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-Jk <.
-^'ISfv'iv'fi''^*
Sr TBS SAMJE AUTHOft.
AN ELEMENTARY TREATISE ON ANALYTIC GEOMETRY,
IMBKACINa
'■ - PLANE GEOMETRY,
AMD AN
INTRODUCTION TO QEOMETRV OF THREE OlMENtlONS.
AN ELEMENTARY TREATISE
DIFFERENTIAL AND INTEGRAL CALCULUS.
WITH NUMEROUS EXAMPLES.
/
J1
«te
cGEOMETRv, I ANALYTIC MECHANICS.
IIMENtlONS.
ISE
:alculus.
WITH NUMEROUS EXAMPLES.
EDWARD A. BOWSER, LL.D.,
riovMKa or mathematics
KUTCnS COLLKW.
NEW YORK:
D. VAN NOSTRAND,
tB McnuT AKD ft WAina Sranr,
1884.
5 "f
..■* 4: X.- ■
T'
OOPTSXOBT, t88k, BT E. A. BOWSXS.
V ■ Jj"v?j''"'/'-:.^^/,*;y:
c^
^te
PREFACE.
"0*^
■♦»»■■
rpHE preoent work on Analytic Mechanics or Dynamics is desired
as a text-book for the mudents of Scientific Schools and Col.
leges, who have received training in the elements of Analytic Oeome-
try and the Calculus.
Dynamics is here used in its tme sense as the science of fortt.
Tlie tendency among the best and most logical writers of the present
(lay appears to l)e to use this torm for the science of Analytic Me-
chanics, while the branch formerly called Dynamics is now termed
Kinetics.
The treatise is intended especially for beginners in this branch of
Hcieuce. It involves the use of Analytic Geometry and the Calculus.
The analytic method has been chiefly adhered to, as being better
adapted to the treatment of the subject, more general in its applica-
tion and more fruitful in results than the geometric method ; and yet
wliere a geometric proof seemed preferable it has been introduced.
The aim has been to make every principle clear and intelligible,
to develop the di£Ferent theories with simplicity, and to explain fully
the meaning and use of the various analytic expressions in which the
|)rinciple8 are embodied.
The book ocmsists of three parts. Part I, with the exception of a
preliminary chapter devoted to definitions and fundamental princi*
pies, is entirely given to Static*.
Part II is occupied with Kinematics and the principles of this
important branch <>f mathematics are so treated that the student may
enter upon the study of Kinetics with clear notions of motion, veloc-
ity and acceleration. Part III treats of the KineUcs cf a particle and
of rigid Imdiee.
tW"
!^^!^'S^ -f_ ' t. ' ; Jnu; 1 ^."V g !"
' W^: :;;^:;,;^i----^. -^ PREFACE. -
In this arrangement of ti.«« work, with the exception of Kine-
matics, I have followed the plan naually adopted, and made the
subject of Statics precede that of Kinetics.
For the attainment of that grasp aX principles which it is the
special aim of the book to impart, numerous examples are given at
the ends of the chapters. The jyreater part of thorn will present no
serious difficulty to the student, while a few may tax his best
efforts.
In prepwring this book I have availed myself of the writings of
many of the beat authont The chief sources frtm which I have
derived assistance are the treatises of Price, Miuchin, Todhunter,
Pratt. Routh, Thomson and Tait, Tait and Steele, Weisbach, Venta-
roli, Wilson, Browne, Gregory, R»nkine, BoucharUt, Pirie, Lagrange,
and La Place, while many valuable hints as well as examples have been
obtained from the works of Smith, Wood, Bartlett, Young, Moeek/,
Tate, Magnus, tioodeve, Parkinson, Olmsted, Gamett, Benwick, Bot-
tomley, Morin, Twisden, Whewell, Oalbraith, Ball, Dana, Byrne, ihe
Elusjyclopedia Britannica, and the Mathematical Visitor.
1 have again to thank my old pupil, Mr. B. W. Pi«ntiss, of the
Nautical Almanac Office, and formeriy Fellow in Mathematics at the
Johns Hopkins University, fbr reading the MS. and for valuable sug.
gestiona Several others also of my friisnds have kindly assisted me
by correcting proofi^beets and verifying copy aad foimule.
E. A. B.
BtrroBRS Collbob, i
Nbw Bbdhbwick, N. J., June, 1884. '
eptlon of Kine-
and made tho
which it is tlio
ilea are given at
will present no
J tax his best
TABLE OF CONTENTS.
' the writings of
I which I have
lin, Todhunter,
eisbacb, Yentu-
'irie, Lagrange,
nples have been
oung, Moselc^ ,
Benwick, Bot.
ma, Bjrrne, the
r.
'rentiss, of the
ematios at the
valuable aug.
aasisted me
vim.
E. A. B.
PART I.
lly
CHAPTER I
FIRST PBINCIPLE8.
m. '^»
1. Definitions— Statics, Kinetics and Kinematies 1
2. Matter *
8. InerJa *
4. Body, Space and Time 8
6. Rest and Motion 8
6. Velocity •• 8
7. Pormube for Velocity *
b. Acceleration ■ *
9. Measure of Acceleration S
10. Geometric Representation of Velocitj and Accelerat'on 6
11. -Mass *
12. Momentum "^
18. Change of Momentum '
14. Force ®
15. Static Measure of Force 8
16. Action and Reaction ®
17. Method of Comparing Forces •
18. Representation of Forces 1®
19. Measure of Accelerating Forces 10
20. Kinetic Measure of Force H
31. Absolute or Kinetic Unit of Force 18
22. Throe Ways of Measuring Force 14
28. Meaning of fli In Dynamics l**
IS'.,
rr. ,, ".i ' W. ^ g!^.
'#■
roirrxyrs.
24. OniTltation Units Df Force and Mam...... ic
25. Gravitation and Absolute Measure 17
Examples 18
■ STATICS (REST). -^
CHAPTER II.
THE COMPOSITION AND BE80LUTI0N OF CONCUBBINO
FOROKS — CONDITIONS OF EQUILIBBIUM.
26. Problem of Statics 21
27. Concurring and Conspiring Forces 21
28. Composition oi Conspiiring Forces 22
29. Composition of Velocities 23
30. Composition of Forces 24
81. Triangle of Forces. 25
82. Three Concurring Forces in Equilibrium 26
83. The Polygon of Forces 27
84 Paralleloplped of Forces 28
85. Resolution of Forces 80
86. Magnitude and Direction of Resultant 81
87. Conditions of Equilibrium 32
88. Resultant of Concurring Forces in Space 84
89. Equilibrium of Concurring Forces in Space 85
40. Tenaion of a String 85
41. Equilibrium of Concurring Forces on a Smooth Plane 39
42. Equilibrium of Concurring Forces on a Smooth Surface 41
Examples 45
CHAPTER III.
COMPOSITION AND BKSOLCTION OF FOBCES ACTING ON A
RIGID BODY.
48. ARlgldBody 57
44. Transmissibility of Force 57
46. Resultant of Two Parallel Forces 58
46. Moment of a Force flo
'.*!a3i-Bia»»»»u.im.'.
-.v\aiifevj^v.a».£?i<ii<»aB«iiii<fe';-fa'n>^>&<fe^
^te
PAO«
It
17
18
CONCUBBIlfO
IIUJI.
21
21
22
28
24
25
26
27
28
80
81
32
84
85
86
lane 30
urface 41
46
iCTINQ ON A.
57
57
60
COm'JBJfTS.
»a
AST. VASI
47. Signs of Moments. , 61
48. Geometric Repreeent&tion of a Moment 61
40. Two £qual anU Opposite Panllel Foroea 61
5U. Moment of a Couple 62
51. Effect of B Couple on a Rigid Body 68
52. Efiect of Transferring Couple to Parallel Plane not altered.. . 64
68. A Couple replaced by another Couple 61
54. A Force and a Couple 06
55. Resultant of any number of Couples 66
56. Resultant of Two Couples. 07
57. Varignon'H Thiorem of Moments 69
58. Varignon'a Theorem for Parallel Foices 71
59. Centre of Parallel Forces 71
6C. Equilibrium of a Rigid Body under Pcrallel Forces. 74
61. Equilibrium of a Rigid Body under Fozoes in any Direction. . 76
62. Equilibrium under Tliree Forces. 77
63. Centre of Parallel Forces in hifferent Planes 86
64. Equilibrium of Parallel Forces in Space. 86
65. Equilibrium of Forces acting in any Direction in Space 88
Examples 00
CHAPTER IV.
OBNTRB OF GRAVITY (CENTRE OP KASS).
66. Centre of Gravity 100
67. Planes of Byiametry — Axes of Symmetry 101
68. Body Suspended from a Point ICl
00. Body Supported on a Surface 102
70. Different Kinds of Equilibrium 102
71. Centre of Gravity of Two Masses 108
72. Centre of Gravity of Part of a Body 108
78. Centreof Gravity of a Triangle 108
74. Centre of Gravity of a Triangul.ir Pyramid 106
75. Centre of Gravity of a Cone 106
76. Centre of Gravity of Frustum of Pyramid 107
77. Investigations involving Integration 100
78. Centre of Gravity of the Ate of a Curve 110
79. Centre of Gravity of a Plane Ai8<i 116
80. Polar Elements of a Plane Area 118
• ••
VUl
CONTENTS.
Ater.
81.
82.
88.
84.
85.
8«.
87.
go.
PAO»
Double Integration— Polar FomiQlae 120
Doiibln Integration — Rectangular Pormulte 122
Centre of Gravity of a Surface of Kevolution. 123
Centre of Gravity of any Curved Surface 126
Centre of (Jruvity of a Solid of Revolution 127
Polar Fonnu'ffl • 130
Centre of Gravity of any I'felid ]3i
Polar Elements of Mass 188
Special Methods isg
Theon-ms of Pappus igg
Examplui 140
CHAPTER V. •
, - '■•■ :*':■' FRICTIOK. ;.' '"•>":'''■ "
91. Friction 149
92. Laws of Friction 150
93. Magnitudes of Coefflcionts of Friction i( i^
94. Angle of Friction ^^.^
95. Reaction of a Rougli Curve or Surface 163
96. Friction on an Inclined Plane 154
97. I'riction or. a Doable Inclined Plane 156
98. Friction on Two Inclined Planes. 159
99. Friction of a Trunnion 159
100. Friction of a Pivot .,,, y-Q
Examples 192
CHAPTER VI.
THE PRINCIPLE OF VIRTUAL VELOCITIE&
101. Virtual Velocity igg
102. Principle of Virtual Velocities 167
103. Nature of the Displacement 169
104. Equation of Virtual Momenta '. , 169
105. System of Particles Rigidly Connected 170
Examples 172
120
122
123
126
127
130
181
183
136
188
140
149
J50
1/-:
ib'a
153
154
156
159
159
ICO
162
coifTEyrs.
CHAPTER VII.
. "■■ „ MACHINES. „ %'
u». - ■■ - ' -"'■•»"
106. Functions of a Machine 177
107. Mechanical / Jvantage 178
108. Simple Machinee , • 1^0
109. Th<i Lever '. !..... 181
110. Equilibrium of the Lever 181
111. The Common Balance 184
112. Chief Reqaisites of a Good Balance 186
113. The Steelyard 188
114. To Graduate the Common Steelyard 188
116. The Wheel and Axle 190
116. Equilibrium of the Wheel and Axle 190
117. Toothed Wheels 192
118. Belation of Power and Weight in Toothed Wheels 198
119. Rtlation of Power to Weight in a Train of Wheels. 194
120. Thfl Inclined Plane IWJ
121. The Pulley 1»7
122. The Simple Movable Pulley 198
123. Firsi System of Pulleys. • 1*8
124. Second System of Pulleys 200
126. Third System of Pulleys 201
126. The Wedge. 20S
127. Mechanical Advantage of the Wedge 202
128. The Screw 304
129. Relation between Power tod' Weight in the Screw 204
129a. Prony's Differential Screw 206
Examples 207
166
107
169
169
170
172
CHAPTER VIII.
THE FtTNICULAR POLYGON — THE CATENARY — ATTRACTION.
190. Equilibrium of the Funicular Polygon 216
131. To Construct the Funicular Polygon 218
182. Cord Supporting a Load Uniformly Distributed 219
188. The Common Catenary — Its Equation 221
188a. Attraction of a Spherical Shell 226
Examples 228
1 1 ^ , , ■■« m f iij n |i , »i f y j ^ v i ^j; i ^ ^i LWL ii _^ wi i r,! ■-- ->, i* i i« i i;i. 'i . y.L . ji •n- it'yjr! >f'ft i; Ti^i''viT-r'yKfr'!! mi ' ^.,J V i ' -!tf 'g ^ '! '^r>l !?*P^'7'iy
coAnuvTS.
PART II.
KINEMATICS (N.OTION).
CHAPTER I.
RECTILINEAB MOTION.
ABT. r^ag
184. DefinltlonB— Velocity 281
185. Acceleration 288
188. Relation lKtwe«n Space and Time when Aooeleration = 0.. . 288
187. Relation when the Acceleration is Constant 284
188. Relation when Acceleration Tariea as the Time 235
189. Relation waen Acnolenttion Vfrico ai> the Distance 285
140. Ekiuations of Motion for Falling Bodies. 287
141. Particle Piv^jected Vertically Upwards 280
142. Oompoeitions of Velodtias 242
148. Resolution of Velodtiea. 248
144 Motion on an Inclined Plane 245
146. Times of Descent down <'^ords of a. Circle 247
146. The atraight Line of Quickest Descent 248
Examples. , 249
CHAPTER II
CCEVItlNEAE MOTHyN.
147. Remarks on Curvilinear Motion 258
J48. Composition of Unifo.m Velocity and Acceleration 258
140. C3m position and Resolution of Acceleration 259
Examples 2fli
160. Motion «♦ Project llos '•» Vacuo 266
161. The Path of a Particle in Vacuo <s k. Parabola 266
162. The Parameter— Range— Greatest Height— Height oi Direc
trix 267
188. Velocity of a Particle at an.- point of its Path 260
164. Time of Flight along a Horisoiitn! Plane 260
166. Poiut at which a Projectile will Strike ac IncUnad PImm. . . 270
Klh
LAWS
VA
166. D
166 N
167. R
168. 'K
169. R
170. T
171. W
172. M
178. M
174. V
175. M
170. \
177. V
178. \.
170. M
E
CONTSKTa.
XI
o.
PAOK
Projection for Grenteat Butg* on a Given Plane 370
The Elevation that the Particle maj pen n Given Point. . . 271
Second Method of Finding E^.nation of Trajectory 273
Velocity of Discharge of Bails and Shells 274
Angular Vilocity and Angular Acceleration 275
Accelerations Along and Perpendicular to RadiuB Vector. . . . 278
Acceleraiions Along and Perp<^ndicular to Tangent 279
When Acceleration Perpendicular to Radios Vector is zero . 281
When Angular Velocity is Constant 282
Examples 284
PART III.
KINETICS (MOTION AND FORCE).
858
258
259
.... 261
206
866
u Direc-
.... 267
sin
.... 869
lane... 870
CHAPTER I.
LAWS OP MOTION — MOTION UNDEB XHF ACTION OF A
VARIABLE FORCE- -MOTION IN A RESISTING MEDIUM.
165. Definitions 289
166 Newton's Laws of Motion 289
167. Remarks on taw I , 890
168. "Remarks on Law II 291
169. Komnrks on Law III 894
170. Two LawD of Motion in the French Treatises 895
171. Motion of Particle under an Attractive Force 295
172. Motion under the action of a Variable Ropnlsive Force 298
178. Motion under the action of an Attractive Force ^1)9
174. Velocity acquired In Falling through a Great Height 800
176. Motion in a Resisting Medium. ;!02
176. Motion in the Air against the Action of Gravity 804
177. M^ion of a Projectile in a Resisting Medium 807
178. Motion against the Resistance of the Afnuitiphere 908
179. Motion in the Atmoephere under a smuii Angle of Elevation 812
Examples 818
zU
CONTENTS.
CHAPTER II.
CENTRAL FORCES.
180. I>?flnition8 321
181. A Particle under the Action of a Central Attraction 821
182. The Sectorial .\rea Swept over by the Radius Vector 825
183. Velocity of Particle at any Point of it« Oruit 826
184. Orbit when Attraction as the Inverse Square of Distance. . . 32 J
186. Suppoee the Orbit to be an Ellipee 833
186. Kepler's L»w8 335
187. Nature of the Force which acta upon the Planetary System. 885
Kxampies .... 338
CHAPTER III.
CONSTRAIKED MOTIOK
188. Deflnitions 848
189. Kinetic Energy or Vis Viva— Work 345
190. To Find the Heaetion of the Constraining Curve 848
101 . Point where Particlj will leave Constri ining Curve 849
192. Constrained Motion Under Action of Oraviiy 350
108. Motion on a Circular Arc in a Vertical Plane 850
194. The Simple Pendulum 353
196. Relation of Time, Length, and Force of OravHy 858
196. Height of Mountain Determined with Pendulum ;t64
197. Depth of Mine Determined with Pendulum 866
198. Centripetal and Centrifugal Forces 866
199. The Centrifugal Force at the Equator 058
200. Centrifugal Force at Diffcn>nt Latitudes 869
901. The Conical Pendiilnm— The Uovemor 861
Examples 992
CHAPTER IV,
IMPACT.
209. An Impulsive Porce 370
208. Impact i)r Collision 371
204. Direct and Central Im|4u;t 872
206. Elaaticity of Bodies— Coefficient of Restitution 878
ABT.
20(5.
','07
208.
2()9.
210.
211.
212
213.
214.
215.
210. ]
]
217. 1
1
218. I
219. i
230. \
231. 1
222. >
238. 6
I
224 1
226. 8
236.
237.
238.
239.
230. I
231. J
iilif I
^
CONTSNTa,
xm
... 821
Ion S21
ector 825
826
Distance. . . 32 J
888
385
aiy System . 885
848
845
848
irvo 849
830
850
, 858
858
, :i54
355
856
aw
859
861
870
871
872
878
ABT. rAOB
30«. Direct Impact of Inelastic Bodies 374
'207. Direct impact of Eliifltic Bodien 875
-m. Loss of Kinetic Energy in Impact of Bodies 878
2()9. Oblique Im|jact of Bodies 880
210. Oblique Impact of Two Smooth Spheres 883
Examples 888
CHAPTER V.
WOEK AND ENERGY.
Definition and Meaanre of Work 880
General Case of Worli done by a Force l>90
Work on an Inclined Plane 891
Examples 808
Horse Power 895
Work of Raising a System of Weights ?0«
Examples 8i»7
Modulus of a Machine 400
Examples 401
Kinetic and Potential Energy— Stored Work 404
Examples 406
Kinetic Energy of a Rigid Body Revolving roond an Axis. . . 403
Force of a Blow 411
Work of a Water Fall 418
The Duty of an Engine 414
Wr.k of a Variable Force 415
Sitiipeon's Rule 415
Exampltw 417
211.
212
213.
214.
215.
210.
217.
218.
219.
230.
231.
223.
233.
234.
225.
226.
227.
238.
229.
230.
231.
CHAPTEK VI.
MOMENT UF INERTIA.
Moment of In»«rtia 480
Moments of Inertia relative to Paral'pl Axes or Planes 438
Radius of Gyration 434
Polar Moment of Inertia 4ii6
Moment of Inertia of a Solid of Revolution 487
Moment of Inertia aliout A: is Perpendicular to Geometric
Axis 438
Moment of Inertia of Various Solid Bodies 440
Moment of Inertia of a Lamina with respect to any Axis. . . . 441
' X-f. ;jy-"'j;i ' v4^; f? g ' ^svj? ! gy ' l^w'^:^M^ l ■fti.^l jija^f[ ^ ^4»^ ,i, \ UAm,v: ' ^> !WSSS^^
1^
lir
CONTMNTB.
232. Principal Axes of E Body.... 448
2S3. Products of Inortia ■■ 446
Examples 447
^
Hi
CHAPTER VII.
ROTATORY MOTION.
284. Imprefwed and Effective Forces 461
285. D'Alembert'B Principle 453
286. RoUtion of a Rigid Body about a Fixed Axis 454
287. The Compound Pendulum 457
28M. Length of Second's Pendulum Determined Experimentally. . 462
289. Motion of a Body when Unconstrained 4t?4
240. Centre of Percussion— Axis of Spontaneous Rotation 404
241. Principal Radius of Gyration Determined Practically 467
242. The Ballistic Pendulum 468
248. Motion of a Body about a Horizontel Axle through its Centre 470
244. Motion of a Wheel and Axle 471
245. Motion of a Rigid Body about a Vertical Axis 472
246. Body Rolling down an ..uclined Plane 478
247. Falling Body under an Impulse not through its Centre 475
Examples 477
CHAPTER VIII.
MOTION OF A SYSTEM OF RIGID BODIES IN BPAOB.
248. Equations of Motion obtfline<l by D' '.Icmbert's Principle 481
249. Independence of the Motions of Translation and RoUtiou. . . 482
250. Principle of the Conservation of the Centre o£ Gravity 486
251. Principle of the Conservation of Areu 486
253. Conservation of Vis Viva or Energy 488
253. Composition of Rotations. 493
254. Motion of a Rigid Body referred to Fixed Axes 494
866. Axis of Instantanaoos Rotation 495
266. Angular Velocity about Axis of Instantaneous Rotation 496
257. Eulor's liquations ^^
258. Motion alx)Ut a Princi|»al Axis through Centre of Gravity. . . 499
269. Vel<,.lty abouta Principal Axis when Accelerating Forces
= Ml
860. The Integral of Euler's Equations 502
Examples "**
, 44S
446
447
451
462
454
457
Imentallv,. 462
4«i4
an 404
illy 467
468
liita Centre 470
471
472
478
^ntre 475
477
N SPACE.
indple... 461
lotation... 482
ivity 485
486
488
498
494
405
iation 400
407
Iravity... 499
ng Forces
601
603
505
ANALYTIC MECHANICS.
PART I.
CHAPTER I.
FIRST PRINCIPLES.
1. Deflzdtioiis. — Analytic Mechanics or Dynamics ia
the science which treats of the equilibrium an<J motion
of bodies under the action of force. It is accordingly
divided into two parts, Statics and Kinetics.
Statics treats of the equilibrium of bodies, and the condi-
tions governing the forces which produce it
Kinetics treats of the motion of bodies, and the laws of
the forces which produce it.
The consideration that the properties of motion, velocity,
and displacement may be treated apart from the particular
forces producing them and independently of the bodies sub-
ject to them, has given rise to an auxiliary branch of Dyna-
mics called Kinematics.*
Although Kinematics may not be regarded as properly
in'^luded under Dynamics, yet this brunch of science is so
important and useful, and its application to Dynamics so
immediate, that u jwrtion of this work is devoted to its
treatment.
* Thli name wm given by Amp4re.
1 MATTES, INERTIA, BODY, MOTION, ETC.
Kinematics is the science of pore motion, withont refer-
ence to matter or force. It treats of the properties of
motion without regard to what is moving or how it ia
moved. It is an extension of pure geometry by introduc-
ing the idea of time, and the consequent idea of velocity.
2. Matter. — Matter is that which can be perceived by
the senses, aud which cia transmit, and be acted upon by
force. It has extension, resistanon, and impenetrability.
A definition of matter which wonld ntiefy the metaphyR'.cian Is ■
not required for this work. It ia Bofflcient for na to oouceive of it aa
capable of receiving and transmitting force : becanoe it la in thia
aspect onlj that it ia of importance in the preaent treatise.
3. Inertia. — By Inertia is meant that property of mat-
ter by which it remains in its state of rest or uniform
motion in a right line unless acted apon by force. Inertia
expresses the iact that a body cannot of itself change its
condition of rest or motion. It follows that if a body
change its state from rest to motion or from motion to rest,
or if it change its direction from the natural rectilinear
path, it must have been influenced by some external cause.
4. Body, Space, and Time.— ^ Body is a portion of
matter limited in every direction, and is consequently of a
determinate form and volume.
A Rigid Body is one in whlvh the relative positions of
its particles remain unchanged by the action of forces.
A Particle is a body indefinitely small in every direction,
and though retaining its material propertias may be treated
as a geometric point
Space is indefinite extension. Time is any limited por-
tion of duration.
5. Rest and Motion.— A body is at rest when it con-
stantly occupies the same place in space. A body is in
i
TO.
VMLOCtTT.
ithont refer-
roperties of
>r how it is
by introduo-
f velocity.
aerceived by
;ed upon by
etrability.
ttaphyo'.cian Is .
uceive at it u
I it is in this
iae.
jerty of mat-
; or uniform
rce. Inertia
If change its
it if a body
otion to rest,
il rectilinear
temal cause.
a portion of
uently of a
positions of
forces.
|ry direction,
Ly be treated
limited por-
rhen it con-
body is in
motion when the body or its parts occupy successively dif-
furent positions in space. But we cannot judge of the stute
of rest or motion of a body without referring it to the
]K)iiition8 of other bodies ; and hence rest and motion must
be considered as necessarily relative.
If there were anything which we knew to be absolately fixed in
space, we mi^t perceira aiieolate motion by change of place with
rcfertince to that object. Bat as we know of no such thing as also-
lute rest, it follows that all motion, as measured 1>t ub, must be
rolatiye ;%.»., most relate to something which we assume to be fixed.
Hence the same thing may often I>e itaid to be at rest and in motion
at the same time : for it may be at rest in regard to one thing, anJ. in
niution in regard to another. For example, the objects on a vessel
may be at rest with reference to each other and to the voseel, while they
are in motion with reference to the neighboring shore. So a man,
I>unting his barge up the river, by leaning against a pole which rests
on the bottom, and walking on the deck, is In motion relative to the
barge, and in motion, but in a different manner, relative to the car-
rui}t, while he is at rest relative to the earth.
Motion is uniform when the body passes over equal spaces
iu equal times ; otherwise it is variable.
6. Velocity. — The velocity of a body is its rate of
motion. When the velocity ic eonstant, it is measured by
the space pas^-ed over in a unit of time. When it is varia-
ble, it is measured, at any instant, by the space over which
the* body would pass in a unit of time, were it to move,
during that unit, with the same velocity that it has at the
instant considered.
The speed of a railway train is, in general, variable. If we were to
Bay, for example, that it was running at the rate of 80 miles an hour,
W6 would not mean that it ran 80 miles daring the last hour, nor that
it would run 80 miles daring the next hour. We would mean that, if
it were to ruu for an hour with the speed which it now has, at the
instant considered, it would pass over exactly 80 miles.
I
In order to have a uniform unit of velocity, it is custom-
ary to express it in feet <uid se&iiule ; and when velocities
4 ACCELBRATtOK
are expressed in any other terms, they should be redtieed to
their equivalent vahie in feet and seconds. Thn unit
velocity, therefore, is the velocity with which a body
describes one foot in one second; other units may be taken
where convenience demands, as miles and hours, etc.
When wo speak of the space passed over by a body, we
mean the path or line which a point in the body or which a
particle describes.
7. Fommlte for Velocity.— If » be the space pa^ed
over by a particle in / units of time, and v the velocity, it is
plain that, for uniform velocity, we shall have
8
(1)
that is, we divide the whole space passed over by the time
of the motion over that space.
If the velocity continually changes, equal increments are
not described in equal times, and the velocity becomes
a function of the time. But however much the velocity
changes, it may be regarded as constant during the
infinitesimal of time dt, in which time the body will
describe the infinitesimal of space ds. Hence, denoting the
velocity at any instant by v, we have
V =
da
dt'
(3)
I-
In this case the velocity is tho ratio of two infinitesimals.
These two expressions for the velocity are true whether the
particle be moving in a right, or in a curved, line.
8. Acceleratioii is the rate of change of velocity. It
is a velocity increment. If the velocity is increasing, the
acceleration is considered positive; if decreasing, it is
negative.
Acceleration is said to be uniform when the velocity
~*im
lid be redtieed to
nds. Thn unit
which a body
ts may be taken
lioure, etc.
er by a body, we
I body or which a
the space passed
the velocity, it is
ive
(1)
ver by the time
[ increments are
jfelocity becomes
ach the velocity
ant during the
the body will
Loe, deuotitig the
infinitesimals,
rue whether the
i, line.
of vehcity. It
1 increasing, the
lecreasing, it is
len the velocity
MBAaURB OF accblbratioh. 5
receives eqcal increments in equal times. Otherwise it is
variable.
9. Measure of Acceleration.— Uniform acceleration
is measured by the actual increase of velocity in a unit of
time. Variable acceleration is measured, at any instant, by
tlie velocity which would be generated in a unit of time,
were the velocity to increase, during that unit, at the same
rate as at the instant considered.
Calling /the acceleration, v the velocity, and t the time,
we have, when the acceleration is uniform.
/
(1)
However variable the acceleration is, it may be regarded
as constant during the infinitesimal of time dt, in which
time the increment of velocity will be dv. Hence, denoting
the acceleration at the time t by/, we have
(2)
We also have (Art. 8)
V =
which in (2) gives
ds
It
-_ dw _ rf ds
^ ~ dt ~ dt ' Jt
d*a
(8)
That is, when the acceleration is variable it is measured, at
any instant, by the derivative of the velocity regarded as a
function of the time, or by the second derivative of the
space regarded as. a function of the time.
From (3) we get, by integration.
da
/* = rf7 = ^'
(♦)
>-!
VMLOCtTT AND AOCELaBATtOH,
i/P
rs <;
a a
HJJ
Si"
and
2/8 = v«,
(5)
(6)
which determine the velocity and space.
10. Gtoometrio Reprosentatioii of Valoeity and
Acceleration.— The velocity of a body may be conveni-
ently represented geometrically in magnitude and direction
by means of a straight line. Let the line be drawn from
the point at which the motion is considered, and in the
direction of motion at that point With a convienient scale,
let a length of the line be cat off that shall contain as many
units of length as there are units in the velocity to be repre-
sented. The direction of this line vnll represent the
direction of the motion, and its length will represent the
velocity.
Also an acceleration may be represented geometrically by
a straight line drawn in the direction of the velocity
generated, and containing as many units of length as there
are units of acceleration in the acceleration considered.
Also, since an acceleration is measured by the actual
increase of velocity in the unit of time, the straight line
which represents an acceleration in magnitude and dii*ec-
tion will also completely represent the velocity genei-atcd in
the unit of time to which the acceleration corresponds.
11. The Mass of a body or particle is the quantity of
matter which it contains; and is proportional to the
Volume and Density jointly. The Density may therefore
be defined as the quantity of m»t>«r in a unit of volume.
Let ilf be the mass, p the density, and V the volume, of
a homogeneous body. Then we have
M^ Vp,
(1)
if we so take our units that the unit of mass is the mass of
the unit volume of a body of unit density.
-■i^-VAiia: %. J.-
•iv.
▼aloeity and
lay be conveui-
le and diroction
be drawn from
red, and in the
ionvenient scale,
contain as many
Mjity to be repre-
reprosent the
11 represent the
^metrically by
of the velocity
lengtli aa there
;ion considered.
by the actual
he straight line
Itude and direc-
ity genei-ated iu
orrespoud*.
the quantity of
>rtional to the
\y may therefore
►it of volume,
the volume, of
(1)
is the muss of
MOMKNTUM. 7
If the density varies from point to point of the body, we
liave, by the above formula, and the notation of the
Integral Calculus,
3f=/odV = f/fpdx dy dt, (2)
wlicre p is supposed to be a known function of x, y, t.
In England the unit of mass is the imperial standard
l)ound avoirdupois, which is the weight of a certain piece of
platinum preserved at the standard office in London. On
the continent of Europe the unit of mass is the gramme.
This is Icnown as the absolute or kinetic unit of mass.
12. The Quantity of Motion,* or the Momentum
of a body moving without rotation is the product of its
mass and velocity. A double mass, or a double velocity,
would correspond to a double quantity of motion, and
so on.
Hence, if we take as the unit of momentum the mo-
mentum of the unit of mass moving with the unit of
velocity, the momentum of a mass M moving with velocity
r is Mv.
13. Change of Quantity of Motion, or Change of
Momentum, is proportional to the mas^ moving and the
change of its velocity jointly. If then the mass remains
constant the change of momentum is measured by the
product of the moss into the change of velocity ; and the
rale of change of momentum, or acceleration of momentum,
is measured by the product of the mass moving and the
rate of change of velocity, that is, by the product if the
mass moving and the acceleration (Art. 9). Thus, ct lling
J/ the mass, we have for the measure of the rale of change
of momentum,
rd*«
M
Ifl'
* This phtwe WM a««d bj Newton in place of Uie more modern term " Momen-
mm."
8
STATIC MSAaURS OF FORCE.
14. Force. — Force is any cause which changes, or tends
to change, a body's state of rest or motion.
A force always tends to produce motion, but may be pre-
vented from actually producing it by the counteraction of
an equal and opposite force. Several forces may so act on
a body as to neutralize each other. When a body remains
at rest, though actcu on by forces, it is said to be in
equilibriam ; or, in other words, the forces are said to
produce equilibrium.
What force is, in its nature, we do not know. Forces
are known to us only by their effects. In order to measure
them we must compare the effects which they produce
under the same circumstances.
15. Static Measnre of Force.— 7%e effect of a force
depends on: Ist, its magnitude, or intensity ; 2d, its direc-
tion; i. e., the direction in which it tends to move the body
on which it acts ; and 3d, its point of application .; t. «., the
point at which the force is applied.
The effect of a force is pressure, and may be expressed by
the weight which will counteract it. Every force, statically
considered, is a pressure, and hence ha« magnitude, and
may be measured. A force may produce motion or not,
according as the body on which it acts is or is not free to
move. For example, take the case of a body restiug on a
table. The ame force which produces pressure on the
table would cause the body to fall toward the earth if the
table were removed.
The cause of this pressure or motion is gravity, or the
force of attraction in the earth. In the first case the attrac-
tion of the earth produces a pressure; in the second case it
produces motion. Now either of these, viz., the pressure
which the body exerts when at rest, or the quantity of
motion it produces in a unit of time, may be taken as a
means of measuring the magnitude of the force of attrac-
tion that the earth exerts on the body. The former is
S^'
V^'/il^^-^t,,^*i":i'«V#i4^,"^^*^.-^-S^:''~<^aSi{^S\
«*
hanges, or tends
but may be pre-
counteraction of
38 may bo act on
a body remains
is said to be in
rces are said to
t know. Forcea
)rder to measure
h thoy produce
effect of a force
y; 2d, its direc-
to move the body
ication ,; i. e., tbo
y be expressed by
Y force, statically
magnitude, and
B motion or not,
i or is not free to
>dy resting on a
pressure od the
the earth if the ~
is graTity, or the
]t case the attrac-
e second case it
iz., the pressure
the quantity of
ay be taken as a
le force of attrac-
The former is
MBTffOD OF COMPABiya FOttCKS,
called the static method, and tUe forces ftrc called ttatie
farces; the latter is called the kinetic method, and the
forces are called kinetic forces. Weight is the name given
to the pressure which the attraction of the earth causes a
l)o(ly to exert Hence, since static forces produce pressure,
we may take, as the unit of force, a pressure of one pound
(Art. 11).
Therefore, the magnitude of a force may be measured
ainfically hy the pressure it will produce "t/on some body,
ami expressed in pounds. This is called the Static measure
of force, and its unit, one pound, is called the Oravitation
unit of force.
16. Action and Roactioii are always equal and
opposite. — This is a law of nature, and our knowledge cf
it comes from experience. If a force act on a body hold by
a fixed obstacle, the latter will oppose an equal and con-
trary rasistance. If the force act on a body free to move,
motion ¥rill ensue ; and, in the act of movitig, the inertia
of the body will oppose an equal and contrary resistance.
If we press a stone with the hand, the stone presses the
liuud in return. If we stiike it, we receive a blow by the
act of giving one. If we urge it so as to give it motion, we
lose some of the motion which we should give to our limba
In the same effort, if the stone did not impede them. In
euch of these cases there is a reaction of the same kind as
tlic action, and equal ♦^" it.
17. Method of Comparing Forces.— Two forces are
orjiial when being applied iu opposite directions to a
particle they maintain equilibrium. If we take two ^qual
forces, and apply them to a particle in the same direction,
wo obtain a force double of either ; if we unite three equal
foices we obtain a triple force ; and so on. So that, in
general, to compare or measure forces, we hare only to
K^iopt the same method as when we compare or measure
JO
REPRBSENTATIOH OP P0SCX8.
any qnantities of the same kind ; that is, we maet take
some known force as the unit of force, and then express, in
numbers, the relation which the other forces bear to this
measuring unit. For example, if one pound be the unit of
force (Art. 16), a force of 12 pounds is expressed by 12 ;
and so on.
18. Rspresentatioii of Forces by SymbolB and
IJneg. — If p. Q. B., etc., represent forces, they are numl)ers
expressing the numl)er of times which the concrete unit of
force is contained in the given forces.
Forces may be represented geomehrically by right lines ;
and this mode of reprci^icntation has the advantage of giving
the direction, magnitude, and point of application of each
force. Thus, draw a line in the direction of the given
force ; then, having selected a unit o2 length, such as an
inch, a foot, etc., measure on this line as many units of
length as the given force contains units of W3ight The
marinittide of the force is represented by the measureti
length of the line ; its direction by the direction in which
the line is drawn; and \t& point of application by the point
from which the line is drawn.*
Thus, let the force Pact at the point * ^
A, in the direction AB, and let AB ^'■- '•
represent us many units of length as P contains units of
force; then tlio force P is represented geometrically by
the line AB ; for the forti act? in the direction from A
to B ; its point of application io at A, and its magnitude iii
represented by the length of the line AB.
19. Measure of Acceleratitig Foices. — From our
definition of force (Art. 14), it is clear that, when a singk
• Forcof", velocities, and accclnratlotiH are direeltd (ptantlUfCy p.nd m majr be
roprewntpd by a line, in direction and magnitnde, and may b« coinpoand«d In the
Mmu way a* vtduri.
If auyUilus bu maRnHude and direction, Uie magoltode and dir«ctlon taken
togetlier constitato a r«clor.
res.
is, we mast take
I then express, in
orces bear to this
nd he the unit of
expreasod by 12;
Symbols and
they are numl)ers
J concrete unit of
ly by right lines;
ivantage of giving
pplication of each
tiou of the given
ength, ftuch as an
as many units of
. of W3ight The
by the measureti
ircction in which
ation by the point
t * «
J Fi9. I.
' contains units of
goonictrically by
direction from A
its magnitude Ia
xces. — From our
lat, when a singlti
uantUitr, Rnd ro tuKj b*
be compoonded In the
tade and diroctton Ukou
MBASURS OF AOCKLSSATmO FOSOXS.
11
iorce acts upon a particle, perfectly free to move, it must
pioduce motion ; and hence the force may be represented
to us by the motion it has produced. But motion is
measured in terms of velocity (Art. 7), and consequently the
\eiocity communicated to, or impressed upon, a particle, in
a given time, may be taken as a measure of the force.
That is, if the same particle moves along a right line so
that its velocity is inorea/jed at a constant rate, i-"" v/ill be
acted upon by a constant force. If a certain uonstani force,
acting for a second on a given particle, generate a velocity
(if 32.2 feet per second, a douWo force, acting ibr one
second on the same particle, would generate a velocity of
'JiA feet per second ; a triple force would generate a
V elocity of 96.0 feet pei* second, and so on.
If the rate of increase of the velocity, (t. e., the accelera-
tion), of the particle is not uniform, the force acting on it
if, not nniform, and the magnitude of the force, at any
imnt of the particle's path, is measured by the acceleration
of the pari.icle at this point Hence, sine* one and the
.«imo particle is capable of moving with all possible accelera-
tions, all forces may be measured by the velocities they
(jennrate in the same or equal particles in the same or equal
times. When forces are so measured they are called
Accelerating Forces.
20. Kinetic or Abftolute Measnre of Forea.*— Let
n equal particles be placed side by side, and let each of them
lie acted on uniformly for the same time, by the same force.
Each particle, at the end of this time, will have the same
velocity. Now if these n separate particles are all united so
as to form a body of n times the mass of each particle, and
If each one of them is still acted on by the same 'orce as
* Arta. M, n, tt, and K, treat of Um Kinetle moarare of force, and may be
omitted till Part til la reached ; but It ii <x)nvenlent to prewnt them once for all,
»ua, for the aake of tefereiicu and comparisou, to place ttiein with the Static
lueaanre of foroo at the beglnuiog of the work.
12 KINBTtO OB ABaOlVTS MBA8USB OP tOMCS.
before, this body, at the end of the time oonsidered, will
have U>e sams velocity that each aeporate partiule had, and
will \ i cted on by n times the force which generated this
velocity in the particle. Comparing a single particle, then,
with tlxe body whose maw is n times the mass of this
particle, we see that, to produce the same velocity in two
bodies by forces acting on them for the same time, the
maf^nitudes of the forces must be proportional to the
masses on which they act.* Hence, generally, since force
vaiies as the velocity when the mass is constant (Art 10),
and varies as the mass when the velocity is constant, we
have, by the ordinary law of proportion, when both are
changed, force varies as the product of the mass acted npon
and the velocity generated in a given time ; that is, it varies
OS the quantity of motion (Art 13) it produces in a given
mass in a given time. If the force bo variable, the rate of
change of velocity is variable (Art 19), and hence the force
varies as the product of the cuiss on which it aots and the
rtUe of chanyi of velocity, ». e., it varies as the acceleration
of the momentum (Art U). Therefore, if any force P act
on a mass M, wo have (Art 10)
P<s: 2H/i
or, in the form of an equation
(1)
(2)
where k is some constant
If the unit of force be taken as that force which, actmg
on the unil of mass for the nnit of time, generates the unit
of velocity, then if Me put iTeaua) to unity, %.»., take the
unit of masj, and /equal to unity, i. «., take the nnii of
acceleration, wo mnst have the force producing the accol-
enition equal to the unit of force, or P equa' to unity.
\F roncB
oonsidered, will
urtiule had, and
I generated this
) particle, then,
B mass of this
velocity in two
same time, the
irtioual to the
illy, since force
istant (Art 19),
is constant, we
when both are
nass acted upoa
that is, it Taries
ices in a given
ble, the rate of
hence the force
it acts and the
the aceeleration
ny force P act
(1)
(2)
e which, acting
lerates the unit
t. 0., take the
ike the nnii; of
icing the accel-
iqual to unity.
TBB ABSOLVTB OK KtNKTlC MSASURB OP FORCB. 18
Hence k must also h^ equal to oinity, ^nd we have the
equation,
P^Mf. (8)
Therefore, the Kinetic or Absolute measure of a force is
the rate of change or acceleration* of momentum it produces
in a unit of lime.
If the force is constant, (3) booomes by (1) of Art 9,
P = — ■•
t
w
And if the force ii variable, (3) beoomes by (3) of Art 9,
P^M
di^
w
21. The AbflolQte or Slttetie Thdt of Force.—
A sec ad, a foot, and a pound being the units of time, space,
and maw, respectively (Art^ 7 and 12), we are required to
find the corresponding unit pf force that the above equation
may be true. The unit of force is that force which, acting
for one second, on the majs of one pound, generates in it a
velocity of one foot per second. Now, from the results of
numerous experiments, it has been ascertained that if a
body, weighing one pound, fall froely for one second at the
seu level, it will acquire a velocity of about 32.2 feet per
second ; t. «., a force equal to the weight of a pound, if
acting on the niaaa of a pound, at the tea level, generates in
it in one second, if free to move, a velocity of nearly 32.2
feet pcT noond. It followa, therefore, that a f^rce ol
^^ of the weight of a pound, if acting on the mass of
a pound, at the lea level, generates in it in one second, if
free to move, a velocity of one foot per second ; and hence
• Dm Tklt and MMto'i DyMndti or • Partlclt, p. «.
14
MBASUBES OF FORCE.
the unit of force is ss-^ of the weight of a pound, or rather
less than the weight of half an ounce avoirdupois ; so thai
half an ounce, acting on the mass of a pound for one
second, will give to it a velocity of onw foot per second.
This is the British absolute kinetic* unit of force.
In order that Eq. 3 (Art. 80) may be universally true
when a second, a foot, and a pound are the units of time,
space, and mass respectively, ail forces must be expressed in
terms of this unit
22. Three Ways of Meaaoring Force.— (1.) If a
force does not produce motion it is measured by the pres-
sure it produces, or the number of pounds it will support
(Art 16). This is the measure of Static Force, and its
unit is the weight of a pound.
(2.) If we consider forces as always acting on a unit of
mass, and suppose tLat there are no forces acting in the
opposite direction, then these forces will be measured
simply by the velocities or accelerations which they generate
in a given time. This is the measure of Accelerating Force,
and its unit is that foroe which, acting on the unit of mass,
during the unit of time, generate the utv", of velocity;
hence (Art 21), the unit of force is the force which, acting
on one pound of mass for one second, generates a velocity of
one foot per second.
(3.) If forces act on different masses, and produce motion
in them, and we consider as before that there are no forces
acting in the opimsite direction, then the forces are meas-
ured by the quantity of motion, or by the acceleration of
momentum generated «« a unit of time (Art 20). This is
the measure of Moving Force, and its unit (Art 21) is the
force which, acting on one pound of mass for one second,
generates a velocity of one foot per second.
* Introduced by OaoM,
JL,
IX.
)f a pound, or rathe
Avoirdupois ; so that
of a pound for one
mw foot per second,
lit of force,
be universally true
e the units of time,
must be expressed in
I Force.— (1.) If a
leasured by the pres-
)unds it will support
Uatic Force, and its
Etcting on a unit of
forces acting in the
I will be measured
which they generate
f Accelerating Force,
on the unit of mass,
le un*", of velocity;
! force which, acting
nerates a velocity of
and produce motion
;here are no forces
he forces are meas-
the acceleration of
(Art. 20). This is
nit (Art 21) m the
lass for one second,
MSAmyO OF O JiV dynamicb.
15
It must be understood that when we speak ui static,
accelerating, or moving forces-, we do not refer to different
kinds of force, but only to force as measured in different
ways.
23. Meaning of gr in Dynamice.— The most impor-
tant case of a constant, or very nearly constant, force is
gravity at the surface of the eaxth. The force of gravity is
so nearly constant for places near the earth's surface, that
fivlling bodies may be taken as examples of motion under a
constant force. A stone, let fall from rest, moves at first
very slowly. During the first tenth of a second the velocity
is very small. In one second the stone has acquired a
velocity of abou* 32 feet per second.
A great number of experiments have been made to ascer-
tain the exact velocity which a body would acquire in one
second under the action of gravity, and freed from the
resistance of the air The most accurate method is indi-
rect, by meaus of the pendulum. The result of pendnlum
experiments made at Leith Fort, by Captain Kater, is,
that the velocity acquired by a body falling unresisted for
one second is, at that phuse, 32.207 feet per sbcond. The
velocity acquired in one second, or the acceleration (Art.
10), of a body f.-'lling freely in vacuo, is found to vary
slightly with the latitude, and also with the elevation above
the sea level. In London it is 32.1889 feet per second. In
latitude 46°, near Bojtleaux, it is 32.1703 feet per second.
This acceleration is usually denoted by g ; and when we
say that at any place g is equal to 32, wo mean that the
velocity generated per second in a body falling freely*
under the action of gravity at that place, is a velocity of
82 feet per second. The averafre value of g for the whole
of Great Britain differs but little from 32.2 ; and hence the
numerical value of g for that country is taken to be 32.2.
• AbodyUi«ldtol)oino¥liig.rt'»«^wh«»ttU«ctodnponby notwrce* except
tlioM nader eoniUenUon.
16
TUX ixyiT or MAaa.
The fonnula, deduced from observation, and a certain
theory regarding the figure and density of the earth, which
may be employed to calculate the moat probable value of
the apparent force of gravity, is
^ = (? (I rf .0061113 8in» A),
where G is the apparent force of gravity on a unit mass at
the equator, and g the force of gravity in any latitude A;
the value of Q, in terms of the British absolute unit, being
32.088. (See Thomson and Tait, p. 82^.)
24. Onivitatlon Unite of Foroe and Mass.— If in
(8) of Art. 20, we put for P, the weight W of the body,
and write g for / since we know the acceleration is o, (3)
becomes " \ '
W=^mg. (1)
W
W
m = — .
9
m
and hence - may be taken as the measure of the mass.
In gravitation memure forces are measured by the pres-
sure they will prodvce, and the unit of force is one pound
(Art. 16), and the unit of mass is the quantity of matter in
a body which weighs g pounds at that place where the accel-
eration of gravity is g.
This definition gives a unit of mass which is constant at
the same place, but changes with the loc Uty ; i «., its weight
changes with the locality while the quantity qf matter in it
remains the same. Thus, the unit of mass would weigh at
Bordeaux 32.1703 pounds (Art. 23), while at Leith Fort it
would weigh 32.207 pounds. Let m be the mass of a body
which weighs w pounds. The quantity of matter ia |hi»
body remains the same when carried from place to place.
If it were possible to transport it to another planet its mass
m, and a certain
' the earth, which
probable value of
),
>n a nnit mass at
any latitude A;
lolute unit, being
id Mass.— If in
W of the body,
sleration i^ g, (3)
m
m
of the mass.
^tred by the pres-
rce is one pound
'My of matter in
where the aceel-
1 is oonstant at
'; i^e.iiU weight
t of matisr in it
would weigh at
iji Leith Fort it
mass of a body
' matter in Ibis
place to place,
planet its mass
OS A vrrATtON MMAStr/fB Of PORCK.
17
would not be altered, but its fimght wonM be very different.
Its weight wherever placed would vary directly m the force
of gravity ; but the aoeeleratioH afeo would vary diivotly as
tlie force of gravity. If placed on the sub, for example, it
would weigh about 5J8 tiraca acr macfe m ©b the surface of
the earth j but the aeceleration on tht ssn would also b«
28 times as much ae on the surface of th© earth ; that i%
the ratio of the weight to^ the acceleration, anywhere in
W
the universe is constant, and henoe — , which is the
9
numerical value of m (Bq. 2), is oonstuit for tiw same
mass at all places.
25. Compsiisoa of Gteavitatioii and Absolnta
Maasnra.— The pound weight has been long used for the
measurement of force instead of mass, and i»the recognized
standard of reference. It came into general use because it
afforded the moat ready and simple method of estimating
forces. The pressure of steam in a boiler is always reck-
oned in pounds per square inch. The tension of a string is
estimated in pounds; the force necessary to draw a train of
cars, or the pressure of water against a look-gate, is
fxnivBsed in pounds. Such expressions as "a force of
10 pounds,^" or " a pressure of steam equal to 50 pounds on
the inch," are of every day occurrence. Therefore this
method of measuring forces is eminently convenient in
practice. For this reason, and because it is the one used
by most engineers and writers of mechanics, we shall adopt
.it in this work, and adhere to the measurement of force by
pounds, and give all our results in the usual gravitation
measme. In this measure it is convenient to represent the
W
mass of a body weighing W pounds by the fraction —
(Art 1^4), so that (3) of Ari 20 becomes
P = If
0)
18
BXAMFLES.
To do so it will only be necessary to assnme that the unit
of mass is the quantity of matter in a body weighing //
pounds, and changes in weight in the same proportion tbut
g changes (Ai-t. 24).
Of course, the units of mass and force in (a) of Art. 30
may be either absolute or gravitation units. U absolute,
the unit of mass is one pound (Art. 12), and the unit of
force is - pounds (Art. 21). If gravitation, the units arc
g times as great; i. «., the unit of mass is g pounds (Art.
24), and the unit of foroe is one pound (Art 16).
The advantage of the gravitation measure is, it enables ns
to express the force in pounds, and furnishes us with a con-
stant numerical representative for the same quantity of
matter ; that is to say, a mass orepresented by 20 ou the
equator would be represented by 20, at the pole or on
the sun. Hence, in (1), P is the static measure of
any moving force [Art 22, (3)], W is the weight of the
body in pounds, g the acceleration of gravity (Art. 24),
— the-mass upon which the force acts [(2) of Art 24], and
which is frf.e to move under the action of P, the unit of
mass being the mass veighing g pounds, and / the
acceleration which the force P produces in the mass.
EXA.MPLBS *
1. Compare the velocities of two points which move
uniformly, one through 5 feet in half a second, and the
other through 100 yards in a minute. Ans. As 2 is to 1.
2. Compare the velocities of two points which move uni-
formly, one through 720 feet in one minute, and the other
through 3Jf yards in three-quarters of a second.
Ans. As 6 is to 7.
3. A railway train travels 100 miles in 2 hours ; find
the average velocity in feet per second. Ans. 73^.
inme that the unit
i body weighing //
ae proportion tbut
B in (3) of Art. 20
mita. If absolute,
i), and the unit of
Hon, the units are
is g pounds (Art.
A.rt 16).
are is, it enables ns
shes us with a con-
same quantity of
ted by 30 on the
at the pole or on
static measure of
the weight of the
gravity (Art. 24),
i) of Art. 24], and
of P, the unit of
unds, and / the
in the mass.
oints which move
a second, and the
ns. As 2 is to 1.
s which move uni-
iite, and the other
econd.
ns. As 6 is to 7.
in 2 hours ; find
Ans. 73f
ItXAMPLBS.
19
4. One point moves nnjfonnly round the circumference
of a circle, while another point moves uniformly along
the diameter ; compare th^ir velocities.
Ahs. As tr is to 1.
5. Supposing the earth to be a sphere 26000 miles in
circumference, and turning round once in a day, deter-
mine the velocity of a point at the equator.
Ans. 1527| ft. per sec.
6. A body has described 60 feet from rest in 2 second^
with uniform acceleration ; find the velocity acquired.
From (1) of Art. 9 we have
and from (4) we have /< = v ;
.'. V = 60.
7. Find the time it will take the body in the last exam-
ple to move over the next 160 feet.
From (6) of Art. 9 we have
etc
Ans. 2 seconds.
8. A body, moving with uniform acceleration, describes
63 feet in the fourth second ; find the acceleration.
Ans. 18.
9. A body, with uniform acceleration, described 72 feet
while its velocity increases irom 16 to 20 feet per second ;
find the whole time of motion, and the acceleration.
Ans. 20 seconds ; 1.
10. A body, in passing over 9 feet with uniform accelera-
tion, has its velocity increased from 4 to 5 fe *: per second ;
find the whole space described from rcsut, and thie accelera-
tion. Ans. 26 feet ; |.
20
MXAMPLMa.
11. A body, aniformly accelerated, is found to be mov-
ing at the end of 10 seconds with a velocity which, if
continued uniformly, would caiJ7 it through 46 mites in
the next hour ; find the acceleration. Am. ^.
12. Find the matis of a straight wire or rod, the d^naity
of which varies directly as the distance fh>m one end.
Take the end of the rod aa origin ; let o = its length ;
let the ''aetance of ary point of it from that end = a; ;
and let u = the area of its transverse section, and k = the
density at the uxut'g distance from the origin. Thea
dV=i<adx', and p=zhx',
and (2) of Art. 11 becomes
M
2
13. J?iiwl aie ma«i of a circular plate of uniform thick-
ness, the density of whiph varies as the distance from the
centre.
Am. |7r*Aa«, where a is the radius, k the density at
the unit's distance, and h the thickness.
14. Find the mass of a sphere, whose density varies
inversely as the distance flrom the centre.
An3. 2jrp«», where p is thje density of the onteide strfttooik
ind to be mov-
locity which, if
igh 45 mileg ia
rod, the d^wity
1 ooe end.
( = its length ;
that end = x ;
a, and k = the
n. Thea
uniform thick-
tanoe from tho
the density at
density y^es
•ntside stratoo^
STATICS (REST)
CHAPTER U.
THE COMPOSITION AND RESOLUTION OF CONCUR.
RING FORCES-CONDITIONS OF EgUILIBRIUM.
26. Probimp of Staftios. — The primary conception of
force is that of a caoae of motimi (Art. 14). If only one
force acts on a particle it is clear that the particle cannot
remain at rest. In statics it is only the tendmey which
for ,$8 have to prodaoe motion that is considered. There
must be at least two forces in statics ; and they Me con-
sidered as acting so as to counteract each other^t tefide^oy
to cause motion, thereby producing a state of equilibriara
in the bodies to which they are applied. The ftjrces which
act upon a body may be in equilibrium, and yet motion
exist; but ia such cases the motion is uniform. Hence
there are two kinds of equilibrium, the one relating to
bodies at rest, the other relating to bodies in motion. The
former is sometimes called Static Equilibriam and the lat-
ter Kinetic (or Dynamic*) Equilibrium. 3%e problem of
atatica ia to detei^ine the conditiona under vohick foreea act
when they keep bodiea at rest.
27. Cdncnrring and Coiwpiring Forces.— Resnlt-
ant.— When sereral forces have a common ryoint of applir
cation they are called concurring forces ; whan they act ivt
the same point and along the same right line they are
called contpirinff forces.
The resultant of two or more forces is that force which
singly wiU produce the same effect as the forces them-
selves when acting together. The individual foroec^ when
considere d with reference to this resultant, are called
" — ■ "" ™^™^^— — ^W*^™^— ^Wl ■ " ■ ■■- I I 11 I 11, MHi ■■II. PM IIHI I H" l
* Oregoiy's XecbMilG*, p. 11
m
coMPoamoN of conspibinq porcbs.
components. The process of finding the resultant of several
forces is called the composition of forces.
28. Composition of Conapiring Forces.— Condi-
tion of Equilibrium. — When two or more conspiring
forces act in the same direction, it h evident that the
resaltant force is equal to their sum, and acts m the same
direction.
When two conspiring forces act in opposite directions
their resultant force is equal to their difference, and acts in
the direction of the greater component
When several conspiring forces act in different directions
the resultant of the forces acting in one direction equals
the sum of these forces, and acts in the same direction ;
and so of the forces acting in the opposite direction.
Therefore, the resultant of all the forces is equal to the
differepce of these sums, and acts in the direction of the
greater sum. Hence, if the forces acting in one direction
are reckoned positive, and those in the opposite direction
negative, their resultant is equal to their algebraic sum ;
its sign determining the direction in which it acts. Thus,
if Pit -Pg, 'Pi, etc., are the conspiring forces, some of
which may be positive and the others negative, and R is
the resultant, we have
/? = P, + P, + P, + eta = £P,
(1)
in which S denotes the algebraic sum of the terms similar
to that written immediately after it«
CoE.— The condition that the forces may be in equilib-
rium is that their resultant, and therefore their algebraic
sum, must vanish. Hence, when the forces are in equilib-
rium we must have i? = ; therefore (l) becomes
■Pi + ^1 + Pi + etc = £P = 0.
(»)
it of several
I.— Condi-
conspiring
tt that the
I the same
directions
ind acts in
directions
bion equals
direction ;
direction,
nal to the
ion of the
9 direction
i direction
raic sum;
ts. Thns,
, some of
and Ji is
(1)
as similar
tt ©qnilib-
algebraie
D eqoilib-
(2)
COMPOSITION OF VSL0CITIS8. 88
t
29. Composition of VelocltieB.— // » particle be
moving with two uniform velocities represented in
magnitude and direction by the two adjacent sides
of a parallelogram, the resultant velo' Uy will be
represented in majnitudi and direction by the
diagonal of the parallelogram.
Let the particle move with a uniform
velocity t', which acting alor:3 will take
it in one i :cond from A to B, and with
a uniform velocity v', which acting
alone will take it in one second from A
to C ; at the end of one second the par-
ticle will be found at D, and AD will represent in magni-
tude and direction the resultant of the velocities represented
by AB and AC.
Suppose the particle to r^oro uniformly along a straight
tube which starts from AB, and moves uniformly parallel
to itself with its extremity in AC. When the particle stary
from A the tube is in the position AB. When the particle
has moved over any part of AB, the end of the tube has
moved oyer the same part of AC, and the particle is on the
line AD. For example, let AM be the - th part of AB, and
AN be the -th part of AC ; while the particle moves from
n
A to M, the end A with the tube AB will move from A to
N, and the particle will be at P, the tube occupying the
position K L, and PM being parallel and equal to AN. F
can be proved to be on the diagonal AD as follows :
AM : MP : : —
AB AC
n
AB : AC (= BD);
therefore P lies on the diagonal AD. Also since
AM : AB : : AP : AD,
fc » W * W.p i»«« i*^ »"*'^*' ■
84
coMPoamoH ow mnomiL
the resultsnt velocity is unifonn. Hence, the diagonal AD
represents iu magnitude and diniotioa the reaaltunt of the
Telocities represented by AB and AC.
This proposition is known as th0 Parallelogram of
Vehciiies.
30. Compositloii of Forces.— From the Parallelo-
gram of Velocities the Parallelogram of Forces follows
immediately. Since two simultaneous velocities, AB and
AC, of a particle, result in a single velocity, AD, nnd since
these three velocities may bo regarded as the measures of
three separate forces all acting for the same time (Art. 10),
it follows that the cifect produced on a particle by the com-
bined action, for the same time, of two forces may be pro-
duced by the action, for the same time, of a single force,
which is therefore called the resultant of the other two
forces ; and these forces ore represented in magnitude and
direction by AB, AC, and AD. (See Minchin, p. 7, also
GametfB Dynamics, p. 10.)
Hence if two concurring farces be represmted in magni-
tude and direction by the adjacent sides of a parallelogram,
their resultant mil be represented in mag7iitude and direction
by the diagonal of the paraiielogram. Care must be token
in oonstructiug the parallelogram of forces that the com-
ponents both aot from the angle of the parallelogram iVom
which the diagonal is drawn.
Thii propoaitioB haa bsen proved in vmtIoim wkju. It wm ennn-
dated !a :t« pnweDt form by Sir laMO Nomumi, and bjr Varisuon, the
oolebnted mathematiciHn, In tlie jmu.* 1067, prcbablj independent of
each other. Since that time vartoiu proofe of it have been given by
difibrent mathematicians. One wcdtk gives a dlacaiBion, mora or leoa
complete, of 4lt other proofs. A noted analytic proof is given by
M. PnisBon. (See Prine'e CaL, Vol. Ill, p. 19). Some authors object
to proving the parallelogram of forces by means of the parallelogram
of velocities. (See Gregory's Meehanlos, p. 14) The student who
wants other proofs is referred to Dochayla's proof ss found in Tod-
hunter's Statics, p. 7, and in Ualbraitb's MeohsAios, p. 7, and in many
) diagonal AD
saltuDt of the
iHehgram of
tie Paralleh-
'oreea follows
ties, AB and
tD, and since
measures of
ne (Art. 10),
by the corn-
may be pro-
single force,
e other two
gnituda and
3> p. 7, also
tl in magni-
ralUhgrmn,
'nd direction
«st be taken
»t the com-
ogram fcom
It WM «nun-
t'^ariguon, the
dependent of
Ben given by
, more or less
1» given by
ithon object
•nllelogram
student who
>and in Tod-
ftnd in niaoy
TMiANOhE or woacsa.
other works ; or to UpUoe-s pnxrf, (Sm M^canicine Celeste, Liv. I,
chap. 1.)
If fl be the angle between the sides of the parallelogram,
AJi and AC (Fig. 2), and P and Q represent the two com-
ponent forces acting at A, and R represent the resultant,
AD, we have from trigonometry,
iZ» = i« -f. ^ -H ^PQ cos 9
m
an equation which gives the magnitude of the resultant of
two forf es in terms of the magnitudes of the two forces and
the angle between their directions, the forces being repre-
sented by two lines, both dra\ni from the point at which
they act.
Cob. —If e = 90^ and « and be the angles which the
direction of R makes with the directions of P and Q, we
have from (1)
C0B« =
COS/3 = ^;
(»)
(K)
from which the magnitude and direction of the resultant
are determined.
31. Triangte of Tonm.—If three oonourring
forces be repreaenied in magnitude and direoiicn
hij the sides of a triangle, taken in order, they will
be in equilibrium.
Let ABO be the triangle who«"
sides, taken in order, represent in
magnitude and direction three foroes
applied at the point A. Complete
%
n*>s
S6
TBlAJiGLM OF FOBCXS.
the pawllelogram ABCD. Then the forces, AB and EC,
applied at A, are expressed by AB and AD (since AD is
e<iual and parallel to BC). But the resultant of AB and
AD is AC, acting in the direction AC. Therefore the three
forces represented by AB, BC, and CA, are equivalent to
two forces, AC and CA, the fomer acting from A towards
C and the latter from C towards A, which, being equal and
opposite, will clearly balance each other. Therefore the
three forces represented by AB, BC, and CA, acting at the
point A, will be in equilibrium.
It should be observed that though BC represents the
magnitude and direction of the component, it is not in the
Une of its action, becanse the three forces act at the
♦ ut A.
The converse of this is also true ; viz.. If three concurrinf^
forces are in equilibrium, they may be represent/nl in m!;g-
nitude and direction by the sides of a triangle, drawn
parallel respectively to the directions of the forces.
Thus, if AB and BC represent two forces in magnitude
and direction, AC will represent the resultant, and henre to
produce equilibrium tbe resultant force AC must be opposed
by an equal and oppoi ite force CA. Therefore, the three
forces in equilibrium will be represented by AB, BC, and
CA.
CoE. — When three concurring forces are in equilibrium,
each is equal and directly opposite to the resultant of the
other tw', •
it. 1«>J:^«< .ns iMtwosn ThrM Oonewring ForoM
in Tic i^■ vm.— Siuce the sides of a plane triangle are
as the Biuo' >/ iie opposite angles, wo have (Fig. 8)
AB : BC (or AD) : AC : : sin ACB : sin BAC : sin ABC
: : sin DAC : sin BAC : sin BAD.
Hence, calling P, Q, and R, the forces represented by AB,
AD, and AC, and denoting the angles between the direo-
lu.
mm
!8,ABand BC,
D (since AD is
tant of AB and
irefore the three
e eqaivaleut to
Tom A towards
being equal and
Therefore t!ie
i, acting at the
represents the
it is not in the
ces act at the
hree concnrrin/-j
sent/nl in m^g-
triangle, drawn
forces.
8 in magnitude
it, and henre to
nust be opposed
(fore, the three
)y AB, BO, and
in equilibrium,
reeuluuit of the
irrinf ForoM
me triangle are
(Fig. 8)
AC:
AC
sin ABO
sin BAD.
aseuted by AB,
woen the direo-
POLTQON OP FORCSa.
tions of the forces P and Q, Q and R, and R and P, by
AAA
PQ> QP> and RP, respectively, we have
P ^ _Q R
. /^ /\ A
em QR sin RP sin PQ
0)
There/ore, when three concurring forces are in equilibrium
they are respectively in the same proportion as the sines of
the angles included between the directions of the other two.
33. The Polygon of Forces.— // amj number of
concurring forces be represented in magnitude and.
direction by the sides of a closed polygon taken in
o: der, they will be in equilibriurrv.
Let the forces be represented in It
magnitude and direction by the lines
AP„ AP„ AP„' AP„ AP,. Take
AB to represent AP,, through B draw
BC equal and parallel to AP, ; the
resultant of the forces AB and BC, or
AP, and AP, is represented bj AO
(Art. 31). Of courae the resultant
acts at A and is parallel to BC. Again through draw CD
equal and parallel to AP„ the resultant of AC and CD, or
AP,, AP„ and AP, is AD. Also through D draw DE
equal and parallel to AP^, the resultant of AD and DE, or
AP,, APg, APj, and AP^ is AE. Now if AE is equal and
opposite to AP, the system is in equilibrium (Art. 18).
Hence the fos-ces represented by AB, BC, CD, DE, EA
wiU be in equilibrium.
Cob, 1. — Any one side of the polygon represents in
magnitude and direction the resultant of all *he forces
represented by the remaining sides.
Cob. 2.— If the lines representing the forces do not form
a closed polygon the forces are not in eqoilibrium ; in this
»8
PARALLMI'OPIPED OP P0K0S8.
case the last side, AE, taken from A to E, or that which is
required to clope up the polygon, represents in magnitude
and direction the resultant of the system.
34. Farallelopijied of Porcea.— // three concur-
ring forces, not ih. the sattve plane, are represented
in magnitude and direction by the three edges of
a parallelopiped, then the resultant will be repre-
sented in magnitude and direction by the diag-
onal; conversely, if the diagonal of a parallel-
opipcd represe.its a force, it is equivalent to three
forces represented by the edges of the parallci-
opiped.
Let the three edges AB, AC, AD of the
parallolopiped represent the three forces,
applied at A. Then the resultant of the
forces AB and AC is AE, the diagonal of
the face ABCE; and the resultant of the
for'^os 4^ and AD is AF, the diagonal of
the parallelogram ADFE. Hence AF represents the
resultant of the three forces AB, AC, and AD.
Conversely, the force, AF, is equivalent to the three
components AB, AC, and AD.
Lot F, Q, S represent the three forces AB, AC, AD ; R,
the resultant ; a, P, y, the angles whicL the direction of B
makes with the directions of P, Q, S, and suppose the
forces to act at right angles with each other. Then aiuce
Sr» = XB* + AC» + AD',
we have R^ = P* + (? + 3*;
, P
alio, cos o = -gi
Q
cos /3 = ^,
8
(1)
m
ta,
\T that which is
I in magnitude
Ihree concur-
s represented
tree edges of
vill be repre-
by the diag-
n parullel-
Zent to three
'ihe parallci-
represents the
D.
; to the three
J, AC, AD; R,
direction of R
kd suppose the
Then aiuce
(1)
(a)
KXAMPtiiS.
»»
from which the magnitude and direction of the resultant
are determined.
EXAMPLES.
1. Three forces of 5 lbs., 3 lbs., and 2 lbs., respectively,
act upon a point in the same direction, and two other forces
of 8 lbs. and 9 lbs. act in the opposite direction. What
single force will keep the point at rest P Atts. 7'lbs.
a. Two forces of 5| lbs. and 3| lbs., applied at a point,
urge it in one dire^'^ion ; and a force of 2 lbs., applied at
the same point, urges it in the opposite direction. What
additional force is necessary to preserve equilibrium ?
Ans. t lbs.
3. If a force of 13 lbs. be represented by a line of 6^
inches, what line nWX represent a force of 7^ lbs.?
Am. 8f inches.
4. Two forces whose magnitudes are as 3 to 4, acting on
a point at right angles to each other, produce a rcuultaut of
20 lbs.; required the component forces.
Am. 12 lbs. and 16 lbs.
5. Let ABO be a triangle, and D the middle point of
the side BC. If the three forces represented in magnitude
and direction by AB, AO, and AD, act upon the point A ;
And the direction and magnitude of the rennltant
Ans. The direction is in the line AD, and the magni-
tude is represented by SAD.
6. When P = Q and fi = 60°, find R.
Ana. R = PVS.
7. When P = Q and 8 = 135°, find R.
Ans. R- P^'z — V^
8. When P = ^ and = 120°, find R.
Ans. R = P.
30
sssoLurrox or roscss.
\
'■
9. If P =: Q, show that their resultant R — ^P cos ?•
10. If P = 8, and e = 10, and (? = 60°, find R.
Ana. R z=2 V^l.
'll. If P = 144, R = 145, and 6 = 90°, find Q.
Ans. Q = 17.
12. Two forces of 4 lbs. and 3 V2 lbs. act at an angle of
45°, and a third force of V42 lbs. acts at right angles to
their plane at the same point ; find their resultant.
Ahs. 10 lbs.
35. Resolntion of Forces.— % the resohition of forces
is meant the process of finding tlie components of given forces.
We have seen (Art 30) that two concurring forces, P and
C = AB and AC, (Pig. 2) are equivalent to a single force
72 = AD ; it is evident then that the single force, R, ucnng
along AD, can be replaced by the two forces, P and Q,
represented in magnitude and direction by two adjacent
sides of a parallelogram, of which AD is the diagonal.
Since an infinite nnmber of pamllelogrnnis, of oac'i of
which AD is the diagonal, can be constructed, it follows
that a single force, R, can be resolved into two other forces
iu an infinite number of ways.
Also, efl 3h of i-ie forces AB, AC, may be resolved into
two others in way similar to that by which ID was
resolved .nto twt ; and so on to any extent. Hence, a single
force nray be resolved into any number of forces, whose
oombinod action is equivalent to the original force.
Cor. — The most convenient compo-
nent into which a force can be resolved
are those whose directions are at right
angles to eacli other. Thus, let OX
and OF be any two lines at right
ujiglea to each other, and P any force acting at in the
Fia.a
i = 2P cos J
9°, find R.
E — 2 V'ai.
\ find Q.
ns. Q = 17.
at an angle of
right angles to
111 taut.
Atis. 10 lbs.
)hition of forces
of given forces,
l forces, P and
a single force
force, it, acting
rcos, P and Q,
' two adjacent
diagonal,
ms, of oac'i of
:ted, it follows
wo other forces
e resolved into
vhich AD was
Hence, a single
forces, whose
force.
a
ig at in the
MAomrmiS and DinEcrtoir or bssultant.
plane XOY. Then completing the rectangle OMPN we
find the ocnponenta of P along the axes OX and OF to be
Oif and ON, which denote by X and Y. Then we have
clearly
X = P cos «, ) /jv
r = P sin «; ) ^ '
where a is the angle which the direction of P makes with
OX. These wmponents X and Y are called the rect-
angular components. The i-ectangular component of a
force, P, along a right lino is Px cosine of angle between^
line and direction of P.
In strictness, when we speak of the component of a given
force along a certain line, it is necessary to mention the
other line along which the other component acts. In thia
work, unless otherwise expressed, the component of a force
along any line will be understood to-be its rectangular
component; le., the resolution will be made along this line
and the line perpendicular to it.
36. To find th« lAagnitode and Direction of the
Resultant of any number of Concurring Forces in
one Plane.— When there are several concurring forces, the
condition of their equilibrium may be expressed as in
Art. 33, Cors. 1 and 2. But in practice we obtain much
simpler results by using the principle of the Resolutwn of
Forces (Art -35), than those given by the principle of
Composition of Forces.
Let be the point at which all
the forces act. Through draw the
rectangular axes XX', YY'. I^t
P,, P„ P,, etc., be the forces and
«,, «„ a J, etc., be the angles which
their directions make with the axis
of*. •
Now resolve each force into its two
components along the axes of x and
Fig.7
Then the com-
88 MAomruDJi avd otRscnoN of sxauLTAjrr.
f
L
ponenta along the axia of x (aHwmponents) ore (Art
35, Cor.), P, COS a,, P, cos a,, P^ cos «,, etc., and thoae
along the axis of y are Pj sin ct|, P, sin «„ P, ain «„
etc. ; and therefore if X and Y denote the algebraic sum of
the ^-components and ^-components respectively, we have
X= P, cos cr^+P, cos «,4-P, cosa, 4-etc. ) .
= 2P coa «, ) ^ '
y = P, sin cci + Pi sin a, + Pj ain a, +etc.
=r iP sin a.
(2)
Let R be the resultant of all the forces acting at 0, and
the angle which it makes with the axis of x ; then resolving
R into its x- and ^-components, we have
i2 COS e = -T = £P coa o,
5 si- 3 3= y =s £P ain a,
'}
/? = X»+r»; tane = -=.
(8)
which determines the magnitude and direction of the
resultant.
ScH. — Regarding OX and OF as positive and OX^ and
OY^ as negative as in Anal. Geom., we see that Oa;,, Oy^,
Oy, are positive, and Oa;,, Oa;„ Oy, are negative. The
forces may always be considered as positive, and hence the
signs of the components in (1) and (2) will be the same as
those of the trigonometric functions. Thus, since a, is
> 90° and < 180° its sine is positive and cosine is negative;
since a, is > 180° and < 270° both its sine and cosine are
negative.
37. Tha Conditions of Eqnlllbrtam for any nnmbor
of ConcTuring ForcM in one Plane.— For the equilibrinm
of the forces we must have R = 0. Hence (4) of Art. 36
becomes
X» -H r» = 0. (I)
Urn
VLVAHT,
s) ore (Art
3tc., and thoae
Kg, P, ain «„
robraic sum of
ely, we have
4- eta)
(1)
+eto.
} <^\
g at 0, and 9
then resolving
(3)
■', (4)
lOtion of the
and OX^ and
hat Oa;,, Oy„
jgative. The
md hence the
) the same as
B, since a, is
ae is negative;
nd cosine are
•nynnmbcr
tie eqnilibrinm
4) of Art 36
(1)
BXAMPLMB.
88
Now (1) cannot be satigfled so long 'as X and F are real
quantities unleae X=:0, F.=:a; therefore,
X=:£i'co8« = and F=SPsino = 0. (2)
Hence these are the two necessary and sufficient conditions
for the equilibrium of the forces; that is, the algebraic sum
of the rectangular components of the forces, along each of
two right lines at right angles to each other, in the plane of
the forces, is eqwA to zero. As the conditions of equilibrium
must be independent of the system of co-ordinate axes, it
follows that, if any number of concurring forces in one
plane are in equilibrium, the algebraic sum of the rectan-
gular components of the forces along every right line in their
plane is zero*
EXAMPLES.
1. Given four equal concurring forces whose directions
are inclined to the axis of z at angles of 16°, 76°, 136%
and 225° ; determine the magnitude and direction of their
resultant
Let each force be equal * > P ; then
X = P ooa 16° + P coe 76 + P cos 136° + P cos 225°
.3* -a
= p:
2*
F = P sin 16° + P sin 76° + P sin 136° + P sin 226"
.'. 7? = P(6-2V^)*-
3*
tan ds=
8
Err
2
2. Giren two equal concurring forces, P, whose direc-
f ions are inclined to the axis of x at angles of 30° and 316°;
tlnd their resultant Ans. B = 1,69 P.
2*
i piW saa'Wiywwi
34
CONCtmRIlfO FOROSa.
i.
3. Given three concurring forces ol 4, 6, and 6 lbs.,
whose directions are inclined to the axis of « at angles of
0°, 60°, and 136° respectively ; fi nd their resnltant.
An$. R= V97 + 16 V 6 - 39 V2.
4. Given three equal concurring forces, P, whose direc-
tions are inclined to the axis of x at angles of 30°, 60°, and
165° ; find their resultant.
Ans. i2 = 1.67 P.
5, Given three concurring forces, 100, 50, and 200 lbs.,
whose directions are inclined to the axis of x at angles of
0° 60° and 180°; find the magnitude and direction of
their resultant Ans. R = 86.6 lbs. ; = 150°.
38. To find the Magnitnde and Direction of the
Resultant of any number of Concnrrlng Forces in
Bpace.-Let P^, P„ P„ etc., be the forces, and the
whole be referred to a system of rectangular co-ordmates.
Let a„ /3t, yj, be the angles which the direction of Pj
makes with three rectangular axes drawn through the point
of appUcation ; let «„ a„ r„ be the angles which the direc-
tion of P, makes with the same axes; aj, Pi, Ts. tne
angles which P, makes with the same axes, etc. Resolve
these forces along the co-ordinate axes (Art. 34) ; the com-
ponents of P, along the axes are Pj cos o„ Pj cos 0,, P,
cos y . Resolve each of the other forces in the same way,
and let X, Y, Z, be the algebraic snms of the components
of the forces along the axes of x, y, and z, respectively ;
then we have
X = P, cos «i + P, cos «, + Pj cos «, -H etc.)
= SP COS a.
Y=P. cos /3j + P, COS /3, + Pj COS 0, + etc.( ^^^
= IP cos |3.
Z = Pi COS yi + Pg cos y, + P» cos y, + etc.]
= SP cos y.
6, and 6 Iba,
jf x at angles of
Bgultant.
V 6 — 39 V2.
P, whose direc-
of 30°, 60°, and
B = 1.C7 P.
50, and 200 lbs.,
f a; at angles of
md direction of
M.;e = 150°.
Irection of the
ring Forces in
forces, and the
liar co-ordinates.
direction of Pi
through the point
s which the direc-
Kj, Ps, Ts. *!»«
ses, etc. Resolve
rt. 34) ; the com-
!„ Pi COS0,, P,
in the same way,
! the components
a z, respectively ;
BOS «j + etc.)
cos 0, + etc.^ ^»
cosy, + etc.]
CONDITIOira OF EQlflUBRIUJl.
3S
Let i2 be the resultant of all the forces; and let the
angles which its direction makes with the three axes be a,
b, c ; then as the resolved parts of B along the three co-or-
dinate axes are equal to the sum of the resolved parts of
the several components along the same axes, we have
^ cos a = X, iJ COB ft = r, i2 cos c -- Z. (2)
Squaring, and adding, we get
B? =z X^ + m -^ Z*',
X Y
cos fl = -p, cos ft = -^,
cos c = -p;
(3)
('4
which determines the magnitude of the resultant of any
system of forces in space and the angles its direction makes
with three rectangular axes.
39. The Conditions of Equilibrium for any num-
ber of Concnrring Forces in Space.— If the forces are
in equilibrium, ^ = ; therefore (3) of Art. 38 becomes
2'» + r«+z» = o.
But as every square is essentially positive, this cannot be
unless X = 0, F = 0, Z = ; and therefore
2:Pco8a = 0, IP 008/3 = 0, SPco8y = 0; (1)
and these are the conditions among the forces that they
may be in equilibrium ; that is, the sum of the components
of the forces along each of the three co-ordinate axes is
equal to zero.
40. Tension of a String. — By the iension of a string
is meant the pull along its fibres which, at any point, tends
to stretch or break the string. In the application of the
preceding principles the string or cord is often used as a
■ 'swviStWMB^.'flKs-wtwsair!'.' ' ■
36
BXAMPLKa.
moiuis of oommnnioating force. A string is said to be per-
fectly flexible when any force, however small, which is
applied otherwise than along the direction of the string,
will change its form. In this work the string will be
regarded as perfectly flexible, inextensible, and withont
weight
If such a string be kept in eqnilibrinm by two forces,
one at each end, it is clear that these forces must be equal
and act in opposite directions, so that the string assumes
the form of a straight line in the direction of the forces.
In this case the tension of the string is the same throngh*
out, and is measured by the force applied at one end ; and
if it passes over a smooth peg, or over any number of
smooth surfaces, its tension is the same at all of its points.
If the string should be knotted at any of its points to other
strings, we must regard its continuity as broken, and the
tension, in this case, will not be the same in the two por-
tions which stext from the knot
BXAMPLES.
1. A and^ B (Fig. 8) are two fixed
points in a horizontal line ; at A is
fastened a string of length h, with a
smooth ring at its other extremity, 0,
through which passes another string with
one end fastened at B, the other end of
which is attached to a given weight W ;
determine the position of C.
Before setting about the solution of statical problems of
this kind, the student will clear the ground before him, and
greatly simplify his labor by asking himself the following
questions : (1) What lines are tbiue in the flgore whose
lengths are already given P (2) What forces are there
whose magnitudes are already given, and what are the
forces whose magnitudes are yet unknown? (3) What
it is required to
mid to be per-
mall, which ia
I of the string,
string will be
, and without
by two forces,
must be equal
string assumes
of the forces.
same through*
one end ; and
my number of
1 of its points,
points to other
roken, and the
in the two por-
is required to
al problems of
i)efore him, and
the following
le figure whose
irces are there
what are the
nP (3) What
mXAMPLKS.
87
variable lines or angles in the figore would, if they were
known, determine the required position of P
Now in this problem, (1) the linear magnitudes which
are given are the lines AB and AC. (2) The forces acting
at the point C to keep it at rest are the weight W, a ten-
sion in the string CB, and another tension in the string
CA. Of these W is given, and so is the tension in
OB, which must also be equal to W, since the ring is
smooth and the tension therefore of WCB is the sanio
throughout and of course equal to W. But as yet there is
nothing determined about the magnitude of the tension in
CA. And (3) the angle of inclination of tbo string OA tc
the horizon would, if known, at once determine the posi-
tion of 0. For if this angle is known, we can draw AC of
the given length ; then joining to B, the position of the
system is completrly known.
Let AB = fl, AC = b, CAB = e, CBA = ^, and the
tension of the string AC = T. Then, for the equilibrium
of the point C under the action of the three forces, W, W,
and T, we apply (2) of Art. 37, and resolve the forces
horizontally and vertically j and equate those acting towards
the right-hand to those acting towards the left ; and those
acting upwards to those acting downwards. Then the
horizontal and vertical forces are respectively
Trsin0 + TsinS = W.
Eliminating T we have
cos d = sin (9 + 0) ;
.-. 25 + = 90°.
(1)
Also, from trigonometry we have
nn (0 -H 0) _ a .
sin </> d '
(2)
r
38
EXAMPLES.
T\%M
from (1) and (3) (? and may be fonn-^ ; and therefore T
may be found; and thus all the circumstances of the
problem are determined.
2. One end of a string is attached to
a fixed point, A, (Fig. 9) ; the string, after
passing o?er a smooth peg; B, sustains a
given weight, P, at its other extremity,
and to a given point, C, in the string is
knotted a given weight, W. Find the posi-
tion of equilibrium.
The entire length of the string, ACBP, is of no conse-
quence, since it is clear that, once equilibrium is estab-
lished, P might be suspended from a point at any distance
whatever from B. The forces acting at the point, 0, are
the given weight, W, the tension in the string, CB, which,
since the peg is smooth, is /*, and the tension in the string
OA, which is unknown.
Let AB = a, AC = 4, CAB = », CBA = ^, and the
tension of the string, AC = T. Then for the equilibrium
of the point Cj we have (Art. 32),
(1)
P
cosfl
>r~
sin (e -1- 0) '
also,
from the geometry of the figure, we ]
ft sin (0 + 0) = a sin 0.
From(l)
and (2) we get
P
dcc«( Q
W~
a sin^'
or
rin^ =
hW .
-p cos tf;
(«)
COS0 =
Vo»/'»-ft»Fr» coB« 9
___
\
c I I I i l i f I jj IWBW
id therefore T
Btances of the
ve
P*
■■*>*^p*
■Hfii
is of no conse-
briam is estab-
at any distance
be point, C, ore
Qg, CB, which,
n in the string
. = ip, and the ^
the equilibrium
(1)
(3)
>B»»
MXAMPLIJS, 99
Expanding sin (fl + ^) in (2), and substitating in it these
values of sin ^ and cos 0, and reducing, we have the
equation
^ =0,
^,,S!!f^^^±D^»,
2W*b
from which 6 may be found. (See Minchin's Statics,
p. 29.)
3. If, in the last example, the weight, W, instead of
being knotted to the string at C, is suspended from a
smooth ring which is at liberty to slide along the string,
AOB, find the position of equilibrium.
Atu. sin = ^.
41. BqniUbrinm of Coneiurrliig Forces on a
Smooth Piano. — If a particle be kept at rest on a smooth
surface, plane or curved, by the action of any number of
forces applied to it, the resultant of these forces must be in
the direction of the normal to the surface at the point
where the particle is situated, and must be equivalent to
the pressure which the surface sustains. For, if the
resultant had any other direction it could be resolved into
two components, one in the direction of the normal and the
other in the direction of a tangent ; the first of these would
be opposed by the reaction of the surface ; the second being
unopposed, would cause the particle to move. Hence, we
may dispense with the t»lane altogether, and regard its
normal reaction as one of the forces by which the particle
is kept at rest. Therefore if the particle on which the
statical forces act b« on a smooth plane surface, the case is
the same aa that treated in Art. 39, viz., equilibrium of a
particle acted upon by any number of forces ; and in vnnt-
ing down the equations of equilibrium, wo merely have to
include the normal i action of the plane among all the
others.
p^
40
JtXAMPLSa.
BXAMPLBS.
1. A heavy particle is placed on a
smooth inclined plan», AB, (Pig. 10),
and is sustained by a force, P, whidi
acts along AB in the vortical plane
which is at right angles to AB ; find
P, and also the pressure on the in-
clined plane.
The only eflfect of the inclined plane is to produce a
normal reaction, R, on the particle. Hence if we intro-
duce this force, we may imagine the plane removed.
Let IT be the weight of the particle, and « the inclina-
tion of the plane to the horizon.
Resolving the forces along, and perpendicular to AB,
since the lines along which forces may be resolved are
arbitrary (Art 37), we have guooeflsively,
P— ITsittft SE 0, or PzszWfAna',
and
R — W'cos « = 0, or R = IT cos a.
If, for example, the weight of the particle is 4 or, and
the in hnation of the plane 30°, there will be a normal
pressure of 2^3 oz. on the plane, and the force, P, wUl
be 2 oz.
a. In the previous example, if P act horizontally, find
its magnitude, and also that of R.
Resolving along AB and perpendicular to it, we have
Bucccsdiveiy,
P cos a— IT sin a = 0, or PssTTtana;
\^
and PBin«+ F oo«« — i? = 0, .•. J? =
W
oobm*
8 to produce a
ce if we intro>
amoved.
a the inclina-
licnlar to AB,
>e resolved are
)OS a.
le ia 4 oz., and
I be a normal
> force, P, will
risontally, find
o it, we have
Ftan «;
V^
coimmoifa op gqpuMnnm.
U
3. If the particle is auatained by a ft)roe, P, mining a
given angle, 6, with the inclined plane, find the mi^itade
of this force, and of the pressure on the plane, ail the forces
acting in the same vertical plane.
Besolving along and perpendicular to the jdaae succes-
sively, we have
PcobO— Wnna = 0,
and JJ + P sine— ff COB « = 0,
from which we obtain
P^W^^,; R
008 $'
_- |^r Coe(« + ^)
COS0
Rem.— The advantage of a judicious selection of direc-
tions for the resolution of the forces is evident. By resolv-
ing at right angles to one of the unknown forces, we
obtain an equation free fh)m that force; whereas if the
directions are selected at random, all of the forces will
enter each equation, which will make the solution less
simple.
The student will observe that these values of P and B
cculd have been obtained at once, without resolution, by
Art. 32.
42. Conditiosui oi SqidUbiiaiii for may number of
Ooncnrring ForoM whoa the perticle on which they
act ia Oeaatrained to Remain on a Qiven Smooth
Bvrfiwe.— If a particle be kept at rest on a smooth sur-
face by the action of any number of forces applied to it,
the resultant of these forces must be in the direction of the
normal to the surface at the point where the particle is
situated, and must be equivalent to the pressure which the
surface sustains (Art. 40). Hence since the resultant is in
the direction of the normal, and is destroyed by the roao-
iSt
CONDinOtfS OF MqmUBRIUM.
tion of the Brrface, we may regard this reaction as an
addildonal force directly opposed to the normal force.
Let N be the normal reaction of the surface, and a, 0, y,
the aiigles which JV makes with the co-ordinate axes of x,
y, and t, respectively. Let X, Y, Z, be the sum of the
comfoneuts of all the other forces resolved parallel to the
three axes respectively. The reaction JVmay i., considered
a ne^r force, which, with the other forces, keeps the parti-
cle im eqoilibriom. Therefore, resolving N parallel to the
three axes, we have (Art 39),
X+ JVcosa = 0,
F-f-JVcos/S
Z+ JVcosy
= 0. >
(1)
Leit u =/(«, y, «) = 0, be the equation of the given
Burfa<3e, and x, y, t the co-ordinates of the particle to
which the forces are applied. We have (Anal. Geom.,
Art. :176),
a'
cos a =
oosjS =
cosy =
y
Vo'« -I- »'» + I'i
1
(2)
where a' and V are the tangents of the angles which the
projections of the normal, N, on the co-ordinate planes xz
and yt make with the axis of tt. Since the normal is per-
pendicular to the plane tangent to the surface at (x, y, t),
t))o projections of the normal are perpendicular to the
traces of the plane. Therefore (AnaL Qeom., Art 37,
Cor. 1), W6 have
and
1 -I- aa' = 0,
1 + W = ;
(8)
(4)
mm/Hmfm^
UM.
s reaction as an
rmal force.
Pace, and a, 0, y,
dinate axes of z,
the sum of the
ed parallel to the
nay <., considered
keeps the parti-
¥ parallel to the
(1)
ion of the given
[ the particle to
e (Anal. Geom.,
(2)
angles which the
rdinate planes xz
he normal is per-
irfaco at {x, y, *),
endicular to the
Qeom., Art. 37,
(8)
(4)
in Thich
a
CONDrnOIfB OF EqUIUBRIUK.
I ' y
dx , daf ,_dy j.__^
(Calculus, Art. 66a.) Substituting in (3) and (4), we have
x.S-^ = o;
and
>-^|-f = <".
from which
du
d^
-§^= ? (OaLArt87) = a',
dx du ^
dt
(5)
du
,„a ^ = -^ = # = *'. <«)
M dy du
dM
Substituting these values of a' and 7/ in (2) and multiply-
du ,
ing both terms of the fraction by ^, we have
008a =
008/3 =
dm
tSi
3.
cosy =
(7)
I
f
M
44
CONDITTOm OP SQVIMBMIXrM.
\
which give the value of the direction cosines of the aonaai
at {x, y, z).
Putting the denominator equal to Q, tot shortness, and
fiubstitating in (1) and transposing, we have
^ N du
r — —— ~
Q' dy'
Z — —— ^
~ Q' dz
(8)
Pi
(10)
X
T Z
du
du du
dm
^ ST
Eliminating N between ihese three equations, we obtain
the two independent equations,
m
which express the conditions that must exist among the
applied forces and their directions in order tiiat their
resultant may be normal to the surface, t. e., that there may
be equilibrium. If these two equations are not satisfied,
equilibrium on the surface cannot exist Hence the point
on a given surface, at which a given particle under the
action of given forces will rest in equilibrium, is the point
at which equations (11) are satisfied.
Cob. 1.— Squaring equations (8), (9), (10) and adding, we
get
/rfM\> /rftt\» (du\
\dxf ]dy}_ \dz}
SW "^ <?» "^ "^J
jv= V Jr» + r» + z*,
IP;
m
■ I. w i wp i y *» i Mf i t n mM mKimtif
^
r.
' shortness, and
(8)
(9)
(10)
ions, we obtain
(11)
ist among the
ler tiiat their
that there may
i not satisfied,
mce the point
icle under the
u, is the point
and adding, ve
»» J
(12)
MXAMPLSa.
W
which is the yalae of the normal reeistanoe of the surface
and is precisely the same as the resultant of the acting
forces, as it clearly .ehoald be ; but this resistance must act
in the direction opposite to that of i^e resultant
Cob. 2.— Multiplying (8), (9), (10) by dx, dy, dz, respeo-
tirely, and adding, and ramembering that the total differ-
ential of « = is zero, we get
Zdx + Tdy + Zdx = 0,
(13)
which is an equation of condition for equilibrium. If (13)
cannot be satisfied at any point of the surface, equilibrium
is impossible.
OoR. 3. — If the forces all act in one plane, the surface
becomes a plane curve ; let this curve be iu the plane xy,
then z = 0; therefore (11) and (13) become
X
du
7_
du'
and
Zdx + Ydy = 0,
(U)
(15)
in which (14) or (IS) may be used according as the equation
of the curve is given as an implicit or explicit function.
EXAMPLES.
1. A particle is placed on the surface of an ellipsoid, and
is acted on by attracting forces which vuy directly as the
distance of the particle from the principal pliraes* of cec-
tion ; it is required to determine the position vf equilibrium.
Let the equation of the ellipsoid be
« =
= /rx,y.,; = ^; + g + ^-l:=0;
• nuM of cy, yti M).
l ipill W I lli Hjn Bl|l u| |l
^'If^fT'^W
\M
I
46 SXAMPLKS.
•'• dx~ a*' dy~ V dt ~ (^*
and let the a;-, y-, and c-components of the foroei be
respectively,
JT = — UiXy r = — t«,y, Z= —Ut»\
then (11) will give
Uid^ = «,{' = UfO*;
which may bo pnt in the form
»l^ _ «£ _ «|^ _ «i + «t 4- «» .
If these conditions are fulfilled, the particle will rest at all
points of the sarface.
3. Again, take the sam;) surfince, and let the forces vary
inversely as the distances of the point from the principal
planes; it is required to determine the position of eqnili-
briom.
Here X=-^, F=-^, Z=-?^;
therefore (11) becomes
3fi
u.,
«8
»1 Mi «8 «1 + «t + **»
by putting u for Ui + u, + tig,
1
It
- ^=«("# » = »(^f. ' = 'fe)*.
2t
J*
the foroee be
-«»*;
!»..
».-«
nrill rest at all
the forces vary
t the principal
ition of eqaili-
11'
_ >
1
-m*-
immm
EXAMPLSa.
47
which in (12) gives
=^ +
y*
= «[^-^^-^}
8. A particle ia placed ingide a smooth spboipe on the con-
cave surface, and is acted on by gravity and by a repulsive
force which varies inversely as the square of the distance
from the lowest point of the sphere; find the position of
equilibrium of the particle. ,
Let the lowest point of the sphere be taken for the origin
of co-ordinates, and let the axis of « be vertical, and posi-
tive upwards; then the equation of the sphere, whose
radius is a, is
Lot TT = the weight of the particle, and r = the distance
of it from the lowest point; then
r« = a^ + ^ + «' = 2o«.
Also, let the -epnlsive force at the unit's distance = « ;
then at the distuioe r it will be
— ^
X =
r =
Sox'
2as
u
w
-If.
ji.
i
mmmmm^i mmsBss
48
MXAMPLXa.
Let iV = the normal pressni-e of the curve ; then (8) and
(10) give
i.t-^+^'-^-o..
from which we have
t =
^a^W^'
whence the position of the particle is known for a given
weight, and for a given value of u. (See Price's Anal.
Mechanics, Vol. I, p. 39.)
4. Two weights, jP and C, are fastened to the ends of a
string, (Fig. 11), which passes over a pulley, 0; and ^
hangs Ireely when P rests on a plane curve, ^P, in a
vertical plane ; it is required to find the position of equili-
brium when the curve is given.
The forces which act on P are (1) the
tensiou of the string in the line OP, which
is equal to the weight of Q, (2) the weight
of P acting vertically downwards, (3) the
normal reaction of the curve R.
Let be the origin of oo-ordinates, and
the axis of x vertical and positive down-
wards. Let OM — X, MP = If, OP = r,
POM =e,OA=a. Then,
Flg.ll
X= P-Qfmd—R
dy
T^-Qmxd+R^',
then <8) and
n for a given
Price's Anal.
the ends of a
jr, 0; and Q
ve, AP, in a
ion of equili-
Fig.ll
MXAMPm.9,
therefore from (15) we have
(P — Q COB 6) dx — Q sin ddy = 0,
4«
or
But since
we have
a* + »• = f4,
zdx -\- ydy =. rdr\
.'. Pdz—Qdr = 0;
(1)
which is the oonditidn that most be aatisfed by P, Q, and
the equation of the curve.
6. Required the equation of the curve, on ^sill points of
which P will rest.
Integrating (1) of Ex. 4, we have
Px-^Qr= O. (1)
But since P is to rest at all points of the curve, this equr
tion must be satisfied when P is at A, from which we get
a; = r = o ; therefore (1) becomes
Pa—Qa=z C;
which in (1) gives
P '
1 — ^ cos
which is the equation of a conic section, of which the focus
is at the pole ; and is an ellipse, parabola, or hyperbola,
according ac P <, =, or > Q.
3
IK>
SXAMPLBa.
EXAMPLES.
1. Two forces of 10 and 20 lbs. act on a particle at an
angle of 60° ; find the resuliont. Ans. 26.5 lbs.
2. The resultant of two forces is 10 lbs.; one of the
forces is 8 lbs., and the other is inclined to the resultant at
an angle of 36°. Find it, and also find the angle between
the two forces. (There are two solutions, this being the
ambiguous case in the solution of a triangle.)
Ans. Force is 2.66 lbs., or 13.52 lbs. Angle is 47° 17'
05", or 132° 42' 55".
3. A point is kept at rest by forces of 6, 8, 11 lbs.
Find the angle between the forces 6 and 8.
Ana. 77° 21' 52".
4. The directions of "iwc forces acting at a point are
inclined to each other (1) at an angle of 60°, (2) at an
angle of^ 120°, and the respective resultants are as
V? : VS ; compare the magnitude of the forces.
Ans. 2 : 1.
5. Three posts are placed in the ground so as to form an
equilateral triangle, and an elastic string is stretched round
them, the tension of which is 6 lbs. ; find the pressure on
each post. ^„«. e Vs.
6. The angle between two unknown forces is 37°, and
their resultant divides this angle into 31° and 6° ; find the
ratio of the component forces. Ans. 4927 : 1.
7. If two equal rafters support a weight, W, at their
upper ends, required the compression on each. Let the
length of each lufter be a, and the horizontal distance
between their lower ends be & , aW
Ans.
V4o»-i»
particle at an
ns. 26.5 lbs.
I.
10
one of the
. reaaltaut at
angle between
this being the
ingle is 47° 17'
' 6, 8, 11 Iba.
77° 21' 52".
at a point are
60°, (2) at an
Itants are as
rces.
Ana. 2 : 1.
> as to form an
TetcLed round
le pressure on
Ana. 6 \/3.
ses is 37°, and
i 6° ; find the
». 4.927 : 1.
;, W, at their
och. Let the
ontal distance
aW
V4o» - V
■m
■■•
MXAMPUa^
wmmm
51
8. Three forces act at a point, and include angles of
90° and 45°. The first two forces are each equal to 2P,
and the resultant of them all is VlOP; find the third
fofce. ^ng. P V%.
9. Find the magnitude, R, and direction, 0, of the
resultant of the three forces, P, = 30 lbs., P, = 70 lbs.,
Pj = 50 lbs., the angle included between P, and P,
being 56°, and between P, and P, 104°. (It is generally
conTenient to take the action line of one of the forces for
the axis of a;.)
Let the axis of x coincide with the direction of P^ ; then
(Art. 36), we have
X = 22.16 ; Y = 75.13 ; B = 78.33 ; « = 73° 34'.
10. Three forces of 10 lbs. each act at the same point ;
the second makes an angle of 30° with the first, and the
third makes an angle of 60° with the second ; find the
magnitude of the resultant. Ana. 24 lbs., nearly.
11. If three forces of 99, 100, and 101 units respectively,
act on a point at angles of 120°; find the magnitude of
their resultant, and its inclination to the force of 100.
Ana. VS; 90°.
12. A block of 800 lbs. is so situated that it receives
from the water a pressure of 400 lbs. in a south direction,
and a pressure from the wind of 100 lbs. in a westerh
direction ; required the magnitude of the resultant pres-
sure, and its direction with the vertical.
Ana. 900 lbs.; 27° 16'.
13. A weight of 40 lbs. is supported by two strings, one
of which makes an angle of 30° with the vertical, the other
45° ; find the tension in each string.
. Ana. 20 {VQ - Vi) ; 40 ( V3 — 1).
tMM
6%
BXAMPLSa.
14. Two forces, P and P\ acting along the diagonalfl of
a parallelogram, keep it at rest in such a position that one
of its sides is horizontal ; show that
P sec «' = P' sec o = W cosec (« + «')>
where W is the weight of the parallelogram, and a and «'
the angles between *;he diagonals and the horizontal side.
16. Two persons pall a heavy weight by ropes inclined
to the horizon ut angles of 60° and 30° with forces »jf
160 lbs. and 200 lbs. The angle between the two veitical
planes of the ropes is 30" ; find the single horizontal force
that would produce the same eflfect Ans. 245.8 lbs.
16. In order to raise vertically a heavy weight by means
of a rope passing over a fixed pulley, three workmen pull at
the end of the rope with forces -of 40 lbs., 50 lb8.,8tul
100 Ibr. ; the directions of these forces being inclined to
the horizon at «n ax^gle of 60". What is the magnitude of
the resultant force which tends directly to riise the weight?
Ans. 164.64 lbs.
17. Three persons pull a heavy weight by cords inclined
to the horizon at an angle of 60°, with forces of 100, 120,
atid 140 lbs. The three vertical planes of the cords are
inclined to each other at angles of .S0°; find the single
horizontal force that would pi-oduoe the same effect
Am. 10 '^^146 + 72 Vl lb«.
18. Two forces, P and Q, acting respectively parallel to
the base and length of an inclined plane, will each singly
sustain on it a particle of weight, W\ to det«rmino the
weight of W.
Let « = inclination of the plane to the horizon ; then
resolving in each case along the plane, so that the normal
pressures may not enter into the equations (See Bern., Ex. 3,
Art. 41), wo have
T
!e diagonals of
sition that one
1, aod a and «'
triiiODtal eide.
' ropes inclined
with forces of
he two veitical
horizontal force
nti. 245.8 lbs.
eight by means
workmen pull at
>8., 50 lbs., and
iug inclined to
e mugnitude of
;J8e the weight ?
s. 164.64 lbs.
r cords inclined
XJ8 of 100, 120,
f the cords are
find the single
le effect
+ 72 VS lb«.
vely parallel to
rill each singly
i determine tho
B horizon ; then
that tlie normal
See Bern., Ex. 3,
MXAMPLBS.
St
Poofla = fTsina; C=lfBina;
jr =
PQ
{p*
W
19. A cord whose length is 21, is faster* • m .■ vnd B, in
the same horizontal line, at a distance i iii .*cli other
oqual to 2rt ; and a smooth ring upon the cortt sustains a
weight W; find the tension of the cord.
Ana. T .: --^^==.
2 VP — fit'
20. A heavy particle, whose weight is W, is sustained on
ii smooth inclined plane by three forceg applied to it, each
equal to ^; one acts vertically upward, another horizon-
O
tally, and the third along the plane ; find the inclination,
«, of the plane.
Ans. tan
1
2'
21. A body whose weight ia 10 Ihs. Is supported on a
smooth inclined plane by a force of 2 lbs. acting along the
plane, and a horizontal force of 5 lbs. Find the inclination
of the plane. ^»«. sin-» |.
22. A body is sustained on a smooth inclined plane (in-
clination a) by a force, P, acting along the plane, and a
horizontal torce, Q. When the inclination is halved, and
the forces, P and Q, each halved, the body is still observed
to rest; find the ratio of P to ^. ^^^
23. Two weights, P and Q, (Fig, 12), rest
on a sipooth doxihle-inolined plane, and are
attiiched to the extremitios of a string
which ijaases over a smooth peg, 0, at a
point vertically over the intersection of the
Hues, the jieg and the weights being in a
Fi|.a
■PBiWiP
M
MXAMPLSa,
vertical plane. Find the position of equilibrium, if Z = the
length of the string and h = CO.
Ans. The position of equilibrium is given by the equa-
tions
p 8in_a _ -, sin /3 ►
cos * ~ ^ cos ^ *
cos « oos_^ _ I
sin sin ~ ^*
24. Two weights, P and Q, connected by a string,
length /, rest on the convex side of a smooth vertical
circle, radius a. Find the position of equilibrium, and
show that the heavier weight will be higher up on the
cirele than the lighter, the radius of the circle drawn to P
making an angle with the vertical diameter.
Ana. P sin 6 = Q sin ( eh
25. Two weights, P and Q, connected directly by a
string of given length, rest on the convex side of a smooth
vertical circle, the string forming a chord of tii«? circle ;
find the position of equilibrium.
Ans. If 2a is the angle subtended at the centre of the
circle by the string, the inclination 0, of the string to the
vortical is given by the equation
P
cot fl = p- ^^ tan o.
26. Two weights, P and Q, (Fi^. 13),
Teat on the concave Hide of a pari< )la
whose axis is horizontal, and are con-
nected by a string, length I, which
]iasse8 over a smooth peg at the focus, F.
Find the {wsition of equilibrium.
A»8. Let — the angle which FP
Fi|.,ll
turn, if 2 r= the
n by the equa-
by a string,
nooth vertical
lilibriuni, and
ler up on the
le drawn to P
wmm-
»n
G'-»)-
directly by a
le of a smooth
of tii'5 circle ;
centre of the
3 string to the
'IfcU
BXAMPLSa.
65
makes with the aria, and 4m = the latus reotnm of the
parabola, then
e Py/l- 2m _
27. A particle is placed on the convex side of a smooth
ellipse, and is acted upon by two forces, F and F', towards
the foci, and a force, F"y towards the centre. Find the
position of equilibrium.
^^ ,. _ __A__, where r = the distance of the par-
Vl — «•
tide ftom the centre of the eUipse ; b = semi-minor was,
F-F'
and n = —^ —
28. Let the curve, (Fig. 11), be a circle in which the
origin and pulley are at a distance, c, above the centre of
the circle : to determine the position of equilibrium.
Q
Ans. r = -pa.
29. Let the curve, (Fig. 11), be a hyperbola in which the
origin and pulley are at the centre, 0, the transverse axis
being vertical ; to determine the position of equilibrium.
30. A particle, P. is acted upoL oy two forces towards
two fixed points, 8 and H, these forces being ^ and ^p
respectively ; prove that P will rest at all points inside a
smooth tube in the form of a curve whose equation is 8P.
PH = ]^,k being a constant
31. Two weights, P and Q, connected by a string, r«8t
on the convex side of a smooth cycloid. Find the position
of equilibrium.
56
aXAMPUUL
Am. If 2 s the length of the string, and « s radios of
generating circle, the position of eqoilibriam is defined by
the equation
.0 Q I
"" a = pTq ' 45'
where B is the angle between the vertioal and the radius to
the point on the generating circle which corresponds to P.
32. Two weights, P and Q, rest on the convex side of a
smooth vertical circle, and are connected by a string which
passes over a smooth peg vertically over the centre of the
circle ; find the position of equilibrinm.
Ana. Let h =. the distance between the peg, B, and the
centre of the circle ; and ^ = the angles made with the
vertical by the radii to P and Q, respectively ; a and j3 =
the angles made with the tangents to the circle at P and
Q by the portions PB and QB of the string ; / =b length
of the string; then
COS a '
_ nsin^
- ^C08/3'
\cos a
Bin *\
cos /)/ - *»
ft cos (fl -f
a) =. a cos a.
A COS (^ + /3) = a cos j3.
« =: radios of
u defined by
. the radias to
esponds to P.
nvex ride of a
a sti-ing which
centre of the
Bg, B, and the
made with the
; a and j3 =
irole at P and
b; ; 2 =B length
CHAPTER III.
COMPOSITION AND RESOLUTION OF FORCES ACTING
ON A RIGID BODY.
43. A Xtigid Body. — In the last chapter wo considered
the action of forces which have a common point of applica-
tion. We shall now consider the action of forces which are
applied at different points of a rigid body.
A rigid body is one in which the particles retain invari-
able positions with respect to one anotbeis so that no
pxternal force can alter them. Now, as a matter of fact,
there is no such thing in nature as a body that is perfectly
rigid ; every body yields more or less to the forces which
act on it. If, then, in any case, the body is altered or com-
pressed appreciably, we shall suppose tiuit it has assumed
its figure of equilibrium, and then consider the points of
application of the forces as a system of invariable form.
The term body in this work means rigid body.
44. TransmisBibility of Force.— When & force acts
at a definite point of a body and along a definite line, the
iffect of the force will be unchanged at whatever point of
its direction we suppose it applied, ])rovided this point be
I'ither one of the points of the body, or be invariably con-
iioctcd with the body. This principle is called the trans-
missibilily of a force, to any point in its line of action.
Now two Ciiual forces acting on a particle in the same
line and in opposite directions neutralize each other (Art.
10); BO by this principle two equal forces acting in the
same line and in opposite directions at any points of a
rigid body in that line neutralize each other. Hence it is
dear that when many forces are acting on a rigid body,
any twc, which aie equal and huvo the same line of action
■k$'f
%
ip!ii)UWU I l ( «WI I UPM. I lllM I WMIlMi l » l l.l i HMM
68
SBSXTLTANT OF PARALLEL FOBCKS.
rHhM
and act in opposite directions, may be omitted, and also
that two equsJ forces along the same line of action and in
opposite directions, may be introduced without changing
the circumstancQi of the qrstem.
46. Rorahast of Two PanlM
Force*.*— (1) Let P and Q, (Fig.
14), be the two parallel forces acting
at the points A and B, in the same
direction, on a rigid body. It is re-
qaired to find the resultant of P
and Q.
At A and B introduce two equal
and opposite forces, F. Th^ introduction of these forces
win not disturb the action of P and Q (Art. 44). P and F
at A are equivalent to a single force, R, and Q and /* at B
are equivalent to a single force, 8. Then let R and S be
snpposed to act at 0, the point of intersection of their lines
of action. At this point let them be resolved into their
components, P, F, and Q, F, respectively. The two forces,
F, at 0, neutralise each other, while the components, P
and Qy act in the line OG, parallel to their lines of action
at A and B. Hence the magnitude of the resultatti is
P-\- Qt (Art S8). To find the point, G, in which its line
of action outs AB, let the eztremitief of P and R (acting at
A) be joined, and complete the parallelogram. Then the
triangle PAR is evidently nmikr to GOA ; therefore,
P GO . ., , <? GO
y=gj; similarly p=gg;
therefore, by division.
P
Q
OB
GA*
(1)
• Miacbte'i BtiUlM, |). ».
8CBS,
Daifcted, and also
)f action and in
itboat changing
£_ o_/
I of these forces
t.44). Pandi^
1 Q and /* at B
let B and S be
ion of their lines
olved into their
The two forces,
components, P
lines of action
the resnltani is
a which its line
ind R (acting at
ram. Then the
therefore,
(1)
RSSULTAJtT OF PAMAJUtML JfOSCSS. fift
(2) When the forces ad in oppwiU dirtctiont,^-At A and
B, (Fig 15), apply two equal and opposite forces F, as
before, and let B, the resultant of P
and Ft and S, the resultant of Q and
F, be transferred to 0, thiir point of
intersection. If at the forces, B
and S, are decomposed into their
original components, the two forces
F, destroy each other, the force, P,
will act in the direetion GO parallel to the direction of
Pftaah:^ and the force Q will act in the direction OG.
Hence the resultant is a force = P — Q, acting in the line
GO. To find the point G, we har^ irom the similar
triangles, PAB andwOGA,
ris.u
F~ QA' "^/'-rin'
GB'
6B
i9A'
m
Hence the resultant of two parallel forces^ aiding in the
same or opposite directions, at the extremities of a rigid
right line, is parallel to the components, equal ta tlisir
algebraic sum, and divides the line or the line produced^
into two segments which are inversely as the forces.
In both cases we have the equation
P X GA = <? X GR (8)
Hence the following theorem :
If from a point on the resultant of two parallel forces a
right line be drawn meeting the forces, whether perpendicu-
larly or not, tfie products obtained by multiplying each force
by its distance from the resultant, measured along the arbi-
traty line, are equal.
ScH.— The point G poMesa^ti this remarkable property ;
'i\m
iF**
mmfimmmmt
-^^f^^^Kl^n^^^S^
60
MOMENT OP A POBCS.
L
that, however P and Q are turned about their points of
application, A and B, their directions remaining parallel,
G, determined aa above, remains fixed. This point is in
consequence called the centre of the parallel forces, P
and Q.
46. Moment of a Force.— 7%e moment of a force with
respect to a point is the product of f lie force and the perpen-
dicular let fall on its line of action from th« point. The
moment of a force measures its tendency to produce rota-
tion about a fixed point or fixed axis.
Thus let a force, P, (Fig. 16), act on
a rigid body in the plane of the paper,
and let an axis perpendicular to this
plane pass through the body at any
point, 0. It is clear that the effect of
the force will be to turn the body round this axis (the axis
being supposed to be fixed), and the turning effect will
depend on the fhagnitude of the force, P, arid the perpen-
dicular distance, p, of P from 0. If P passes through 0,
it is evident that no rotation of the body round can take
place, whatever be the magnitude of P ; while if P
vanishes, no rotation will take place however great p may
be. Hence, the measure of the power of the force to
produce rotation may be represented by the product
P-P,
and this product has received the special name of Moment.
The unit of force being a pound and the unit of length a
foot, the unit of moment will evidently Imj & foot-pound.
The point is called the origin of moments, and may or
may not be chosen to coincide with the origin of co-
oi-dinates. The solution of problems is often greatly sim-
plified by a projwr selection of the origin of moments. The
perpendicular from the origin of moments to the action line
of the force is called the arm of the force.
their points of
aining parallel,
his point is in
allel forces, P
of a force mth
nd the perpen-
h^ point. The
I produce rota-
axis (the axis
ing effect will
nd the perpen-
eo through O,
nd can take
while if P
great p may
the force to
rodact
e of Moment.
lit of length a
)ot-pound.
s, and may or
origin of co-
greatly sim-
oments. The
he action line
siaifa OF MOHEifra.
61
47. Signs of Moments. — A force may tend to turn a
body about a point or about an axis, in cither of two direc-
tions; if one be regarded as positive the other must be
negative; und hence we distinguish between positive and
negative moments. For the sake of uniformity the moment
of a force is said to be positive when it tends to turn a body
from left to right, t. e., in the direction in which the hands
of a clock move ; and negative when it tends to turn the
body from right to left, or opposite the direction in which
the hands of a clock move.
48. Oeometrlc Representation of the Moment of
a Force with respect to a Point— Let the line AJ}
(Fig. 16), represent the force, P, in magnitude and direc-
tion, and p the perpendicular OC ; then the moment of P
with respect to is AB xp (Art. 46). But this is double
the area of the triangle AOB. Hence, the moment of a force
with respect to a point is geometrically represented by double
the area of the triangle whose base is the line representing
the force in magnitude and direction, and whose vertex is
the given point,
49. Case of Two Equal and Opposite Parallel
Forcea — If the forces, P and Q, in Art. 45, (Fig. 16) are
equal, the equation
P X GA = ^ X GB
gives GA = GB, which is true only when G is at infinity
on AB ; also the resultant, P—Q, is equal to zero. Such a
system is called a Couple.
A Couple consists of two equal and opposite parallel forces
acting on a rigid body at a finite distance from each other.
We shall investigate the laws of the composition and
resolntion of couples, since to these the composition and
62
MMIHI
MOMSNT OF A COUP LB.
resolution of forces of every kind acting on a rigid body
may be reduced.
Fi|.l7
V
50. Moment of a Conpla— Let ^
(Fig. 17) be anj point in the plane of the
couple; let fall the perpendiculars Oa
and Ob on the action lines of the forces
P. Then if is inside the lines of actio.i
of the forces, both forces tend to produce
rotation round O in the same direction, and therefore the
sum of their moments is equal to
P{Oa + Ob), or P X e*
*
If the point chosen is O', the sum of the moments is
evidently
P (O'a - 0'*), or P X oi,
which is the same as before. Hence the moment of the
couple with respect to all points in its plane is constant.
The Arm of a couple is the perpendicular distance
between the two forces of the ooaple.
- The Moment of a couple is the product of the arm and
ono of the forces.
The Axis of a couple is a right line drawn from any
chosen point perpendicular to the plane of the couple, and
of such length as to represent the magnitude of the mo-
ment, and in such direction as to indicate the direction in
which the couple tends to turn.
As the motion, in Statics is only virtual^ and not actual,
the direction of the axis is fixed, but not the position of it;
it may be any line perpendicular to the plane of the couple,
and may be drawn as follows ; imagine a watch placed in
the plane in which several couples act. Then let the axes
of those coaples whifih tend to prodoce rotation in the
n a rigid body
a o b d
I
d therefore the
he moments is
noment of the
is constant.
colar distance
>f the arm and
»wn from any
le couple, and
udr of the mo-
le direction in
ad not actital,
position of it ;
of the couple,
itch placed in
m let the axes
tation in the
COUfLKB.
6S
direction of the motion of the hands be drawn upward
through the face of the watch, and the axes of those which
tend to produce the contrary rotation be drawn tlowntmrd
through the back of the watch. Thus each couple is oom-
plotoly represented by its axis, which is drawn upward or
downward according as the moment of the couple is posi-
tive or negative ; and couples are to be resolved and
compounded by the same geometric constructions performed
with reference to their axes as forces or velocities, witii
reference to the lines which diredly represent them.
Wo shall now give three propositions showing that the
effect of a couple is not altered when cu-tain changes are
made with respect to the couple.
51. Tlie Effect of a Couple on a Rigid Body is not
altered if the arm- be turned through an;/ angle
about one extremity in the plane of the Couple.
Let the plane of the paper be the
plane of the couple, AB the arm of
the original couple, AB' its new posi-
tion, and P, P, the forces. At A
and B' respectively introduce two
forces each equal to P, with their
action lines perpendicular to the arm
AB', and opposite in direction to
each other. The effect of the given
couple is, of course, unaltered by the introduction of those
forces. Let BAB' = W ; then the resultant of P acting at
B, and of P acting at B', whose lines of action meet at Q,
is 2F sin 6, acting along the bisectw A^ ; and the result-
ant of P acting at A peipendicular to AB and of P per-
pendicular to AB', is %P sin 9, acting along the bisector
AQ in a direction opposite to the former resultant Hence
these two resultants nentjalire each other; and there
remains the couple whose arm is AB', and whose forces aro
P, P. Hence the effect flf the couple is not altered.
,.--m':f.x»mum\uimm
ummmmmmi:'.
64
COUPLES.
4
\
^B
Fig. 19
52. The Effect of a Couple on a Rigid Body is
not altered if we transfer tlie Cojtple to any other
Parallel Plane, the Arm remaining parallel to
itself.
Let AB be the arm, antl P, P, the
forces of the given couple; let A'B'
be the new position of the arm par-
allel to AB, At A' and B' apply two
eqaal and opposite forces each equal
to P, acting pi^rpendicular to A'B',
and in a plane parallel to the plane of
the original couple. This will not alter the effect of the
given couple. Join AB', A'B, bisecting each other at ;
then P at A and P at B', acting in parallel lines, and in
the same direction, are equivalent to 2P acting at ; also
P at B and P at A', acting in parallel lines and in the
same direction, are equivalent to 2/* acting at 0. At O
therefore these two resultants, being equal and opposite,
neutralize each other ; and there remains the couple whose
arm is A'B', and whose forces are each P, acting in the
same directions as those of the original couple. Hence the
effect of the couple is not altered.
53. The Effect of a Couple on a Rigid Body is
not altered if we replace it by another Couple of
which the Moment is the same ; the Plane remain-
ing the same and the Arms being in the same
straight line and having a
common extremity.
Let AB be the arm, and P, P, the
forces of the given couple, and sup-
pose P=Q+R. Produce AB to C
Bo that
AB : AC :: Q : P{=Q + B),
and therefore AB : BO :: Q : B',
tP=Q+R
F{gM
♦ P=Q+R
(1)
igid Body is
to any other
parallel to
\/
7'
Fifl.l9
I effect of the
!i other at O ;
1 lines, and in
]g at ; also
es and in the
: at O. At O
and opposite,
couple whose
acting in the
Hence the
^,id Body is
r Couple of
ne rcniain-
the same
mmm
.30
Q
P=Q+R
(1)
(2)
FOSCS AND A COUPLB.
66
at introduce opposite forces each equal to Q and parallel
to P ; this will not alter the effect of the couple.
Now E &t A and Q at C will balance Q + li at B from
(3) and (Art. 45); hence there remain the forces, Q, Q,
acting on the arm, AC, which form a couple whose moment
is equal to that of P, P, with arm, AB, since by (1) we
have
P X AB = Q X AC.
Hence the effect of the couple is not altered.
Rem. — From the last three articles it appears that we
may change a couple into another couple of equal moment,
and transfer it to auy position, either in its own plane or
in a plane parallel to its own, without altering the effect of
the couple. The couple must lomain unchanged so far as
concerns the direction oj rotation which its forces would
tend to give the arm, i. e., the axis of the couple may be
removed parallel to itself, to any position within the body
acted on by the couple, while the direction of the axis from
the plane of the couple is unaltered (Art 50).
54. A Force and a Couple acting in the same
Plane on a Rigid Body arc equivalent to a Single
Force.
Let the force be /'and the couple {P, a), that is, P is
the magnitude of each force in the couple whose arm is a.
Then (Art 53) the couple (P, o) = the couple \F, ^.
Let this latter couple be moved till one of its forces acts in
the same line as the given force, F, but in the opposite
direction. The given force, F, will then be destroyed, and
there will remain a force, F, acting in the same direction
as the given one and at a perpendicular distance from it
aP
= -F'
S
• i
v8
RBSULTANT OF COUPLXa.
Cob.-— J force and a couple acting on a rigid h}dy cannot
produce equilibrium. A couple can be in equilibrium only
with an equivalent couple. Equivalent couples are thotie
whose moments are equal.*
The resultant of several couples is one which will produce
tk'j safne effect singly as tfie component couples.
55. To find the Resultant of any number of
Couples acting on a Body, the Planes of the
Couples being parallel to each other.
Let P, Q, R, etc^ be the forces, and a, b, c, etc., their
arms respoctively. Suppose all the coaples transferred to
the same piano (Art. 52) ; next, let them all be transferred sa
as to have their arms in the same straight line, and one
extremity common (Art. 51) ; lastly, let them bo replaced
by other couples having the same arm (Art. 53). Let » be
the common arm, and P,, Qi, R^, etc., the new forces,
80 that
P^a ::=: Pa, Qia = Qb, Uia — Re, etc.,
then P| = P-, Qi = e-, Rt=R*-, etc.,
i.e., the new forces are P , Q , R-, etc., actlug on the
common arm a. Hence their resultant will be a ooaple of
which each force equals
p? 4. ^* 4- i?*: ^. etc.,
a a a
•nd the arm ~ n, or tho moment cquaU
Pa+ Qb + Re + etc
If one of tho oonplcs, as Q, act in a direction opposite to
• The nioDteaU nroqiUmlent cou|iit« iiiajr >wt(i like or uullke ilgiM.
^?«?? '
id body cannot
mlibrium only
pUa are thoKC
h will produce
number of
ines of the
», c, etc., their
transferred to
} transferred sa
', line, and one
;m bt) replaced
•3). Let a be
10 new forces,
c, etc.,
actliig on the
le a couple of
11
m
1 opponito tt
iliko llgM.
aSSULTANT OF TWO COUPLES.
•Y
the other couples its sign will be negative, and the foroe at
each extremity of the nrm of the resultant couple jv ill be
p?_^* + i?- + etc.
Hence the moment of the resultant couple is equal to the
algebraic sum of the momentfi of the component couples.
S6. To Find the Resultant of two Couplet not
acting in the same Plane,*
Let the planes of the con pies b^
inclined to each other at an
angle y ; let the couples be trejis-
ferred in their pianos so as to
have the same arm lying along
the line of intersection of the two
planes ; and let the forces of the
couples thus traiisfbn-ed Ixi P and Q. Let AB hk\ the com-
mon arm. I^et R be the rosnltant of the forces J"' and Q at
A acting in the ^i! rection AJS ; and of P and (? at B acting
in the direction Jtit. Then einoe P and Q at A are parallel
to P and Q at B respectively, tlierefore £ at A is parallel
to i2 at B. Hence the two couples are equivalent to the
single couple R, R, acting on the arm AB ; and since
PA.Q =. y, V.0 LiTO
i? = P« + g» + iPQ cofi Y (Art 30).
(1)
Draw A«, Bi perpendicular to the planes of the conples
/•, /', and Q, Q, respectively, and proportional in length to
thflir moments.
Draw Ac perpendicular to the plane of R, R, and in the
same proportion to Aa, Bb, that the moment of the couple,
R, R, is t<^) those of P, P, and Q, Q, respeotively. Then
Art, Kb, Ac, nuiy hi' Inkoii bb the axes of P, P ; Q, Q; and
* tMhoDlar't autlM, p. «. AtM PraU't UmIuuiIo*, p. K
■HMMMRMP
G8
RESntTAXT OF TWO COUPLES.
R, R, respectively (Art. 50). Now the three strai^^^ht linos.
Art, Ar, Kb, make the same angles with each other that
AiF^ A^, A^ make with each other; also they are in the
euine proportion in which
or in which
AB • P, AB . i2, AB . ^ are,
P, /?, ^ are.
But R is the resultant of P and Q ; therefore Ac is tho
diagonal of the parallelogram on Ao, Ai (Art. 30).
Hence if two straight lines, having a common extrnmitif,
represent the axes of two couples, that diagonal of the
parallelogram descril/ed on these straight lines as adjacent
sides which passes through their comn*on extremity repre-
sents the axis of the resultant couple.
Con. — Since R • AB is the axis or moment of tho reeui'-
ant couple, we have from (1)
/p. All' = /". Al?+ Q»- A^+iP- AB- Q- ABcos y. (2)
If X and M represent the axes or moments of the com-
ponent couples and G, that of the resultant couple, (2)
Dtcones
CP = L* + M' + 2L • M COB y.
(3)
Pen. 1. — \{ L, M, N, are the axes of three component
couples which act in planes at right angles to one a.iother,
and G tho axis of the resultant couple, it may easily be
shown that
C*= D + M*^ N\
(4)
If A, n, V be the angles which the axis of tho resultant
makes with those of the componont8, we have
009 A =
^'
M
cos /* = ^-,
y
COS V = ^.
vs.
B straight linos,
ach other that
;hoy are in the
;fore Ac \a tho
rt. 30).
mon extremity,
iagonal of the
Ines as adjacent
xtremity repre-
it of tho resuit-
AB-cosy. (3)
ta of the com-
iit couple, (2)
(3)
ree component
o one aiiothcr,
may easily bo
(*)
tho rosnltanf
o'
^ggggggm*«*- 1 1 , 111 , 1 1 iiiii« || p | i i i. ii .
VARroifoy's theorem of moments.
iBil"i»SBS.r
6tf
Sen. 2. — Hence, conversely any couple may be replaced
by three coupIos acting in planes at right aiigloa to ono
another ; their moments being G cos k, G cos /u, G cos v ;
where G is the moment of the given couple, and A, p,, v tho
angles its axis makes witli the axes of the three couples.
Thus the composition and rcaolution of covplcs follow
laws similar to those which apply U) forces, tho ••xis of the
couple corresponding to tho direction of the force, and the
moment of the couple to the magnitude of tho force.
57. Varignon's Theorem of Moments.— TTta mo-
rri'CMt of the resultant of two conifjonen!, forces
with respect to any point in their pic %e is equal
to the algebraic sum of the mnmenis of the two
components with respect to the same point.
Let A P and .4 Q represent two com-
ponent forces ; complete the parallelo-
gram and draw the diagonal, Ali,
representing the resultant force. lict
O be tho origin of moments (Art. 46).
Join OA, OP, OQ, OR, and draw PC
and QB parallel to OA, and let p = tho perpendicular let
fall from to All.
Now tho moment of A P about is the product of A P
and the per|)endicular let fali on it from (Art. 40), which
IS double the area of the triangle, AOP (Art. 48). But
the area of the triangle, AOP, = the area of tlie triangle,
AOC, since these triangles have tho same base, AO, and
are between tho same parallels, AO and CP. Hen^o the
moment of AP about = the moment of vlC about
O = AC -p. Also the moment of AQ v^^.cni is dooblo
tho area of tho triangle, A OQ, = double the area of tho
triangle, A OB, since the two triangles hovo the same baso,
AO, end are between thn same parallels, AO and QB.
Hence tlie moment of AQ about — Uu' moment of AU
Fig.Zl
70
VASIOjrON'S TBBOREX OF MOMEKTS.
about = AB • p. Therefore the sura of the moments of
AP and AQ about = tlie sum of the moments ot AC
and AB about = {AC + AB)p, = {A3 + £i?)p,
{auci AC = fift firom the equal triangles ^PCand QBR)
= AR ' p = the moment o£ the resulttmt.
If the origin of momente fall between AP and AQ, the
forces will tend to produce rotation in opposite directions,
and hence their moments will have contrary signs (Art
4t). In this caae the moment of the resultant = the dif-
ference of the moments of the components, as the student
will find no difficulty in showing. Hence in either case
the moment of the resultant is equal to the algebraic sum
of the moments of the components.
Con. 1.— If there are any number of component forces,
wo may compound them in order, taking any two of them
first, then finding the resultant of these two and a third,
and so on ; and it follows that the sum of their momenta
(with their proper signs), is equal to the moment of the
redultojit.
OoR. 2.— If the origiv of moments be on the line of
action of the reroltant, p = 0, and therefore the moment
of the resultant = ; hence the sum of the moments of
the componeuta is ecjual to isero. In this case the moments
of the forces in one direction balance those in the opposite
dirpction ; t. e,, the forces that tend to produce rotation in
one direction Mf^ counteracted by the forces that tend to
produce rotation in the opposite direction, and there is no
tenden'jy to rotation.
Oo». 3.— If all the forces are in eqailibriam the resultant
J? = 0, and therefore the moment oi R — Q; henoe the
sum of the momontfl of the conponents is equal to aewj,
and there is no teudeacty to motion either of tnuiulatiou or
rottktion.
VlfTS.
the moments of
loments ot AC
{A3 + BB)p,
IPC and QBJi)
iP and AQ, the
>aite directions,
rary signs (Ai-t.
[«nt = the dif-
I as the student
i in either case
3 alj/ebratc sum
nponent forces,
,ny two of them
,\vo and a third,
their moments
moment of the
on the line of
re the moment
he moments of
se the moments
in the opposite
uoe rotation in
[>s that tend to
md there is no
im the resultant
= 0; hence the
equal to serri,
trauulatiou or
p
R
Q
1
(
1
B
VARIGNOIf'a TBEORMM FOR PAKALLSL FORCES. 71
CoK. 4. — Therefore when the moment of the resultant
= 0, we conclude either that the resultant — (Cor. 3),
or that it passes through the point taken aa the origin of
moments (Cor. 2).
58. Varignon'B Tlieoram of MomamtB for Parallel
Forcea. — The aum of the rnamients of two parallel
forces abotit any point is equal to the moment of
their resultant about the point.
Let P and Q be two paraliel forces
acting at A and B, and R their result-
ant acting at 6, and let be the point
about which moments are to be taken.
Then (Art. 45) we have
P X AG = Q X BG,
.•. P(OG - OA) = C (OB - OG),
.-. (P + C) OG = P X OA + C X OB,
/Z xOG = PxOA + G xOB;
that is, the earn of the moments = the moment of the
resultant
OoB.— It follows that the algebraic sum of the momenta
of any number of parallel forces in one plane, with respect
to a point in their plane, is equal to the moment of their
resultant with respect to the point
69. Centre of Parallel Forcea. -To iind ttte mag-
nitude, direction, and point of application of the
resultant of any number of parnllel forces acting
on a rigid body in one plane.
kHH
iT'lMMBtmill
72
CBNTRV OF PARALLEL FORCES.
Let P., P,, P., etc., denote the
ZM
fonjM, if,, Jf,, M3, etc., their points ^ -. >■''•
of ufplicution. .Take any point in
the plane of the forces as origin and
draw the rectangular axes OX, OV.
L«t («i. yi). (««» 1/%)f etc., be the
points of application, if,, i/,, etc.
Join M^M^; and take the point if on M^M,, so that
^.
/•
6 «&
Fig.24
if, if
(1)
then the rcsnitant of P, and P, is P, + Pg, and it acts
through M parallel to P, (Art 46).
Draw M^a, Mb, M^c parallel, and Mie perpendicular to
the axis of y. Then wo have
p
Mb-yy= -jr—j- iy, - Vi) '.
, , Mb- p^^p^ ,
(2)
which gives the ordinate of the point of application of the
rcEultant of P, and Pg.
Now since the resultsat of P, and P,, which is
P, 4- P,, acts at M, the resultant of P, + P, at M, and
P, at M^, is P, + P, -f- P, at g, and substituting in (8)
P, + /'„ P„ ifi, and y, for P,, P„ y,, and y, rcgpec-
tiywiy, we have
mm^
mm
mmt
m.
7m ^
i^
g-^U,
a c
» b b
Fig.24
r,, BO that
0)
Pg, and it acts
erpendicalar to
« -yi);
y».
(2)
ieation of the
Pg, which is
I\ at if, and
jtnting in (3)
md yg ro§pec-
ItJ- P^Vi .
'• + ^3
;(3)
CBHTRS or PARALLSL P0BCE8,
T3
and this process may be extended to any number of parallel
forces. Let R denote the resultant forCe and y the ordi-
nate of the point of applicjation ; then we have
i? = Pi + P, + ^3 + etc.
SP.
7, - P^Vx + ^«y« + ^sy._+_.etc. _ I.Py
y - p, + p, + P3 + etc. ^P
IP
Similiirly, if x bo the abscissa of the point of application of
the resultant, we have
X =
XP'
The values of zl y are indeponth^nl of the angles which
the directions of the forces muko wllh the axcM. Hence
t' these directions be turned about the points of application
(f the forces, their parallelism being preserved, tlie point of
ii|)pliciition of the resultant will not move. For this reason
tlie point (i, y) is called the centre of parallel forces. We
shall hereafter have many applications in which its position
is of great importance.
ScH. 1. — The moment of a forcf. toifh respect to a plane
to which it is parallel, is the product of the force into ♦ho
perpendicular distance of its point of application from the
plane. Thui, Pi^i is the moment of the force P,, iu
reference to the plane through OX perpendicular to OV.
This must be carefully distinguished from the moment of a
force with respect to a point. Hence the Aquations for
determining the posit'on of the centre of pfprailel forcet
show that the sum of tlie mmufnts of the parallel fofoen icitti
respect to any plane parallel to tfifun, i« eq</0X to the moment
of their resultant.
SoH. 2. — The moment of a force with respect to any line
is the product of the component of the force |.Hirj)endicalar
■^
74
comurroxa of xquiLiasivx.
♦Pt
FlgJI
♦r
to the line into the shortest distance between the line and
the line of action of the force.
60. Conditions of Equilibrium of a Rigid Body
acted on by Pandlal Foroes in one Plane.— Ut
P,, P„ P„ etc., denote the forces. Take
any point in the plane of the forces as
origin, and draw rectangular axes, OJT,
OV, the latter parallel to the foroes. Let
A bo the point where OJT meets the direc-
tion of P,, and let OA = x^.
Apply at two opposing forces, each
equal and parallel to P,; this will not disturb the equili-
brium. Then P, at A is replaced by P, at along OV,
and a couple whose moment is P^ ■ OA, i. e., P,a;,. The
remaining forces, P„ />„ etc., mr.y be treated in like man-
ner. We thus obtain a stt o1 forces, P^, P„ P„ etc.,
acting at along OY, and a sot of couplr.., P^r^, P,x,,
/'jij, etc., in the plane of the forces tending to turn the
^fcbdy from die axis of x to the axis of ;/. These forces are
etjuivalcnt to a single resultant force /'^ f\ + /*, + etc.,
and the couples iirc equivalent to a s^gji i^ciilt&n! rioaj^
Pjjr, + P,x, + P,«, + etc. {Art 6^.
Hence denoting the resultant force by 72, and tlie mumout
of the resultAiil couple by 0, we have
il = P, -I- P, 4- /», -H eta =r SP;
= P,*, + P,a!, + P,x, 4- eta .-= S.PTi
that is, a system of parallel forces can be reduced to a
single fiitj* Hiiij a couple, which (Art. fl4. Cor.) (annot
pioduoe equilibrium. Heuoo, for equilibrium, the force
and the couple must vanish ; or
XP = 0, and LPx = a
TM.
^n the line and
B Rigid Body
18 Plane.— Let
♦P.
turb the equili-
it along OF,
e., P,a;,. The
ied in hko man-
' >» ■* S' 6tc.,
rig 10 turn (he
Phese forces are
/% + /*, + etc.,
■oopte,
uU the momt)ut
IP;
reduced to n
Cor.) runnnf
uin, the force
V
1
^1 >R
H
'^ ,yi
-A
'
M'
F
i8.a«
coifDmoira ojc squiLiBBiuM. 75
Hence the ooinditions of eqailibriam of a system of pur-
iillel forces acting on a rigid body in one pkue are :
The sum of the forces must = 0.
The suin of the momenta of the forces about every point in
their plane mttst = 0.
61. Conditioiift of EqaililMiiiin of a Rigid Body
acted on by Forces in any direction in one Plane.—
Let Pj, Pg, Pj, etc, be the forces acting at the points
(^i» yi)> (a^8. y«)« i^i' ys)> etc., in the
plane xy. Resolve the force P, into t>Yo
components, X,, F,, parallel to OJT
and OY respectively. Let the direc-
tion of Kj meet OX at M, and the
direction of X, meet OFat iV. Apply
at two opix)sing forces each equal and parallel to Xj,
and also two opposing forces each equal and parallel to F, .
Hence Fj at .4,, or M, is equivalent to Y^ at O, and a
louplo whoso moment is F, • OM; and X^ at J,, or iV, is
equivalent to Xj at 0, and a couple whose moment is
V, ov.
Ifenc. Fj is replaced by F, at 0, and the couple F,.t, ;
and A", is voplanul by Xj at (), and the couple X|//i (Art.
47). Therefore the force P, may bo replaced by the com-
ponents Xj, F, acting at 0, and the couple whose
moment is
and which equals the moment of P, about (Art. 67).
By a similar resolution of all the forces we shall have
them replaced by the forces (X,, F,), (Xj, Fj), etc.,
acting at along the axes, and the couples
TtX, — X,y„ Fa^s — Xj^,, etc
Adding together the couples or moments of P,, P,, etc.,
■i;
76
SQVILIOBr'M VNDEH THREE FORCES.
and denoting by G the moment of tho resultant conple, wc
get the total moment
If the sum of the cumponents of the forces along OX is
denoted by SX, and the sum of the components along OV
by £F, the resultant of the forces acting at is given by
the equation
i? = (SX)» -h (£F)«.
If a be the angle which R makes with the axis of X, wc
have
R cos a = IX, Ji Bin a = I.V;
tan a =
iX'
Therefore, any system of forces acting in any direction
in one plane on a rigid botly may be reduced to a single
force, li, and a single couple whose moment is O, which
(Art, 64, Cor.) cannot produce equilibrium. Hence for
equilibrium we must have R = 0, and G' = 0, which
requires that
tx=o, ir=o,
^irx — Xy) = 0.
Hence the conditions of equilibrium for a system of
forces acting in any direction in one plane on a rigid body
arc :
Tlfte sum of the components of the forces parallel to each of
two rectangular axes ..ivM = 0.
The sum of the mofir its of the forces round ever tf point in
their plane must = 0.
rt*i
FORCES.
ultant coaple, wc
EXAMPLES.
77
roes along OX is
inents along OV
at is givon by
;he axis of X, we
in any direction
iced to a single
ent is O, wliich
Hence for
O — 0, wliich
m
3r a system of
m a rigid body
rallel to each of
i every point in
Cor. — Convoraely, if the forces are in equilibrinni tho
sum of the com{K>nents of the forces parallel to any diroc-
tioii will = l>, and also the sum of the moments of the
forces about any point will = 0.
62. Condition of Equilibrium of a Body under the
Action of Three Forces in one Plane -// thn
forces jtiaintain a hi'dy in equilibrium,, their
directions must meet in- a point, or be parallel.
Suppose the directions of two of the forces, P and Q, to
meet at a point, and take moments round this point ; then
tlio moment of each of these two forces = 0; thcrefon> ilio
moment of the third force li = (Art. 61, Cor.), which
requires either that Ji — 0, or that it pass through the
point of intersection of P and Q. If R is not = 0, it must
jtiiss through this jmint. Hence if any two of the forces
meet, tiie third must pass through their point of intersec-
tion, an keep it at rest, and each force must be equal and
opposite to the resultant of the other two. If the angles
])etween them in pairs I c p, q, r, tho forces must satisfy the
conditions
r : Q '. R = «inp : sin & : sin r (Art. 32).
If tro of the forces are parallel, the third must bo
parallel o them, aud equal md directly opposed to their
resultant.
EXAMPLES. .
1. Suppose six parallel forces proportional to the numbers
1, 2, 3, 4, 5, 6 to act at points (—2, —1), (—1, 0), (0, 1),
(1, 2), (2, 3), (3, 4) ; find the resultant, R, and the centre
of parallel forces.
By Art 69 we have
J2 = SP = l + 2 + ...6 = 21;
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■- f .-.sii^irTp tf?
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•ft XXAMPLMS,
SP* =-2-2 + 4 + 10 +18 = 28;
XPy = - 1 + 3 + 8 + 15 + 24 = 49.
. i-^Z^i - 2S. -_2Py _ 49
'' ~ £-" ~21' ^~"£F~2i'
2. At the three vertices of a triangle pnrallel forces are
applied which are proportional respectively to the opposite
Bides of the triangle; f nd the centre of these forces.
Let(»i» !ft)> (»i, y»), («„ Jf,) be the vertices, and let a,
i, c be the sides opposite to them; then
;_ aa;,+te,+cr , -_ ay, +» »-,+ gy,
a+4+c ' » a+i+c
3. If two parallel foroeu, P and ^, act in the same direc-
tion at A and 5, (Pig. ji), and make an angle, B, with
^5, find the moment o/ each about the point of epplicur
tion of their resnltant
The moment of F with respect to (7 is
P-^(?8iu(!»(Art.46).
But from (1) of Art 4ft, we have
P+ Q
.'. AG =
4B
AG*
P + Q
aB,
which in P - AG an gives
for the moment of P which also e<]uals the moment of (j.
lOfflHP^:-
( = 49.
y
49
21*
■nrallel forces are
J to the opposite
Bse forces,
iiidoe^ and let a,
k the Mune direc-
u ao^, 0, with
point ol epplicur
Qoment of (j.
MXAMPtML
n
I
Fit-V
4. Two paiallel foroet, aotiBg in ike nm» direction;
liave their magnitndes 5 and 13, and th«r points of appUcar
tion, J and B, 6 feet tipmti. Find the nu^iind« of their
resultant, and the point of application, 0.
Ant, B = 18, J.0 rts 4|, BG = If
6. On a Ftraif^ht rod, AF, there are snspended a weights
of 6, 15, 7, 6, -uid 9 poands respectively at the points A, B,
D, E, F; AB = 3 feet, BD = Q feet, DE = & feet,
EF = 4 feet l^'ind the magnitnde of the resultant, and
the distance of its point of application, O, from A.
Ana. Ji = i2 pounds. AG = 6f feet
6. A heavy nniform beam, AB, rests
in a vertical phuae, with one end. A, on a
emooth horizontal plane and the other
end, B, against a smooth vertical wall ;
the end. A, is prevented from sliding by
a horizontal string of given length ii£-
tened. to the end of the beam and to the vail ; determine
the tension of the string and the preesurss against the
horizontal plane and the wall
Let ito =: the length of the beam, and let IF be its weight,
which as the beam i> uniform, wn may suppose to act at its
middle point, G. Let B be the vertical pressure of the
hori«>ntal plauvj against the beam ; '\nd R the horizoot'U
pressure of the vertical wall, and T the tension of the hor-
izontal string, AC ; let BAG = o, a known angle, since
the lengths of the boaa and th« string are given. Then
(Art 61), we have
tot horiaontal forces, T=s R;
for vertioal forces, 17 = JK ;
for moments abont A (Art 47), 2/2*^ dn « = TTa cos •> ;
80
XXAMPLSa.
i
7. A heavy beam, AB = 2a, rests
on two givea amooth planes which are
inclined at angles, a and /3, to the
horizon ; required tho angle which
the beam makes with the horizontal
plane, and the presfiores on the
planes.
Let a and b be the segmeats, AG and BO, of the beam,
made by its centre of gravity, G; let E and R' be tho
pressures on the planes, AC and BC, the lines of action of
which are perpendicular to the planes since they are smooth,
ard let W be the weight of the beam. Then we have
for horizontal forces, Rsin a = R' mnP; \ (1)
for yertical f oroM, R coa a + R" coa p = W; (8)
for moments about G, ^ cos (a—0)=:R'b cos {li+0)\ [ (3)
pi.riding (3) by (1), we have
aoot a -{- atanO = bcotp — hta,n6;
a cut a — 6 cot i?
therefore,
tan9 =
a + b
and from (1) and (2) wo have
Wain (3
R =
sin {a + 13)'
and R'
IT sing
sin (a 4-0)'
Otherwise thus : since the beam is in equilibrium under
the ftotion of only three forces, they must meet in a point 0,
(Art. 62), and therefore we obtain immediately from the
geometry of the figure^
P
w
sing
siu (c+/3)'
« =
Wainfi
mi (a + 0)'
IG, of the beam,
t and R' be the
lines of action of
they are smooth,
;n we have
ramU; \ (1)
ip= W; {%)
Aoo8(/3-fe), (3)
tan $;
Bin a
lilibrium under
Bet in a point O,
lately fh)m the
sin /3
« + /3)*
■^ipBHMK^'^
SXAMPLSa.
81
ana -n-, =
R
W
sin a
R' =
IT Bin a
sin (a + (iy • " — gin („ ^ py
Also since the angles, GOA and OOB, ai-e equal to a and (i,
respectively, and AGO = 5 — (?, we have
«
(a + ft) cot AGO = o cot GAO — b cot GOB;
a cot a — ft cot /3
therefore,
tan0 =
a + b
Hence, if x = K^r-gi the beam will rest in a horizontal
position.
8. A heavy uniform beam, AB, rests with
one end, A, against a smooth vertical wall,
and the other end, B, is fastened by a string,
BO, of given length to a point, C, in the
wall ; the beam and the string are in a vertical
l)lane ; it is required to determine the pressure
against the wall, the tension of the string, and
the position of the beam and the string.
Let AG = GB = a, AC = x, BC = ft.
F)|.M
weight of beam = W, tension of string = T, prousare of
waU = Ji,
BA£ = 0, BOA = t.
Then wo have
for horiiontttl forces, R =: Tain ^; (1)
for vertical forooB, If = 7* cos ^; (2)
for momenta about A, Ifa sin 9 = T- AD = Tz sin ^; (%)
. ' . a sin 9 = a; tan 9 ; (4)
89
SXAMPLSa,
and by the goometr/ of the figure
sin B
2a
sin ^ '
z _ sin (O—ji)
Ha" sin ^
(8)
(6)
Solving (4), (6), and (6), we get
. . 1 ri6o» - ft»"]*
nn
from which ^ and T become known. (Price's Anal.
Moch'a, Vol. I, p. 69).
To determine M tho anknown qoantitles many problems in Stat cs
require eqoatlona to be formed bj geometric relations as well as ttatic
relations. Thus (1). (S), (8) are stntic equations, and (5) is a geometric
equation.
0. A uniform heavy beam, AB = 2a,
rests with one end, A, against the inter-
nal surface of a smooth hemispherical
bowl, radius = r, while it is supported
at some point in its length by the edge
of the bowl ; find the position of equili-
brium.
The beam is kept in equilibrium by three forces, vias., the
reaction, R, at A perpendicular to the surface of contact,
(Ari 42) and therefore perpendicular to the oowl, the
reaction, R', at C which, for the 8an)e reason, is perpen-
dicular to the beam, and the weight IK acting at 0.
rtg.so
ti
w
01
(1
k
.S(
li
t1
E
(5)
(6)
(Price's Anal.
problems in Stat cs
ins M well as tttUic
id (6) is a ge(HU«tric
na.30
forccB, via!., the
irface of contact,
the bowl, the
Mon, is perpon-
ng at O.
MXAMPLSa.
83
Let = the inclination of tho beam to the horizon
= <AGD. The solution will be most readily effected by
resolving the forces along the beam and tbking moments
about 0, by which we shall obtain equations free from the
unknown reaction, R. Then we have
for f 01*068 along AB, £ cos d = TF sin 0,
for moments about 0,
R'2roofi9ma.e = IF (2r cos — a) cos 0.
From (1) we have
/2=irtanfl,
which in (2) gives, after reducing,
8r sin* — 2/* cos* tf 4 o OM = 0,
4r cos* — a cos — 2r = 0,
(1)
(2)
(ir,
(3)
COS* =
a ± \/32r» + a *
8r
Otherwise thns: since the beam is In equilibrium under
I lie action of only three forces, they must meet in a point
() (Art. 62). Draw the three forces AO, CO, GO, which
keep the beam in equilibrium. Let the line, 00, meet the
semicircle, DAO, in the point, Q. Then AQ is a horizontal
line. Also
<QAG = <DCA = 9,
therefore
<OAQ = 29.
llenoe
AQ = AO cos 20,
:ind also
AQ = AO cos 0;
M
MXAMPLXS.
therefore
or
Sr COB 20 = a ooa d,
4r COB* 6) — a COB — 2r =: 0,
which is the same as (3) obtained by the other method.
The student may prove that the reaction, B', at G
a
= »F
2r
n^
10. Find the position of equilibrium of
a uniform heavy beam, on<} end of which
rests against a smooth vertical plane, and
the other against the internal surface of a
smooth spherical bowL
The beam is in equilibrium under the
action of three forces, the weight, fF,
acting at G, the reaction, R, at A, perpen-
dicular to the surface and henoe passing through the centre,
C, and the reaction, R', of the vertical piano perpendicular
to itself and hence horizontal
Let the length of the beam, AB, = 2fl, r = the radius
of the Bphere, d = CD, the distance of the centre of the
sphere from the vertical wall, W = the weight of the beam ;
and let 6 = the required inclination of the beam to the
horiion, and ^ = the inclination of the radius AO to the
horixon. Then we have
for rertioal forces, /2 sin ^ =. fT ; (1)
for moments about B, R- 2a sin (^—9) = W-aootO; (8)
Dividing (2) by (1) we have
a sin (» — 0)
or
BlU ^
tan ^ = 2 tan #.
008 9,
(3)
mrmi
0,
her method,
ction, B', at G
n«.ii
rough the centre,
10 perpeudicnlar
, r = the radius
he centre of the
ht of the beam ;
le beam to the
adius AO to the
W; (1)
W-ao(me; (3)
(3)
CSlfTBS OF PARALLEL FORCKS.
85
Then we ha'^e, from the geometry of the fignre, the
liorizontal distance from A to the wall — the horiiontal
projection of AO 4- CD, that is.
fta coa 9 z= r COS <f> ■{• d.
(4)
From (3) and (4) a value of 6 can be obtained, and hence
tlie position of equilibrium.
Other^'ise thus : since the beam is in equilibrium under
tiie action of only three forces they must meet in a point, 0.
Geometry then gives us
2 cot 0GB = cot AOG — cot GOB = cot AOG,
or
2 tan 9 = tan ^
which is the same as (3).
63. Centre of Parallel Forces in Different Planes.
— To find the magnitude, direction, and point of
application of the resultant of any number of
parallel forces dieting on a rigid body.
The theorem of Art. 59 is evidently true also in the cose
in which neither the parallel forces nor their fixed points of
application lie in the same plane, hence, calling $ the third
co-ordinate of the point of application of the resultant, we
have for the distance of the centre of parallel forces from
tlie planes yz, zx, and xy,
m =
:lPx - _ I;Py . _ 1P«
iF
y =
ii*
£P'
Hence (Art. 59, Sch.) the equations for determining the
position of the centre of parallel forces show that the sum
of the f I' omenta of the parallel forces with respect to any
plane paraiUi to litem is equal to the mommt of their
reanUani.
li'i
»^^
ii i iiJMia i iili i iiui i ^^
•^ V
Fis*31
86 squtLTBRum of parallsl forcbb m spacxs.
64. Conditioiu of Equilibrinm of a Sjrstem of
Parallel Forces Acting upon a Rigid Body in
Space. — Let P,, P„ P,, etc., denote the forceF, and let
them be referred to three rectangular axes,
OX, or, OZ; the last parallel to the
forces ; let (a;,, y„ «,), (z„ y,, «,), etc.,
be the points of application of the forces,
/•,, Pj, etc. Ijet the direction of P^
meet the plane, ay, at Af,.
Draw if,i\r, perpendicular to the axis
of X meeting it at JV,. Apply at O, and also at JV,, two
opposing forces each equal and parallel to P^. Then the
force Pj at Mi is replaced by
(1) P, at along OZ;
(2) a couple formed of P, at Jf, and P, at JV, ;
(3) a couple formed of P, at JV, and P, at 0.
The moment of the first couple is Pijft, and this couple
may be transferred to the plane ye, which is parallel to its
original plane, without altering its effect (Art. 52). The
moment of the second couple is P,a;,, and the couple is in
the plane xz.
Replacing each force in this manner, the whole ejstem
will be equivalent to a force
■Pi +P, + Ps +etc., or SP at along OZ,
together with the couple
Ayi +-P»yt +^»y8 +eto., or I,Py, in the plane yz,
and the couple
P,a:i + Pt^g+P^x, +etc., or SPx in the plane xz.
The first couple tends to turn the body from the axis of y
to that of z round the axis of x, and the second couple
8 IN BPACMa.
r a System of
Xigid Body in
be forces, and let
Plfl<32
1 also at JVj, two
» P,. Then the
»,atJV,;
1 at 0.
and this conplc
is parallel to its
(Art. 62). The
the couple is in
he whole system
ong OZ,
the plane yt^
he plane xz.
am the axis of y
second couple
MQVTLIBRIPS Of PARALLEL FORVBa IN SVACB. 87
tends to turn the body from the avis of a; to that of t
round the axis of y. It is customary to consider those
couples as positive which tend to turn the body in tlie
direction indicated by the natural order of the letters, i. «.,
positive from x to y, round the «-axi8 ; from y to « round
the z-axis ; and from z to x round the y-axis ; and
negative in the contrery direction.
Hence the moment of the first couple is +^Py, and
therefore OX is its axis (Art. 60) ; and the moment of
the second couple is —I,Px, and therefore OV \» its axis.
The resnltant of these two couples is a single couple whose
axis is found (Art. 66) by drawing OL (in the positive
direction of the axis of x) — ^Py, and OM (in the nega-
tive direction of the axis of y) = ZPx, and completing the
parallelogram OLOM. If 00, the diagonal, is denoted by
0, we have
and
i2 = 2:P;
R being the resultant force.
Now since this single force, R, and this single couple, O,
cannot produce equilibrium (Art 64, Cor.), we must have
R = 0, and = 0, and O cannot be = unless I,Px =
and £Py = ; the conditions therefore of equilibrium are
R = 0,
IPx = 0, SPy
0.
Hence, the conditions of equilibrium of parallel forces in
space are:
7%0 sum of the forces must = 0.
I%e sum of the moments of the forces mth respect to
every plane parallel to them must = 0.
wm
88
SqUrLtBRWM OF FORCXa.
i
KM.
Fig. 33
65. ConditioiiB of EqtdUbriiim of a System of
Forces acting in any Direction on a Rigid Body Sn
Space. — Let P,, P,, Pj, etc., denote the forces, and let
them be referred to three rectangular axes, OX, OY, 02;
let (a;,, yi» «i), (*i» y». 2|). etc., be the pointa of applica-
tion of P,, Pf, etc.
Let J, be the point of application of
Pjj resolre Pj into components X,,
J",, Zj, parallel to the co-ordinate axes.
Lot the direction of Z^ meet the plane
xy at if J, and dr»»w if,iVi perpendicu- ^
lar to OX Apply at iV\ and also at
two opposing forces each equal aud par-
allel to Z,. Hence Z^ at J, or Jf, io equivalent to Z, at
O, and two couples of which the former has its moment =
Z, X iV",iIf, r= Z,yj, and may be supposed to act in tho
plane yz, and the latter has its moment = Z, x ONi =
— Z^x^ aud act* in the plane zar.
Hence Z, is replaced by Z, at 0, i. couple Zi^j in the
plane yz, and a couple — Z,a;j (Art. 64) in the plane zx.
Similarly X, may be replaced by X, at 0, a couple X,«,
in the plane zx, and a couple — X,y, in the plane xy.
And F, may be replaced by T, at 0, a couple JT^a;, in the
plane xy, and a couple — Y^Zi in the piano yz. Therefore
the force P, may be replaced by X,, F,, Z,, acting at 6>,
and three couples, of which the moments are, (Art. 54),
Z,y, — Fi«, in the plane yz, around tho axis of a?,
X,«, — ZiX^ in the plane zx, around the axis of y,
I'jo;, — X^y, in the plane xy, around the axis of «r.
By a similar resolution of all the forces we shall Lave
them replaced by the forces
sx, sr, iz,
acting at along the axes, and the couples
a SyBtein of
Rigid Body ^
>e forcoa, and ht
, OX, or, OZ;
tointfl of applica-
,f
4
*z.
r^tT'
!J-X
ivalent to Z, at
KB ita moment =
ed to act in the
Z, X ON^ =
iple Ziff^ in the
n tho plane zx.
a conple X,«,
a the plane xy.
pie J",*, in the
yz. Therefore
,, acting at 0,
■e, (Art. 54),
tho axis of x,
the axis of y,
tho axis of z.
!s we shall Lave
tiimuMm ■MMmn)'^
BqviLiBntmt of forces. 89
£ (Zy — F«) = L, suppose, in tho plane yt,
S (Xz — Zx) = M, suppose, in the plane tz,
S ( Fa; — Xy) = N, suppose, in the plane xx.
Let U be the resultant of the forces which act at 0; a,
b, c, the angles its direction makes with the axes ; then
(Art. 38),
R* = (1X)» + (2F)» + (^Zy»
sx , sr xz
coc a = -jT-> cos — -g- » cos c = -pT •
Let G be the moment of the couple ^"rhich is the result-
ant of the three couples, L,M,N; A, ^<, v, the angles its
axis makes with the co-ordinate axes ; then (Art. 5G, Sch.),
<]h = D + M^ + IP,
, L M
cos A = -^, cos /i = -g,
COS V =
N
Therefore any system of forces acting in any direction on
a rigid body in space may always be reduced to a single
force, R, and a single couple, 0, and cannot therefore pro-
duce equilibrium (Art 64, Oor.). Hence for equilibrium
we must have ^ = and Q = Q; therefore
ijLXf -H (sr)» + {-LZf = 0,
and Z» -H iP + ^ = 0.
These lead to the six conditions,
SX=0, 2r=0, SZ=0,
£ (% - r«) = 0, 1 (Xz — Zx) =r 0,
l,{Tx^Xy) = 0.
fel-
90
KXAMPLXa.
EXAMPLES.
1. If the weights, 1, 2, 3, 4, 6 lbs., act pevpendicalarly
to a stniight line at the reep^ctive distances of 1, 2, 3, 4,
5 feet from one extremity, find the resultant, and tho dis-
tance of its point of application from the first extremity.
Ans. i? = 15 lbs., x — ^ feet.
2. Four weights of 4, —7, 8, —8 lbs., act perpendicularly
to a straight line at the points A, B, C, D, so that AB =
6 feet, BC = 4 feet, CD = 2 feet ; find the resultant and
its point of application, G.
Ana. i? = 2 lbs., AG = 2 feet.
3. Two parallel forces of 23 and 42 lbs,, act at the points
A aud B, 14 inches apart; find GB to three places of
decimals. -Ans. 4.954 ins.
4. Two weights of 3 cwts. 2 qrs. 15 lbs., and 1 cwt, 3 qis.
25 lbs. are su; ported at the points A and B of a straight
line, the length AB - 3 feet 7 inches ; find AG to tlirfco
places of decimals offset Ans. 1.268 ft.
0. A bar of iron 15 inches long, weighing 12 lbs., and of
uniform thickness, has a weight of 10 lbs. suspended from
one extremity ; at what point must the bar be supported
that it may jnst balance.
Tbe weight of the bar acts at its centre.
An«. 4-^ in. from the weight.
6. A bar of uniform thickness weighs 10 lbs., and is
6 feet long ; weights of rf lbs. and 5 lbs. are suspended from
its extrumitios ; on wlut point wiU it balance ?
Am. b in. from the centre of the bar.
7. A beam 30 feof long balances itself on a point at one-
third of its length fcom the thioker end ; but when a weight
of 10 lbs. is suspemtod from tbe smaller end, tho prap must
Mte
pei'pendicalarly
»8 of 1, 2, 3, 4,
ut, and tho dis-
■st extremity.
« = 3| feet.
perpendicularly
), 80 that AB =
e resultant and
A.G = 2 feet.
ict at- the points
three places of
ru. 4.954 ins.
nd 1 cwt. 3 qid.
B of a straight
id AG to thrfco
Ins. 1.268 ft.
12 lbs., and of
suspended from
ix be supported
n the weight.
10 lbs., and is
mspendod from
0?
re of tho bar.
a point at one-
when a weight
tho prap must
n
XIXAMPLSa.
n
be moved two feet towards it, in order to maintaiu the
equilibrium. Find tho weight of the beam. Am. 90 lbs.
8. A uniform bar, 4 feet long, weighs 10 lbs., and weights
oi 30 lbs. and 40 lbs. are appended to its two extremities ;
where must the flilcrum* be placed to produce equilibrium P
An». 8 in. from the centre of the bar.
9. A bar of iron, of uniform thickness, 10 ft long, and
weighing 1^ cwt, is supported at its extremities in a hori-
zontal position, and carries a weight of 4 cwt suspended
from a point distant 3 ft from one ex^^^mity. Find the
pressures on the points of support
^ns. 3.55 cwt, and 1.05 owt
l'^ \ bar, each foot in length of which weighs 7 lbs.,
rests upon a fulcrum distjint 3 feet ttoTa one extremity ;
whttt must be its length, that a weight of 71^ 'Hs. sus-
pended from ihat extremity may just be balai i?ed by
20 lbs. suspended from the other r Ans. 9 ft
11. Five equal parallel forces act at 6 angles of a regular
hexagon, whose diagonal is a ; find the point of application
of Lheir resultant
Ana. On the diagonal passing through the sixth angle, at
a distance from it of \a.
12. A body, /*, suspended from ona end of a lever with-
out weight, is balanced by a weight of 1 lb. at the other
end of the ie^ er ; and when tho fulcrum is removed
through naif the length of the lever it requiiea 10 lbs. to
balance P ; find the weight of P. Ana. 5 lbs. or 2 lbs.
13. A carriage wheel, whoM weigbi is W and radius r,
rests upon a level road ; show tha*; the force, F, necessary
to draw the wheel ever an obstacle, of height A, is
„V2rA-A»
F=z W
-T"
* nto ■wpo't oil whieh U N^s.
0S
BXAMPLEa.
14. A beam of nniform thickoesB, 5 feet long, weighing
10 lbs., ia Bupported on two props at the ends of the beura ;
iind where a weight of 30 lbs. must be placed, so that the
pressures on the two props may be 15 lbs. aud ib lbs.
Ana, 10 ins. from the centre.
15. Forces of 3, 4, 5, lbs. act at distances of 3 ins.,
4 ins., 5 ii s. 6 ins., from the end of a rod ; at what distance
from the same end does the resultant act ?
Ans. 4| inches.
16. Four vertical forces of 4, 6, 7, 9 lbs. act at the four
corners of a squai'e ; find the point of application of the
resultant. Ana. -f^oi middle line from one of the sides.
17. A flat board 12 ins. square is suspended in a hori-
zontal position by strings attached to its four comers, A,
B, C, D, and a weight equal to the weight of the board is
Inid upon it at a point 3 ins. distant from the side AB and
4 ins. from AD ; find the relative tensions in the four
strings. An$. As | : | : | : ^.
18. A rod, AB, moves freely about the end, B, as on a
hinge. Its weight, W, acts at its middle point, and it is
kept horizontal by a string, AG, that makes an angle of 45"
with t. Fina the tension in the stiinr. . W
" Ana. — — •
19. A rod 10 inches long can turn freely about one of
its ends ; a weight of 4 lbs. is slung to a point 3 ins. from
this end^ and the rod is held by a string attached to its free
end and inclined to it at an angle uf 120° ; find the
tension in the striug when the rod is horizontal.
Ana. \ v/S lbs.
20. Two forces of 3 lbs. and 4 lbs. act at the extremities
cf a straight lever 12 ins. long, and inclined to it at angles
of 120° aud 136° respectively ; find the position of the
fulcrum. j^na. (8-3 V6) x 9.6 ins. from one end.
Ji
long, weighing
i of the beuni ;
id, 80 tliat the
J 25 lbs.
1 the (cntre.
ices of 3 ins.,
: what distance
>. 4| inches.
ct at the foar
lication of the
of the sides.
ied in a hori-
ur corners, A,
f the board is
side AB and
I in the four
id, B, as on a
)int, and it is
angle of 45"
Ans. — — •
V2
abont one of
nt 3 ins. from
led to ita free
0° ; find the
al.
t \/3 lbs.
le extremities
} it at angles
sition of the
m one end.
MXAMPLKS. ^
21. Find the true weight of a body which is found to
weigh 8 ozs. and 9 ozs. when placed in each of the scale-
pans of a false balance. ^,„. eVaozs.
22. A beam 3 ft. long, the weight of which is 10 lbs.,
and acts at its middle point, rests on a rail, with 4 lbs. hang-
ing from one end and 13 lbs. from the other ; tind the point
at which the beam is supported ; and if the weights at thu
two ends change plaoee, what weight meat be added to thn
lighter to preserve equilibrium ?
Ans. 12 ins. from one end ; 27 Ibe.
23. Two forces of 4 lbs. and 8 lbs. act at the ends of a
bar 18 ins. long and make angles ot 120° and 00° with it;
find the point in the bar at which the resultant acts.
An*. H (* — VS) ins. from the 4 lbs. end.
24. A weight of 24 lbs. is suspended by two flexible
strings, one of which is horizontal, and the other is inclined
at an angle of 30° to the vertical. What is the tension in
each string ? ^^s. 8 V3 lbs. ; 16 ^3 lbs.
25. A pole 12 ft long, weighing 25 lbs., rests with one
end against the foot of a wall, and frotn a point 2 ft. from
the other end a cord mns horizontrtlly to a point in the
wall 8 ft. from the ground ; find the tension of the cord and
the pressure of the lower end of the pole.
Ans. 11.25 lbs.; 27.4 lbs.
20. A body weighing 6 lbs. is placed on a smooth piano
which is inclined at 30° to the horizon ; find the two direo-
tionn in which a force equal to the body may act to produce
equilibrium. Also find vha*. is the pressure on the plant
in each case.
Atu. A force at 60° with the pUne, or vertically upwards ;
R = ey/3, or 0.
27. A rod, AB, 5 ft. long, without weight, is hnng fn>m
a point, 0, by two strings which are attached to its ends
i
94
SXAMPLSa.
and to the point ; the atring, AG, is 3 ft^ and BO is 4 ft in
length, and a weight of 2 lbs. is hang from A, and a weight
of 3 lbs. from B ; find the tensions of the strings.
Ana, V5lbs.; 2 V5 lbs.
28. Find the height of a cylinder, which can just rest on
an inclined plane, the angle of which is 60°, the diameter
of the cylinder being 6 ins. and its weight acting at the
middle point of its axis. Am. 3.46 ins.
29. Two equal weights, P, Q, are connected by a string
which passes over two smooth pegs, A, B, situated ii> a
horizontal line, and supports a weight, W, which hangs
from a smooth ring through which the string pusses ; find
the position of equilibrium.
Atu. The depth of the ring below the line
W
AB =
2 ^4/^ - W
AB.
80. The resultant of two forces, P, Q, acting at an angle,
$, is = {2m + 1) \//* + Q^; when they act at an angle,
a — 6, it is = (2f» — 1) VP*+ Q*; show that tan e =
«
tn — l
m + 1
81. A uniform heavy beam, AB = 2a,
rests on a smooth peg, P, and against a
smooth vertical wall, AD ; the horizontal
distance of the peg from the wall l)cing
h ; find the inclination, 0, of the beam to
the vertical, and the pressnrcs, R and S, on the wall and peg.
Ans. e = s.n-. (*)*; S = w(^f; R = ff^^^.
83. Two equal smooth cylinders rest in oortaot on two
smooth planes inclined at angles, a and (i, to the horizon ;
Pig.M
■■- 9. mtr s fmmvmt
1 BO is 4 ft in
i, and a weight
ings.
1.; 2 V5 1b8.
an just rest on
, the diameter
: acting at the
ns. 3.46 ins.
;d by a string
situated in a
which hangs
g passes; find
g at an angle,
at an angle,
that tan 6 =
rig.J4 i
wall and peg.
A*
ntaot on two
tlie horizon;
MXAMPLHa.
90
find the inclination, 9, to the horizon of the line joining
their centres. Ans. tan =- 1 (cot a — cot j3).
33. A beam, 6 ft. long, weighing 5 lbs., rests on a ver-
tical prop, CD = ^ ft. ; the lower end, A, is on a hori-
zontal plane, an-l is prevented from sliding by a string,
AD = 3^ f t. ; find the tension of the string.
Ans. T—\ lbs.
34. A uniform beam, AB, is placed with one end, A,
inside a smooth hemispherical bowl, with a point, P, rest-
ing on the edge of the bowL If AB = 3 times the radius
R, find AP. Ans. AP = 1.838 R.
35. A body, weight W, is suspended by a cord, length I,
from the point A, in a horizontal plane, and is thrust out
of its vertical position by a rod without weight, acting at
another point, B, in the horizontal plane, such that
AB = rf, and making the angle, 6, with the plane ; find
the tension, T, of the cord.
Am. r = IF ^ cot <».
36. Two heavy uniform bars, AB and
CD, movable in a vertical plane about
their extremities. A, D, which rest on a
horizontal plane and are prevented from
sliding on it; find their position of
equilibrium when leaning against each
other.
Let the bars rest against each other at B, and let
AD = o, AB = h, CD = «?, BD = a;, W and }\\ = the
weighte of AB and CD, respectively acting at thoir middle
points ; then we have
2«« »r (a» -h 4» - a^) = c FT, (a« -H *• - «») (i« + a^ - ««),
which is an equation of the fifth degree, and hence always
has one real root, the value of which may be determined
when numbers are pat for <t, b, and c
96
BXAMPLMa.
87. A parabolic curve is
placed in a vertical plane with
itfl axis vertical and vertex
downwards, and inside of it,
and against a peg in the focas,
and against the concave arc, a
smooth uniform and heavy
beam rests ; required the posi-
tion of equilibrium.
Let PB be the beam, of
length /, and of weight W,
resting on the peg at the focus,
F ; let AF =: ;> and AFP = ft
♦w
rrs.M
Awi. = 3co8-»(|)*
38. Find the form of the curve in a vertical plane such
that .1 heavy bar resting ou its concave s'de and on a peg at
a given point, say the origin, may be at rest in all
positions.
Ans. r = y + k eeo e, in which I = the length of the
bar, k an arbitrary constant, and d the inclination of the
bar to the vertical. It is the equation of the conchoid of
Nicomedes.
39. A rod whose centre of gravity is not its middle point
is hung from a smooth peg by means of a string attached
to its extremities ; find the position of equilibrium.
A)U. There are two positions in which the rod hanga
vertically, and there is a third thus defined :— Let F be the
extremity of the rod remote {torn the centre of gravity, k
the distance of the centre of gravity from the middle point
of the rod, 2a the length of the string, and 2c the length of
the rod ; then measure on the string a length FP from F
equal to o (l + - 1, and place the point P over the peg.
This will define a third poutiou of equilibrium.
3 co«r« (^f.
ical plane such
jnd on a peg ut
it rest in all
length of the
ination of the
le conchoid of
middle point
itring attached
brium.
he rod hangs
-Let F be the
of gravity, k
middle point
the length of
FP from F
)Ter the peg.
a.
MXAMPLMa.
W
40. A smooth hemisphere is flxdd on a horizontal plane,
with its convex side turned upwards and its base lying in
the plane. A heavy uniform beam, AB, rests against the
hemisphere, its extremity A being just out of contact with
the horizontal plane. Supposing that A in attached to a
rope which, passing over a smooth pulley placed vertically
over the centre t^t the hemisphere, sustains a weight, tind
the position of tqnilibrium of the beam, and the requisite
magnitude of the snspended weight
Ans. Let W be the weight of the beam, 2o its length, P
the suspended weight, r the radius of the hemisphere, h
the height of the pulley above the plane, 6 and ^ the
inclinations of the beam and rope to the horizon ; then the
position of equilibrium is defined by the equations.
r oosec = A cot ^,
r cosec* = a (tan ^ -f- cot 6),
which give the single equation for 6,
r (r — a sin 9 cos 6) = ah sin* B.
Also
W
sin 9
cos {p — (?)
= W
a sin» e Vr* + A» sin* (9
(1)
(«)
(8)
(4)
41. If, in the last example, the position and magnitude
of the beam be given, find the locus of the pulley.
Ans. A right line joining A to the point of intersection
of the reaction of the hemisphere and W.
42. If, in the same example, the extremity. A, of the
beam rest against the plane, state how the nature of the
problem is modified, and find the position of equilibrium.
Ans. The suspended weight must be given, instead of
being a result of calculation. Equation (1) still holds, but
BXAMPLSa,
li ■'
not (2) ; and tho position of equilibrium is defined by the
equation
Ph* co8^ = War Bin» 0.
43. If the fixed hemisphere be replaced by a fixed sphere
or cylinder resting on the plane, and the extremity of the
beam rest on the ground, find the position of equilibrium.
Ans. If A denote the vertical height of the pulley above
the point of contact of the sphere or cylinder with the
plane, we have
r cot 5 = A cot ^,
Pr (1 + cot 5 cot 0) cos = FTa cos 0.
44. One end, A, of a heavy uniform beam rests against a
smooth horizontal plane, and the other end, B, rests against
a smooth inclined plane ; a rope attached to B passes over
a smooth pulley situated in the inclined plane, and sustains
a given weight; find the position of equilibrium.
Let 6 be the inclination of the beam to the horizon, a the
inclination of the inclined plane, W the weight of the beam,
and P the suspended weight ; then the position of equili-
brium is defined by the equation
cos « ( ITsin « - 2P) = 0. (1)
Hence we draw two conclusions: —
(o) If the given quantities satisfy the equation IF sin a
— 2P = 0, the beam will rest in all positions.
(*) There is one position of equilibrium, namely, that in
which the beam is vertical.
This position requires that both planes be conceived as
prolonged through their line of intersection.
45. A uniform beam, AB, movable in a vertical plane
about a smooth horizontal axis fixed at one extremity, A, is
li
•taatV fttatrar^ i::tl,
defined by the
r a fixed sphere
tremity of the
t equilibrium,
iie pulley above
uder with the
cos 0.
rests against a
B, rests against
> B passes oyer
le, and sustains
um.
horizon, a the
\t of the beam,
ition of equili-
(1)
nation TFsin «
IB.
lamely, that in
)e conceived aa
vertical plane
xtremity. A, is
MXAMPLES.
99
attached by means of a rope BC, whose weight is negligible,
to a fixed point in the horizontal line through A, such
that AB =: AC; find the presau.^ on the axis.
Am. If d = <CAB, TF — weight of beam, the re-
action is
iTr|/4 8in«s + Bec»*
2
2
CHAPTER IV.
CENTRE OF GRAVITY* (CENTRE OF MASS).
66. Centre of Gravity. — Gravity i» the name given to
the force of attraction which the earth exerts on all bodies ;
the effects of this force are twofold, (1) statical in virtuo of
which all bodies exert pressure, and (3) kinetical in virtue
of which bodies if unsupported, will fall to the ground
(Art. 15). The force of gravity varies slightly from pluce
to place on the earth's surface (Art. 23) ; but at each phu;o
it is a force exerted upon every body and upon every
particle of the body in directions that are normal to the
earth's surface, and which therefore converge towards the
earth's centre ; bat as this centre is very distant compared
with the distance between the particles of any body of
ordinary magnitude, the convergence is so small that the
lines in which the force of gravity acts are sensibly parallel.
The centre of gravity of a body is the point of application
of the resultant of aU the forces of gravity which act upon
every particle of the body; and since these forces are
practically parallel, tlis problem of finding its position may
be treated in the same way as that of finding the centre of a
system of parallel forces (Arts. 46, 59, 63). The centre of
gravity may also be defined as the point at which the whole
ireighl of a body acts. If the body be supported at this
point it will rest in any position whatever.
The weight of a bod« is the resuUant of all the forces of
gravity which act upon every particle of it, and is equal in
magnitude and directly opposite to the force which mil just
support the body. Since the centre of gravity is here
* CsUed also OntAw of Man and Cmtrt ot tntrUa ; ud the tenn OtntroU hM
Utelf come into qm .o destgnate It.
iF MASS).
name given to
I ou all bodies ;
5al in virtue of
stical in virtue
to the {^round
itly from pluce
t at each plii«o
nd ujjon every
normal to t lie
re towards the
tant compared
any body of
small that the
insibly parallel.
t of application
which act upon
ese forces are
position may
the centre of a
The centre of
^hich the whole
)ported at this
n the forces of
nd is equal in
vhich mil just
rarity is here
e term OenlnM hu
CMyTRH OF aSAVTTT.
101
regarded as the centre of parallel forces, it is more t'mly
conceived of as the "cdntre of mass;" yet m deference to
usage we shall call the point the ''centre of gravity."
67. Planes of Symmetry.— Asm of Symmetry.— If
a homogeneous body be symmetrical with reference to any
plane, the centre of gravity is in that plane.
If two or more such planes of symmetry intersect in
one line, or axis of symmetry, the centre of gravity is in
that axis.
If three or more planes of symmetry, intersect each other
in a point, that point is the centre of gravity.
By observing these principles of the symmetry of the
figure there are many cases where the centre of gravity is
known at once ; thus, the centre of gravity of a straight
line is its middle point. The centre of gravity of a circle
or of its circumference, or of a sphere or of its surface, is its
centre. The centre of gravity of a parallelogram or of ita
perimeter is the point in which the diagonals intersect.
The centre of gravity of a cylinder or of its surface is the
middle of its axis; and in a similar manner we sh^
frequently conclude from the symmetry of the figure, that
the centre of gravity of a body is in a particular line which
can be at once determined.
When we speak of the centre of gravity of a line, we
are really considering a material lino of the same density
and thickness throughout, whose section is infinitesimal ;
and when we consider the centre of gravity of any surface,
wf are really considering the surface as a thin uniform
lamina, the thickness of which, being uniform, can be
neglected.
68. Body Stupended from a Voiat— When a body is
suspended from a point about which it can turn freely, it
will rest with its centre of gravity in the vertical line pass itig
•Qugh the point of suspension. For, if the point of sus-
1»
BODY aVPPORTKD ON A SURPACB.
peiision and the centre of gravity are not in a vertical line,
the weight acting vertically downwards at the contro of
gravity and the reaction of the point of suspensiou vert'cally
upwards form a statical couple and hence there will be
rotation.
69. Body Supported on a Sufkce.— When a body is
placed on a surface it will stand or fall according as the
vertical line through the centre of gravity falls within or
without the base. For if it falls within the base the reaction
of the surface upward and the action of the weight -down-
ward will be in the same vertical line, and so there will be
equilibrium. But if it falls without the base the reaction
of the surface upward and the action of the weight down-
ward form a statical couple and hence the body will rotate
and fall.
7D. Different Kinds of Sqnilibrinm.— According to
the proposition just proved (Art. 69) a body ought to rest
upon a single point without falling, provided that its cent^
of gravity is placed in the vertical line through the point
which forms its base. And, in fact, a body so situated
would be, mathematically speaking, in a position of equili-
brium, though practically the equilibrium would not sub-
sist The body would be moved from its position by the
least force, and if left to itself it would depart further from
it, and never return to that position again. This kind of
equilibrium, and that which is practically possible, are
distinguished by the names of unstable and stable. Thus
an egg on either end is in a position of unstable equilibrium,
but when resting on it« side it is in a position of stable
equilibrium. The distinction may be defined generally as
follows :
When the body is in such a position that if slightly dis-
placed it tends to reftim to its original position, the equili-
brium is stable. When it tends to move further away from
cs.
emitKg or oravttt of a trijuiols. 103
ft vertical lino,
) tiie contra of
nsiou vert'cally
tbore ivill be
When a body is
wording as the
falls within or
»8e the reaction
I weight -do wn-
> there will be
le the reaction
weight down-
ody will rotate
-According to
ought to rest
that its cent;^
i]gh the point
iy 80 situated
ition of equili-
onld not snb-
)8ition by the
; further from
This kind of
possible, are
stable. Thus
'e equilibrJum,
tion of stable
generally as
f slightly dis-
)n, the equili-
er away from
its orifiutf J^sition, its equilibrium is unstable. When it
reinaim in its new position, its equilibrium i . neutral. A
sphere or o^ndrical roller, resting on a horizontal surface,
{equilibrium. In stahk equilihrium the centre
IS m
of grwntp occupies the lowest possible position; and in
unstable it occupies the highest position.
Wc shall first give a few elementary examples.
71. Oiven the Centres of Qravlty of two Mauea,
Ml and M„ to find the Centre of Qravity of the two
Masses as one System.— Let gi, denote the centre of
gravity of the mass if,, and g, the centre of gravity of the
mass Mg. Join g^ g, and divide it at the point, O, so that
Og.- M\'
ma6ses as one system (Art 46).
then G is the centre of gravity of the two
72. Oiven tbe Centre of Gravily of a Body of
Mass, M, and also the Centre of Ghnvity of a part
of the Body of Mass, m, to find the Centre of
Qravity of the remainder. — Let denote the centre of
gravity of the mass, M, and jr, the centre of gravity of the
mass, nt|. Join Og^ and produce it through O tog,, so tliat
^^ =r ^_* — , then g, is the centre of gravity of the
remainder (Art 45).
73. Centre of Gravity of a Triangnlar Figure of
Uniform Thickness and Density. — Let ABO be the
triangle; bisect BC in D, and joiu AD;
draw any line bdc parallel to BC ; then it
is evident that this line will be bisected by
AD in d, and will therefore have its centre
of gravity at d ; similarly every line in the
triangle parallel to BO will have its centre
of gn^vity in AD, and therefore the centre of gravity of the
triangle must be somewhere in AD.
Frg.37
■ ?>*/■'
104
CXXTBB OF ORAVnr or il TRZAJteLM.
In like )> .annor the pentre of gravity must lie on the line
BE which joiuB B to the middle point of AC. It is there-
fore at t le intersection, G, of AD and BE.
Join DE, whih wiU be parallel to Afi ; then the triangle^
ABO, DEG', are similar : therefore
AG
GD
AB_ BO
DE~DO
i'
or GD = iAG = ^AD.
Hence, to find ih« centre of gravity of a triangle, bisect any .
side, join the point of bisection with the opposite angle, the
centre of gravity Ivss one third the way up this bisection.
Cor. 1.— If three equal particles be placed at the vertices
of the triangle ABC their centre of gravity will coincide
wich that of the triangle.
For, the centre of gravity of the two equal particles at B
and is the middle point of BO, and the centre of gravity
of the three lies on the line joininjj this point to A.
Similarly, it lief> on the line joining B to the middle of AC.
Therefore, etc.
Cob. 2.— The centre of gravity of any plane polygon may
be found by dividing it into triangles, finding the centre of
gravity of each triangle, and then by Art 69 deducing the
cenb'c of gravity of the whole figure.
Let the coK)rdinate8 of A, referred to any axes,
«. ; those of B!, x,, y„ «, ; and those of 0, ar„
*t »
Cob. 8.
be a;,, yj, .. ,
y„ «, ; then (Art 69), the co-ordinates, 5, y, t, of the centre
of gravity of three equal particles placed at A B, 0, respec-
tively, are
• =
3
* ~ 3
+ y.
; = 'jL±_«ijtii;
M..
i«ta
AjreLK.
t lie on the line
0. It is there-
en the trianglcSy
%ngle, Usect any .
poaite angle, the
'tis bisection.
d at the Tcrtices
ty will coincide
1 particles at B
entre of gravity
point to A.
middle of AC.
18
ms polygon may
ig the centre of
9 dedacing the
>ed to any axes,
i>
those of C, X
•, of the centre
A. B, 0, respec-
+ y.
CJWVTBi OF ORAVTTr OF A PTRAXID,
which aro also the co-ordinates of the<»ntre of gravity «f
the triangle ABO (Oor. 1).
74. Cemtre of Qncvity of a Trlangnlar Pyramid of
Unifonn Density.— Let D-ABO be a triangular pyramid;
bisect AO at E; join BE, DE; take EP
= ^EB, tLan P is the centre of gravity of
ABC (Art 73). Join FD ; draw ab, h;-, ea
parallel to AB, BC, CA respectively, and
let DF meet the plane, abc, at /; join bf
and produce it to meet DE at e. Then ^^
since in the triangle ADC, ac is parallel ^*''* °
to AO, and DE bisects AC, e ip tne middle point of «c; also
but
'herefore
K-W - sL.
BF ~ DF ~ EF •
EF = IBF,
therefore /is the centre of gravity of the triangle a Jc (Ari
73). Now if we suppope the pyrnmid to be divided by
planes parallel to ABC into an indefinitely great number of
triangular laminae, each of these laminae has its centre of
fjravity in DF. Henoe the centre of gravity of the pyramid
if^ in DF.
Again, take EH r= JED ; join HB cutting DF at G.
Then, as before the centre of the pyramid must be on BH.
It is therefore at the intersection, G, of the liues DF
and BH.
Join FH ; then PH is parallel to DB. Also, EP = pB,
therefore FH = JDB ; and in the similar triangles, FGH
and BGD, we nave
therefore
PG _ FH _ 1
M ~ D B "~ 8 '
FG = JDG = iDF.
106
CMSTRM OF QRA. VrrT OF A CONS.
\.s'
3
u
m
Eli'
I
nonce, the centre of gravity of the pyramid ia one-fourth
of the way up ike line joining the centre of gravity of the
base with the vertex. (Todhuni a Statics, p. 108. Also
Pratt's Mechanics, p. 63.)
Cor. 1.— The centre of gravity of four equal particles
placed at the vortjices oi the pyramid coincideo with the
centre of gravity of the pyramid.
Cob. 2.— Let (x,, yj, «i) be one of the vertices ; («„ y„ «,)
a second vertex, and so on ; let (i, y, i) bto the centre of
gravity of the pyramid ; then (Art 59)
• = J (*, + «, + ;r, + «4). •
» = i (yi + yi + y» + ^t).
; = J («, + «,+«, + «4).
Cor. 3.— The perpendicular distance oi' t'le centre of
gravity of a triangular pyramid from the base is 64031 to {
of the height of the pyramid.
75. 0«ntn of Qnvitj cX a Cone of Uniform
Denaity havjog mnf Plana Baaa.— Consider a pyramid
whose base ij a polygon of any number of sides. Divide
the base into triangles ; join the vertex of the pyramid with
the vertices of all the triangles ; ihen we may consider the
pyramid as composed of a number of triangular pyramids.
Now the centre of gravity of each of these triangular
jiyramids lies in a plane whose distance fr&m the base is
one-fourth of the height of the pyramid (Art 74, Cor. 3) ;
tf ,refore the centre of gravity of the whole pyramid lies in
this plane, ». u, i*« perpendicular distance from the base is
one-fourth of the height of the pyramid.
Again, if we suppose the pyramid to be divided into an
i?»deflnitely great yiumber of laminu. as in Art 74, each of
these lamina!) has its centre of gravity on the right UiM
nid is om'fourih
of gravity of the
:8, p. 108. Also
r equal particles
incidef) with the
tice«;(a;„y„«,)
to the centrt) of
)i' the centre of
Nisc is e4a3l to \
M of Uniform
laidcr a pyramid
of sides. Divide
he pyramid with
nay consider tho
ovular pyramidfi.
Iicse triangnlar
f^m tho base is
Art. 74, Cor. 3) ;
pyramid lies io
from the base is
divided iato an
Art 74, each of
the right liiM
csNTRs OF Qturmr.
107
joining the vertex to the centre of gNTity of the base ; and
hence the centre of gravity of the whole pyramid lies on
tills Une, and hence it mast bo one-fourth tiie way up this
line. There is no limit to the number of sides of the poly-
gun which forms the base of the pyramid, and henoe they
may form a continnous curve.
Therefoie, the centre of gravity of a cone whose base is
any plane curve whatever is found by joining the centre of
gravity of the base to the vertex, and taking a point one-
fourth of the tvay up this line.
76. Contre of Ckavity of tho Fnutom of a Pyra-
mid. — Let ALC-abc (Fig. 38) be the frustum, formed by
ilie removal of the pyramid, D-abc, from the whole pyramid,
D-ABO ; let A, and ff be the perpendicalar heights of these
])yramids, respectively; let m and if denote their masses;
and let Xi, «,, denote the prpendicular distances of the
centres of gravity of the pyramids D-ABO, and D-abe, and
tho frustum, from the base \ then we have (Art 60, Soh. 1)
Jf«j = i ( J/" — m) + ««, ;
or
Rat
mz
M-m
». = -r;
i.
(1)
4'
= (ir-»,) + V = ^-**»-
Also, the maasea of the pyramids are to each other an the!.
M)lumos* by (1) of Art \%, and therefore as the oubsa of
tlieir heights. Henoe (1) becomes
* irtiM bodiM Mw iMMMRenmut, tlw roiniiiM or Ike weigtato m« proportkwal to
I 111- oMwaM, and m»f be mibi>tlwia4 kn tlHM.
108
SXAMPLSa.
H-
H* + iTA, + A, *
(2)
Instead of the heights we may use any two corresponding
lines in the lower and npper baaes, to which the heights are
proportional, as for example AB and <dt. Denoting these
lines by a and h, and the altitude of the firostum by h, {i)
becomes
This is true of a fnistum of a pynunid on any base, a
and b being homologons sides of the two ends, and hence it
is true of the frustum of a cone standing on any plane baae.
EXAM PL.ES.
1. Find the centre of gravity of a t/apeioid in terms of
the lengths of the two parallel sides, a and b, and of the
line. A, joining their middle points.
Take moments with raferanoe to the longer pnimllol tide.
An$. On the line bisecting the parallel sides and at a
A a + 2i
distance fh>m its lower end =
3 a-\'b
2. If out of any cone a similar cone ic cut so that their
axes are in the same line and their bases in the same plane.
And the height of the centre of gravity of tha remainder
above the base.
Take momenta with refefenoe to the
VLV
+JV
(2)
'0 oorresponding
I the heights are
Denoting these
ostum hy A, (2)
(3)
on any base, a
ds, and hence it
I any plane base.
oid in t«rm8 of
d b, and of the
lol tide.
sides and at a
t so that their
the same plane,
thd remainder
llfTSOBATIOJf FORMULA.
109
Ana, J . TTZTTt* "^^^^ *' " *^® height of the"^)^iginal
cone, and h', the height of that which is cut out of it.
3. If out of any cone another cone is cut having the
same base and their axes in the same line, find the height
of the centre of gravity of the remain«ier above the base.
Ans. J(A + Ai), where h and A, are the respective
heights of the original cone and the one that is cut out
uf it.
4. If out of any right cylinder a cone is cut of the same
base and height, find the centre of gravity of the remainder.
Ans. Iths of the height above the base.
77. iDvestigKtions Iiivolvivf Integration.— The
general formula for the co-ordinates of the centre of gravity
vary according as we consider a material line, an area or
thin lamina, or a solid ; and assume different forms accord-
ing to the manner in which the matter is supposed to be
divided into mfinitesimal elements.
In either case the principle is the same ; the quantity of
matter is divided irto an infinite number of inficitesimal
elements, the mass of the element being dm ; multiplying
the element by its co-ordinate, x, for example, we get
X ' dm, which is the moment of the element* with respect to
the plane yt (Art. 63) ; and /x • dm is the sum of the
moments of all the elements with respect to the plane yx,
and which corresponds to SPz of Art 63. Also, /dm is
the sum of the nuases of all the elements which correspond
to £P of the come Article. Hence, dividing the former by
the latter we have
* The momsnt of th« ft>TC* Mtlng on elemAnt dm U Mriotlj dm -gic, bnt cinM
the eooittnt g appean In botb terma of expreeilnn ror co-ordinate* of centre of
Kravlty, it may be omitted and It becomet more conrenlent to ipaak of the ptomtnt
ot the timtmt, mtaniag by It the product of the maM of the al«awF>t dm, and it«
arni.z. Tha momant of an ato a a i it maaa nw a lt« «g» a dateraUUng the poaltfcm
of the oantrt of gmTUj.
110
CKIfTKS OF QBA VITT OF A LiytU
m
fx • dm
/dm
Similarly
i = ^
dm
t =
/dm *
/t'dm ^
~/dm'*
(1)
(2)
(3)
the limits of integration being determined by the form of
the body ; the sign, /, is used as a general symbol of sum-
mation, to be replaced by the symbols of single, doable, or
triple integration, according as dm denotes the mass of an
elementary length or surface or solid. Hence, the co-or-
dinate of the centre of gravity referred to any plane is i tal
to the gum of the moments of the elements of the mass
referred to the same plane divided by the sum of the elements,
or the whole mass. If the body has a plane of symmetry
(Art. 67), we may take it to be the plane xy, and only (1)
and (2) are necessary. If it has an axis of symmetry we
may '^ke it to bo the axis of z, and only (1) is necessary.
79. Centn of Qn^ity of the Are of a Conro.— If
the body whose centre of gravity we want is a material line
in the form of the arc of any curve, dm denotes the mass of
an elementary length of the carve.
Let ds =■ the length of an element of the curve ; let
h = the area of a normal section of the curve at the point
ix, y, z), and let p = the density of the matter at this
point Then (Art. 11), we have dm = kpds, which is the
mass of the element ; multiplying this mass by its co-or-
dinate, X, for example, we have the moment of the element,
(kftxds), with respect to the plane, ? ».
Hence, substituting for dm in (1), (2), (8), of Art. 77,
the linear element, kpds, we obtain, for the position of the
centre of gravity of a b«jdy in the form of any curve, the
oquutioni
u,
(2)
(3)
1 by the form of
I symbol of Bum-
single, doable, or
I the mass of an
Hence, the co-or-
\ny plane is t lal
ents of the mass
m of the eUmenis,
Hie of symmetry
ty, and only (1)
of symmetry we
) is necessary.
9f aCnnr*.— If
s a material line
lotea the mass of
' the curve; let
rte at the point
matter at this
ds, which is the
ass by its co-or-
of the element,
(8), of Art. 77,
position of the
any cnrre, the
MXAkPLJSH.
fkpxdi
C :=
-_/kpyds
^ - J'kpds '
• =
/kptd$
fkpds'
(1)
(8)
(8)
The quantities h and p mast be given as functions of the
position of the point {x, y, «) before the integrations can
be performed.
If the curve is of doable corvature all three equations
required. If it is a plane curve, we may take it to be
are
in the plane xy, and (1) and (2) are sufficient to determine
the centre of gravity, since i = 0. If the curve has an axis
of symmetry, the axis of * may be made to coincide with
it, and (1) is sufficient
«
KXAMPL.ES.
1. T find the centre of gravity of a circular 4ro of uni-
form thickness and density.
Let BC be the arc, A its middle point,
and the centre of the circle. Theti as
the arc is symmetrical with respect to OA
its centre of gravity must lie on this line.
Take the origin at 0, and OA as axis of x.
Then, since k and p are constant, (1) be-
comes
fxds
m
X being the co-ordinate of any point, P, in the arc. Let 9
be the angle POA, and a the radius of the circle, and let
=: the angle BOA. Then
II
m
113
and
Hence m =
/!-
BXAMPLSa.
2 = a COS 9,
da = ade.
J COB Od$
COB e dO
= a
f*a do PdO
= a
Bin a
Therefore, the distance of the centre of gravity of the arc of
a circle from tJte centre is the product of the radius and the
chord of the arc divided by the length of the arc.
Cob.— The distance of the centre of gravity of a semi-
circle* from the centre is
2a
3. Find tb« centre of gravity of the quadrant, AT/, (Fig.
89), referred to the co-ordinate axes OZ, OF.
The equation of the circle is
a* + y« = a».
. ^ <?y _ Vda? + dt^ _ds _
J axdxy
.'., xde =z /
y
and
yds sz adx,
J ache
ds = ;
y
/
which in (1) and (3), after canceling h and p, give
rxdx r -la
r dx
h-sl
f;»)
= a
Bin a
\y of the are of
radius and the
\rc.
ivity of a semi-
tint, AT/, (Fig.
r
th
give
8a
iza\
SXAMPLBS.
/** H
y =
r__dx_
-v/o»-a?
hil
2a
3. Find the centre of gravity of the arc of a cycloid.
Take the origin at the starting point of the cycloid, and
let the base be taken as the axis of x. The eqaaUou of
the curve is
X = a vers-» ^ — (2ay — j/*)* j
&! _ dy __ ds ,
y*~(2a-y)*~(2a)**
it is evident that the cenire of gravity will be in the axis of
the cycloid ; therefore i=:na; and as k and p are constant,
(2) becomes
l^_ffdj/_
c/q (2a-y)* _
•^« (2a -y)*
OoB. — For the aio of a semi-cycloid, we get
5 = fi, jr = |a.
4. Find tbe centre of gravity of a circular arc of uniform
section, the density varying as the length of the arc from
one extremity.
Let AB (Fig. 39), be the arc ; let ft be the density at the
units distance from A, then fts will be the density at the
distance « from A ; let OA be the axis of x, and a the
Z. AOB. Then, putting fu for p, and a cos 9, a sin 0, a (J9,
and ofl, for x, y, dt, and a, in (1) and (2),
wm
Bin edB
Odd
of a loop of a
W, I being the
to.
= o»
2*-l
rod, the den-
hc distance of
to cdndde with
CSlfTXE OF OMA VTTT OF AN ARSA.
116
7. Find the oentro of gravity of the arc of a semi-car-
dioid, its cqaation being
r = o (1 + cos fl)'
Am. The co-ordinates of the centre of gravity referred
to the aiis of the curve and a perpendicalar mrongh the
cnsp, as axes of x and y, are
^.
'i^
« =
y = ^«-
c
^
79. Centra of Qravity of a Plane
Area. — Let ABCD be an area bounded
by the ordinates, AC and BD, the carve
AB whose equation is given, and the axis
of a; ; it is required to find the centre of
gravity of this area, the lamina (Art 67)
being supposed of uniform thickness and density. We
divide the a.ea into an infinite number of infinitesimal
elements (Art. 77). Suppose this to be done by drawing
ordinates to the curve. Let PM and QN be tw6 consecu-
tive ordinates, let (x, y) be the point, P, and let g be the
centre of gravity of the trapezoid, MPQN, whose breadth is
dx and whose parallel sides are y and y + dy. The area of
this trapezoid \Aydx, (CaL, Art 184).
Let p be the density and k the thickness of the lamina.
Then (Art 11) we have dm — kpy dx, which is the mass
of the element MPQlJ ; multiplying this mass by its co-or-
dinate, X, for example, we have tuo moment of the element
{lepxydx), with respect to OY, and multiplying by the
other co-ordinate, \y, we have the moment with respect to
OX. Hence, substituting for dm in (1) and (2) of Art. 77,
the surface element, hpy dz, and remembering that k and p
are constants, we obtain, for the position of the centre of
gravity of a body in the form of a plane area, the equations,
116 MXA.MPLX&
yydx.' *-*jf^* (1)
the integrations extending over the whole area CABD.
EXAM Pt.ES.
1. Find the centre of gravity of the area of a semi-parab-
ola whose equation is y* r= 2px.
Let a = the axis, and * the extreme ordinate, then we
have from (1)
J^V2p z* dx J*^ dx
— _L
_ *'o
J V^x^dx Jx^dx
y = i
pa
f 2pzdx
J V2px^dx
= a/P:^
X"'^
/ x^dx
= |4.
2. Find the centre of gravity of the area of an elliptic
quadrant whose equation is
y = ^ Va» — a^.
Here
. - 4«
(1)
rea CAfiD.
)f a semi-parab-
linate, then we
xdx
— =1*.
*dx
% of an elliptio
^xdx
)^dx '
MXAMPLtCa.
4»
117
.-. y =
3^'
Hence for the centre of gravity of the area of a circnlar
quadrant we have
- - 4<i
3.. Find the centre of gravity of the area of a semi-
cycloid.
Take the axis of the cnrve as axis of ar, and a tangent at
the highest point as axis of y ; then the equation is (AnaL
Geom., Art 167),
y = vers-*- + 's/'Hax — a*]
where a is the radius of the generating circle. Prom (1) wo
have
CD =
J^ydx [yx-fxdy'^
\yx - /(2ax - 3?)*dx'^ * TO. 2a - 4»ra» '
since when x = and 2a, y = Oanina.
Also,
■i ,
Z'
118 POLAB ELSMElfTS 01^ A PLANE AJIBA.
[y»a:-3 J^y (^csr - «»)* dxT
37ra*
[fx-^J*{2ax~a?')i yevr^-ch—%f{^tax-7?) dx'f
37r««
[fx-%a3?4-\:t» — %nj{%ax — a*)i verg-' - dxT
arrfa»
3 2
3»ra»
3ira»
a
which the student can verify by asBuming
vers"* - = fl.
a
(See Todhunter's Statics, p. 118.)
80. Polar Btomenta of a Piano
Area. — Let AB be the arc of a curve,
and let it be required to find the centre of
gravity of the area bounded by the arc
AB and the extreme radii-vecton*, OA
and OB, drawn from the pole, 0, to the
extremities of the arc.
Divide the area into infinitesimal trianj^les, such as POQ,
included between two oontrtLfCutive radii-vectors, OP and
0(j. I^t {r-9) be the point, P, then the area of the
element, POQ =:ft^c^ (Cal., Art. 191) ; and if the thick-
ness and density of the lamina are uniform, the centre of
K19.41
UtX—7?) d£
a Jo
r
such as POQ,
itoni, OP and
e area of the
if the thiok-
the centre of
MXAMPtm.
in
gravity of this elementary triangle will be on a straight lino
drawn from to the middle of PQ, and at a distance of
two-thirds of this straight line from (Art 73). Hence
the co-ordinates of the centre of gravity, g, of POQ, are
OM and Mi/, or,
fr cos 0, and \r sin 6.
Hence, (Art 77),
- /|rcosgjJfJ<?9 /f*oo80<l9 .
, /frsing-jH ^W „ , /r'singtJ9 .
""" J-\f*dO ""* J'r*dd '
the integrations extending over the whole area, AOB.
(1)
(2)
EXAMPLB.
Find the centre of gravity of the area of a loop of Ber-
nonilli's Lemniscate whose equation is r* = «^ cos 5W.
As the axis of the loop is symmetrical with respect to
the axis of a?, y = 0, and the abscissa of the centre of
gravity of the whole loop is evidently the same as that of
the half-loop above the axis. Substituting in (1) for r its
value a cos^ 20, we have
• = fa
X"'
cost 20 cos 9 (i9
X'
cos 20 do
s= \aj{) — 2 sin* 0)< d siii 0.
Put sin = -7?, then
lao
DOUBLB IXTKOBATTOy.
_ Ja n^^^^ ^ J^ . I J (CaL, Art. 167).
3^/2
• =
Tta
4V^
81. Double Integrattoa— Polar Fonniil«.— When
the density of the lamina varies from point to point, it may
be BeoesBary to divido it into el. met ' the isecond order
instead of rectangular or triar^tu;.> aements of the first
order (Arts. 79 and 80).
Suppose that the density of the hunina AOB (Fig. 41),
is not nniform. If we divide it into triangular eletnents,
POQ, the element of mass will be no longer proportional to
the element of area, POQ = ^i*d9', nor will the centre of
gravity of the triangle, POC* ^ ^ distant from 0.
Let a series of circles be described with as a centre,
the distance between any two successive circles being dr.
These circles will divide the triangle, POQ, into an infinite
number of rectangular elements, abed = rdBdr. If k is
the thickness and p is the density of the lamina at this ele-
ment, the elemeii' of mass will hedtn — kprdSdr; and
the coH>rdinates of its centre of gravity will b '' ri?9 6 and
r sin 0, Hence, from (1) and (3) of Art '/7. f n'*
m =
and
r fk pr done rdBdr f fkpr* cop & n'J - r
ffhr de dr ff^ ^ *''*
V —
/A'*
Bin e de dr
//^
d»dr
(1)
(•)
In each of these integrals the values of k and p are to be
snbstitnted in terms of r and 6, and the intQgrstiona taken
between proper limits.
3al.,Art.l57).
nnlaB.— When
I point, it may
Hocond order
ts of the first
OB (Fig. 41),
ular eletaientfl,
>roportional to
the centre of
>mO.
as a centre,
^les being dr.
tto an infinite
ddr. If k is
oa at this ele-
prdBdr; and
>
K
' fi?» 8 and
6 rrj r.r
- , (1)
)dr
(2)
d p are to be
rations taken
MXAMPLM.
EXAMPLB.
m
Find the centre of gravity of the area of a cardioid in
which the density at a point increases directly as its distance
from the cusp.
Let ft = the density at the unit's distance from the
cusp, then p = (ir, is the density at the distance r from
the cusp.
As the axis of the curve h an axis of symmetry (Art 67),
y = 0, and the abscissa ol the whole curve is the same as
for the halfabove the axis ; then (1) becomes
w =
= }
f /f*cmedddr
r rt*dedr
Jr*0Med6
fr*dB
by performing the reintegration.
The equation of the curve is
d
r s a (1 + COB 9) = 2a oo*» 5
Substituting this value for r, and putting ^ = 0, we have
2 s: la
J 00^ ^ (2 cos* ^ — 1) «^
/ oo^^d^
= H«.
in
RXCTJLNaVLAB FORMULAE.
82. DonUe Integntioii.— Reotugiilar Fommte.—
Let » aeries of consecative straight lines be drawn parallel
to the axea of x and y respectively, diTidiog the area, ABOD,
(Fig. 40), into an infinite nnmber o<' rectangular elements
of the second order. Then the area of each element, as
abed, — d^dy; and if k and p are the thickness and density
of the lamina at this element, the element of man will be
dm s= kpdxdy, and the co-ordinates of its centre of gravity
will be X and y. Uenoe from (1) and (2) of Art 77, we
have
• =
9 —
J jk (Kcdxdif
yykpdxdy
JJhpydxdy
J I lepdxdy
the integrations being taken between proper limita
(1)
(»)
EXAMPLE
Find the centre of gravity of tVa area of a cycloid the
density of which varies as the nth power of the distance
from the base.
Take the base as the axis of x and the starting point as
the ori^!ii. Then the equation of the curve is
• s a vers-*^ - (Say - y»)* ;
dx-
V2ay-
r Fonnnlie.—
drawn parallel
le area, ABOD,
^lar elements
ich element, as
ess and density
f man will be
sntre of gravity
of Art 77, we
(1)
(2)
limits.
a cycloid tbe
the diatance
rting point as
aUBFACa OP RSVOLUnON.
123
Let p = ^y" = density at the distanoe y from the base.
It is evident that tbe centre of gravity will be in the axis of
the cycloid ; therefore i = rra ; and as it is constant (3)
becomes
y =
J^J^trdydx
n + I t/p *
_n + l ^o V2ay — y «
«-»-2* n + 3 "
'v/2ffy — y»
rtf*** dy
V2ffy — y«.
r
V2«y — p
y =
n_-f_l 2n + 5
n + 2" n + 3 '
83. Centre of OniTitjr of • Bnrfiuse of Revola-
tion. — Let a surface be generated by tbe revolution of the
curve, aB (Fig. 40), round the axis of x. Then the
elementary arc, PQ, (= ds), generates an element of the
nirf«o« whose aree = Stry di (Gal., Art 1.93). It k ia the
tbicknoB8 and p the density of the lamina or shell in this
elementary zone, the element of mass will be dm = %nkpy ds.
Also the centre of gravity of this zone is in the axis ul x at
134
SXAMPLSa.
the point M whose absoiaaa is x and ordinate 0. Hen<» (1)
of Art. 77 becomes, after cancelling *in,
(1)
Jkpxy da
X := — — __
Jkpydt
(.he integrations being taken between proper limits.
BXAMPLSS.
1. Find the centre of gravity of the surface formed by
the revolution of a semi-cycloid round its base.
The equation of the generating curve is
a
» =z a vers-* * — \^^y — y*;
d»
or
V3a-y
which in (1) gives, after cancelling 's/'ia kp.
2. Find the centre of gravity of the sarteoe formed by
the revolution of a semi- cycloid round its axis.
It is clear that the centre of gravity lies on the axis of
the curve ; hence y = 0. '
0. Hence (1)
(1)
[imits.
ace formed b;
e.
e formed by
D the axis of
MXAMPLM8.
The equation of the generating carve is
135
y = a yerr-> - + V-Sox — a».
a
Here
which in (1) gires
ds = VZax'^dx,
J^ifxidx
m =
J^y7r\ix
[lya;* — \fx's/%a — x dxT
\%yx^ — 2 /V2a — xifoT*
_ |ff<i(2a)*-- AW*
"" »Tr (2o)* — |(3a)»
= A«
15 tr-8
8. Find the centre of grayity of the sarfaoe formed by
the revelation of the semi-cycloid round the axis of y in the
last example, t. e., round the tangent to the carve at the
highest point
Ah8. P = jg (ISrr
8).
i iSii
tse
ANT CUBVBD BVaFACK.
M. Centre of Ghra-vity of Any Onnred Sarface.—
Ixjt there be a shell having any given curved surface for
one of its boandaries; and let * = tb? thickness, p = the
density, and ds — the area of an element of the surface at
the point {x, y, z); then (1) of Art. 83 becomes
/ kpxda
(1)
and similar expressions for y and i.
Substituting the value of ds (CaL, Art 201) and cancel-
ling k and p, we have
r r I, , difi , rfa»\i
X =
ds*
ffh%-'4f-^y
BX AM PLE8.
1. Find the centre of gravity of one-eighth of the
surface of a sphere.
Here
«» + y* + «^ = a«.
V "^ «te» "^ dyV - (o» - «» - y»)**
rr ^dxdy
- — ^ ^ ( < i» - ;>^ - fy
**'•"/* r <^dy *
J J {a*-^ r* - y«)*
I BnifAoe.—
!<i surface for
ness, p = the
he surface at
(1)
I and caucel-
\>
;hth of the
30i«/D or aUVOLOTION.
137
First perform the y-inljegn-
tion, X being constant, firom
y ^0 to y = -W = Sfi =
\/«* — a^ ; the effect will be
to sum up all the elements
similar to pq from H to I.
The eflbct of » subsequent
a;-integration will be to sum
all these elemental strips that
are comprised in the surface
of which OAB is the projec-
tion, and the limits of this iutegratiou are x = and
X = OA =z a. Hence t
Fig. ♦?.
a =
Jo Jn (a'
xdxdy
»»)*
P» /f' dxdy
Jn Jq (a» - a» - y»)*
J^rxdz
id.
Similarly
P = fi, i =r i<i.
2. Find the centre of gravity of onn-eighth of the sorfiMM
of the sphere if the density varies as the «-ordinate to any
point of it. Here p = fix.
. . 4a . 4a . 3a
J,«. . = 3-; , = _;. = -.
85. Centre of Qtmvity of * Solid of Revolution. —
Let a solid be generated by the revolution of the curve, AB»
(Fig. 40), round the axis of x. Then the clemeutury
rectangle, PQNM, {=yflx), gen^raitea m element q£ th»
198
SOLID or RBVOLVnON.
solid whose volnmc = n^dx (Oal, Art. 908). Henoe if the
density of the solid is aniform, we have for the position of
the centre of gravity ^whlch evidently is in the axis of x),
I iTi/hi dx I ^xdx
^dx
>dz
(1)
the int^rations being extended over the whole area,
CABD, of the bounding curve.
If the density varies, the element of mass may require to
be taken differently. If the density varies with z alone, t. c,
if it is uniform all over the rectangular strip, PQNM, the
volume may be divided up as already done, and the element
of mass = Ttfyj^ dx. Hence, we shall have v\ this case,
m =
Jpj/*xdx
— — — ■ — ■ ' ■ - — - •
fpfdx
(2)
If the density varies as y alone, we may take a rectangular
element of area of the second order, dx dy, at the point
(x, y) ; this area will generate an element of volume
= 2ny dx dy ; therefore the element of mass = 2irpy dx dy,
and we have
» =
JJpxydx :y
fj'pydxdy
(8)
the y-integrations being performed first, from to y, the
ordinate of a point P, on the bounding curve ; and then
the x-integrations from OC to OD.
mm'
8). Henoe if the
»r the position of
I the axis of x),
(1)
the whole area,
18 may require to
with « alone, t. 0.,
rip, PQIf^M, the
and the element
in this case,
(«)
ke a rectangular
iff, at the point
lent of volame
= 2iTpy dx dy,
(8)
■om to y, the
iirve ; and then
BXAMPLSa.
SXAMPLE8.
189
- 1. Find the centre of gravity of the hemisphere generated
by the revolution of the quadrant, AD, (Fig. 39), round OA
(taken as axis of x), (1) when the density is uniform ; (2)
when it is constant over a section perpendicular to OA and
varies as the distance of this section from OD ; (3) when
it is constant at the same distance from CA And varies as
this distance.
(1) From (1) we hare
X =
jj/hidx
J y*dx
Putting 35 = r cos 6, and y = r sin e, where r is the
i-adius of the circle and integrating hetween = and
e
2'
we have
(2) Since p = fia;, we have from (2)
faN/»dx
w =
which gives
(3) Since p
» =:
~m - tV-
Hy, we have from (3)
r Ixj^dxdy Cxf^dx
ffy*dxdy fi^dx'
180
axAMPLm.
■•, If
aatl the previons Bnbstituv^ons for x and y give
16r
u =
ISir
2. Find the centre of gravity of a paraboloid of revolu-
tion, the length of whose uxis is A. Ant. i = \h.
3. Find the centre of gravity (1) of a portion of a prolate
spheroid, the length of whose axis measured from the vertex
is c, and (2) of a heini-spheroid.
An». (1) i = I -g— ~; (2) * = |a.
66. Polar Fonnnlie.— a solid be generated by the
revelation of AB, (Fig. 41, 1 the axis of x. Then the
elementary rectangle, ahcd, whose mass = pr dd dr, (Art
81), the thickness being omitted, generni'ts a ring which is
an element of the solid whose volume = %nr sin pr dd dr ;
and the abscissa of the centre of gravity of the ring is
r cos 0. Hence (1) of Art. 77 becomes
r jp^ sin cos (/d dr
~7P-
sin eMdr
(1)
in which p mnst be a function of r and 6 in order that the
integrations may be effected.
If the density depends only on the distance from a fixed
point in the axis of revolution, this point may be taken as
origin, and p will be a function of r ; if the density depends
only on the distance from the axis of revolution, p will
be a function of r sin 6.
IXAMPLK.
The vertex of a right circular cone is in the sur&ce of a
sphere, the axis of the cone coinciding with a diameter :>f
ygire
rsboloid of revolu-
Ang. i = |A.
wrtion of a prolate
rod from the vertex
I generated bj the
18 of X. Then the
= pr dO dr, (Art
5^8 a ring which is
Inr sin 6 pr dS dr ;
ty of the ring is
dr
(1)
in order that the
nee from a fixed
may be taken as
density depends
evolution, p will
the surface of a
th a diameter c»f
OENTRK OF ORAVtTT Of AITT 80UO.
131
the sphere, the base of the cone being a portion of the sur-
face of the sphere. Find the distance of the centre of
gravity of the cone from its vertex, 2a being its vertical
angle, and a, the radius of the sphere.
Here the r-limits are and 2a cos ; the 0-limits are
and « ; p is constant ; hence from (1) wo have
a =
*'0 «^0 y
sin 6 cos dS dr
r Ht* sin e dd dr
= 1
r(2fl (,08 (?)
* ein cos d$
= |«
r(2a cos 6)* sin d9
/"co8^ e sin 9 d9
/■
cos* sin d do
1 — coal* a
1 — cos*
a.
87. Centre of Gkavity of any Solid.— Let (x, y, z)
and {x •{■ dx, y ■{■ dy, z + dz) be two consecutive points E
und F, (Fig. 42), within the solid whose centre of gravity is
to be found. Through E, pass three planes parallel t<o the
co-ordinate planes xy, yz, zx ; also through F pass three
planes parallel to the first. The solid included by these six
planes is an infinitesimal parallelopiped, of which £ and F
are two opposite angles, and the volume = dx dy dz. If p
is the density of the body at £, the element of mass at E
= pdT dy dz. Hence the co-ordinates of the centre of
gravity ot the solid are given by the equations
IM
MXAMPLKS.
m =
t =
J J J pxdxdydt
fffpdxdydz '
J jfpydxdydt
/ffpdxdydz
fffptdxdydB
the integrations being extended over the whole solid.
(1)
(^)
(3)
EXAMPLES.
1. Find the centre of gravity of the eighth part of an
ellipsoid included between its three principal planes.*
Let the equation of the elliiMSoid be
^ + |i + -^^i-
Here the limits of the ^-integration are
which call e, and ; the limits of y are
which call y, and ; the z-limits are a and 0.
* FlMMt of xf . y*. w.
1, jjjgiwWJWKTww
mm
POLAR KLKMSlfTS OF " .J.
188
(1)
(2)
(3)
irhole solid.
ighth part of an
pal planes.*
0.
First integrate with respect to z, and we obtain the
infinitesimal prismatic colnmu whose base is PQ. (Fig. 42),
and whose height is Pp. Then we integrate with respect
to y, and obtain the sum oi* all the columns which form
the elemental slice Kplinq. Then integrating with respect
to as, we obtain the snm of all the slices included in the
solid, OABO. Hence (1) becomes, since the density is
uniform.
rrr^'^^^y^
m :=
/"y.'X'*"'**
^^^•-^^*''^*
n-^),>.
.'. »=s^a.
Similarly y = fft, ■ = |«.
2. Find the centre of gravity of the solid bonnded by th '
planes » =r (ix, n = yx, and the cylinder y» = 2a« — i*.
88. Polar El«m«iit» of MMm.— Lei Fig. 43 repre-
sent the portion of the volume of a solid included between
its bounding surfaov; and three rectangular oo-ordinato
planon,
'Ul
134
POLAR MLEMSSTB OF MASS.
mi
(1) Throngh the axis of z draw
a aeries of consecutive plaues, divid-
ing the soUd into wedgenshc^d
slices such as OOBA.
(2) Round the axis of t describe
a series of right cones with their
vertices at O, thus dividing each
slice into elementary pyramids like
0-PQST.
(3) With as a centre describe
a series of consecutive spheres;
thus the solid is divided into elementary rectangular par-
allelopipeds similar to abpt, whose volume — ap-ps' at.
Let XOA = <l>, COP = e. Op =2 r,
AOB = dit, rOQ =:de, pa- dr.
Then pq is the arc of fe circle whose radius is r, and the
angle is de ; therefore
pq = rd9.
Also pa is the furc of a circle in which the angle is d<p,
and the radius is the perpendicular from p on OZ, or
r sin 0; tharefore
ps — r sin e d^.
Therefore the volume of the elementary parallelopiped =
>^ sin 6 dr dd dp ;
and if p is the density of the solid at p, the element of
mass is
pr* »iu e dr de dp.
Also the oo>ordiDates of the centre of gravity of this
element are
r sio ooe ^, r ein 9 aiu <P, and r cos 9 ;
ii ' SM ' f f
y rectangttlar par-
' ~ ap'ps' at.
= r,
-dr.
ins is r, and the
the angle is d<l>,
'm p on OZ, or
>araIlelopiped =r
the element of
rwvity of thii
rco8ff;
mmm
HXAMPLSa. 135
hence for the centre of gravity of the whole solid we have
/ / fpf* sin* Ocoifclrddd^
J'fjp** Bin edrdddit
a =
fffpr* »in* e iia ift dr de d^
Jjffpf* sin edrded^
r r ipr* sin COB dr dO d^
ti =
fffpf* sin Bdrded^
the limit! of integration being determined by the figure of
the solid considered.
The angles, 9 and ^, are sometimes called the eo4at%tud»t
and longitude, respectively.
BXAMPLB8.
1. Find the centre of gravity of a hemisphere whose
density varies as the nth power of the distance from the
centre.
Take the axis of f perpendicnlar to the plane bMC of the
hemisphere. Let a = the radios of the sphere, and
p = fir*, where n is the density at the nnits distance from
the centre. First integrate with respect to r from to a,
and we obtain the infinitesimal pyramid 0-PQST. Then
integrate with respect to from to ^n, and we obtain the
sum of all the pyramids which form the elemental slice,
CODA. Then integrating with respect to ^ from to arr,
we obtain the sum of all the slices included in the hemi-
sphere. Heuce,
/ / Bin (9 COS rfd d^
y / sin rf0 0?^
i = P = 0.
2. Find the centre of gravity of a portion < f a solid
sphere conlaired in a right cono whose vertex is the centre
of the sphere, the density of the solid varying as the «th
power of the distance from the centre, the vertical angle of
the cone being = 2«, and the radius = a.
Take the axia of the cone as that of t, and any plane through it as
that from which longitude is nieaaured.
^*^- • = ^ ^ ^(1 + cos e), and i = y = 0.
89. 8p«oi«l Metfiods.— In the preceding Articles we
have given the usual forraulw for finding the centres of
gravity of bodies, bnt particular cases may occur which may
be most conveniently treat U by special methods.
BXAMPLBS.
1. A circle revolves round a tangent line through an
angle of 180° find the centre of gravity of the solid
generated.
drd$dift
iedi»
i6d<tt
r^
rtion ( f a solid
tex is the centre
ying as the «th
vertical angle of
lane through It aa
I * = y = 0.
ng Articles we
the centres of
icur which may
uds.
e through an
of the solid
BXAXPLKB,
137
Let 07 be the tangent line about
• which tb<) circle revolves, and let the
plane of tbe paper bisect the solid ; the
centre of gravity will tberefora lie in
the axis of x. Lot P and Q bo two
consecutive points : and let OM = x,
and MP = y = V2«.c — »«. The
elementary i-ectangle, PQy;?, will gen-
crate a semi-cylindrical shell, whose volume = 2ffnxdx,
the centre of gravity of which will be in the axis of re at a
distanoe — ftom (Art 78, Ex. 1, Oor.). Hence,
r~-'>
TTxdx
JT'y
nx dx
„ / 3^ V2ax — afidx
A I'D
ir /**« ~
/ X V a«w — s^dx
6a
2^*
3. Find the centre of gravity of a right pyramid of uni-
form density, whose base is any regular plane figure.
Let the vertex of the pyramid be the origin, and the axis
of the pyramid the axis of x; divide the pyramid into slices
of the thickness dx by planes perpendicular to the axis.
Then as the areas of Ihnse sections are as the squares of
their homologous sides, and as the sides are as their dis-
tances from the vortex, so wiU the areas of the sections be as
the squares of their distances from the vertex, and therefore
the masses of the slices arc as ^ho squares of their distances'
from the vertex. Now imi^ne each slice to be condensed
into its centre of gravity, which point is on the a^ic of x.
Then the problem is reduced to finding the centre of grav>
1S8
TBXORKMS OF PAPPUS.
Jty of a material line io which the density varies as the
square of the distance from one end, and which may be
found as in Ex. 6, (Art 78). Calling a the altitude of the
pyramid, we have
a^dz
X ■■=
_ fa,
which if the same as in Art. 75.
■I
I
90. TlMormiui of Pappus.*— (1) If a plans curve
revolve rourd any axis in its plane, the area of the
surface generated is equal to the length of the
revolving curve multiplied by the length of the
path described by its centre of gravity.
Let « denote the length of the carve, x, y, the co-ordinates
of one of its points, i, y, the co-ordinates of the centre of
gravity of the curve; then, if the curve is of constant
thickness and decity, we have from (3) of Art. 78,
y- —71—;
J"
in pa = Zn / yds;
(1)
the second member of which is the area of the surface
generated by the revolution of the curve whose length is 8
about the axis of x, (GaL, Art. 193) ; and thf first member
is the length of the rovolviug curve, s, multiplied by the
length of the path described by its centre of gravity, 2ffy.
• tJnkllr oafled Onldin'i Tbeormm, bnt origlDally •■randated Ity n^vpns. 0m
Watton** MeduBloai Protdnaa, p. 48, Sd W.)
lity varios aa the
d which may be
le altitade of the
a plane eurve
the area of the
length of the
length of the
'ty.
the co-ordinatea
of the centre of
e is of constant
Art. 78,
(1)
of the surface
lose length is a
u first member
iltiplied by the
gravity, 2»ry.
Med Iqr Fipptu. ifim
wmm.
wm
wammm
TBSORBMB OF PAPPUS.
139
(2) // a plane area revolve round any axis in its
plane, the volume generated is equal to the area of
the revolving figure multiplied by the length of the
path described by its centre of gravity.
Let A denote the plane area, and let it be of constant
thickness and density, then {%) of Art 82 becomea
V =
jfy dx djf
ffdxd/
or 2?ry y fdA ■= %ir J jy dx dy,
(substituting dK for dx dy),
(»)
the integral being taken for every point in the perimeter of
the area; but the second member is the volume of the
solid generated by the revolution of the area (Oal., Art.
208) ; and the first member is the area of the revolving
figure, A, multiplied by the length of the path described
by its centre of gravity, 2ny.
Cob. — If the carve or anaa revolve throngh any angle, 0,
instead of 2it, (1) and (2) become
and
6i8 = efyd$,
9yK = ^fi/»dx.
itnd the theorem* are still true.
SoH. — If the axis cuts the revolving curve or area, the
theorems still apply with the convention that the surface
or volume generated by the portioni of the curve or area on
opposite sides of the axis are affected with opposite rignc.
i*fl3!
m
140
JSXAMPLMS.
EXAMPLES.
1. A circle of radios, a, revolves round ua axis in its own
plane at a distance, c, from its centre; find the surface of
the ring generated by it.
The length (circumference) of the revolving carve =
2na; the length of the path described by its centre of
gravity = inc;
.'. the area of the surface of the ring = in*ac.
2. An ellipse revolves round an axis in its own plane,
the perpendicular distance of which from the centre is c ;
find the volume of the ring generated daring a comph t€
revolution.
Let a and b be the semi-axes of the ellipse ; then the
revolving area = nab ; the length of the path described by
itf( centre of gravity = 2nc ;
. • . the volume of the ring = in'abc.
Observe that the volome is the same for any positioii of the axes
of tlie ellipse with respect to th« axis of revolution, provided the per-
pendicular diiitance from that axis to the centre of the ellipse is the
same.
3. The surface of a sphere, of radius a, = inefi; the
length of a somi-circumference = na ; find the length of
the ordinate to the centre of gravity of the arc of a semi-
circle. ^ - 2a
Ans.
y = __.
4. The volume of a sphere, of radius a, = fTro* ; the
area of a semicircle = rra* ; find the distance of the centre
of gravity of the semicircle from the diameter.
Aru. y =
40
37r*
6. A circular tower, the d''uneter of which is 20 ft., is
being bnilt, and for every foot it rises it inclines 1 in. from
I axis in its own
i the Bur&ce of
)lving carve =
ly its centre of
its own plane,
the centre is c ;
ing a complete
lipse; then the
th described by
idtion of the axes
provided the per-
the elllpoe ia the
, = 4rro*; the
the length of
arc of a ecmi-
- 2a
ns. y = — .
It
= ^ira' ; the
of the centre
4tf
"*• ^ = rn
h is 20 ft., is
nes 1 in. from
JSXAMPLSa.
141
the vei:tical ; find the greatest height it can reach without
falling. Ans. 240 ft.
6. A circular table weighs 20 lbs. and rests on four tegs
in its circumference forming a square ; find the least ver-
tical pressure that must be applied at its edge to overturn it.
Ans. 20 {V2 + 1) = 48.28 lbs.
7. If the sides of a triangle be 3, 4, and 5 feet, find the
distance of the centre of gravity from each side.
Ans. I, 1, f ft.
8. An equilateral triangle stands vertically on a rough
plane ; find the ratio of the height to the base of the plane
when the triangle is on the point of overturning.
Ans. V3 : 1.
9. A heavy bar 14 feet long is bent into a rigbc angle so
that the lengths of the portions which meet at t.ie angle
are 8 feet and 6 feet respectively ; show that the distance
of the centre of gravity of the l»r so bent fh)m the point
of the bar which was the centre of dravity when the bar
9 V^
was straight, is — ^ — feet.
10. An equilateral triangle rests on a sqnare, and the base
of the triangle is equal to a side of the square ; find the
centre of gravity of the fligure thus formed.
Ans. At a distance from the base of thn triaugle equal to
7= of the base.
8 + 2^3
11. Find the inclination of a rough plane on which half
p. i^gnkiT hexagon can just rest in a vertical position with-
out overturning, with the shorter of its parallel sides in
contact with the plane. Ans. 3 y/Z : 5.
12. A cylinder, the diameter of which is 10 ft, and height
60 ft, rests on another cylinder the diameter of which is
■i*ii
143
EXAMPLMS,
18 fL, and height 6 ft ; and their axes coincide ; find their
common centre of gravity. Ans, 273|f ft. from the base.
13. Into a hollow cylindrical vessel 11 ins. high, and
weighing 10 lbs., the centre of gravity of which is 5 ins.
from the bai>e, a uaiform sohd cylinder C ins. long and
weighing 20 lbs., is just fitted ; find their common centre of
gravity. Ans. 3f ins. from base.
14. The middle points of two adjacent sides of n square
are joined and the triangle formed by this straight line and
the edgep is cut ofF; find the centre of gravity of the
remainder of the square.
Ans. ^ot diagonal from centre.
15. A trapezoid, whose parallel sides are 4 and 12 ft.
long, and the other sides each equal to 5 ft., is placed with
its plane vertical, and with its shortest side on an inclined
plane ; find the relation between the height and base of the
plane when the trapezoid is on the point of falling over.
Ans. 8 : 7.
16. A regular hexagonal prism is placed on an inclined
plane w:th its end faces vertical ; find the inclination of
the plarie so that the prism may just tumble down the plane.
Ans. 30".
17. A regular polygon just tumbles down an inclined
plane whose inclination is 10° ; how many sides has the
polygon ? Ans. 18.
18. Prom a sphere of radius R is removed a sphere of
radius r, the distance between their centres being c; find
the centre of gravity of the remainder.
Ans. It is on the line joining their centres, and at a dis-
cr*
tance
/2«-r»
from the centre.
19. A rod of uniform thickness is made np of equal
lengths of three sabstauoes, the densities of which taken in
icide; find their
from the base.
1118. high, and
which is 5 iiis.
6 ins. long and
foimon centre of
na. from base.
des of a square
traight line and
gravity of the
from centre.
e 4 and 12 ft.
, is placed with
on an inclined
and base of the
falling over.
J««. 8 : 7.
on an inclined
i inclination of
iown the plane.
Ans. 30°.
vn an inclined
sides has the
Ans. 18.
'ed a sphere uf
i being e ; find
, and at a dis-
• tip of equal
rbich taken in
tXAMPLWa.
143
order are in the proportion of 1, %, and 3 ; find the position
of the centre of gravity of the rod.
Ans. At -^ of the whole length from the end of the
densest piu*t.
20. A heavy triangle is to bo suspended by a string pass-
ing through a point on one side ; determine the position of
the point so that the triangle may rest with one side
vertical.
Ans. The distance of the point from one end of the side
=r twice its distance from the other end.
21. The sides of a heavy triangle are 3, 4, 5, respectively ;
if it be suspended- from the cetitre of the inscribed circle
show that it will rest with the shortest side horizontal.
22. The altitude of a right cone is h, and a diameter of
the base is i ; a string is fastened to the vertex and to a
point on the circumference of the circular base, and is then
put over a smooth peg ; show that if the cone rests with its
axis horizontal the length of the string is ^/{Ifl + ¥).
23. Find the centre of gravity of the helix whose equa-
tions are
X =■ a cos ^ ; y = a sin ; « = ha^.
Ans. X
, y . J a—x - %
ka^; y = ka -— -; z = ^y
Z z a
24. Find the distance of the centre of gravity of the
catenary (Cal., Art 177), from the axis of x, the curve
being divided into two equal portions by the axis of y.
Ans. If 2Z is the length of the curve and (/*, k) ig the
extremity, the centre of gravity is on the axis of y at a
distance — ^ — trom the axis of x.
KXAMPLBS.
25. Find the centre of gravity of the area included
between the arc of the parabola, y» = 4aa:, and the straight
line y = kx
Arts, i =
?1
y =
2rt
26. Find the centre of gravity of the area bounded by
the cissoid and its asymptote, the equation of the cissoid
being y» = --^--. ^^. ^ ^ j^,
27. Find the centre of gravity of the area of the witch
of Agnesi.
Afis. At a distance from the asymptote equal to ^ of the
diameter of the base circle.
28. Find the centre of gravity of the area included be-
tween the arc of a semi-cycloid, the circumference of the
generating circle, and the base of the cycloid, the common
tangent to the circle and cycloid at the vertex of the latter
being taken as axis of z, the vertex bein^r origin, and a the
radius of the generating circle.
. .37r» — 8 _
Ans. x= —^--— a; y == in.
39. Find the centre of gravity of the area contained be-
tween the curves y' = ax and f = 2ax — a^, which is
above the axis of a;. , . 15 n- — 44
y =
Atis. i = o
a
IStt — 40'
3n - 8
30. Find the centre of gravi>;^ of the area included by
the curves y* = ax and a^ = ly.
Ana. i: = ^ijl ; y — ^ibK
31. Find tno distance of the centre of gravity of the area
of the circular sector, BOCA, (Fig. 39), from the centre.
Let 20 = the angle included by the bounding radii
.„„ - .sin ©
Ans. « = f* -^-.
SXAMPLSa.
146
area inclndod
ind tho straight
8a _ _ 2rt
ea bonnded by
I of tho cissoid
ins. i = \a.
a of the witch
[ual to I of the
[. included be-
ference of the
, tho common
of the latter
^n, and a the
y — {a.
contained be-
■ «*, which is
— ^
included by
= A«***.
_, of the area
:he centre.
g radiL
, sin 9
32. Find the distance of the centre of gravity of the
circular segment, BCA, (Fig. 39), from the 'centre.
. - , a sin' B
Ana. « = f
BC'
» — sin «• cos e ~ 12 urea of ABC
33. Find the centre of gravity of the area bounded by
the cardioid r = o (1 + cos 6). Ana. i = ^a.
34. Find the centre of gravity of the area included by a
loop of the curve r == a cos 20. _ 28a Vi
Ans. X =
105n-
35. Find the centre of gravity of the area included by a
loop of the curve r = a cos 3 . _ gja y^
Ana. te =
80rr
36. Find the centre of gravity of the area of the
sector in Ex. 31, if the density varies directly as the dis-
tance from the centre.
Ana. X =
3a sin 6
e
37. Find the centre of gravity of the area of a circular
sector in which the density varies as the «th power of the
distance from the centre.
Ana. — ^t_ . _, where a is the radius of the circle, I the
fi + o I
length of the arc, and c tue length of the chord, of the
sector.
88. Find the centre of gravity of the area of a circle in
which the density at any point varies as the nth power-of
the distance from a given point on the circumference.
Ana. It is on tho diameter passing through the given
2(n +
point at a distance from this point equal to
a being the radius.
« + 4
<h
149
SXAMPLSa.
39. Find the centre of gravity of the area of a quadrant,
of an ellipse in which the density at any point yaries as
tiie distance of the point from the major axis.
Ans. ii = |a ; y
3?r.
40. Find the distance of the centro of gravity of the sur-
face of a cone from the vertex.
Let a = the altitude.
Ans.
K
41. Find the centre of gravity of the surface formed by
revolving the curve
r = o (1 -f- cos 6),
round the initial I'ne.
. - 60a
Ans. • = -gg .
42. A parabola revolves round its axis ; find the centre
of gravity of a portion of the surface between the vertex
and a plane perpendicular to the axis at a distance from
the vertex equal to J of the latus rectum.
Ans. Its distance from the vertex = f | (latns rectum).
43. Find the centre of gravity of a cone, the density of
each circular slice of which varies as the nth power of its
distance from a parallel plane through the vertex.
Let the vertex be the origin and a the altitude.
. . n 4- 3
Ant. m = — i— J a.
n + 4
44. Find the centre of gravity of a ccae, the density of
every particle of whicli increases as its distance from the
\a, whore the vertex is the origin and a tha
Ans. w
altitude.
46. Find the centre of gravity of the volume of uniform
density ounlained between a hemisphere and a cone whoso
vertex is the vertex of the hemicphcre and base is the bast)
of the hemisphere.
3?ji^«««f*«w»tsw,'aaH«»»MS««if«ii^^
3a of ft qaadrant
y point varies as
xis.
■avity of the sur-
Ans. i = ^a.
rface formed by
ins. i =
50a
63*
find the centre
;ween the vertex
a distance from
(latns rectum).
the density of
th power of its
ertei.
itude.
n + 3
n + 4
a.
the density of
ituDce from the
igin and a r.he
me of aniforni
a cone whoso
Hvse is the baso
XXAMPLSSL
147
a
Ans. * = 2> ^^^^ the vertex is the origin and a the
altitude.
46. Find the distance of the centre of gravity of a henii-
aphere from the centre, the radius being a,
Ans. i = |fl.
47. Find the centre of gravity of the solid generated by
the revolution of the cycloid,
y = V2ax — a» + a vers"*?,
a
(1) round the axis of z, and (2) round the axis of y.
A /i\- (637r»_64)« .... /16 Tr»\ 2a
Am. (1) . = V(9^;ii)-5 (2) » - (y + i) IT-
48. Find.the centre of gravity of the volume formed by
the revolution round the axis of x of the area of the curve
f — axf + X* — 0.
^
Ant. i
3an
32 '
49. Find the centre of gravity of the volume generated
by the revolution of the area in Ex. 2U round the axis of y.
. - 8a
^'"- >' = aTr5inri4)-
60. Find the centre of gravity of a hemisphere when
the density varies as the square of the distance from the
ooutro. . . 5a
''12'
Ant.
51. Find the centre of gravity of the solid generated by a
scmi*pai-abola bounded by the latns reotum, rsvolviug
round the lutus reotum.
Ant. Distance from focus = /^ of latus reotum. ^
. (I'fc'fi
148
SXAMPLBS.
62. The vertex of a right circular cone is at the centre of
a sphere ; find the centre of gr&vity of a body of uniform
density contained within the couo and the sphere.
Ans. The distance of the centre of gravity from tue ver-
tex of the cone = ~ (1 + cos «), where a = the semi-
o
vertical angle of the cone and a = the radius of the
sphere.
53. Find the distance firom the origin to the centre of
gravity of the solid generated by the revolution of the
cardioid round its prime radius, its equation being
r = a (1 + cos 6).
Ana. X = fa.
54. Find by Art. 90 (1) the surface and (2) the volume
of the solid formed by the revolution of n cycloid round
the taigent at its vertex.
Ans. Surface = ^na'; Volume = rrV.
55. Find (1) the surface and (2) the volume of the solid
formed by the revolution of a cycloid round its liaso.
Am. (1) *^na* ; (2) orV.
56. Au equilateral triangle revolves round its base,
whose length is a ; find (1) the area of tho surface,
and (i) the volume of the figure described.
Ana. (1) na* a/3 ; (2) -^.
67. Find (1) the surface and (2) the volnmo of a ring
with a circular section whose intomal diameter is 12 ins.,
and thickness 3 ins.
Am. (1) 444.1 sq. in.; (2) 833.1 cub. in.
at the centre of
•ody of uoifonu
phere.
y from tue ver-
a = the Bemi-
radiua of tbe
the centre of
relation of the
1 heing
ins. X = |a.
(2) the volume
cycloid round
ume = tV.
me of the Rolid
its liasc.
»» ; (2) 5rW.
und its base,
tho surface,
\/3 ; (2)
iTrt»
nniu of a ring
iieter is 12 ins.,
33.1 cnb. in.
CHAPTER V.
FRICTION.
91. Friction. — Friction is that force which acts between
two bodies at their surface of contact, and in the direction
of a tangent to that surface, so as to resist their sliding on
each other. It depends on the force with which the bodies
are pressed together. All the curves and surfaces which we
have hitherto considered were supposed to be smooth, and,
as such, to offer no resistance to the motion of a body in
contact with them in any other than a rormal dircctiou.
tSuch curves and surfaces, however, are rot to be tbund in
nature. Every surf two is capable of destroying a certain
amount of force in its tangent plane, i.e., it possesses a certain
degree of roughness, in virtue of which it resists tho sliding
of other surfaces upon it. This resistance is called friction,
and is of two kinds, viz., sliding and rolling friction. The
rst is that of a heavy body dragged on a plane or other
surface, an vie turning in a fixed box, or a vertical shaft
turning on a narizo'zU! plate. Friction of the Be<,ond kind
is that of a v;beel rolling along a plane. Both kinds of
friction are governed by the same laws ; the former is much
Teater than the latter under the same circumstances, and
< I he only one that we shall consider.
.'\. smooth surface i i one which opposes no resistance to
the motion of a body upon it A rough surface is one
which does oppose a resistance to the motion of a body
u])ou it.
TIiP fliirfkcM of all bodiM eonslBt of rerj Rintll elevations and
deproMlons, ho that if thejr are prMsed againut vacli (ith«r, the
elevatlona of ooo fit, mom or lesa, into the deprpwions of thv other,
and tbe Burfaoea iitt«rp«netrate each otkor ; and the- mutual penetra-
\ht
160
LAWS OF FRIcnON.
tion is of ooane greater, if the presBing force is greater. Henoe,
when a force is applied bo as to cause one liody to move on another
with which it is iu contact, it is necessary, lieforo motion can take
place, either to break off the elevations or compress them, or force tlio
iKxliea to separate far enough to allow them to pass each other.
Much of this roughneM may be removed by polishing ; and the effect
of much of it may be destroyed by lubrication.
Friction always vets along a tangent to the surfiuje at the point of
contact ; and its direction is opposite to that of the lice of motion ; it
presents itself in the motion of a body as a passive force or resistance,*
since it can only hinder motion, but can never produce or aid it. In
investigations in mechnnius it can be considered as a force acting in
opposition to erery motion whose direction lies in the plane of contact
of the two bodies. Whatever may be the direction in which we move
a body resting upon a boriaontal or inclined plane, the friction will
always act in the opposite direction to that of the motion, i. «., when
we slide a body down an inclined plane, it will appear as a force up
the plane. A surface may also resist sliding motion by means of tha
adhetion > ^ween its substance and that of another body in contact
with it.f
Tho friction of a body on a surface is measured by the
leaxt force which will put the body in motion along the
surface.
92. Laws of FriotioB. — In onr ignorance of the
ooudtitutiun of bodies, the laws of friction must be deduced
from experiment. Experiments made by Coulomb and
Morin have established the following laws of friction :
(1) The friction varies as the normal pressure when the
materials of the surf..^js in contact remain the same. Subse-
quent experiments have, however, considerably modified
this law, and shown that it can be regarded only as an
approximation to the truth. When the pressure is very
great it is found that tho friction is less than this law
would give.
* WalHbwh, p. «».
t 8e« Haaklue'i ApplM MsetMDks, p. au».
I greater. Henoe,
movA on another
3 motion can take
them, or force tlie
paw each other,
ag i and the efl'cxrt
aoe at the point of
line of motion ; it
brce or resistance,*
duee or aid it. In
ts a force acting in
lie plane of contact
in which we move
le, the friction will
motion, i.e., when
>pear as a force up
a by means of tha
ler body in contact
leasured by the
)tion along the
loranoe of the
ust be deduced
Coulomb and
frictiou :
|e«jiur(? wJien the
same, Subsc-
Irsbly modified
only as nn
assure is very
than this law
ijiwa or rRicnoN.
(2) Tht frictioH is independent of the extent of the aur-
facee in contact so hng as the normal pressure remains the
same. When the surfacec iu contact are very small, as for
instance a cylinder resting on a surface, this law gives the
friction much too great.
These two laws are true when the body is on the point of moying,
and also when it is actually in motion ; but in the case of motion the
magnitude of the friction is not always the same as when the body is
beginning to move ; when there is a difference, the fHction is greater
in the state bordering on motion than In actual motion.
(3) The friction is independent of the velocity when the
body is in motion.
It follows from those laws that, if .R be the normal
pressure between the bodies, F the force of friction, and ft
the constant ratio of the latter to the former when slipping
is about to ensite, we have
F= fiR.
(1)
The fraction n is called the co-efficient qf friction ; and if
the first law were true, fi would be strictly constant for the
■ame pair of bodies, whatever the magnitude of tho normal
pressure between them might bo. This, however, is not
tho case. When the normal pressure is nearly equal to that
which would crush either of the surfaces in contact, the
force of friction increases more rapidly than the normal
pressure. Equation (1) is nevertheless very nearly true
when the differences of normal pressure are not very jfr<?at ;
und in what follows we shall assume this to be the case.
Rbmabk. — The laws of friction were established by Coulomb, a
distinguished French officer of Engineers, and were founded on
expeHments made by him at Rochefort. The results of these experi-
ments were presented in 1781 to the French Academy of ScienreH, and
in 178A his Memoir on Friction was published. A very full nbHlract-
of this paper is given in De Young's Natural PhUoKtphy, Vol. II,
p. 170 (1st Ed.). Further ex|it>rinienM wert; made at Mi-ts by Morin,
1881-S4, by direction of the French military authorities, the raialt of
I f .
153
AlfOLK OF FRICTION.
i
which has been to oonfirm, with slight exeeptions, all the results of
Coulomb, and to determine witli conBldersble precision tlie numerical
values of the coefficients of friction, for all the substances usually
employed in the construction of machiaes. (8ee Oalbraith's Me-
chanics, p. 68, Tiwisdeu's Practical Mechanics, p. 188, and Weisbach's
Mecbauica. VoL I, p. 817.)
93. Magnitndes of CoeffloientB of Friction.— Prac-
tically there is no observed coeflRcient much greater than 1.
Most of the ordinary coefficients are less than ^. The fol-
lowing results, selected from a table of coefficients,* will
afford an idea of the amount of friction as determined by
experiment : these results apply to the friction of motion.
For iron on stone i* varies between .3 and .7.
For timber on timber " " " .2 and .5.
For timber on metals " " " .2 and .6,
For metals on metals " " '* .15 and .25.
For full particulars on this subject the student is referred
to Rankine's Applied Mechanics, p. 209, and Mosilcy's
Engineering, p. 124, also to the treatise of M. Morin, where
he will find the subject investigated in all its completeness.
94. Angle of Friction. — The angle at which a rough
plane or surface may be inclined so that a body, when acted
upon by the force of gravity only may just rest upon it with-
out sliding, is called the Angle of Friction, f
Let a bo the angle of inclination of
the piano AB just as the weight is on
the point of slipping down; W the
weight of the body ; R the normal pres-
sure on the plane ; F the force of fric-
tion acting along the plane = fiR (Art.
92). Then, resolving the forces along and perpendicular to
the plane we have for equilibrium
* Ranklne'B AppHod Mechautcs, p. m.
t SoiuetimM called " Ibe aogto of ropoae; " al»o caUad " the Umltlng asglo of
resUtaiMe."
na-45
all the rasolts of
ion tlie namerical
ubatauces nsnalljr
Galbnith's Me-
B, and Weisbach's
fiction.— Prac-
greater than 1.
an f The fol-
efficiente,* will
determined by
on of motion.
.3 and .7.
.2 and .5.
.2 and .6.
15 and .25.
dent is referred
and Mosiley'fl
Morin, where
) completeness.
which a rough
ody, tvhen acted
upon it with'
erpendicnlar to
the limiting angle of
BK ACTION or A SOUOH CVBVK.
uB = ffaina; R — ffoosa;
tan a = /»,
a)
which gives the limiting valae of the inclination of the
plane for which equilibrium is possible. The body will rest
on the plane when the angle of inclination is less than the
angle of friction, and will slide if the angle of inclination
exceeds that angle ; and this will be the case however great
W may be ; the reason being that in whatever manner
we increase W, in the same proportion we increase the
friction upon the plane, which serves to prevent IT from
sliding.
From (1) we see that the tangent of the angle of friction
is equal to the coefficient of friction.
95. ReactiOB of a Rough Cnrvo
or Surface. — Let AB be a rough cuive
or surface ; P the position of a particle
on it ; and suppose the forces acting on
P to be confined to the plane of the
paper. Let R^ = the normal resistance of the surface,
acting in the normal, PN, and F = the force of friction,
acting along the tangent, PT.
The resultant of if, and F, called the Totai Resistance*
of the surface, is represented in magnitude and direction by
the line PR = R, which is the diagonal of the parallelo-
gram determined by R^ and F. We have seen that the
total resistance of a smooth surface is normal (Art. 41) ; but
this limitation does not apply to a rough surface. Let ^
denote the angle between R and the normal £, ; then ^ is
given by the equation
* tan = ^•
* MlBcitia'p Btatici, p. (M.
ris.4e
154
WBiCTION ON Air INCLiySD PLAKIl,
Hence, ^ will be a maximum when the force of friction,
F, bears the greatest ratio to the normal pressnre B^. But
this greatest ratio is attained when the body is just on the
point of slippin/; along the surface, and is what we called
the coefficient of friction (Art 92), that is
Tr = ^;
. • . t«m ^ = ^.
Therefore the greatest angle by which the Total Reaietanc^
of a rough curve or surface can deviate from the normal is
the angle whose tangent is the coefficient of friction for the
bodies in contact ; and this deviation is attained when slip-
ping is about to commence.
Cob.— By (1) of Art 94, tan « = f» ;
ft.
hence, the direction of the total resistance, R, is inclined at
an angle a to the normal ; i.e., the greatest angle that the
Total Resistance of a rough curve or surface can make with
the normal is equal to the angle of friction, correspmiding
to the two bodies in contact.
96. Friction on an Inclined Plane. — A body rests on
a rough inclined plane, and is acted on by a given force, P,
in a vertical plane which is perpendicular to the inclined
plane ; find the limits of the force, and the angle at which
the least force capable of drawing the particle up the plane
must act
Let t = the inclination of the plane to the horiison ; 6 =
the angle between the inclined plane and the line of action
of P; (I =■ the coefficient of friction ; aud let us first snp-
pose that the body is on the point of moving down the
force of friction,
)Tes8are B^. But
»dy is just on the
is ivhat we called
i Total Betistance'
om the normal is
'' friction for the
tained when slip-
R, is inclined at
st angle that the
ce can make with
m, correspotuling
A body rests on
a given force, P,
to the inclined
angle at which
cle up the plane
le horizon ; =
le line of action
let us first snp-
loving dowti the
mmmmmmmBm^':.
FRICTION ON AN INCLINMD PLANS.
156
plane, so that friction is a force acting up the plane^ then
resolving along, and perpendicular to, the plane, we have
F + Pco&e =z W fan if
B + Pmie= Wooai,
F^fiB;
. sin t — ^ cos t
P=W
cobO — fianO
(1)
And if P is increased so that motion up the plane is just
beginning, F acu in an opposite direction, and therefore
the sign of /u must be changed and we have
p — j|r »^P*'4-^C0St
COS $ + n Bind'
(2)
Hence, there will be equilibrium if the body be acted on by
a force, the magnitude of which lies between the values of
P in (1) and (2). Substituting tan ^ for /« (Art. 95) ; (2)
becomes
n_ nr wn(t-f »)
COB
(«)
To . determine 6 in (2) so that P shall be a minimum we
must put the first derivative of P with respect to fl = 0,
therefore
dP nr / • • ■ ^ and — fi COB 6
,*. tan 9=:^;
that is, the' force P necessary to draw the body up the plane
will be the least possible when =. the angle of friction.
166
DODBhE-UrCLINMD PLANE.
Henoe we infer that a given force acts to the greatest
advantage in dragging a weight up a hill, if the angle ut
which its line of action is inclined to the hill is equal to
the angle of friction of the hilL Similarly, a force acts to
the greatest advantage in dragging a weight along a hori-
zoatal plane if its line of action is inclined to the plane
at the angle of friction of the plane. We may also deter-
mine from this the angle at which the traces of a drawing
horse should be inclined to the plane of traction.
These results are those which are to be expected, becanse
some part, of the force ought to be expended in lifting the
ight from the plane, so that friction may be diminished.
V oe Price's Anal. Mech's, Vol. I, p. 160.)
97. Friction on a Doubla-Inclined Plane.— Two
bodies, whose weights are P and Q, rest on a rough double-
inclined plane, and are connected by a string which passes
over a smooth peg at a point. A, vertically over the intersec-
tion, fi, of the two planes. Find the position of equili-
brium.
Let « and /3 be the inclinations of
the two planer ; let 2 = the length of
the string, and h = AB; and let
and 0' be the angles the portions of
the string make with the planes.
Suppose P is on the point of
ascending, and Q of descending.
Then, since the motion of each Inxiy is about to ensue, the
total resistances, R and ^S', must each make the angle of
friction with the corresponding normal (Art 95, Oor.) ; and
since the weight, P, is about to move upwards the friction
must act downwards, and therefore R must lie below the
normal, while, since Q is about to move downwards, the
friction must act upwards, and therefore 8 must be above
the normal.
Fig.4y
&ilKiS^&-
i to the greatest
, if the angle at
) hill is equal to
a force acts to
it along a hori-
led to the plane
may also deter-
ces of a drawing
;tion.
tpeeted, because
;d in lifting the
' be diminished.
1 Plana.— Two
a rough double-
g wliich passes
ver theintersec-
ition of equili<
F(s.47
i to ensue, the
e the angle of
95, Oor.) ; and
ds the friction
t lie below the
ownwards, the
must be above
DOUBLE-mcUNSD PLAIflS.
187
If T is the tension of the string, we hare for the eqni-
libriam of P, (Art. 32),
y^j> «»(« + »)
cos (d — 1^)
And for the equilibrium of Q,
~ ^co8(e' ■\-<p)'
Equating the v^ues of ^ we get
pS in(«-|-^) _ sin (0 -0)
C08((? — 0) ^co8(e'4-<>)'
and if P is about to move down the piano, the friction acts
in an opposite direction, and therefore the sign of must
be changed and we have
„ sin (« — 0) _ si n (/? + 0)
cos {0 + <p) ^ cos (»' — 0)'
(2)
(1) or (2) is the only statical equation connecting the
given quantities.
We obtain a geometric equation by expressing the length
of the string in terms of h, «, /3, 6, and B', which is
/
Vsin e ^ sin Bf
(3)
From (1) or (2) and (3) the values of 6 and 0' can be found,
and this determines the positions of P and Q.
Otherwise thus :
Instead of considering the total resistances, R and S, we
may consider two iwrmai resistances, Ji^ and Si, and two
158
DOUBLE-INCLISSD PLANK,
forces of friction, /»iJ, and /u/S,, acting respectively down
the plane « and up the plane /3. In this case, considering
the equilibriam of P, and resolving forces along, and per-
pendicular to, the plane a, we have
P sines + M^, = TcmB,
P cos a = i?i + T sin e,
:■]
(4)
and for the eqnilibriam of Q,
C8in/3 = |ii/S, + TcosffA
Cco8/3= /S, + Tsmff. I
(5)
Eliminating i?,, S^, and T from (4) and (5) we get (1),
the same statical equation as before.
The method of considering total resistances instead of
their normal and tangential components is usually more
simple than the sep8,rate consideration of the latter forces.
(See Minchin's Statics, p. 60.)
CoE. — If Q is given and P be so small that it is abont to
ascend, its value, P^, will be given by (1),
P _ ft ^inj^^^l^ii?:!!^ ,R^
^^-^ flfn (a + ^) COS {ff + 0)' ^'
nnd if P is so large that it is abont to drag Q up, its value,
P,, will be given by (2)
P, = Q
sin (/3 + «f>) COB (0 + ^)
sin (o — 0) cos ((?' — 0)
(7)
the angles and 6' being connected by (3).
There will be equilibrium if ^ be acted on by any force
whose magnitude lies between P| and Pg.
fc —
espectively down
case, considering
along, and per-
(*)
(6)
i (5) we get (1),
ances instead of
is usually more
;he latter forces.
lat it is about to
ho" <*^>
Q up, its value,
n by any force
Kiii iaiiilliMWIM I WipWIWI'l
fmm
ut'iiimnmHmmimmMmixiKn'
FRICTION OF A TRUNNIOie.
159
Pi8.4S
98. FrictioB on Two XncUned Planes.— A beam
rt'sts ou two rough inclined planes; find the position of
equilibrium.
Let a and b be the segments, AG
and BG, of the beam ; let be the
inclination of the beam to the hori-
zon, a and /3 the inclinations of the
planes, and R and S the total resist-
ances. Suppose that A is on the
point of ascending; then the total
resistances, R and 8, must each
make the angle of friction with the correeponding normal
and act to the right of the normal.
The three forces, W, R, S, must meet in a point (Art.
62) ; and the angles GOA and GOB are equal to a -f- 0,
and (3 — (p, respectively.
Hence {a + b) cot BGO = o cot GOA — b cot GOB,
or (a + ft) tan e = o cot (a + 0) — i cot (/3 — 0). (1)
Cob. — If the planes are smooth, ^ = 0, and (1) becomes
{a + b) taxi6 = acota — b cot j3.
(See Ex. 7, Art 62.)
99. Friction of a Trunnion.* — Trunnions are the
cylindrical projections from the ends of a shaft, which rest
on the concave surfaces of cylindrical boxes. A shaft rcEits
in a horizontal position, with its trunnions ou rough
cylindrical surfaces; find the resistance due to friction
which is to be overcome when the shaft begins to turn
about a horizontal axis.
« BometimM called " JonmaL"
160
mWTrON OP A PIVOT.
FiR.4»
Let IM and BAED be two right
flections of the trunnion and its box;
the two circles are tangent to each
other internally. If no rotation
takes place the trunnion presses
upon its lowest point, H', through
which the direction of the resulting
pressure, R, passes ; if the shaft
begins to rotate in tiie direction AH, the trunnion ascends
along the inclined surface, EAB, in consequence of the
friction on its bearing, until the force, ^S, tending to move
it down just balances the friction, /'. Resolving R into a
normal force AT and a tangential one, 8, we have, since the
t<<ngential coini>onent of R in urging the trunnion down
the burface = tlie Mction which opposes it.
•8=F-(iN; but R* ^ S* + N*;
or
therefore
and the friction
R* = ix*2P + I^;
R
N-
fiR
Vi +^«'
Rtantft
or
vT-f n* Vl + tanS
P =r Ram (p.
(Art. 96),
Ilonco, fo fiiv^ (he friction upon n (rtinnioti, muUiply the
resultant of the forces which act upon it by the sine of the
angle of friction.
100. Friction of • Pivot—A heavy circular shaft
rests in a vortical position, with its end, which is a circular
Fip.49
trunnion ascends
isequcnce of tho
, tending to move
aohing R into a
e have, since the
>o traimion down
t.
+ N*',
rt. 95),
)//, multiply the
y the sine of the
circular shaft
til is a circular
SSfi^^
FBicnoN Of A prroT.
IGI
section, on a horisontal plate; And tho resigtimoe due to
friction which is to be overcon^e, when tho shaft begins to
revolve about a vertical axis.
Let a be the radius of the circular section of the shaft;
let the plane of ( •. $) be the horizontal one of contact
between the end of the shaft and the plate; and let tho
centre of the circular area of contact be the pole. Let
W = the weight of the siiaft, then the vertical pressure on
W
each unit of surface is — j; and therefore, if rdrd6 is the
area-element, we have
the pressure on the element
W
rra-
-, r dr de ;
the friction of the element sz n~- rdr dO.
no'
The friction is opposed to motion, and the diroc- jii oi its
action is tangent to the circle descril)ed by the element ;
the moment of the friction about the vertical axis through
the centre
^Wt^d rde
■^ "~ ita*" '
therefore the moment of friction of the whole circular end
«/o t/fl
(iWr^drde
2ftWa
8 '
(1)
and consequently varies aa the radius. Hence arises the
advantage of reducing t« tho smallest possible dimensions
llio area of the ba«o of a vertical shaft revolving with its
end reiiting on a horizontal be<l.
From (1) wn -nay regard the whole friction due to the
pressure oa acting at a single point, and at a distance from
the centre of motion equ«l to two-thirds of the nulins o(
m
BXAMPLES.
the base of the shaft,
lever of fi-iction.
This distance is called the mean
; ' r
When the shaft is veriicai, and rests upon its circular end
in a cylindrical socket the cylindrical projection is called a
Pivot,
EXAMPLES.
1. A mass whose weight is 750 lbs. rests on a horizontal
plane, and is palled by a force, P, whose direction mal es
an angle of 15" with the horizon ; determine P and the
total resistance, B, the coefficient of friction being .62.
Ans. P = 413.3 lbs.; iJ = 756-9 lbs.
2. i> lermine P in the last example if its direction is
hoiizontal. Ans. P = 465 lbs.
8. find the force along the piano required to draw a
weight of 25 tons np a rough inclined plutte, the coefficient
of friction being -j^, and the inclination uf the plnno being
snch that 7 tons acting along the plane would siip|)ort the
weight if the plane were smooth.
Ans. Any force greater than 17 tons.
4. Find I'lo force in the preceding example, supposing
it to act at the most advantageous inclination to the plaice.
Ans. ]5t^ tons.
5. A ladder inr-Hned at an angle of 60° to the horizon
rests between a rouyh pavement and the smooth wall of a
house. Show that if the ladder kgin to slide when a man
hAB osoopdod so that his centre of gravity in half way up,
then the coefficient of friction between the foot of the
ladder and the pavement is \ VS.
6. A body whose weight is 20 lbs. ip just sustained on a
rough inclined plane by a horizontal force of 2 iba., and a
force of 10 lbs. along the piano ; the coefficient of friction is
I ; find the inclination uf thu plane. Ans. tan~' (||).
lalled the mean
iis circular end
Hon is called a
on a horizontal
iircctiou mal es
dine P and the
I being .62.
= 756-9 lbs.
its direction is
P = 465 lbs.
irod to draw a
!, the coefficient
the plnne l)eing
Id support the
Lhau 17 tons.
i])Ic, supposing
to the plaiiC.
}5-j^ tons.
to the horizon
rnootfi wall of ii
c when a mun
is half way up,
foot of the
sustained on a
1 2 lt)0., and ti
it of frictioa is
tan-'(il).
i-'mmmmmmm
BXAMPLBS.
163
7. A heavy body is placed on a rough pljine whose
inclination to the horizon is sin "* (f), and is counected by
a string passing over a smooth pulley with a body of equal
weight, which hangs freely. Supiwsing that motion is on
the point of t-nsu'ng up the plane, find the inclinatiop of
the striag to the plane, the coeCaoient of friction being \.
Ana. ^ = 2 tan"' (i).
8. A heavy body, acted uiwn by a force equal in magni-
tude to its weight, is just about to ascend a rough inclined
plane under the influence of this force ; find the inclination,
d, of the force to the inclined plane.
Am. e = ^ — *■» or 2(;fr + t - 1 where t = inclination
2 *
of the plane, and <> = angle of friction, {d is here sup-
posed to be measured from the upper side of the inclined
plane). If 5 > 2^ -f », * is negative and the applied force
will act towards the under side.
9. In the first solution of the last example, what is the
magnitude of the pressure on the plane ?
Ana. Zero, Explain this.
10. If the shaft, (Art. 100), is a square prism of the
weight W, and rotates about an axis in its centre, prove
that the moment of the friction of the square end varies as
the side of the square.
11. If the shaft is composed of two equal circular
cylinders placed side by side, and rotatcH about the line of
contact of the two cylinders, show that the moment of the
friction of the surface in contact with the horizontal piano
_ 3a/mlF
-~ 9n '
12. What is the least coefficient of friction that will
allow of a heavy body's being just kept from sliding down
EXAMPLES.
an inclined piano of given inclination, the body (whoso
weight is W) being sustained by a given horizontal force, P ?
An,, l^'^P.
»K + /' tan i
13. It is obsoi-ved that a body whose weight is known to
be fKf'flin bo juet sustained on a rough inclined plane by a
liorizontil force P, and tlmt it can also be just sustained on
the same plane by a force Q up the plane; express the
angle of friction in terms of these known forces.
Am. Angle of friction = cos~* •
14. It is o' ved that a force, ^,, acting up a rough
inclined pL .11 ju«t sustain on it a body of weight W,
and that a for-^e, Q^, acting up the piano will jest drag tho
same body up ; find the angle of friction.
Ans. Angle of friction — sin"' ^l -T-jiJ
15. A hea>> uniform rod rests with its extremities on
the interior of a rough vortical circle; find the limiting
position of equilibrium.
Ans. If 2« is the angh subtended at tho centre by the
rod, and A the angle of friction, the limiting inclination of
the rod to the horizon is given by the equation
tan 6
sin 2A
cos 2A -f- cob 2a
16. A solid triangular prism is placed, with its axia
horiisonUil, on a rough inclined i)lane, the inclination of
which is gradually increased ; determine tho nature of the
kititil motion of the prism.
Am. If the ta-ianglc ABO is the section iwrpondionlar to
the uxis, aud the gide AR is isi contact with the plane, A
B body (whoso
zontal force, P ?
rtan i — p
V + .Tii^'
^ht is known to
led plane by a
i8t sustained on
s; express the
ces.
PW
ig up tt rough
7 of weight W,
jest drag tho
xtreniities on
the limiting
centre by the
inclination of
ith its axis
U'linution of
atiire of the
pudionlar
he plane,
EXAMPLES.
165
being the lower vertex, the initial motion will be one of
tumbling if
5» + 3<? - o«
l^>
4A
the sides of the triangle being a, b, c, and its area A. If ft
is less than this value, the initial motion will be one of
slipping.
17. A frustnaa of a solid right cone is placed with its
base on a rough inclined plane, the inclination of which
is gradually increased ; determine the nature uf the initial
motion of the body.
Am. If the radii of the larger and smaller scctionB arc R
and r, and h is the height of the frustum, the initial motion
will be one of tumbling or slipping according as
'*><T
4R R* + Rr + t*
R* + 2Rr + 3r»
18. An elliptic cylinder rests in limiting equilibrium
between a ro'igh vertical and an equally rough horizontal
plane, the nxis of the cylinder being horizontal, and the
major axis of the ellipse inclined to the horizon at an angle
of 46°. Find the coefficient of friction.
Ans. ft = — ■ ■ - _"~ — ^-, e being the eccentricity
of the ellipse.
P
ffifiiiinn
CHAPTER VI.
THE PRINCIPLE OF VIRTUAL VELOCITIES*
101. Virtual Velocity.— If the point of application of
a force be conceived as displaced through an indefinitely small
spacs. tho resolved part of the displaceiHcnt in the direction
theforct^, is calkd the Virtual Velocity of the force,' and
the product of the force into the virtual velocity has been
called the virtual moment\ of the force.
"J hue, let O be the original, and A
the new point of application of the
force, i*, acting in the direction OP,
and let AN be drawn perpendicular to
it. Then ON is the virtual relocity of
P, and P • ON is the virtual moment,
virtual displacement of the point.
If the projection of the virtual displacement on the line
of the force lies on the side of O toward which P acts, as in
the figure, the virtual velocity is considered positive; but
if it lies </fi the opposite side, i. «., on the action line pro-
longed through 0, it is negative. The forces are always
n>garfied as positive ; the sign, therefore, of a virtual mo-
ment will be the same as that of tho virtual v«'locity.
Cob. — If d be the angle between the force and the virtual
dibpiacement, we have for the virti?al moment,
P . ON = P • OA cos « = P co£ e . OA.
OA is called the
* Thf- rHw)tpN> 3t Vinnil VeloctUwi ww dlfoovered by Otllleo, mi wm rmj
Ailly d'-" ''ped by BnrtiouiUi and La^gnngt.
t Soiiu tlniM oatM '
bj UuluuotJ.
VirtiuU Work." Tte ovaa " Virtiul Momaat " <
i^ren
.'JiM**»
^■^
OCITIES*
f application of
idefinitely small
n the direction
the force ; and
tlocity has been
Fig.50
. is called the
it on the line
P acts, ofi in
positive; but
tion line pro-
08 are always
a virtual mo-
looity.
nd the virtual
OA.
lleo, and ww very
omont " WM giTen
'■ il l Wg ff'WIII I IIIIIII I IWWIU I llll l
VIRfUAL VELOCITIES.
107
Now P cos 6 is the projection of the force on the direction
of the displacement, and is equal to OM, OP being the
force and PM being drawn perpendicular to OA. Hence
we may also define the virtual moment of a force as the
product of the virtual displacement of its point of applica-
tion into the projection of the force on the direction of this
displacement; and this definition for some purposes is
more convenient than the former.
Rbmahk.— A forca is said to do work if It movM the body to wbidi
it is applied ; and tlie work doDu by it is measuriMi by tlie product i>{
tlie force into tiie space through wliich it moTce the body. Generally,
the work done by any force during an infinitely tuBall dkaplacemeut uf
its point of application is tlie product of the rettolved paii of tbc force
in the direction of the displacement into tlie displaomnent . hmI thia
is the same as the virtutM^ «»>«M«Mt of Um force.
102. Principle of ▼ht—l "W^HocStx^m. — (1) The
virtual moment of a force m e'^ttui iv the xum of the virtual
moments of its components.
Let OR represent a force, ft, act-
ing at 0, and let its components be
/' and Q, represented by OP and
OQ. Let OA be the virtual dis-
placement of 0, and let its projec-
tions on R, P, and Q, be r, p, and
q, respectively. Then the virtual
moments of these forces are R • r, P
Pm, and Qo, perpendicalar to OA.
Fia.M
p, Q • q. Draw /?>/,
Then On, Om, and Oo
(= mn), are the projections of R, P, and Q, on the dirr-c-
tion of the displacement ; and henoe (Art. lOX, Cor.) we have
i2 . r = OA On ;
P.;>=OA.Om;
^ - q ss OA • mn.
168
rinvuAL VELoctTtsa.
Hence P • ^ + Q • y = OA (Om + mn)
= OA ' On = E-r.
(See MinohiD'a Statics, p. 68.)
(2) If there are any number of component forces we may
compouud them iu order, taking any two of them first, and
finding the virtual moment of their resultant as above, then
finding the virtual moment of the resultant of these two
and a third, likewise the virtual moment of the resi!i>«nt of
the first three and a fourth, and bo on to the last ; or we
may use the polygon of forcew (Ai-t. 33). The sum of the
virtual moments of the forces is equal to the virtual dis-
placement mullipliod by the sum of tiic projci (ions on the
displacement of the sides of the polygon Avhich rt'pre.^ent
the forces (Art 101, Cor.). But the sum of these projec-
tions is cyual to thv projection of the remaining side of the
polygon,* and this side represents the resultant, (Ar!. 3'6,
Cor. J). Therefore, the sum of the virtval moments (if any
number of concurring forces is equal to the virtual moment
of the resultant.
(3) If the forces are in equilibrium, their resultant is
equal to zero ; hencA , it follows that ivhen any number of
concurring fnrren ur» in equilibrium, the sum of their
virtual moments =■ 0.
This principle la gon(*rully lin<»wn as the Principle of
Virtual Velocities, and Ih of great use in the BoluUon of
practical problems in Statics.
* Krom Uis lUture of projectiuiia (Anal. Ocom., Art. laB), It |8 Oleir llilt in any
terleit of polnU the projection (on a glTen line) of tlio lliio n lili.li Jlijna ilio flret and
l8Bt, Ir eq;ukl to the oum of the projection)! of the Ilium which Join the pointH. iwa
and two. Thim, If the eldes of a cloxed polfKun. taken In order, be morkcd witli
arrows pointing from each vertns to the next one ; and If their projettlon-j be
marked with arrows In the same dln^ctlong, then, lined mcasorsd from left to right
being oonsldered pocltive, and linos from right to left negative, the swn (ff tKr pro-
Jtetionn <if the Hdet nf a dvMd pohfgon on any rl(f/ii Rm U atro.
8
n)
nt forces we may
»f them first, and
kiit as above, then
ant of these two
' the rest! Hunt of
the last ; or we
The sum of the
the virtual dis-
ojoc/iujis on the
Avhich represent
of those projec-
ining side of the
iUant, (Art. 33,
moments (if any
virtual moment
eir resultant \»
any number of
sum of (heir
tl
Principle of
10 solution of
I- iHf.v tint in aiiy
jiilnt Die flrst and
jiilu the potnti), two
be marked with
tlii'lr projeitlons bo
rsd IW>m left to right
the sum of the pro-
VntTDAL MOMMJfTS.
169
103. Natnre of the Sisplaeemttnt.— It must be care-
fully observed that the displacement of the particle on
which the forces act ia virtual and arlilrarv. The word
virtual in Statics is used to intimate tha^. tbo displacements
are not really made, but only supposed, i. e., they are not
actual but imagined displacements ; but in the motion of a
particle treated of in Kinetics, the dJsplacemcni is often
taken to be that which the particle actually undergoes.
In Art. 101, the displacement was limited to an infiniteai-
mal. In some cases, however, & finite displacement may be
used, and it may be even more convenient to consider a
finite displacement. But in very many cases any finite dis-
placement is suflScient to alter the amount or direction of
the forces, so as to prevent the principle of virtual velocities
from \mug applicable. This diflSculty can always bo avoid-
ed in practice by assuming the displacement to be infinitesi-
mal ; a»d if the virtual disphujement is infinitesimal the
virtual volocitieci are all inflnitcsimaL
104 Equation of Virtual Moments. —Let P^, P,,
Pf, < !. iiofr the fonos, and ip^. (5jt>,, '5/;,, etc., the vir-
taai I lien the priut>ipUs of virtual velocities is
expressed (Art 102) by the etiuation
^i • ^Pi + Pt • ^P% + i\ ' ^P, + etc. = 0;
or ^P6p = 0, (1)
which is called tbo equation of virtu/il moments.*
Hon. — If the virtual diHplacemont is at right angles to
the direction of any force, it is clear that dp, the virtual
velocity, is rijual In m-ro. Hence, when the virtual dia-
placement is at ripht angles, to the direction of the force,
• Or virtual wort (Set Art. 101, Rem.). Tbl* eqiutlon IiM been in&de hy La.
itranKe tho fiiaudation of hia Kreat work on Xeclivilo, " MiJbaahtno /jialytlque,'*
(I'rlce'H AiuO. Uocb., ~.'ol. I, p. 14t.)
170
SrSTBM or PAItTICLXa.
m.
m
»M
the virtual moment of the force = 0, and the force will not
enter into the equation of virtual moments.
Such a virtual displacement is always a convenient one
to choose when we wish to get rid of some unknown force
which acts upon a pari:icle or system.
105. System of Particlea Rigidly Connected.— (1)
If a particle in equilibrium, under the action of any forces,
be constrained to maintain a fixed distance from a given
fixed point, the force due to the constraint (if any) is
directed towards the fixed point
Let B be the particle, and A the fixed point. Then it is
clear that we may substitute for the string or rigid rod
which connecta B with A, a smooth circular tube enclosing
the particle, with the centre of the tube at A. Now, in
order that B may be in equilibrium inside the tube, it is
necessary that the resultant of the forces acting upon it
should be normal to the tul)e, t. e., direct^ towards A.
(2) Let there be any number of
particles, mi, m,, 7n,, et^., each
acted on by any forces. P,, P,, Pg,
etc., and connected with the others
by inflexible right lines so that the
figure of the system is invariable.
Then eacli particle is acted on by all
the external forces applied to it, and
by all the internal forces proceeding from the internal con-
nections of the particle with the other particles of the
system. Thus tlie particle, m, is acted on by Pj, Pg, etc.,
and by the internal forces which proceed from its connec-
tion with m^, «!,, «t,, etc., and which act along the lines,
mm I, mm^, etc., by (1) of this Article. Denote the forces
along the lines »iw,, «i»i„ mm,, etc., by /j, t^, t^, etc.,
and their virtual velocities by dt^, rf^,, <J/,, etc. Now
^ force will not
convenient one
unknown force
innected.— (1)
of any forces,
from a given
int (if any) in
it. Then it is
I or rigid rod
tube enclosing
t A. Now, in
the tnbe, it is
«ting npon it
>wards A.
internal oon-
icles of the
P,, Pj, etc.,
its connec-
ig the linos,
e the forces
/,, <s, etc.,
etc. Now
BrSTSM OV PART. US8.
171
imagine that the system is slightly displaced so as to
occupy a new position. Then (1) of Art. 104 gives us
for n»,
I\6p^ + P,dp, + etc. + ti&l, + ttStt + etc. ^0, (1)
for nil,
t'JPi. + Pi^Pt, + etc. + /i<J<i + <,<»<, + etc s= 0, (a;
proceeding in this way as many equations may be formed as
there are particles in the system.
Now it is clear that /,d/,, and /,«J^j, in (1) have contrary
signs from what they have in (2). Thus if the system is
moved to the right in its displacement, t^Sti, and t^it^ will
Ix positive in (1) and negative in (2) (Art. 101), and hence,
if we add (1) and (2) together, thoije terms will disappear ;
in the same way, the virtual moment of the internal force
along the line connecting m with any other particle disap-
pears by addition, and the same is true for the interii^^l
force between any two particles of the system. Henco,
adding together all the equations, the internal forces
disappear, and the resulting equation for the whole
system is
SPdp = 0, (1)
A
■>
and the same result is evidently true whatever be the num-
ber of particles forming the system. Hence, if any num-
ber of forces in a system are in equilibrium, the sum of
their virtual moments = 0.
The converse is evidently true, that if the sum of the
virtual moments of the forces vanishes for every virtual
displacement, the system is in equilibrium.
The following are examples which are solved by the
principle of virtual velocities.
^m
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SMAGE EVALUATION
TEST TARGET (MT-S)
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microfiches.
Canadian Institute for Historical Microraproductiont / Institut Canadian da microraproductiona historiquaa
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I
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lis*
17a
MXAMPLXS,
EXAMPLES.
1. Determine thv> condition of equilibrium of a heavy
body resting on a smooth inoline^l plane under the action
of given forces.
Let W bo the weight of the body
sustained on the plane BC by the
force, P, making an angle, 0, with the
plane. To avoid bringing the un-
known reaction, R, into our equation,
we make the displacemeLt of it« point
of application perpendicular to its line of action, (Art. 104,
Sch.); hence we conceive as receiving a virtual dis-
placement, OA, at right angles to R, the magnitude of
which in the present case is unlimited. Draw Am and
A» perpendicular to W and P respectively, Om and 0»
are the virtual velocities of W and P, (Art 101) ; and
W • Om and P • On are their virtual moments. Horce (1)
of Art. 104, gives
W • Of» - - P . On = 0.
*
^H. But
Om = OA sia a,
^^H
On = OA cos & ;
^^BE therefore
W sin a — Poos* s= 0;
m
which agrees with Ex. 3, Art. 41.
:h
If the Toroo acts parallel to the plane, 9 = 0, aud (1)
becomes
P = W sin « ;
which agrees with Ex. 1, Art. 41.
mmemirwai
ibrinm of a heavy
ander the action
action, (Art. 104,
Dg a virtual dis-
thfl magnitude of
Draw Aw and
vely, Om and 0»
(Art. 101) ; and
ents. Horce (1)
(1)
« = 0, aud (1)
MXAMPLga.
178
2. Sappoae the plane in Ex. 1 to be rough, and iha« the
body is ou the point of being draggsd up the pUine, find
the condition of equilibrium.
The normal resistance will now be
replaced by the total resistance, C,
inclined to the normal at an angle
= <f>, tht- argle of friction (Art. 95,
Cor. ). Let the virtual d itplacemeut,
OA, take pL>i0e perpendicularly to
R, then (1) of Art. 104, gives
WOm — P.On = 0.
But Om = OA sin (o -f ^),
i>nd On = OA cos (^ — 6) ;
therefore W aic (a + 0) =: P oos (^ — 9) ;
which agrees with (3) of Art. 96.
?,. Determine the horisontal force
which will keep a particle in a given
position inside a circular tube, (1)
when the bibe is smooth end (3)
when it is rough.
(1) Lot the virtual displaoament,
OA, be an infinitesimal, = tis, aloiig
the tube. Then since dt is infinites*
imal the virtual velocity of B = 0. Then the equation of
virtual moments ij
— W • Om 4- P • On = 0.
Wi§,»
But
Om = da ■ sin 6,
and
On =z Ha -co* 8]
therefore
W«in« = PooiS;
or
P = W An e.
EXAMPLSa.
{%) Snppoao tho force, P, jnst sustains the particle ; the
normal rcsiatanoe must now be replaced by the total resist-
ance, making the angle, ^, with the normal at the right of
it. Take the virtual displacement, OA', at right angles to
the total resistance (Art 105, Sch.), and let it be as before,
an infinitosimal d*. Then (1) of Art 104, gives
--W-Om + P.0«'=0.
But Om = <fo • sin (0 — <»),
aud On' = d» ' QOi (0 — ^),
therefcrs W • sin (* — 0) = P • cos (0 — ^) j
or P = W . t»n (fl — ^).
Similarly, if the force, P, will just drag the particle up
toe tube we obtain
P = W • tan (© + <•).
4. Solve by virtual velocities Ex. 6, Art. 62.
Let the displacement be made by diminishing the angle
«, which the beam makes with the horiaontal plane, by da,
the ends of the beam still remaining in contact with the
horizontal end vertical planes. Then the virtaal velocity of
T s d • 2a cos a = ~ 2a sin a dm',
aud that of
W = (2aBinn = aco8a da,
and those of the leaotinns, R aUd R', yaoish. Then the
equation of viitnal momenta if
the particle; the
by the total resist-
aal at the right of
at right angles to
let it be as before,
.gives
0.
-f);
1 the particle up
62.
liehing the angle
al plane, by da,
ontact with the
irtaal velocity of
dm;
isb. Then th«
MXAMPMCa. 179
— T2asina(;«.fWaco8«c^=rO;
.*. 2T sin a =r W cos a.
6. Solve Ex. 8, Art 63, by virtual velocities.
Let the displacement be made by increasing the angle
d by d9, the point, A, remaining in contact with the wall ;
the virtutil displacement of B is at right angles to the
direction of the tension, T, and h'ince the virtual moment
of T is zero ; the virtual velocity of W is
d (5 cua -- a cos 0) — amxO dO — b wa^ d^.
Then (1) of Art 104, gives
yi {a «in e de — b An ift d«l>) — Q\
.'. & siu ^ (^^ = a sin 9 dd.
But from the geometry vt the figuro we have
d sin ^ = 3a sin 9;
.'. 6 cos ^ rfi^ = 2a COS 9 <fO;
.-. 2tan^ = tant;
which, combined with (6) of Ex. 8, Art. 63, gives us the
values of sin and cos ^ ; and these in (6) of that Ex.
give us the value of x.
6. Solve Ex. 38, Art. 65, by virtual velocities.
■ Slow the bar Is to rest in all positioiui on the curve and the
peg, its centre of gnrity will neither rise nor fall when the
bar reoeivee a displaouuent, hence Its virtual Telocity hIU s ;
• t MA*
176
EXAMPLEB.
7. In Ex. 4, Art 42, prove that (1) is the equation of
virtual moments.
8. Find the inclination of the beam to the vertical in
Ex. 31, Art. 66, by virtual velocities.
9. Deduce, by virtual velocities, (1) the formula for the
triangle of forces (see 1 of Art 32), and (2) the formula for
the parallelogr&m of forces (See 1 of Art. 30).
the equation of
D the vertical in
fonnola for the
I the formula for
0).
CHAPTER VII.
MACHINES.
106. Fnnetioiu of a Machine.— Ji maehiue, Statically,
is any instrument by means of which we may change the
direction, magnitude, and point of application of a given
force ; and Kinetically, it is any instrument by means tf
which we may change the direction and velocity of a given
motion.
In applying the principle of virtual velocities to a system
of connected bodies, odvantige is gained by choosing the
virtual displacements in certain directions (Art. 104, Sch).
When we use this principle in the discussion of machines
the displacements which we shall choose will be those which
the different parts of a machine actually undergo when it
is etni)loyed in doing worilc, and instead of equations of
virtual work we shall have equations of actual work; and
in future the principle of virtual velocities will often be
referred to as the Principle of Work. (See Minchin's
Statics, p. 383.)
Every machine is designed for the purpose of overcoming
ceitain forces which are called resistances; and the forces
which are applied to the machines to produce this effect are
called movitig forces. When the machine is in motion,
every moving force displaces its point of application in its
own direction, while the point of application of a resistance
is displaced in a direotion opposite to that of the resistance.
Ilenoe, a moving force is one whose elementary work * i>
positive, and a resistance is one whose elementary work is
negative. The moving force is, for convenience, called the
* See Art. 101, Ran.
Is,
178
MECHANICAL ADVANTAOE.
power ; and becanse the attraction of gravity is the most
common form of the force or resistance to bo overcome it is
usually called the weight.
The weight or radstanoe to be overcome may be the enrth's attrac-
tion, aa in raising a weight ; the molecular attractiona between the
particles of a body as in stamping or catting a metal, or dividing
wood ; or fristlon, aa in drawing a heavy body along a rough road.
The power may be that of men, or horses, or the steam engine, etc.,
and may be jost sufficient to overcome the resistance, or it may be in
excess of what is necessary, or it may be too small. If just sufficient,
the machine, if in motion, will remain uniformly so, or if it be at rest
it will be on the point of moving, and the power, weight, and friction
will be in equilibrium. If thr power be in excess, the machine will
be set in motion and will continue in accelerated motion. If the
power be too small, it will not be able to move the marJiine ; and if
it be already in motion it will gradually come to rest.
The geueral problem with regard to machines is to find
the relation between the power and the weight. Some-
times it is most convenient that this relation should be one
of equality, t. e., that the power should equal the weight
Generally, however, it is most convenient that the power
should be very different from the weight Thus, if q man
bus to lift a weight of one ton hanging by a rope, it is clear
that he cannot do it unless the mechanical contrivance
provided enable him to lift the weight by exercising a pull
of very much less, say one cwt When the power is much
smaller than the weight, as it is in this case, which is a
very common one, the machine is said to work at a mechan-
ical advantage. When, as in some other cases, it is de8irul)le
that the power should be greater than the weight, there is
said to be a mechanical disadvantage of the machine.
107. Mechanical Advantage.— (1) Let P and W he
the power and weight, and p and w their virtual velocities
respectively ; and let friction be omitted. Then fh>m the
equation of virtual work (Art 104), we have
Pp-Ww=iO, OT ~-y
mum
OE.
gravity is the most
,0 be overcome it is
be the eitrth's attrac-
itmctions between the
: « metal, or dividing
y along a rough road,
the Bteam engine, etc.,
lance, or it may be in
all. If just sufficient,
f BO, or if it be at rest
3r, weight, and friction
cesB, the machine will
rated motion. If the
re the machine ; and if
> rest.
machines ia to find
e weight. Some-
Ition should be one
equal the weight
t that the power
Thus, if % man
a rope, it is clear
nical contrivance
exercising a pull
he power is much
|s case, which is a
work at a mechan-
laees, it is de8irul>le
e weight, there ia
e machine.
iLet P and »r be
virtual velocities
Then from the
^ve
10
y
pir'j ,11m
MSCHANICAL ADVANTAQX.
\n
which shows that the smaller P is in comparison with W,
the smaller w will be in comparison with p. But the
smaller P is in comparison with W, the greate; is the
mechanical advantage. Hence, the greater the mechanical
advantage is the lefc3 will be the virtual velocity of the
weight in comparison with that of the power. Now, if
motion actually takes place the vi/tual velocities become
aciual velocities ; and hence we have the principle what is
gained in power is lost in velocity.
(3) There are no cases in which the weight and power
are the only forceo to be considered. In every movement
of a machine there will always be a certain amount of fric-
tion ; and this can never be omitted from the equation of
virtual work. There are cases, however, as that of a balance
on a knife-edge, where the frict'on is very small ; and for
these the principle, what is gained in jwwer is lost in
velocity, is very approximately true. Where the friction is
considerable this is no longer the case.
Lot F and / bo the resistance of tiiction and its virtual
velocity, then the equation for any machine will take the
form
Pp — Ww — Ff = 0,
which shows us that although P can be made as small sta we
wish by taking p large enough, yet the mechanical
advantage of diminishing P is restricted by the fact that /
increases with/>; and therefore as P diminishes there is' a
corresponding increase of the work to be done against fric-
tion. Hence if friction be neglected, there is no practical
limit to the ratio of P to fF ; but if the friction be con-
sidered, the advantage of diminishing P has a limit, since
if Pp remains the same, Ww must decrease as /y increases;
i. c, the work done against friction increases with the^
complexity of the machine ; and thus puts a practical limit
to the mcchanicAl advantage which it is possible to obtain
by the use of machines.
180
aiXPLB MACBlNSa.
108. Sln^te MacMnea.— The simple machines, some-
times called the Mec/ianical Powers, are generally enumei-
ated as six in number ; the Lever, the Wheel and Axle, tho
Inclined Plane, the Pulley, the Wedge, and the Screw.
The Lever, the Inclined Plane, and the Pulley, may bo
considered as distinct in principle, while the others are
combinations of them.
llie efficiency* of a machiuQ is the ratio of the useful
work it yields to the whole amount of work performed by
it The use/til work is that which is performed in over-
coming useful resistances, while losi work is that wiiich is
spent in overcoming wasteful resistances. Useful lesist-
anccs are those wliich the machine is specially designed to
overcome, while the overcoming of wasteful resistances is
foreign to its purpose. Friction and rigidity of cords are
wasteful resistances while the weight of the body to be
lifted is the useful resistance.
Let W be the work done by the moving forces, W^ the
useful and Wi the lost work when the machine is moving
uniformly. Then
If = r. + Wi,
and if Jf denote the efficiency of the machine, we have
W
In a perfect machine, whore there is no lost work, the
efficiency is unity ; but in every machine some of tho work
is lost in overcoming wasteful resistances, so that the
efficiency is always less than unity ; and the object of all
improvements in a machine is to bring its efficiency as near
unity as possible.
The most noticeable of the wasteful resistances are fric-
tion and rigidity of cords ; and of these we shall consider
* HonmimM ckUad mo^Mn.
«*
KnviLianitTM OF raw lkvsk.
lU
bines, eome-
illy enumer-
<id Axle, tho
the Screw,
ley, Tuay bo
otliers arc
f the useful
erformed by
ned in over-
lat which is
'»eful lesist-
dedigned to
eaistances is
of cords are
body to be
ces, Wm the
is moving
re have
work, the
tho work
that the
hject of all
ncy ns near
68 are fric-
II consider
only the first The student who wants informauon on the
experimental laws of the rigidity of cords is referred to
Weisbach's Mechanics, VoL I, p. 363.
109. The Levor.—A lever is a rigid bar, straight or
onrved, movable about a fixed axis, which is called the
fulcrum. The parts of tho lever into which the fulcrum
divides it are called the arms of the lever. When the arms
are in a straight line it is called a straight lever ; in all
other cases it is a bent- lever.
Lovers are divided, for convenience, into three kinds,
according to the position of the falorom. In the first kind
the fulcrum is between the power and tho weight ; in tho
second kind the weight acts between the fulcrum and tho
power ; in ihe third kind the power acts between the ful-
crum and the weight. In the last kind the power is always
greater than the weight
A pair of scissors fumislies an example of a pair of levers
of the first kind ; a pair of nut-crackers of the second kind ;
and a pair of shears of the third kind.
110. Conditions of Eqnilibrinm
of the Lever. — (1) Wilh<mt FVictton.
Let AB be the lever and its fulcrum ;
and let the two forces, P and W, act in
the plane of the paper at the points, A
and B, in the directions, AP and BW.
From C draw CD and OE perpendicular to the directions
of P and W. Let a and P denote the angles which the
directions of the forces make with the lever. Then, taki.ig
moments around C, w^ have
P.CD = rCE,
;i
ii
or
P _ perpendicular on direction of W
W " perpendicular on dirootiou of P'
(1)
^^_.v k:_ (^:j
I
I
^i
i
188
EqUILIBRIUU OF TBS LSVJSS.
That is, tho condition of equilibriam requires that the
power and weight should he to each otlter inversely as the
length of their respective arms (Art. 46).
To find the pressure on tho fulcrum, and its dii'cctiou ;
let the directions of the pressures, P and W, intersect in
F; join and F; then, since the lover is in equilibrium
by the action of the forces, P and W, and the reaction of
the fulcrum, the resultant of P and W must be equal and
opposite to that reaction, and hence must prss through
and be equal to the pressure on the fulcrum. Denott) this
resultant by R, tho angle which it makes with the lever by
; and the angle AFB by <•> ; then we have by (1) of Art. 30
or
It* = P* + W» + 2PW COB AFB ;
IP = P» + W^ + 2PW cos fc).
(8)
ic?iich gives the pressure, R, on thefukrum.
To find its direction resolve P, W, und R parallel and
perpendicular to the lever, and we have
for parallel forces, P cos a — W cos fi—R cos = 0;
for perpendicular forces, P sm a + W sin (i—R sin = 0;
by transposition and division we get
P sin « -f- IF sin /3
tan e =
F cos a — W cos /3'
(8)
which gives the direction of the pressure.
Cob. — When the lover is bent or curved the condition of
equilibrium is the same.
Solution by the principle of virtual velocities.
Suppose the lever to be turned round in the direction
of P through the angle dO, into the position ab; let p and
[aires that the
iversely as the
its dii-oi^tioii ;
V, intersect in
Id cquilibrinm
he reaction of
be e<]Tia!l and
R88 through
Denoto this
!i the lever by
(1) of Art. 30
^;
(2)
! parallel and
^ cos = 0;
i? sin © = 0;
(3)
condition of
he direction
let p and
BQUILIBRIUM OF TBB LJSVXR.
"*1*pi
188
q be the perpendiculars CD and G£ respectively, then the
virtual velocity of P will be (Art. 101),
Aa sin a — AC-dO-aln a ■=. pd&.
Similarly, the virtual velocity of TF is — qdB.
Hence, by the equation of virtual work we have
P'P-dd-^ W-qdd = 0;
.-. P p= W-q.
which is the same as (1).
(*)
(3) With Friction. -In the above wo have supposed fric-
tion to be neglected ; and if the lever turns round a sharp
edge, like the scale beam of a balance, the friction will be
exceedingly small. Levers, however, usually consist of flat
bars, turning about rounded pins or studs which form the
fulcrums, and between the lever and the pin there will of
coni-se be friction. To find the friction let r be the radius
of the pin round which the lever turns ; then the friction
on the pin, acting tangentially to the surface of the pin
and opposing motion, = ^ sin ^ (Art. 99) ; and the virtual
velocity of the point of application of the friction = rdO ;
and hence the virtual work of the friction = £ sin ^- rdO,
Hence the equation of virtual work is
p.pd6 — W-qM — ^ sin ^ r<W = 0.
Substituting the value of R from (2), and omitting d6, we
havo
Pp —Wq = r sin fjt ^/WT~W* + 2Pr cos w ; (5)
solving thie quadratic for P we have
X84
TES COMMON BA/.AJ/CS.
P=W
pq + r' cos u) giu* ^
/J* — r* sin*
± Wt sin ^
vV>' 4- ^j?? COS w 4- g* — »•* sin* (p sin* ta
p^ — t^ siu* flk
.(6)
which gives the relation between the power and the weight
when friction is considered, the upper or lowei sign of
r sin <) being taken according as P or W ie about to pro-
ponderate.
Cob. — If the friction is so small that it nay be omitted,
r sin ^ = 0, and (6) becomes
w~ p
m
112.. Th« Common Balano. — In machines generally
the object is to produce motion, not rest ; in other words
to do woik. The statical investigation shows only tlie limil
of force to be a'-plied to put tiio machine on the point of
motion, or to give it uniform motion. For any work to bo
done, the force applied must exceed this limit, and the
grealor the excess, the greater the amount of worL done.
Thei-e 's, however, one class of applications of the levt/
where the object is not to do work, but to produce equi-
librium, and which !>,re therefore 8))eL'iRlly adapted for treat-
ment by statics. This is the class of measuring machines,
where the object is not to overcome a particular resistance,
but to measure its amount. The t'Osting machine is a gotnt
example, measuring the pull which a bar of any material
will sustain Ixifore breaking. The common balance and
steelyard for weighing, are familiar examples.
The common balance is an instrument for weighing ; it
is a lever of the hrst kind, with two equal anns, with a
Bcale-pr.u suspended from each extremity, the fulcrum
being vertically above the centrn of gravity of the borm
when the latter is horiscnt«l, and therefore vertically above
lin' sin* u» ,„.
, (C)
EUid the weight
lower sign of
about to pro-
ay be omitted,
(7)
hines generally
n other words
3 only the limit
n the point of
my work to bo
limit, and the
jf work done.
3 of the lever
produce equi-
pted for treat-
ing machines,
ar ix^sistancc,
ine is a go<Hl
any mittorial
balance and
weighing ; it
arms, with a
the fulcnim
of the borra
rtically abovo
TBB COMMON BALAlWt.
185
the centre of gravity of the system formed by the beam, the
scale-pans, and the weights of the acate^paus. The sab-
stunce to be weighed is placed in one ecalc-pan, and weights
of known magnitude are placed in the other til) the beam
remains in equilibrium in a {)erfectly horizontal position,
in which case the weight of the substance is indicated by
the weights which balance it If these weights differ by
ever so little the horizontality of tbo beam will be disturbed,
and after oscillating for a short time, in consequence ot
the fulcrum being placed above the centre of gravity of the
system, it will rest in a position inclin -i to the horizon at
an a.igle, the extent of which is a measure of the sensibility
of the balance.
Dte preceding explftnatioc raprpaenta tlie balanoo in ita simplflet
fom; in practice there aru manj modiflcationa and roDtrivances
introduced. Sfacli akill hso been expended upon the cch mrction of
balincea, and great delicacy haa be.ia olitainr'J. Tfaoa, the lieam
ehouid l>e auapended hj me&na of a Icnifeedge, i. e. , a projecting
metallic edge transverse to its len){th, which reatfl open a plate of
agate or othei hard sulMtance. The chains which rapport the acaln-
pauB should be suspended tiom the extremities of the beam in the
same manner. The point of support of the beam (folerum) should be
at equal distances from the points of suspension of the scales ; and
when the bnlance is not loaded the Iteam s.iould be horisontal. We
can ascertain f these conditions are satisfied by olwerving whether
tliere is still 'quillbrium when the snbntanoe is transferred to the
scale which th ) weight originally occupied and : he weight to thai
which the subst inoe originally oocnpind.
The chief requisites of a good balance are :
(1) When equal weights are placed in the sca1e-3Hui8 the
beam should be perfectly horizontal.
(2) The balance should poss*;?'^ great misibilitif ; x. «., if
two weights which are very nearly equal be placed in the
Bcale-pana, the beam should vary t$n«ibly from its horizontal
position.
ij-i
186
aSQUIBITSa OP a qood balancm.
(8) When the bala..^ is disturbed it should readily
return to its state of rest, or it should have staMity.
L
112. To DetenniiM the
Chief RaqniBites of a Good
Balanpe. — Let P and W be
the weights iu the scale-pans ;
O the fliicrum ; h its distance
from the straight line, AB,
which joins the points of at-
tachtuent of the sc»le-parf> to
the beam ; O the centre of grayity of the beam ; and let
AB be at right angles to 00, the line joining the falcmwt
to the centre of gravity of the beam. Let AC = OB = a;
OG — k\ w = the weight of the beam ; and 6 = tho
angle which the beam makes with the horizon when there
is equilibrium.
Now the perpendicular from
on
the direction of P
—
a cos
e-h
sin
0',
it
((
((
W
=
a cos
e + h
sin
9\
<<
«
«
to
—
k sin
e;
therefore tak'ng moments round we have
P (a COS 6~h sin 5)— If (a cos e+A sin 6)—wh sin d = ;
tan =
(P- W)a
{P + W)h + wh'
a)
This equation determines the position of equilibrium. Tho
first requisite — tho horizontality of 'the beam when P and
W are equtd — is Ba^^isfied by making the arms equal.
The second requisite [(2) of Art. Ill], requires that, for
a given value of P — W, the inclination of the beam to the
horisoii must be «s great as possible, and therefore the sen-
sibility is greater the greater tan is for a given value of
P — W'y and for a given value of tan 9 the sensibility is
- fietwassMsw-v-jMiMrm
mam*m f^ff ii H Ki ! m^ii>slsm9S» i' 9 ^wiv* l
^■IIWi'fcl|Wfl<WW*^j|g_
rCA
RBQUJSITS8 Of A GOOD BALANCB.
187
should readily
\tability.
beam ; and let
ig the fulgrnui
C = CB = rt;
and B — tho
pn when there
sin 9;
gin 9;
tPit ain = ;
(1)
ibrinm. The
when P and
equal.
lires that, for
e beam to the
Bfore the son-
iven value of
sensibility is
greater the smaller the valae of P — W 'a\ hence the sen-
sibility may be measured by -5- — ^, which requires that
be as small as possible. Therefore a must be brge, and to,
h, and k must be small ; t. «., the arms must be long, tuo
beam light, and the distances of the fulcrum from the
beam and from the centre of gravity of the beam must be
small.
The third requisite, its stability, is greater the greater
the moment of the forces which t«nd to restore the beam to
iti former position of rest when it is disturbed. 1( P=W
this moment is
[{P +W)h + wk] sin 6,
which should be made aa large as possible to secnre the
third requisite.
This 'M)ndition is, to som<> extent, at variance with the
second requisite. Tlicy may both b .• satisfied, however, by
making {P + W)h -\- wk large and a large also ; t. e., by
increasing the distances of the fulcrum from the beam and
from the centre of gravity of the beam, and by lengthening
the arms. (See Todhunter's Statics, p. 180, also Pratt's
Mechanics, p. 78.)
The comparative importance of these qualities of sensi-
bility and ntability in a balance will depend upon the use
for which it is intended ; for weighing heavy woightA,
stabilitif is of more importance; for use in a chemical
laboratory the balance must possess great BMmbiUty ; ar d
instruments have been constructed which indicate a varia-
tion of weight less than a miUionth part of the whole. In
a balance of great delicacy the fulcrum is made as thin as
possible ; it is generally a kni/t-tdge of hardened steel or
agate, resting on « polished agate plate, which is supported
on a strong vertical pillar of brass,
t^:m^' -1^ jvn
188
TBS STEBLTASD.
113. The Ste«l]rarcL— This if a kiud of balance in
which the arras &ve unequal in length, the longer one beiuti
graduated, along which a poist may be moved in order to
balance different weignts which 01*6 placed in a scale-pan on
the short-arm. While the moment of the substance
weighed is changed by increasing or diminishing its quan-
tity, its arm remaining constant, that of the poise is
changed by altering its arm, the weight of the poise
remaming the same.
114. To Qradnate the Common Steelyard.— (1)
When the point of suapeiunon is coincident with the centre
of gravity.
Let AF be the beam of the steel-
yard suspended about an axis pass-
ing through its centre of gravity,
C ; on the arm, CF, place a mov-
able weight, P ; then if a weight,
W, equal to P, is suspended from
A, the beam will balance when P
on the long arm is at a distance
from C equal to AO. If W equals twice the weight of P,
the beam will balance when the distance of P from is
twice AO ; and so on in any proportion. Hence if W is
successively 1 lb., 2 lbs., 3 lbs., etc., the distances of the
notches, 1, 3, 3, 4, etc., where P is placed, are as 1, 2, 3,
etc.. I. 0., the arm CF is divided into equal divisions, begin-
ning at the fulcrum, 0, as the zero point
(2) When the point of suspension is not coincident toith
the centre of gravity.
Let C b<» the fulcrum, W the substance to be weighed,
hanging; at the extremity. A, and P the movable weight.
Suppuse that when W is removed, the weight, P, placed at
B will balance the long arm, CF, and keep the eleelyard in
a horizontal position ; then the moment of the instrument
WWi M *w r>n »-<|«ttgw>ya
of balance in
•nger one beiui--
ved in order to
1 a scale-pan on
the substance
hing its quon-
f the poise is
t of the poise
teelyard.— (1)
vith the centre
p-r^
n|.M
weight of P,
P from is
lence if IF is
lances of the
re as 1, 2, 3,
risions, begin-
nncident mth
be weighed,
table weight.
P, placed at
aleelyard iu
instrument
MXAMPLXa.
189
itself, aboBt 0, is on the side, CF, and is equal to P' CB.
Hence, if W hangs from A, uid P from any point E, then
for equilibrium we must have
PCE + PBC = r. AC; •
or P-BB=fr-AO;
BE = -^ . AO.
If we make W successively equal to P,2P, SP, etc., then
the values of BE will be AG, 2AC, SAC, etc., and these
distances must be measured off, commencing at B for the
zero point, and the points so determined marked 1, 2, 3, 4,
etc. Such a steelyard catinot weigh below a certain limit,
corresponding to the first notch, 1.
To find the length of the divisions on the beam, divide
BE, the distance of the poise from the zero point, by the
weight, W, which P balances when at the point E. The
steelyard often has tivo fulcrums, one for small and the
other for large weights.
EXAMPLKS.
1. What force must be applied at one end of a lever
12 ins. long to raise a weight of 30 lbs. hanging 4 ins. from
the fulcrum which is at the other end, and what is the
pressure on the fulcrum f Ans. 10 lbs. : 20 lbs.
2. A lever weighs 3 lbs., and its weight acts at its middle
point ; the ratio of its arras is 1 : 3. If a weight of 48 lbs. ,
be hung from the end of the shorter arm, what weight
must be suspendeu from the other end to prevent motion ?
Ans. 15 lbs.
f
HI
IW
WHEEL Am) AXLM.
8. The arms of a bent lever are 3 ft. and S ft. and inclined
to each other at an angle B = 150°. To the short arm a
weight of 7 lbs. is applied and to the long osm a weight of
6 lbs. is applied. Required the inclination of each arm to
the horizon when there is equilibrium.
Ans. The short arm is inclined at an angle of 18° 32'
above the horizon, and the long arm is inclined at an angle
of 48° 22' below the horizon.
1.15. The Wheel and Axle.— This
machine consists of a wheel, a, rigidly
connected with a horizontal cylinder,
b, movable round two tmnnions (Art
99), one of which is shown at c. The
power, P, is applied at the circumfer-
ence of the wheel, sometimes by a cord
coiled round the wheel, sometimes by
handspikes as in the capstan, or by
handles as in the windlass ; the weight, W, hangs at the
end of a cord Listened to the axle and coiled round it
116. Conditions of EqcUibrinm of tiie Wheel and
Axle. — (1) Let a and b be the radii of the wheel and axle
respectively ; P and W the power and weight, supposed to
act by strings at the circumference of the wheel and axle
perpendicular to the radii a and b. Then either by the
principle of virtual velooiti < or by the principle of momenta
we have
Pa = Wby
or
P _ radius of axle
W ~ radius of wheel'
(1)
It is evident that, by increasing the radius of the wheel
or by diminishing the radius of the axle, any amount of
mechanical advantage may be gained. It will also be seen
5 ft. and inclined
the short arm a
ium a weight of
D of each arm to
angle of 18° 22'
lined at an angle
V, hang8 at the
round it
he Wheal ftud
wheel and axlo
;ht, sapposed to
wheel and axle
m either by the
pie of moments
(1)
ioB of the wheel
any amount of
rill also be seen
DTFTERSIfTIAL WBJBSL AlfD AXLB.
191
that thi" machine is only a modification of the lever ; the
peculiar advantage of the wheel and axle being that an end<
lc'88 scries of levers are brought into play. In this respect,
then, it surpasses the common lever in mechanical advan-
tage.
In the above we have supposed friction to be neglected,
or, what amoui.f« to the same thing, have assumed that the
trunnion is indefinitely small In practice, of course, the
trunnion has a certain radius, r, and a certain coefficient of
friction. Calling R the resultjint of P and If, and taking
into account the friction on the trunnion we have for the
relation between P and W •
Pa = ITJ + r sin ^ V/^lPrMP 1Fcm"w, (2)
u being the angle between the directions of P and W
exactly as in Art 110.
(2) Differential Wheel and
Axle. — By diminishing b, the radius
of the axle, the strength of the
machine is diminished ; to avoid this
disadvantage a differential wheel and
axle is sometimes employed. In this
instrument the axle consists of two
cylinders of radii h and b' ; the rope
is wound round the former in one
direction, and after passing under a
movable pulley to which the weight
is attached, is wound round the latter in the opposite direc-
tion, so that as the power, P, which is applied ae before,
tangentially to the wheel of radius, a, moves in its own
direction, the rope at h winds up while the rope at b'.
unwinds.
For the equilibrium of the forces (whether at rest or in
uniform motion), the tensions of the rope in hn and b'n
Pia.eo
■:>vaiisimmtmg^
192
TOOTHED WBSBLa.
are each equal \Ai\W. Hence, taking moments ronod the
centre of the traniiion, e, we have
Pa ■{- \Wh' - \m ^ Oi
(3)
hence by making the difference, b — b', small, the power
can be made as small as we please to lift a given weight.
Let the wheel turn through the angle dd; the point of
application of P will describe a space := add, and the
weight will bo liftad through a space = ^ (6 — b') de,
which latter will be very small if b — b' is very small.
Therefore, since the amount of taork to be done to raise the
weight to any given height, is constant, economy of power
is accomplished by a loss in the time of performing the
work.
117. Toothed Whoels. — Toothed or cogged toheeh arc
wheels provided on the circumferences with projections
called teeth or cogs which interlock, as shown in the figure,
and which are therefore capable of transmitting force, so
that if one of the wheels be turned round by any means,
the other will be turned round also.
When the teeth are on the aides of the wheel instead of
the circumference, they are oalldd croton wheels. Wlicn
the axes of two wheels are
neither perpendicular nor
parallel to each other, the
wheels take the form of
frustums of cones, and are
called beveled wheels. When
tliero is a pair of toothed
wheels on each axle with the
teeth of the large one on one
axle fitting between the teeth . ifq Fi».M
InaiA
^^^^MmMMm
j.i;j«a:n«tPn*r.rf-i'
.:^^^R~^^l^-^
nomeDts roaod the
0;
(3)
, small, the power
ifb a given weiglit.
<W; the point of
3 ::= add, and the
9 == ^ (A _ b) 60,
- V is very small,
e done to raise the
Bconomy of power
>f performing the
cogged wheels arc
with projectiona
awn in the figure,
smitting force, so
id by any means,
wheel instead of
wheels. When
f\%.U
TOOTHED WHEELS,
193
of the small one on the next axle, the larger wheel of each
pair is called the wheel, and e smaller is called the pinion.
By means of a combination of toothed wheels of this kind
called a train of wheels, motion may be transferred from
one point to another and work done, each wheel driving
the next one in the aeries. The discussion of this kind oic
machinery possesses great geometric elegance ; but it would
be out of place in this work. We shall give only a slight
sketch of the skimpiest case, that in which the axes of the
wheels are all parallel. For the investigation of the proper
formfe of teeth in order that the wheels when made shall
run truly one upon another the student is referred to other
works,*
118. To Find the Relation of the Power find
Weight in Toothed Wheels.— Let j* and B be the fixed
centres of the toothed wheels on the circumferences of
which the teeth are arranged ; QCQ a normal to the sur-
faces of two teeth at their point of contact, C. Suppose an
axle is fixed on the wheel, B, and the weight, W, suspended
from it at E by a cord ; also, suppose the power, P, acts at
D with an arm DA; di-aw Aa and BA perpendicular to
QCQ. Let Q be the mutual pressure of one tooth upon
another at C ; this pressure will be in the direction of the
normal QCQ. Now since the wheel, A, is in equilibrium
about the fixed axis. A, under the action of the forces,
P and Q, we have
P.AD = ^.Aa; (1)
and since the wheel, B, is in equilibrium abont the fixed
axis, B, under the action of the forces, Q and W, we have
Fr.BE=: ^B*.
(2).
• 8m Ouodere'i OimmU tf MtOmUm ; Rankice's AppHtd MetAanie* ; Mom.
ley's Oiglnetrbtg; WUlls'* PHmetplm tf UtehamUm; CoUlgnon's SloMfiM,- and
• Paf$r qf Mr. Mtf* <« M< Om*. mi. Trmt., Vol. n, p. fn.
9
"F
194
TRAIN OP IT WaSBLa.
Dividing (1) by (2) we have
P
W
AD
BE
Aa
B4'
or
mome nt of P _ Aa
moment of Jf ~ B4*
If the direction of the normal, QCQ. at the point of con«
tact, C, changes ae the action passes from one tooth to the
succeeding, the relation of P to TT becomes variable. But,
if the teeth are of such form that the normal at their point
of contact shall always be tangent to both wheels, the lines
Aa and Bd will become radii, and their ratio constant.
And since the number of teeth in the two wheels is propor-
tional to their radii, we have
mome nt of P _ number of tee th on the wheel P
moment of fT "" number of teeth on the wheel W'
(3)
119. Relation of Power to Weight in a Train of n
Wlieela — Let R^, R,, R^, etc., be the radii of the suc-
cessive wheels in such a train ; Ti, r„ r„ etc., the radii of
the corresponding pinions; and let P, P,, P,, P,, . . . W,
be the powers applied to the circumferences of the successive
wheels and pinions. Then the first wheel is in equilibrium
about its axis under the action of the forces Pand P.^,
since the power applied to the circumference of vhe second
wheel is equal to the reaction on the first pinion, therefore
Similarly
p
X R^
=
Pi
X r,.
Pt
A R,
=
P,
X r, ;
p.
X /?,
=
P,
X r, ;
etc
=
etc.;
Pn-l
X Rn
=3
w
X r»
•■ . ■ jpaiMiy^jiiJMLwiMiik^mi!^^
mxisenmmsswf&'i^sirx:
m.
EXAMPLES.
195
the point of con-
Q one tooth to the
IBS variable. But,
mal at their point
) wheels, the lines
'ir ratio constant.
I wheeUi ia propor-
khe wheel P . .
he wheel W' ^ '
in a Train of n
radii of the sac-
!., the radii of
Pi. Ps, . . . W,
of the successive
in equilibrium
jrcea Pand Pj,
ce of the second
inion, therefore
is
Multiplying these equations together and omitting common
factors, wo have
P
W
_ '1
X r, X r. X
• ■ a •
Jii X R, X Jti X
(1)
It will be observed, in toothed gearing, that the smaller
the radius of the pinion as compared with the wheel, the
greater will be the mechanical advantage. There is, how-
ever, a practical limit to the size that can be given to the
pinion, because the teeth must be large enough for strength,
and must not be too few in number. Six is generally the
least number admissible for the teeth of a pinion. Equa-
tion (1) shows that by a train consisting of a very few pairs
of wheels and pinions there is an enormous mechanical
advantage. Thus, if there are three pairs, and the ratio of
each wheel to the pinion is 10 to 1, then P is only one
thousandth part of W; but on the other hand, W will only
make one turn where P makes one thousand. Such trains
of wheels are very useful in machinery such as hand cranes,
where it is not essential to obtain a quick motion, and
where the power available is very small in comparison to
the weight (See Browne's Mechanics, p. 109.)
EXAMPLES.
1. What is the diameter of a wheel if a power of 3 Ibe.
is just able to move a weight of 12 lbs. that hangs from the
axle, the radius of the axle being 2 ins.? Arts. 16 ins.
2. If a weight of 20 lbs. be supported on a wheel and
axle by a force of 4 lbs., and the radius of the axle is
f in., find the radius of the wheeL Ana. 3^ ins.
3. A capstan is worked by a man pushing at the end of
a pole. He exerts a force of 50 lbs., and walks 10 ft.
round for every 2 ft. of rope pulled .in. What is the
icsistauce overcome ? Ans. 250 lbs.
196
tNCUi^SD PLAlfE.
4. An axle whose diameter is 10 ine., iias on it two
wheels the diameters of wliich are 2 ft and 2| ft. respew-
tively. Find the weight that would be suppoi ted on the
axle by weights of 25 lbs. and 24 lbs. on the smaller and
lurgor wheels respectively. Ana. 264 lbs.
120. The Inclined Plane.— This has already been
partly considered (Art 96, etc.). Let the power, P, whose
direction makes an angle, B, with a rough inclined plane,
be employed to drag a weight, W, up the plane. Then if
is the angle of friction and i the inclination of the plane,
we have from (3) of Art. 96,
P _ w wn (t 4- ») •
"^ cos (« - e)'
If P acts along the plane, = 0, and (1) becomes
p - |yB' P(»' + ^) ,
cos ^
If P acts horizontally, 9 = — t, and (1) becomes
P= ff' tan («• + 0).
(1)
(2)
(8)
Cor.— If we suppose the friction = 0, (1), (3), and (3)
become respectively
rsin t
P= W
COS0*
<*)
P= TTsini, (5)
P = fF tan i. (6)
SoH.-— It follows from (4), (5), and (6) that the smaller
Mituitttsus»<>iititmmmm
ms., lias un it two
ft aud 2| ft. respac-
*o Buppoited on the
on the smaller uud
Ann. 264 lbs.
} has already boon
the power, P, whose
agh inclined plane,
he plane. Then if
oation of the plane,
(1)
(1) becomes
(2)
(1) becomes
(3)
(1), (2), and (3)
(4)
(6)
(6)
that the smaller
— S!
w^mmmmm.
TBE PULLEY.
mr-
197
the inoKtiation* of the plane to the horison, *he greater will
be the mechanical advanta^. If we take in friction there
exception to this rule wnen i > ^ — 0. The
18 an
2
gradients on railways are the most common examples of
the use of the inclined plane ; these are always made au low
iis is convenient in order to enable the engine to lift the
lieavieat possible train.
121. The Pnlley. — The pulley consists of a grooved
wheel, capable of revolving freely abo><t an axis, fixed into
a framework, called the block. A cord passes over a por-
tion of the circumference of the wheel in the groove.
Wlien the axis of tlie pulley is fixed, the pulley is called a
Jixed pulley, and its only eflTect is to change the direction
of the force exerted by the cord ; but where the pulley can
ascend and descend it is called a movable pulley, and a
mechanical advantage may bo gained. Combinations of
pulleys may be made in endless vaxnety; we shail consider
only the simple movable pulley and three of the more
ordinary combinations. No account will be here taken of
the weight of the pulleys or of the cord, or of friction and
stiflfness of cords. The weight of a set of pulleys is gener-
ally small in comparison with the loads which they lift ;
aud the friction is small. The use of the pulley is to
diminish the effects of friction which it does by transferring
the friction between the cord arid circumference of the
wheel to the axis and its supports, which may be highly
polished or lubricated. The mechanical principle involved
in all calculations with respect to the pulley is the constancy
of the force of tension in all parts of the same string
(Art. 40).
• To And the incIitiAtion of the plane for a maxlniiim value of P when It acts
parallel to the plane we pnt Ihe derivative of P with respect to J = 0, and get
!''* = IF '^°' ^' *-♦]! = 0, .-. I = " - *. Hence whilfl the incllnatton of the plane
dl coa^ %
18 dimlnisbiug from ' to ^ - ^, mociumical advantoie U diministaiug.
198
riRST araTBM of pullets.
12?. The Simple Movable Pulley.— Let be the
centre of the pulley which is supported by a cord passing
un ^er it with cue end attached to a beam at A and the
other end stretched by the force /.
Now since the tension of the string,
ABDP, is the same thoughont, and the
weight, W, io supported by the two
strings at B and D, in each of which
the tension is P, re have
2P= W',
P
W
1
2*
The same result follows by the prin-
ciple of virtual Yelot^ities. Suppose the
pulley and tbe weigbt, W, to rise any
distance. Then it is, clear that both halves of the string
must be shortened by the same distance, and hence P
must rise double the distance ; and therefore the equation
of virtual work gives
P 1
The mechanical advantage with a single movable poUoy
is 2.
123. First Bjwxma of PnUeye, in which
the same curd passes round all the Fal-
leya. — In this systcitn there arc two blocks, A
and R, the upper of which is fixed and the
lower movable, and each containing a num^^r
of pulleys, each pulle> being movable round
the axis of the block in v/hioh it is. A single
cord is attached to the lower block and passes
alternately round thi pulleys in the np|)er and
lower hlooks, the portions of the cord l)otwoon
successive pulleys being iHtrallul. The (Kirtiou
ijfrjs.
»y.— Let be the
L by a cord passing
beam at A and the
^
1
©
Fig.e2
Hives of the string
nee, and hence P
efore the equation
lo raorable pulley
IIS8T arsTMM or pullbts.
199
of cord proceeding from one pulley to the next is called a
ply; the portion at ivhioh the power, P, is applied is
called the taokli-faU.
Since the cord passes round all the pulleys its tension is
the rame throughout and equal to P. Then if n be the
number of plies at the lower b^ock, nP will be the resultant
upward tension of the cords at the lower block, which
must equal W ;
.-. nP = W,
or
P
W
1
n
This result follows also by the principle of virtaal yeloci-
ties. Let^ denote the length of the tockle-fall and x the
common length of the plies ; then since the length of the
cord is constant, we have
/» + »*»:
— cor
istant
•
• •• dp +
ndx :
= 0.
But the equation
of virtual work
is
Pdp +
Wdx
= 0;
•
• •
--1
or
P
w~
1
n
This system it* most commonly used on aooonnt of itR
superior portability and is the only one of practical impor-
tance. The several pulleys are usually mounted on a com-
mon axis, as in the figure, the cord being inclined slightly
asido to rmss from one pair of pulleys to the next.
This forms what is called a set of BloJcs and Falh. It
is very commonly used on shipboard and wherever weight*
have to be Hfted at irregular times and places. The weight
of the lower set of pulleys in this caae merely forms part of
the groM weight W.
300
SBCOND SrSTSX OF PULLMTS.
Tho friction on the spindle of any particnlar pulley is
proportional to the total pressure ^n the pulley, which is
clearly %P. Hence, if fi is the coefficient of faction, the
resistance of iViclion on any pulley = 2P/i; and the
amount of its displacement, when W is raised, will be to
the displacement of W in the ratio of the radius of the
epindle to that of the pnlley.
124 Second Bywltmm of Fnlloys,
in which each Pnlley lumffi from a
fixed block by a sepanite SKring.— <
Let A be the fixed pulley, n the number
of movable pnileys ; each cord has one
end attanb<Ml to a fixed point in the beam,
and all except the lust have the other end
attached to a movable pulley, the por-
Fi9.M
first
tions not in contact with any pulley being all ]>a-^lle1.
Then the tension of the cord passing under the
W
(lowest) pulley = -^ (Art. 128) ; the tension of the cord
W
passing under tho second pulley = -^, and so on ; and the
W
tension of the cord passing uiider tho nth pulley = ^,
which must equal the power, P\
P _ 1
0)
i
The same result follows by the principle of work. Sup-
pose tho first pnlley and the weight W to rise any distan*^
« ; then it is clear ^hat both portions of the oord passing
round this pulley will be shortened by the same distance,
and hence tlio second ])nIloy must rise double this dlst^moe
or 9-", and the third pulley must rise double the disfjtnce of
the second or 2*x, and so on ; and the nth pulley must rise
%*-''^ and P must dvsvcud "if^x ; therefore the work of P
Mk
ticnlar pulley iu
iallej, which is
of friction, the
IPft ; and the
used, will be to
) radins jf the
/\^
under tho first
on of the cord
so on ; and the
pulley = ^,
(l)
)f work. 3up-
le any distame,
e oord paasing
dame diatance,
this distimoe
the distance of
ulloy muHt rise
the work of P
THIRD arSTSM Of PULLKY8.
201
is P2»a;, and the work to be done on fF is W-x. Hence
the equation of work gives
P.2-«=FF.V ... |,= |,.
125. Third SyBtem of Pnlleya, in which each cord
ia attached to the weight— In this aystem oue end of
each cord is attached to the bar from which the weight
hangs, and the other supports a pulley, the cords being all
])arallel, and the number of movable pulleys one less than
the numboi of cords.
Let n bo the number of cords; then the
tension of the cord to which P is attached is
r ; the tension ot the second cord is 2P (Art
122) > t^^at of the next 2>P, and so on; and
the tension of the »th cord is 2»-»P. Then
the sum of all the tensions of the cords
attached to the weigl t must equal W.
Hence
/» + 2P + 2»P +
2»-»P =
2" — 1
In thifl system the weights of the movable pulleys assist P ;
in the two former systems they act against it.
(CXAMPL.ES.
1. What force ia necessary to raise a weight of 480 lbs.
by an arrangement of six pulleys in which the same string
passes round each pulley ? Ana, 80 lbs.
*
8. Find the power which will support a weight of
800 lbs. with three movable pulleyai, arranged as in the
second system. An*. lOOlbs.
inOS
TWf WMDOU,
3. If thero b« eqoiUbriam between P and W with three
pulleys in the third syBtem, what ftdditional weight can be
raised if 2 lbs. be added to P? Ana. 14 lbs.
126. The Wedg*. — The wedge w a triangnlar prism,
usually isosceles, and is used for separating bodies or parts
of the same body by introdacing its edge between them and
then thrusting the wedge forward. This is effected by the
blow of a hammer or other snoh moaii^ which produces a
▼iolent pressure, for a short time, in a direction perpen-
dicular to the back oi the wedge, and the resistance to be
overcome consists of friction and a reaction due to the
moleoalar attractions of the particles of the body which
are being separa>>ed. This reaction will be in a direction
perpendicular to the inclined surface of the wedge.
137. The lfeet4aio«l Ad-
vantage of the Wedge.-^Let
ACB represent a section of the
wedge perpendicular to its in-
clined fitces, the wedge having
been driven into the material a
distauce equal to DC by a force,
P, acting in the direction DO.
Draw DE, DF, perpendicular to
AC, BC, and let B denote the
reactions along ED and FD ; then liR will he the frictitm
acting at E and F in the directions EA and FB. Let the
angle of the wedge or ACB = %a.
Resolve the foroos which act on the wedge in directions
perpendioalar and parallel to the back of the wedge, then
we have for peqiendioular forces
n9.N
P = 2R sin a + 2ftR co» a.
(1)
7%M equaiioH may ako i» oHmntd from Uie prinoiple of
work m follows : If the wedge has been driv«B inW the
W with tbree
weight ean be
Ana. 14 lbs.
ingubir prism,
odiee or parts
reen them and
iflected by the
ch produces a
action perpen-
sistauoe to be
n due to the
ic body which
in a direction
redge.
I9.M
the frictioB
i'B. Let the
in directions
wedge, than
(1)
fmnoiple tf
MBOHANICAL ADVANTAatt OP WJBDGX.
203
material a distance eqnal to DO by a force, P, acting in
the direction DO,, then the work done by i* is P x DO
(Art 101, Rem.); and since the points £! and F were
originally together, the work done against the resistance
i? is 22 X DE -f /? X DP = a^ X DE ; and the work
done against friction is inR x EC. Hence the equation
cf work is
P X DO = 2i2 X DE + 2^i2 x EC, (?)
which reduces to (1) by substituting sin a and cos a for
DE - EC
]T0"*'^D0*
Cob. — If friction be neglected, (2) becomes
P
2DE
DC
AB
AC
that is -i: =
P
R
back o f the we dgg
length of one of the equal sides'
It follows that the narrower the back of the wedge, the
greater will be the mechanical advantage. Knives, chisels,
and many other implements are examples of the wedge.
In the action of the wedge a great part of the power is
employed in cleaving the material into which it is driven.
The force required to effec^: this is so great that instead of
applying a continuous pushing force perpendicular to the
back of the wedge, it is driven by a series of blows. Be-
tween the blows there is a powerful reaction, B, acting to
puBh the weJge back again out of the cleft, and this is
resisted uj the triction which now acts in the directions
EC and FO. tience when the wedge is on the point of
starting back, between the blows, t.a equation of equi-
librium will be from (1)
2S sin a — 2ftR cos a =r ;
. • . as tAn~' I*.
S04
THM 8CBBW.
fi N
And tho wedge will fly back or not according as a > or
< tan~* /«. (See Browne's Mechanics, p. 117. Also Magnus's
Mechanics, p. 157.)
128. The Screw.— The screw consists of a right cir-
cular cylinder, on the convex surface of which there is
traced a uniform projecting thread, abed .... inclined at
L nstant angle to straight lines parallel to the axis of the
cylinder. The path of the thread
may be traced by the edge AG of
an inclined plane, ABC, wrapped
ronnd the cylinder; the base of
the plane corresponding with the
circumference of the cylinder, and
the height of the plane with the
distance between the threads which
is called the pitch of the screw.
The threads may be rectangular or
triangular in section. The cylinder
fits into a blqck, ou the inner sur-
face of which is cut a groove which is the exact counterpart
of the thread. The block in which the groove is cut is often
called the nut. The power is generally applied at the end of
a lever fixed to tho centre of tlie cylinder, or fixed to the nut.
It is evident that a screw never requires any pressure in the
direction of its axis, bnt must ^ made to revolve only ;
and this can be done by a force acting at right angles to
the extremities of its diameter, or its diameter produced.
129. The Relation between the Fewer end Jie
Weight in the Screw. — SuppiiM) the TH)wcr, P, to act in
a plane por{)ondicular to the axis of the cylinder and at the
end of an arm, DE = a, and suppose the screw to have
made one revolution, the power, P, will have moved
through the circumference of which a, is the radius, and
tlie work done by P will be Py.%ita. During the same
Fia.«7
I
^te
MM
ng as a > or
k.180 Magnus's
r a right cir>
bich there is
. iuclined at
le axis of the
J
•7
ooanterpart
cat is often
»t the eod of
i to the nut.
ssuro in tUa
)volve only ;
ht angles to
troduced.
and Jh.9
P, to act in
:* and at the
ew to have
ave moved
radius, and
the same
I
TBS SCREW.
205
time the screw will have moved in the direction of its axis
t'.roug 1 the distance, AB = Sfrr tan a, r being the radius
of the cylinder, nnd n the angle which the thread of the
screw makes with its base. Then as this is the direction in
which the resistance is encountered, the work done against
the resistance, W, is Winr tan a. Hence if no work is lost
the equation of work will be
P X 2ffa = ff X 2rrr tan «.
(1)
That is the power is to the weight as the pitch of the screw
is to the circumference described by the power.
If there is friction between the thread and the groove, let
B bo the normal pressure at any point, p, of the thread,
and nR the friction at this poinf^, iihen the work done
against the friction in one revolution is ftI,R 2nr sec a, I.R
denoting the sum of the normfd reactions at all points of
the thread. Hence the equation of work is
P %na = W2irr tan « -f ^ 2nr sec «li?.
(2)
But, for the oqnilibrinm of the screw, resolving parallel
to the axis, we have
therefore
W = I. (R cos a— nJt sin a),
W
ZR =
cc3a — (i sm a
which in (2) gives
Pa ^ WV tan « H : — ;
cos a — /u sm n'
or Pa = Wr tan (a -f 0),
being the angle of friction.
(3)
306
psoyr'a D/jvmsgfmAL sossw.
129a. Fxonj'u DifEurential 8or«w.— If h denote the
pitch of a screw (1) becomes
aPira = Wh,
which expresses the relation between P and W, when fno-
tion is neglected. Therefore the mechanical advantage is
gained by making the pitch very small. In some cases,
however, it is desirable that the screw should work at fair
speed, as in ordinary bolts and nuts, and then the pitch
must not be too small. In oases where the screw is used
specially to obtain pressure, as in screw-presses for cotton,
etc., we do not care for speed, but only for pressure. But
in practice it is impossible to get the pitch very small from
the fact that *' the angle of inclination is very flac, the
threads run so near each other as to be too weak, in which
case the screw is apt to " strip its thread," that is, to tear
bodily out of the hole, leaving the thread behind.
Where very great pressure is required a difierential nut-
bole is resorted to. Let the screw work in two blocks,
A and B, the first of
which is fixed and the
second movable along a
fixed groove, n ; and let
A be the pitch of the
thread which works in
K SOS SSS|
ns-M
the block, A, and h' the pitch of the thread which works
in the block B. Then one revolution of the screw impresses
two opposite motions on the block, B, one equal to h in the
dii-ection in which the screw advances, and the other equal
to /(' in the opposite direction. If then the block, B, is
connected with the resistance W, we have by the principle
of work
8P»T<i=: »F (*-*');
and the requisite power will be diminished by diminishing
h denote
W, when fric-
advantage is
1 some cases,
work at fair
iien the pitch
I screw is used
es for cotton,
iressnre. But
ty small from
yery flac, the
eak, in which
hat is, to tear
ad.
erential nat-
two blocks,
which works
ew impresses
to h in the
other equal
block, B, is
he principle
diminishiDg
aXAitPLKa.
80t
h — h\ By means of this torew a oompantiVBly small
pressnro may be made to yield a preasnm enormously
greater in magnitude.
EXAMPLES.
1. A lever 10 ini. long, the weight of which is 4 lbs., and
acts at its middle point, balances about a certain point
when a weight of 6 lbs. is bung from one end; find the
point Ant. 2 ins. from the end where the weight is.
2. A lever weighing 8 lbs. balances at a point 3 ins. from
one end and 9 ins. from the other. Will it continue to bal-
ance about that point if equal weights be suspended from
the extremities ?
3. A beam whose length is 12 ft balances at a point 2 ft
from one end ; but if a weight of 100 lbs. be hung from the
other end it balances at a point % ft ftam th&t end ; find the
weight of the beam. Am. 2S lbs.
4. A lever 7 feet long is 6up|K>rted in « horisontal posi-
tion by props placed at its extremities : find where a weight
of 28 lbs. must be placed so that the pressure on one of the
props may be 8 lbs. Ana. Two feet from the e.nd.
6. Two weights of 12 lbs. and 8 Ib& respectively at the
ends of a horisontal lever 10 feet long balance : find how
far the fulcrum ought to be moved for the weights to bal-
ance when each is increased by 2 lbs. Ann. Two inches.
6. A lever is in equilibrium under the action of the forces
P and Q, and is also in equilibrium when P is trebled and
Q is increased by 6 lbs.: find the magnitude of Q.
Ans. 3 lbs.
7( In a lever of the first kind, let the power be 217 lbs.,
the weight 726 lbs., and the angle between them 196°.
Find the pressure on the falcmm. Am. eS2.7 lbs.
ao8
BXAMPhEA
8. If the power and weight in a straight lever of the
first kind be 17 lbs. and 32 Ib&, and make wi^'a each other
an angle of 79° ; find the pressure on the fulcrum.
An8. 39 lbs.
9. The length of the beam of a false balance is 3 ft.
9 ins. A body placed in one scale balances a weight of
9 lbs. in the other ; but when placed in the other scale it
balanceb 4 lbs.; required the true weight, W, of the body
and the lengt,hs, a and b, of the arms.
Ans. F = 6 lbs.; « = 1 ft. 6 ins.; i = 2 ft 3 ins.
10. If a balance be false, having its arms in the ratio of
15 to 16, find how much per lb. a customer really pays
for tea which is sold to him from the longer arm at 3s. 9d.
per lb. ^n«. 4s. per lb.
11. A straight uniform lever whose weight is 50 lbs. and
length 6 feet, rests in equilibrium on a fulcrum when a
weight of 10 lbs. is suspended from one extremity : find the
position of the fulcrum and the pressure on it.
Ana. 2^ ft from the end at whioh 10 lbs. is suspended ;
60 lbs.
12. On one arm of a false balance a body weighs 11 lbs.;
on the other 17 lbs. 3 02.; what is the true weight ?
Ans. 13 lbs. 12 oz.
13. A bent lever is oompoeer of two straight uniform
rods of the same length, inclined to each other at 120°, and
the fulcrum is at the point of intersection : if the weight of
one rod be double that of the other, show that the lever will
remain at rest with the lighter arm horizontal.
14. A uniform lever, / feet long, has a weight of W lbs.,
suspended from its extremity ; find the position of the ful-
crum when the long end of the lever balances the short
•
t lever of the
ta each other
rum.
Ins. 39lbe.
lance is 3 ft.
s a weight of
other scale it
, of the body
2 ft 3 ins.
in the ratio of
;r really pays
arm at 3s. 9d.
4s. per lb.
is 50 lbs. and
mm when a
lity : find the
i suspended;
ighs 11 lbs.;
jht?
lbs. 12 oz.
fht uniform
at 120°, and
le weight of
.he lever will
of W lbs.,
of the ful-
iis the short
9
SXAMPLKS.
209
end with the weight attached to it, supposing each unit of
length of the lever to be w lbs.
■^nt. ^T-n-- , J - \ is the short arm.
15. A lever, I ft long, is balanced ,. her it 's placed upon
a prop i of its length from the thick end ; when a weight
of W lbs. is suspendofl from the small end the prop must
be shifted j ft. towards it in order to maintain eqailibiium ;
required the weig.t of the lever. Ana. ^W.
16. A lever, I ft. long; is balanced on a prop by a weight
of W lbs.; first, when the weight is suspended from the
thick end the prop is a ft. from it; secondly, when the
weight is suspended from the small end the prop is b ft.
from it ; required the weight of the lever.
. W{a + b) ,.
l — {a -1,-0)
17. The forces, P and W, act at the arms, a and b,
respectively, of a straight lever. When P and W make
angles of 30° and 90° with the lever, show that when equi-
fibW
librium takes place P = •
a
18. Supposing the beam of a false balance to be nniform,
a and b the lengths of the arms, P and Q the apparent
weights, and IV the true weight ; when the weight of the
beam is taken into account show that
a
b
P -W
19.
W-Q
If a be the length of the short arm in Ex. 14, what
must be the length of the whole lever when equilibrium
takes place P /2aW
Ana. a + a/ —^^ + aK
w
20. A man whose weight is 140 lbs. is just able to snp-
port a weight that hangs over an axle of 6 ins. radius, by
SIO
MXAMTLMS.
hanging to the rope that passea over the corresponding
wheel, the diameter of which is 4 ft; find the weight sup-
ported. Ans. 560 lbs.
21. If the difference between the diameter of a wheel and
the diameter of the azte be aiz times the radius of the axie,
find the greatest weight that can be sustained by a force of
60 lbs. Ana. 240 lbs.
22. If the radius of the wheel is three times that of the
axle, and the string round the wheel can support a weight
of 40 lbs. onlj, find the greatest weight that can be lifted.
Am. 120 lbs.
23. What force will be required to work the handle of a
windlass, the resistaDce to be overcome being 1156 lbs., the
radius of the axle being six ins., and of the handle 2 ft.
Sins.? Ana. 216.75 lbs.
24. Sixteen sailors, exerting each a force of 29 lbs., push
a capstan with a length of lever equal to 8 ft, the radius of
the capstan being 1 ft. 2 ins. Find the resistance which
this force is capable of sustaining.
Ana. 1 ton 8 cwt. 1 qr. 17 lbs.
25. Supposing them to have wound the rope round the
capstan, so that it doubles back on itself, the radius of the
axle is thus increased by the thickness of the rope. If this
be 2 ins. how much will the power of the instrument be
diminished. ' Ana. By \, or 12f per cent.
26. The radios of the axle of a capstan is 2 feet, and six
men push each with a force of one cwt. on spokes 5 feet
long ; €nd the tension they wiU be able to prodnce in the
rope which leaves the axle. Ans, m cwt.
27. The difference of the diameters of a wheel and axTe
is 2 feet 6 inches ; and the weight is equal to six times the
power ; find the radii of the wheel and the axle.
Ant. 1%\xul; 3 ina.
I
sorresponding
s weight Bup-
18. 560 lbs.
tf a wheel and
IS of the axle,
by a force of
na. 240 lbs.
E>8 that of the
wrt a weight
»n be lifted.
w. 120 lbs.
e handle of a
1156 lbs., the
handle 2 ft.
216.75 lbs.
29 Iba, push
the radins of
stance which
qr. 17 lbs.
)e round the
radius of the
•ope. M this
istmment be
per cent.
eet, and six
spokes 5 feet
oduce in the
IS. 15 cwt.
eel and axle
six times the
e.
ns.; dina.
I '
itAJtPLSa.
an
28. If the rsdins of a wheel is 4 ft., and of the a tie
8 ins., And the power that will balance a weight of
500 lbs., the thickness of the rope coiled round the axle
being one inch, the powe; acting without a rope.
Ana. 88.54 lbs.
29. Two given weights, P and Q, hang vertically from
two points in the rim of a wheel turning on an axis;
find the position of the weights when equilibrium takes
plaoe, supposing the angle between the radii drawn to
the points of suspenrijn to be 90°, and that 6 is the
angle which the radius, drawn to /**& point of sus-
pension, makes with the Tertical.
Ana. tan d =
Q
30. What weight can be supported on a plane by a hori-
zontal force of 10 lbs., if the ratio of the height to the base
isfP Ana. IS^lbs.
31. The inclination of a plane is 30°, and a weight of
10 lbs. is supported on it by a string, bearing a weight at
its extremity, which passes over a smooth pulley at its
summit ; find the tension in the string. Ana. 5 lbs.
32. The angle of a plane is 45° ;. what weight can be
supported on it by a horizontal force ot 3 lbs., and a force
of 4 lbs. parallel to the '^Ivxe, both acting together.
Ana. 3 -H 4 V2 lbs.
33. A body is supported on a plane by a force parallel
to it and equal to | of the weight of the body ; find the
ratio of the height to the base of the plane.
Ana. 1 : 2 -v/e.
34. One of the longest inclined planes in the world is
the road from Lima to Oallao, in S. America ; it is 8 miles
long, and the faU is 511 ft. Calculate the inclination.
Am. 88' 27", OT 1 yard in 62.
212
SXAMPLBB.
3?
35. If the force required to draw a wagon on a horizontal
road be ^th part of the weight of the wagon, what will be
the force required tc draw it up a hill, the elope of which
is 1 in 43. Ans. -rrVitl' pa^t of the weight.
36. If the force required to draw a train of cars on a
IcTcl railroad be yf^th jmrt of the load, find the force
required to draw it up a grade uf 1 in 5(3.
Ans. xrS-jth part of the load.
37. What force is required (neglecting friction) to roll a
ciisk weigliing 964 lbs. into a curt 3 St. high, by means of a
plank 14 ft long resting against the cart.
Arts. The force must exceed 206 lbs.
38. A body is at rest on a smooth inclined plane when
the power, weight and normal pressure are 18, 26, and
12 lbs. respectively ; find the inclination, a, of the plane to
the horizon, and the angle, 6, which the direction of the
power makes with the plane.
Ana. a = 37° 21' 26"; 6 = 28° 46' 54".
39. If the power which will support a weight when act-
ing along the plane be half that which will do so acting
horizontally, find the inclination of the plane. Ana. 60°.
40. A power P acting along a plane can support W; and
acting horizontally can support x ; show that
P»= »'»-««.
41. A weight W would be supjwrted by a power P act-
ing horizontally, or by a power Q acting parallel t<j the
plane \ show that
42. The base of an iooliucd plane is 8 ft., the height
6 ft , and TF = 10 tons ; required P and the uormal
pressure, N, on the plane.
Ana. P = Q tons ; JV = 8 tona.
>n a horizontal
, what will be
lope of which
the weight.
I of cars on a
Qnd the force
of the load.
;ion) to roll a
by means of a
eed 206 lbs.
id plane when
) 18, 26, and
i the plane to
•ection of the
58° 46' 54".
ht when act-
do so acting
Ana. 60°.
)port W; and
(owor P nct-
rallel t<j the
the height
the uormal
= 8 tons.
SXAMPLEA
Sid
43. A weight is supported on an inclined plane by a
force whoso direction is inclined to the plane at an angle
of 30" ; wlieu the inclination of the plane to the horizon ia
30°, show that IT = P V'S-
44. A man weighing 150 Iba raises a weight of 4 cwt. by
a system of four movable pulleys arranged according to the
second system ; what k his pressure on the ground ?
Ans. 122 lbs.
45. What power will be required in the second system
with four movable pulleys to sustain a weight of 17 tons
12 cwt. Ans. 1 ton 2 cwt.
46. Two weights hang over a pulley fixed to the summit
of a smooth inclined plane, on which one weight is sup-
ported, and for every 3 ins. that one descends the other
rises 2 ins.; find the ratio of the weights, and the length
of the plane, the height being 18 ins. Ana, 2 : 3 ; 27 ins.
47. U W = 336 Iba. and P = 42 lbs. in a combination
of pulleys arranged according to the first system, how many
.novable pulleys are there ? Ana. 4.
48. In a system of pulleys of the third kind in which
there are 4 cords attached to the weight, determine the
weight, W, supported, and the strain on the fixed pulley,
the power being 100 lbs., and th>. weight, W, of each
pulley 6 lbs.
Ans. W — 15P •^- llw = 1666 lbs.; Strain = 16P + 16u>
= 1675 lbs.
49. In a system of pulleys of the third kind, there arc
2 movable pulleys, each weighing 2| lbs. What power is
required to support a weight of 6 cwt. ? Aiu. 94.67 Ibc.
60. Find the power th»t mill support a weight of 100 lbs.
by means of a system of 4 pulleys, the strings being all
attached to the weight, and each pulley weighing 1 lb.
Ana. 5\^ lbs.
814
sxjZii'Lsa.
i
51. The cinmmfeMnoe of the circle oorrestwiulhig to tibe
point of applicatioii of P is 6 feet ; find how many tunui
the screw must make on a cylinder 2 feet long, in order
that IT may bo equal to 144P.. An». 48.
62. The distance between two consecutive threads of a
screw is a quarter of an inch, and the length of the power
arm is 5 feet; find what weight will be sustained by a
power of 1 lb. Ann. 480Tr Iba.
53. How muiy turns must be giten to h screw formed
upon a cylinder whose length is 10 ins., and circumference
6 ins., that a power of % ozs. may overcome a pressnre of
IOO0Z8.P Ana. 100.
54. A screw is made to '.evolve by a force of 2 lbs.
applied at the end of a le.or 3.5 ft long; if the distance
between the threads be \ in., what pressure can be pro*
duced r Ant. 9 cwts. 1 qr. 20 lbs.
56. The length of the power-am is 16 inches ; find the
distance between two consecutive threads of the screw,
that the mechan ol advantage may be 30. Ana. it ins.
66. A weight of IF pounds is suspended f • +^i' btflok
of a single movable pulley, and the end f't .: ■ c**vii hi
which the power acts, is fastened at the d'!ut<.uii < h ft
from the fulcrum of a horizontal lever, a ft long, «.f the
second kind ; find the force, P, which must be applied per-
p«Qdioakirly at th« extremity of the lever to initain W.
Ana. P = -^-.
57. In a steelyard, the weight of the bean is 10 lbs., and
the distance of its centre of gravity from the fulcrum it
2 ins., find where a weight of 4 lbs. most be phioed to bal-
ance it Ana. At 6 ini.
'^nmmm^mmtmmgmm
ipoo^hig to tile
low many tnnifl
i long, in ordef
Atu. 48.
ve threads of a
h of the power
eostained by a
rm. 4807rlba.
a Borew fonned
ciroiimferenoe
e a pretsare of
Ana. 100.
force of a lbs.
if the distance
re can he pro-
1 qr. 20 Ibfl.
ches; find the
of the Bci-ew,
Aru. IT ins.
Ir* fh' block
Ui ■ c<»rd in
:i' • 5 ft
long, *A the
>e applied per-
luttain W.
P = I*.
ia
B 10 lbs., and
le fnlcnim it
ihioed to bal-
At Sim.
MXAMPLM8.
iUS
68. A body whose weight is V2 lbs., is placed on a rongh
plane inclined to the horizon at an angle of 46°. The co-
efficient of friction being -p, find in what direction a force
of (V3 — 1) lbs. must act on the body in order jnst to
support it Am. At an angle of 80° to the plane.
59. A rongh piano is inclined to the horizon at an angle
of 60° ; find the magnitni'e and the Jireotion of the least
force which will prevent e body weighing 100 lbs. fiom slid-
ing down the plane, the coefficient of friction being — •
Am. 60 lbs. inclined at 30° to the plane.
U^WIV' ■«'««■
IlLJIBUIIIl IMI^U.
■W^
i
CHAPTER VIII.
THE FUNICULAR* POLYGON— THE CATENARY
ATTRACTION.
130. fiqiiilibfilun of the Fnniciilar Poljrgon.— If a
cord whose weight is neglected, is saspended From two fixed
points, A and B, and if a series of weights, Pj, P„ P„
etc., be saspended from the given points Q^, Q„ Q^, etc.,
the cord will, when in equilibrinn:, form a polygon in a
vertical' plane, which is called the Funicular Polygon.
Let the tensions along
the successive portions
of the cord, AQ I, Qi Q^,
QtQ»> ®^ ^ respec-
tively r,, T„ r,, etc.,
and let 9,, 9„ 9„ eta,
be the inclinations of
these portions to the
horizon. Then Qi is
in equilibrium under the action of three forces viz., P,,
acting vertically, 7",, die tension of the cord AQ^, and 7,,
the tension of Qi Q,. Resolving these forces we have,
for horizontal foroes, jT, cos 0, — T, cos 9,-0, (1)
for vertical forces, P, + T, sin 0, — T, sin 9, = 0, (2)
In the same way for the point Q, we hare,
for horizontal forces, T, cos 0, — T, cos 9, = 0, (8)
for vertical forces, P, + T, sin 9, — T, sin 9, = 0, (4)
• The term, ^mtou l w, bM rafcrwM* Alone to Um oord, ud kaa bo ■«A«iilMd
■ignttOHMe.
Fia-M
Mvomwii .wTRf^f^w"
I.
CATENARY
Poljrgon.— Ifa
)d from two fixed
^tfl, P„ P„ />„
^xj Qt> Qi, etc.,
a polygon in a
%T Polygon.
i.n
foHMS via., P,,
AQ„ and T^,
8 we have.
.0, =.
:0,
(1)
«» =
0,
(2)
». =
0,
(8)
(?, =
0,
(4)
hM no BcchMilMl
MQUIUBRIUM or TBtt WUmCULAS POLTOON. 217
Henoe from (1) and (8) we have
T^ cog 0, = JT, COB ©, = J*, cos 0, = etc. ,
that is, the horitontal components of the tensions iv the dif-
ferent portions of the cord are constant. Lot this constant
be denoted by T; then we have
T,^
T»
^.=
COB 0,
cos tfj * ■*• — cos e, '
which in (2) an ^ (4) give
P, + rtan (J, — rtan fi, = 0,
P, + r tan e, — rtan 8, = 0,
and from (5) and (6) we have
P.
; etc..
(6)
tan e, = tan e, 4- -^
and
Similarly
and
P,
-f"
P,
T'
p
tan 0^ = tan fl, 4- -^i
etc., etc.
tan 6, = tan 0, +
tan 0, = tan e^ +
(n
If we suppose the weights P|, P,, etc., each equal to TT,
(7) becomes
tan ^i — tan 0g = tan 0, ~ tan 9, = tan 0, — tan 04
- _ FT
(8)
Hence, /A0 tangents of the successive inclinations form a
series in Arithmetic Progression. In the figure 0, = 0,
10
M^IWI IWIKIH II— I IHWI>W
818 coNSTancnotr or rat rmncvLA» polygon.
tand, =-^; tan 0, = ^;
T
*»«»*• = -ST ; tan e^ --fgr\ etc.
(9)
131. To Oonstrnct the Fanicnlar Polygon Trhen
the Horisontal Progectioiui of the euocesajve Por-
tions of the Cord are all eqnaL— Let ^,^4, Q^q^, q^q^,
q,qi, etc., be all of constant length = a, and let Q^q^ ■=. c.
Then since by (9) of Art.
130, the tangents of 9^, d„
0,, di, etc., are as 1, 2, 8,
4, etc., we have
Q^n — 3Q,qt = 3c; etc.
Hence, taking the middle point, 0, of the honzontal
portion, QiQ^, as origin, and the horizontal and vertical
lines through it as axes of x and y, the co-ordiiateu of Q,
are (fa, c) ; those of Q, are ({a, 30) ; those of Q^ .%-<^ (fa,
6c), and those of the nth vertex from Q^ are e ^ idently
o 2v 9i Q'
X =
2n + 1 » (« + 1]
— s a; if g= A 'ft
S
Eliminating n from these equations we get
■'"4
(1)
which, being independent of ft, is ntisfled by all the ver-
tices indifi'erently, and is therefore the equation of a oarre
passing through all the vertices of the polygon, and
denotes a parabola whose axis is the vertical line, OY, and
whose vertex is vertically below at a distance = x*
The shorter the distances ^4^1, QtQw ®tc., the more
nearly does the funiouiar polygon ooinoide with the para-
bolic curve.
I'OLrodfr.
tc.
(9)
olygon T7hen
tceMdive Por-
I let Q^q^ = c.
the honzontal
il and vertical
rdiiatei) of Q,
of Qt *"» (K
e ^idently
(1)
tj all the rer-
ion of a onrre
polygon, and
ine, OY, and
c
8
to., the more
th the para-
|e =
O0M9 aVPFOMVmO LOAA
319
:i32. Cord Bnppoiting a Load TJnUemalj Sis-
tribnted owr th* HoriscwtaL— If the number cf vertices
of the polygon be very great, and the suBpended weights all
equal so that the load is distribated uniformly along the
straight line, FE, the parabola which passes through all the
vertices, virtually coincides with the cord or chain forming
the polygon, and gives the figure of the Suspension Bridge.
In this bridge the weights suspended from the successive
portions of the chain are the weights of equal portions of
the flooring. The weight of the chain itself and the
weights of the sustaining bars are neglected in comparison
with the weight of flooring and the load which it carries.
Fit.7l
Let the span, AB, = 2a, and the height, OD, = h.
Then the equation of the parabola referred to the vertical
and horizontal axes of x and y, respectively, through 0, is
y* = 4mx,
(1)
4m being the parameter.
Because the load between O and A is uniformly dis-
tributed over the horizontal, 0£, its resultant bisects OB
at 0; therefore the tangents at A and intersect at
(Art 63).
From (1) we have
^ _ 8» _ y.
m
I
220
CORD aUPPORTINO LOAD^
which is the tangent of the inolination of the ottxve at any
point (a;, y) to the axis of x. Hence the tangent at the
p6int of support, A, makes with the horizon an angle, a,
whose tangent is — , which also is evident from the tri-
angle ACE.
Let If be the v^eight on the cord ; then ^ IF is the weight
on OA, and therefore is the yertical tension, V, at A. Then
the three forces at A are the verticcJ tension V ■= \W, the
total tension at the end of the cord, acting alonj^ who
tangent AG, and the horizontal tension, T, which is every-
where the same (Art. 130). Hence, by the triangle of
forces (Art 31) these forces will be represented by tiie
three lines, AE, AC, CE, to which their directions are
respectively parallel ; therefore we have for the horizontal
tension
r = AE cot o = W^,
and the total tension at A is
4A
EX AMPLB,
The entire load on the cord in (Fig. 71) is 320000 lbs.;
the span is 150 ft. and the height is 15 ft.; find the tension
at the points of support and at the lowest point and also the
inclination of the curve to the horizon at the points of
support.
tan « = -— = .04 ;
a
a ~ 2V 48'.
The yertical tension at each point of support is
r = i weight = 160000 lbs. ;
c
t
it
t
tl:
cc
B
tb
ca
or
lei
th
ax
lii
P,
hi
th
W(
!:he ottxve at any
tangeut at the
ZOQ an angle, a,
t from the tri-
r W is the weight
F, at A. Then
nV=^W, the
;ting aloni^ k.uo
, which is every-
the triangle of
resented by the
• directions are
r the horizontal
is 320000 lbs.;
tnd the tension
Ut and also the
the points of
48'.
Irtif
raw COMMON CATSXASr.
221
the homontal tension is
Tis »r^ = 400000 lbs.;
and the total tension at one end is
.. VV*+ T* = 430813 lb&
133. The Common Catenary.— Its Equation.— A
catenary is the curve assumed by a perfectly flexible cord
when its ends are fastened at two points, A and B, nearer
together than the length of the cord. When the cord is of
constant thickness and density, t. e., when equal portions of
it are equally heavy, the carve is called the Common
Catenary, which is the only one we shall consider.
Let A and B be the fixed
points to which the ends of
the cord are attached ; the
cord will rest in a vertical
plane passing through A and
B, which may be taken to be
the plane of the paper Let
be the lowest point of the
catenary; take this as the
origin of co-ordinates, and
let the horizontal line
through be taken for the
axis of X, and the vertical
line through for the axis of y. Let {x, y) be any point,
P, in the curve ; denote the length of the arc, CP, by « ;
let c* be the length of the cord whose weight is equal to
the tension at ; and T the length of the cord whose
weight is equal to the tension at P.
V
H'
X .
\ "
J
* 4
o
A
X
T'
n
»
* Tb* weif^t of a nuU of ImgUi of tbe conl balog here Uken m the nuit o(
TMa COMMON OATUNART.
Then the arc, CP, after it has assumed its permanent
fonn of equilibrium, may be considered as a rigid body
kept at rest by three forces vis.: (1) T, the tension, acting
at P along the tangent, (2) c, the horizontal tension at the
lowest point C, and (3) the weight of the cord, CP, acting
vertically downward, and denoted by a. Draw PT' the
tangent at P, meeting the axis of y at T'. Then by the
triangle of forces (Art 31), these forces may be represented
by the three lines PT, NP, T'N, to whici* their directions
are respectively parallel. Therefore
or
T'N _ weight of CP
~NP ~ tension at C *
^ — i.
tke
(«)
Differentiating, substituting the value of ds, and reducing,
we have
d
(I)
y/Mtf
e '
Integrating, and remembering that when x = 0, ^ = 0,
we obtain
where « is the Naperian base. Solving thif equation for
2 = »('--0' (•>
^, we obtain
/
ed ita pennanent
1 as a rigid body
he tension, acting
tal tennion at the
cord, CP, acting
Draw PT' the
T'. Then by the
ay be represented
:u their directions
is. and reducing,
. = 0,g = 0.
X
_ *
c'
ii« eqaation for
THM COMMON CATMNAST.
283
and by integration, observing that y = when » = 0,
we have
9 = 1(^ + 0"')-^' (8)
which is the equation required. We may simplify this
equation by moving the origin to the point, 0, at a dis-
tance equal to c below 0, by putting y — c for y, so that
(2) becomes,
=i(.%4
(8)
which is the equation of the catenary, in the usual form.
The horizontal line through O is called the directrix* of
the catenary, and is called the origin.
e r
Cob. 1.— To find the length of the arc, AP, we have
= Y I + i(e^ - e'*} dx, from (1),
= iK+e"7<^; (4)
.-. « = |(<l-«"^ (6)
the constant being = 0, since when x = 0, s = 0.
This equa^ioQ may also be found immediately by equa*
dv
ting the values of ^ in (o) and (1).
• 9m Mo*'* An«L Mtebt., Tol. I, p. tl«.
8M
THE COMMON CATENART.
Cor. 2.— Since c = OC is the length of the cord whoae
weight is equal to tha tension of the curve at the lowest
point, 0, it follows that, if the half, BO, of the curve were
removed, and a cord of length c, and of the same thickness
and density as the cord of the curve, were joined to the
arc CP, and suspended over a smooth peg at C, the curve
would be in equilibrium.
Cob. 3.— We have from the triangle, PNT,
tension at P __ PT
tension at C ~ PiV '
or
!*_- *L — if
c dx c
m (3) and (4),
that is, the tension at any point of the ^■'tenary is equal to
the weight of a portion of the cord whose length is equal to
the ordinate at that point.
Therefore if a cord of constant thickness and density
hangs freely over any two smooth pegs, the vertical por-
tions which hang over the pegs, must each terminate on
the directrix of the catenary.
Cob. 4.— From (3) and (5) we have
y» = «» + (?,
and from (6) we have
dy
(7)
At the point, P, draw the ordinate, PM, and from M,
the foot of the ordinate, draw the perpendicular MT. Then
pr = y cos MPT = y
dy
of the cord whose
mrve at the lowest
of the curve were
the same thickness
vere joined to the
!g at C, the curve
'NT,
(4),
'tenary is equal to
length is equal to
ness and density
the vertical por-
ach terminate on
(6)
(7)
M, and from M^
icularirr. Then
TBJS COMMON CATMUtABT.
which in (7) givQp
PT =: 9 = the OK, CP, (8)
and since j/t ^ PT» + TM*, we have from (6) and (8)
TM=c.
(9)
Therefore the point, T, is on the involute of the oitenary
which originates from the curve at C, TM ib n tangent to
this involute, and TP, the tangent to the catenary, is
normal to the involute, (Seo alculus, Art. 124). As TM
is the tangent to this last v nrve, and is equal to the con-
stant quantity, c, the involute is the eq^itangential Curve,
ortractrix(SeeOalcaJus, p.^67).
By means of (8) and (9) we may construct the origin &nA
dtrecinx of the catenary as follows : Oh the tangent at anv
point, P, measure off a Ungth, PT, equal to the arc, OP;
at T erect a perpendicular, TM, to the tangent meeting the
ordinate of P at M; then the horizontal line through M is
the diredtnx, and its intersection with the axie of tlie (mm
is the origin.
OoB. 6.— Combining (%) and (5) we obtain
(y + c)» = 8» +' c».
therefore
«• = »» + 2<?y.
(10)
The catenary possesses many interesting geometric and
mecnanical properties, but a discussion of them would
carry us beyond the limits of this treatise. The student
who wishes to pursue ihe subject further, is referred to
Prices Anal. Mecha, Vol. I, and Miucbin's Statics.
m^tffiiiiifiiniifrSniiiif
M
I
326
BPSTBRWAL 8ESLL.
133a. Attraction of a Sphwieal SholL-By the law
of umverBal gravitation every particle of matter attracts
every other particle with a force that varies diredly as the
mass of the attracting particle, and inversely as the square
of the distance between the particles.
To find tJ^ temltant attraction of a spherical shell of
um/orm densttv and small uniform thickness, on a par-
(1) Suppose the paiticle, P,
on which the value of the
attraction is required, to be
outside the shell.
Let p ana * be the density
and thickness of the shell,
its centre and M apj particle of it. Let 03f = a
FM = r, OP = c, the angle MOP = 0, the angle which
the plane MOP makes with a fixed plane through OP ^
^ , " IL .""^ **^ *^^ «'«'"«"* »t ^ (Art. 88) is
pka^,m9dBd^ The attraction of the whole shell L-ts
along OP) the attraction of the elementary mass at M on
P in the direction PM
— P^ <** **» " ^ M d<p
therefore the attraction. of i/' on P, resolved along OP,
_ pkcfi Bin 6 de d<p c — a cos
^
We shall eliminate from this equation by means of
t* =: a» + d> - 2ac (m e ;
.*. rdr sz ac Bin 6 d9;
(1)
Shell— By the law
of matter attracts
lies directly an the
rseltf aa the square
I spherical shell of
tckness, on a par-
t. Let OM = o,
; the angle which
I through OP.
M (Art. 88) is
whole shell sots
ary mass at M on
ed along OP,
y moaoa of
(1)
ac
substituting these values in (1), the attraction ot JM on P
along PO
= g.-(' + -;-)** w
To obtain the resultant attraction of the whole shell, we
take the ^integral between the limits and 2n, and the
r-integral between c — a and c + a.
Hence the resultant attraction of the shell on P along PO
npka
1 +
«»-
^dr.
^npkn* mass of the shell
(3)
Since e is the distance of the point P from the centre this
shows that the attraction of the shell on tlie paVticIo at P
is the same as if the mass of the shell were condensed into
its centre.
It follows iTom this that a sphere which is either homo-
geueous or consists of concentric spherical shells cf uniform
density, attracts the particle at P in the same manner as if
the whole mass were collootcd at its centre.
(2) lict the particle, P, be inside the sphere. Then wo
proceed exactly as beioro, and obtain equation (2), which is
true whether the particle be outside or inside the sphere ;
■
h
mmm
MXAKPLMB.
but the r-limits in thia eaae gre « — c and a + c. Hence
from (2) we have, by performing the ^integration,
attraction of sheU = ^ J^ (l - ^-^^r) ^^'
= !:^(2c-ac) = 0.
therefore a particle within the ehell is equally attracted in
every diiection, t. «., is not attracted at all.
CoE.— 11 a particle be inside o homogenons sphere at Uhe
distance r from its centre, all that portion of the sphere
which is at a greater distance ftrom the centre than the
particle produces no effect on the particle, while the re-
mainder of the sphere attracts the particle in the same
manner as if the mass of the remainder were all collected
at the centre of the sphere. Thiw th. attraction of the
sphere on the particle
|7rpr»
_ i-T— or
4Trpr
Honoe, toiihin a homogeneous sphere the attraction varies
as the dittoes from the centre.
The propositious respecting the attraction o. » uniform
Bphericd lihell on an external or internal particle were
given by Newton (Principia, Lib. I, Prop. 70, 71). (See
Todhunter's Statics, p. 276, also Pratt's Mechs., p. 187,
Price's Anal. Mechs., Vol. I, p. «66, Minohin's Statics,
p. 408).
BXAM PLBS.
1. The span Ali =mO foet, and CO = 1600 feet, And
Uu' length of the curve, CA, the height, CH, and the
nda + c. Hence
tegration,
««-e»
)dr.
= 0,
[aally attracted in
JL
noQB sphere at Hbe
tion of the sphere
9 centre than the
icle, while the re-
ticle in the same
were all collected
attraction of the
attraction varies
ioD Oi a nniform
lal particle were
). 70, 71). (See
Mechs., p. 187,
inohin's Statics,
= 1600 feet, find
t, CH, and the
MXAMPLMtH
229
iaoliaation, B, id the ewfe te the horisoai ii eithsr point of
siupension.
(1) Here - = f, and < ==: 3'7183d,
c
therefore
^ =: (a-71828>* = 1-2840,
and • = (2. 71828p = 0- 7788.
Substitnting these values in (5) we get
i$ = 800 X 0.5052 = 404-16.
CA = 404-16 feet.
Hence
(2)
= 800 X 2-0628 — 1600
= 60-24 r^ot
(«)
therefore
tan
« = ^? = t(^*---*).
cte
flrom (1),
= 0-2526,
e = 14° 11'.
8 404-16
Otherwise tan » = -, from (a), = -YaST = 0-2526, as
before.
1600
2. The entire load on the cord in Fig. 71 is 160000 lbs.,
the span is 192 (t, and the height is 15 ft; find the tension
at the points of support, and also the tension at the lowest
point Ana. Tension at one end = 268208 lbs.
Horizontal tension = 256000 "*
Ii :
^lr*
no
MXAMPLXa.
3. A chain, AOB, 10 feet long, and weighing 30 lbs., is
aaspended so that the height, CH, = 4 feet ; find the
horiaontal tension, and the inclination, 9, of the chain to
the horizon at the points of support
Am. Horizontal tension = 3| lbs., 9 = 77° 19'.
4. A chain 110 ft long is suspended from two points in
the same horizontal plane, 108 ft. apart; show that the
tension at the lowest point is 1.477 times the weight of the
chain nearly.
weighing 30 lbs., ig
4 feet; find the
0, of the chain to
M., 9 = 77° 19'.
rom two points in
t; show that the
the weight of the
PART II.
KINEMATICS (MOTION).
CHAPTER I.
RECTILINEAR MOTION.
134. Dsflnitioiia.— Velocity. — Kinematics is that
branch of Djrnamics which treats of motion without refer-
ence to the bodies moved or the foree* producing the mo-
tion (Art. 1). Although we do not know motion as free
ttom force or from the maiter that is moved, yet there are
cases in which it is advantageous to separate the ideas of
force, matter, and motion, and to study motion in the
abstract, t. «., without any reference to what is moving, or
the cause of motion. To the study of pure motion, then,
we devote this and the following chapter.
The velocity of a particle has been defined to be its rate
of motion (Art. 7). The formulae for uniform and variable
velocities are those which were deduced in Art. 8. From
(1) and (2) of that Art. we have
,1
ds
di'
(1)
(»)
in which v is the velocity, s the space, and t the time.
■MtJMo.
888
EXAMPLB8.
:X AM PI.E8.
1. A body moyes at the rate of 754 yar«l8 per hour. Find
ihe velocity in feet per second.
Since the velocity is uniform we use (1), hence
*' = 7 = H7^ STi = 0.628 ft. per sec, Ans.
t bU X oU
2. Find the position of a particle at & given time, /,
when the velocity varies as the distance from a given point
on the rdctiUnear path.
Here the velocity being variable we have from (2)
(to ,
*' = rf-< = *''
where /b is a constant ;
d$
therefore
= kdt;
logs s= kt + e,
(1>
where c is an arbitrary coustant.
Now if we suppose that s^ is the distance of the particle
from the given pomt when < = we have c = log »,,
which in (1) gives
log — = ife/ ; ' or s = s,e^.
3. A railway train tra' Is at the rate of 40 mile«per
hour ; find its velocity in feet per second.
Ans. 58.66 ft per ieoond.
4. A train takes 7 h. 31 m. to travel 300 miles ; find its
velocity. Ans. 39.02 ft per eec.
5. If « = 4^, find the velocity at the end of five seconds.
Ans. 300 f t^ per sea
6. Find the position of the particle in Ex. 2, when the
Telocity varies as th« time. Ans. s sx s^ + ^kfi.
fl per hour. Find
, bence
per sec, Ans.
I giren time, t,
•m a given point
I from (2)
he,
{1}
of the particle
m c = log «„
>f 40 milMper
peraeoond.
miloB ; find its
ft per eec.
of five seconds.
it. per 860.
X. 2, when the
AOCMLMSATnur MMMO.
233
7. Find tb« diertiance the particle will more in one
minute, when the Telocitj is 10 ft. at the end of one
second and Ti<jries as the time. Aiu. 18000 ft.
135. Acceleration. — Acceleration has been defined to
be the rate of change of vOocity (Art. 9). It is a wioeily
increment. The formula for acceleration are £rom (1), (3),
and (3) of (Art. 10),
f-dV
/ =
M"'
(1)
(2)
(8)
(1) being for uniform, and (2) and (3) for variable,
acceleration.
If the velocity decreases, f is negative, and (2) and (3)
become
^^ - /^. *? - _ f .
di~ ~^' dfl ~ ■'*
and the velocity and time are inverse functions of each
other.
136. The Relation between the Space and Time
when the Acceleration = 0.
Here we have
*"-0
so that if V, is the constant velocity we have
da
s=»
• >
« = «,< + «•»
.1 r
'f*
.rt'2]
I' i
tt4
AOOELMnATIOIf CONSTAltT.
1<
3
i
i
in which 9, is the space which the body has passed over
when ^ = 0. If < is computed from the time the body
stairtj from rest, then a = v^t The student will observe
that this is a case of uniform velocity.
137. Th« Relation (1) between tiie Space and
Time, and (2) between the Space and Veloeity,
when the Acceleration ia Oonatant
(1) Let A be the initial position of cT
the particle supposed to be moving
p
Fig.73
toward the right, P its position at any time, t, from A, v
its velooity at that time, and / the constant acceleration c*
its velocity. Take any fixed point, 0, in the line of motion
as origin, and let OA = «, ; OP = s. Then the equation
of motion ia
••• rf/ =/" + «•
Suppose the velocity of the particle, at the point A to be
V,, then when ^ = 0, v = Vs;* hence e ■=■ v„ and
.«. « = !//« + v^t + c'..
But when / = 0, « = «,; hence c' = «,, and
» = !//» + v,< + «„
(2)
(3)
Hence if a particle moves from rest from the origin O, with
a constant acceleration, we have
* OOM MIM ▼•loelty and ipaoe rwpMtlTclr, or the velocity the pwtide lia%
•nd epMe It hM moved over at Uie Inatant ( begini to be nekosed.
T.
iy has passed over
te time the body
dent will observe
tlM Space and
and Velosity,
Flo.73
ime, /, from A, v
it acceleration c*
;he line of motion
hen the equation
(1)
point A to be
v„ and
(8)
and
(8)
origin 0, with
tdty tbe pwtlele haM,
MMd.
ACCMJJISATION VABtABLM.
s:Aj,,-jr--';J;55.i;
235
(4)
and thus the space described varies as the square of the
time.
(2) From (1) we have
... g = 2/, + 0.
But when « = #„ t> = t», ; hence t) = t>,« — 2/5jo, and
therefore
v» i:^ 2/« + Vo" - 2/8,. (6)
Equations (2) and (3) give the velocity and position of the
particle in terras of t ; and (5) gives the velocity in terms
of «.
138. When the Acceleration Vaxiea dirjcily aa
tiie Time from a State of Rest, find the Velocit7
and Space at ttie end of the Time t.
Here
' ' dt
where «, is the initial velocity ;
.-. » r= !«<• + V,
the initial space being since / is estimated from rest.
139 When the Acceleration ▼ariea direotiiy aa
the IMatance ftam a giwen Point bt the line of Mo-
tion, and ia negattve, find the Relation between
the Space and Time.
1 :|;:vj
-'. ?■
m
Jl
itL.y.iy«iWiiiJiuii,v
286
Here
J
BXAMPLBS.
==-ks;
by calling «, the valae of $ when the particle is at rest
da
V^o* - «•
ifcirf/.
the negative sign being taken since the particle is moving
towards the origin ;
. • . cos-' — = kU,
if « = «, when / = ;
i^t.
EXAMPLES.
1. A body commences to move with a velocity of 30 ft.
per sec, and its velocity is increased in each second by
10 ft. Find the spoce de«cribed in 5 seconds.
Here / = 10, v, = 30, s^ = 0, and / = 6, thei^fore
from (2) we have
« = |.10.a6 + 30.5 = 275, Ans.
2. A body starting with a velcoity of 10 ft per sec, and
moving with a constant acceleration, describes 90 ft. in
4 sees.; find the acceleration. Ana. 6^ ft. per sec.
3. Find the velocity of a body which starting fipom reit
with an acceleration of 10 ft per sec., has described a space
of 20 a Ana. Wtt,
;le ifl at rest
rticle is moying
locity of 30 ft
lach second by
: 5, therefore
Ins.
per sea, and
jbea 90 ft. in
1^ ft per sec.
ing from rest
bribed a space
Ana. 30 ft
FALLtJtB BODina.
4. Throngh what space mast a body pan nnder an accel-
eration of 6 ft per sec, so that its velocity may increase
from 10 ft to 20 ft per sec. ? Ans. 30 ft
5. In what time will a body moving* with an cxxselera'
tion of 25 ft per sec., acquire a velocity of 1000 ft per
second? ' Ana. 40 sees.
6. A body starting flrort rest has been moving for 5 min-
utes, and has acqaired a velocity of 30 miles an hour;
what is the acceleration in feet per second ?
Ana. W ft. per sec.
7. If a body moves from rest with an acceleration of { ft
per sec, how long mast it move to acquire a velocity of
40 miles an hour?
Ans. 88 sees.
140. Equations of Motion for Falling Bodies.—
The most important case of the motion of a particle with a
constant uccoleration in its line of motion is that of a body
moving under the action of gravity, which for smtdl dis-
tances above the earth's surface may be considered constant.
When a body is allowed to fall freely, it is found to acquire
a velocity of about 32.2 feet per second during every second
of its motion, so that it moves with an acceleration of 32.2
feet per second (Art. 21). This acceleration is less at the
summit of a high mountain than near the surface of the
earth ; and less at the equator than in the neighborhood of
the poles ; i. e., the velocity which a body acquires in falling
freely for one nocond varies with the latitude of the place,
and vrith its aliitueh above the sea level ; but is independ-
ent of the site of the body and of its masa. Practically,
however, bodies do not fall J>eely, as the resistance of the
air opposes their motion, and therefore in practical cases at
high speed (0. g., in artillery) the resistance of the air must
be taken into account But at present we shall neglect
* In each cate tbe body ia «nppoMd to etart from rect nnleH otherwise stated.
iij
!
II
238
FALLTNO BODIES.
this resistance, and consider the bodies ad moving in vacuo
under the action of gravity, t. e., with a constant accelera-
tion of about 32.2 feet per second.
As neither the substance of the body nor the cause of
the motion needs to be taken into consideration, all prob-
lems relating to falling bodies may be reganled as cases of
accelerated motion, and treated from purely geometric
considerations. Therefore if we denote the acceleration by
g, as in Art. 23, and consider the particle in Art. 137 to be
moving vertically downwards, then (2), (3), (5) of Art 137
become, by substituting jr for/,
s = yp + vj + So,
v* = 2flf« 4- Vo' — 2jrso»
(A)
s being measured as before from a fixed point, 0, in the
line of motion.
Suppose the particle to be projected downward from O,
then A commences with and «, = 0. Hence (A) be-
comes
V =zgt + v^, (1)
8 =yi* + v^t,
i^ = 2g& + «,».
(2)
(3)
As a particular case suppose the particle to bo dropped
from rest at (Fig. 71). Thro A coincides with 0, and
s, = 0, Vj = 0. Hence eqiov-ions (A) become
v = gt.
(4)
» = igi',
(6)
tf> = 2ga.
(6)
WsmKiasmr-
'jn-rn
ovmg m vacuo
istant accelera-
or the cause of
ation, all prob-
rckd as cases of
rely geometric
acceleration by
I Art. 137 to be
(5) of Art. 137
(A)
oint, 0, in the
iward from O,
Hence (A) be-
(1)
(«)
(8)
to be dropped
with 0, and
me
(»)
(«)
PARTIOLX PROJECTED UPWARDS.
239
141. When the Particla im Projected Vextioally
TTpwards. — Here if we measure s upwards from the point
of projection, 0, the acceleration tends to diminish the
space and therefore the acceleration is negative, and the
equation of motion is (Art 135)
= -9-
In other respects the solution is the same. Taking
therefore «, = in (A) and changing the sign of g,* we
obtain
t> = Vo — gt, (1)
t)» = Vo» - ^gif.
(2)
(3)
Gob. 1. — The time during which a particle rises when
projected vertically upwards.
When the particle reaches its highest point, its velocity
is zero. If therefore we put v = in (1), the corresiwnd-
ing value of t will be the time of the particle ascending to a
state of rest.
9
Cor. 2. — The time qf flight hefwe returning to the dart-
ing point.
From (3) we have the distance of the particle from the
starting point after ^ seconds, when projected vertically
upwards with the velocity v,. Now when the particle has
risen to its maximum height and returned to the point of
projection, « = 0. If, therefore, we put « = in (2), and
solve for /, we shall get the time of flight. Therefore,
• g in poaltlve or mgaUye Moordin^ aa the particle is deaoendlng or aa-
cendingr.
■^ r i
240 FABTICLM PROJSCTKD UPWAMl^H.
which givM
/ = 0, or
tv.
I
The first yaluo of t shows the time before the part'cle
starts, the latter shows tine time when it has returned.
2r
Hence, the whole time of flight is -^, which is just double
the time of rising (Cor. 1) ; that is, the time <if riniiig equals
the time of falling.
The final velocity, by (1) of Art. 140, = y/ = ^ x *^
(Cor. 1) = Wo ; hence a body returns to any point in its
path with the same velocity at which it left it. In other
words, a body passes each point in its path with the same
velocity, whether rising or falling, since the velocity at any
point may be considered as a velocity of projection.
•
Cob. ^.—The greatest height to which the partich tuiU
riM,
At the summit w = 0, and the corresponding value of «
will be the greatest height to which the particle will rise ;
when V = 0, (3) becomes
o.t —
Co*. 4.— Since v^* — 2g8, where « is the height frum
which a body fiills to gpin the velocity w„ it follows that »
body will rise through the same space in losing a velocity
V, as it would fall through to gain it
(fore the part^'cle
it has returned.
cb is just double
» <^ rining equaU
any point in its
3ft it. In other
t with the same
n velocity at any
ectiou.
Ihe particle toiu
ding value of «
ftiole will rise ;
le height frum
follows that a
ling a velocity
1. A body projected vertically downwards with a velocity
of 20 ft. a see. from the top of a tower, reaches the ground
in 2.5 sees.; find the height of the tower.
Here t = ^, and v^ ~%0', assume g = 32. Then
from [i) of Art. 140 we havo
, = it^UL + 20 X f = 150 ft
2. A body is projected vertically upwards with a velocity
of 200 ft. per second ; find the velocity with which it will
pass a point 100 ft. above the point of projection.
Here t>, = 200, « = 100 ; therefore from (3) we have
»8 = 40000 - 6400 = 33600 ;
.-. v = 40'v/2T.
3. A man is ascending in a balloon with a uniform
velocity of 20 ft. per sec., when he drops a stone which
reaches the ground in 4 sees.; find the height of the
balloon.
Here v, = 20, and f := 4 ; therefoi-e from (2) we have,
after changing the sign of the second mcrabor to make the
result positive.
fl = - (80 - 256) = 176,
which was the height of the balloon.
4. A body is projected upwards with a velocity of 80 ft. ;
after what time will it return to the band ?
Ana. 5 second*.
8. With what velocity must a body be projected ver-
tically upwards that it may rise 40 ft. P
Atts. 10 VlO ft. per SHJ.
a
,? . !'
-.^>J^.,-Jtf.,........:.|^.«^
'mmmamm
242
COMPOSITION or VBLOCITISa.
6. A body projected vertioallj upwards paBses a oertain
point with a velocity of 80 fl. per sec. ; how much higher
will it ascend P Ans. ICJ ft.
7. Two balls are dropped from the top of a tower, one of
them 3 sees, before the other ; how far will they be apart
5 sees, after the first was let £eiI1 ?
Am. 336 ft.
8. If a body after having fallen for '3 sees. breakF a pane
of glass and thereby loses one-third of its velocity, find the
entire space through which it will have fallen in 4 sees.
Ans. 224 ft.
142. Composition of Velocities.— (1) From the Par-
allelogram of Velocities, (Art. 29, Fig. 2), we see that if AB
represents in magnitude and direction the space which
would be described in one second by a particle moving with
a given velocity, and AC represents in magnitude and
direction the space which would be described in one second
by another particle moving with its velocity, then AD, the
diagonai of the parallelogram, reprenenta the resultant
velocity in magnitude and direction.
(2) Hence the resultant of any two velocities, as AB, BD,
(Fig. 2), is a velocity represented by the third side, DA, of
the triangle ABD; and if a point have simultaneously,
velocities represented by AB, BC, CA, the sides of a trian-
gle, taken in the name o-der, it is at rest.
The linos which are taken to represent any given forces
may clearly be taken to represent the velocities which
measure these forces (Art, 19), therefore from the Polygon
and Parallelopiped of Forces the Polygon and ParaUel-
ojtiped of Velocities follow.
(3) Hence, if any number of velocities be represented in
magnitude and direction by the sides of a closed polygon,
taken all in the same order, the resultant is tsro.
(4) Also, if three velocities be represented in magnitude
passes a certain
w much higher
Ana. ICt) ft.
' a tower, one of
I they be apart
Atu. 336 ft.
. breake a pane
ilocity, find the
m iu 4 sees.
Ana. 224 ft.
Prom the Par-
) see that if AB
le space which
3le moving with
magnitude and
1 in one second
then AD, the
the reaullant
es, aa AB, BD,
side, DA, of
nmullaneoualy,
ka of a triau-
ly given forces
tlocitiea which
the Polygon
and Parallel-
repreaented in
ihaed polygon,
ro.
in magnitude
■W'^^>:iS?^^J!jf»'A*^*fi9ft!j¥JV»>:|^
BEaOLUnON OF VBLOCPtlKS.
243
and direction by the thre* edges of a parallelepiped, the re-
aultant velocity will be repreaented by the diagonal.
(5) When there are two velocities or three velocities in
two or in three rectangular directions, the resultant is the
square root of the sum of their squares. Thus, if
-£, ^-, ^j, -^, are the velocities of the moving point and
its components parallel to the axes, we have from (2) of
Art. 30,
and from (1) of Art. 34,
(1)
da
dt
=V6?)V(I)^(I)* <')
143. Resolution of Velooitie>.— Af the diagonal of
the parallelogram (Fig. 2), whose sides re])re8ent the com-
ponent velocities was found to represent the resultant
velocity, so any velocity, represented by a given straight
line, may be resolvrd into component velocities represented
by the sides of the parallelogram of which the given lino
\ the di'\gonaL
It will b< rtasily seen that (2) of Art. 134 is equally
applicable whether iiie point bo considered as moving in a
straiglit line or in a curved line ; but since in the lalter
ase the direction of motion continually changes, the mere
liount of the velocity is not sufficient to describe the
ii 'tion completely, so it will be necessary to know at every
instant the direction, &b well as the magnitude, of the point's
velocity. In such oasec as thie Ihe method commonly cm-
ployed, whether we deal with velocities or accelerations,
consist.) mainly in studying, not the velocity or acceleration,
directly, but its components parallel to any three assumed
rectangular axes. If the {mrticle be at the point (x, y, «),
' , i
n
1 ■
a: )
" f 1 i 1-
i I
I'-
Jj!l
if}
. i
-J
I
tfiiMMn
^jMW imitn iili Hiiii'ftyiJiiMiMHiiiri ■ n
iii_rnm-i Ti-Tn-iiriT-TT'iir""'V iiirir-rfT""
i-
2U
BXAMPLS9.
at the time /, and if we denote its relocitiet parallel
respectively to the three axes by «, v, w, we hare
dx dy dz
dt
dt
dt
Denoting by v the velocity of the moving particle along
the curve at the time t, we have as above
and if o, /3, y be the angles which the direction of motion
along the curve makes with the axes, we have, as in (2) of
(Art 34), •
dx ds
^ =s ^ oca a = V 008 « = u;
dt da
^ = J-. cos y = » cos y = «>.
m dt
. , . d!* rfv rf' • 1. u
Hence c«*ch of the components ^, ^, j^ is to be
found from the whole velocity by retolving the velocity,
I. e., by multiplying the velocity by coeim of the angle
between the direction of motion and that of the compo-
nent.
EXAMPLES.
1. A body moves under the influence of two velocities,
at right angles to each other, equal respectively to 17.14 ft.
and 13.11 ft. per second. Find the magnitude of tb"
resultant motion, and the angles into which it divides
right angle.
Ans. 81.679 ft. per sec. ; 37° «ft' and 68°
ities parallel
ITC
•article along
I
(1)
)n of motion
> as in (2) of
h
if
is to be
the velocity,
of the angle
the compo-
▼elocities,
to 17.14 ft.
ode of tb"
divide*
dSJ"
MOnON ON AH tlfCUNSD PLANB,
%. A ship Bails due north at the rate of 4 knots per
hour, and a ball is rolled towards the east, across her deck,
at right angles to her motion at the rate of 10 ft. per
second. Find the magnitude and directicoi of the real
motion of tlie ball.
Am. 12.07 ft per soc.; and N. 66° E.
3. A boat moves N. 30° E., at the rate of 6 miles per
hour. Find its rate of motion northerly and easterly.
Ans. 5.2 miles per hour north, and 3 miles per hour
east.
144. Motion on an ZncUnod Plane.— By an exten-
sion of the equations of Art. 140, we may treat the case of
a pai-tiole sliding from rest down a smooth inclined plane.
As tb>is is a very simple case in which an acceleration is
resolved, it is convenient to treat of it in this part of our
work ; yet as it properly belongs to the theory of con-
strained motion, we are unable to give a complete solution
of it, until the principles of such motion have been ex-
plained iu a future chapter.
Let P bo the position of the particle at
any time, t, ou the inclined plane OA, OP
sr «, its distance from a fixed point, O, iu
the line of motion, and let « be the inclina-
tion of OA to the horizontal line AB. Let
Pi represent g, the vertical acceleration with rfj-M
which the body would move if free to fall. Resolve this
into two components, Po = ^ sin a along, and Vc — g
cos a perpendicular to OA. The component g cos a pro-
duces pressure on the plane, but does not affect the motion.
The only acceleration down the plane is that component of
the whole acceleration which is parallel to the plane, vie.,
g sin «c. The equation of motion, therefore, if
^ = ^ sin «,
(1)
— f-
j».t*
^^..^^■liiHrWBIIilllttfi
MIMIMM
240
DBCOENT DOWN CHORDS OF A OIBCLX.
the Bolation cf which, aag nintt is constant, is incladed in
that of Art. 140; and all the resalts for particles moving
vertically aa given in Arts. 140 and 141 will be made to
apply to (1) by writing g sin et for g. Thns, if the particle
be projected down or up the plane, we get from (1), (2), (3)
of Arts. 140 and 141, by this means
V = Vo ± jf sin u't,
s = Vft ± \g sin a- 1,
V* = v,» ± 2^ sin «•«,
(3)
(4)
in which the -f or — sign is to be taken according as the
body is projected down or up the plane.
If the particle starts from rest from 0, we get from (4),
(5), (6) of Art. 140
V = ^ sin a* t, (6)
a = 4jr sin a- ^,
»» = 2jf sin «•«.
(8)
(7)
CoE. 1. — Tk« velocity acquired by a particle in falling
down a given inclined plane.
Draw PO parallel to AB (Fig. 74), then if » be the
velocity at P, wo have from (7)
1^ = 2g sin a*«
Hence, from (6) of Art. 140 the velocity is the same at P
as if the particle had fullen throngh the vertical space OC ;
that is, tlte velocity acquired in falling down a smooth
inclined plane is the same as would be acquired in falling
freely through the perpendicular height of the plane.
tCLg.
is incladed in
rticles moving
ill be made to
if tbe {Hirticle
>m(l),(a),(3)
(2)
(3)
(4)
;ording as the
} get from (4),
(ft)
(8)
(7)
cfo in falling
if V be the
le same at P
kl space OC ;
V« a stnoolh
in falling
\lan0.
DBaCSNT DOWN CBOBDS OF A CtBCLX.
CJOB. 2. — When the particle is projected up the plane teith
a given veloef^y, to find how high it will ascend, and the time
ofascetU.
From (4) we have
1^ = w,' — 2^ sin a'S.
When V = the particle will stop ; henoe, the distance it
will ascend will be given by thb equation
= r,» — Zgaina'S,
8 =
2g sin «
To find the time we have from (2)
V = v^ —g aina-t;
and the particle stops when v = 0, in which case we have
i =
g ana
From (6) we derive the following canons and nseful
result
145. The Times of Descent down
all Chords drawn from the Highest
Point of a Vertical Circle are eqnaL—
Let AB be the vertical diameter of the
circle, AG any cord through A, a its
inclination to the horizon ; join BO ; then
if ^ be the time of descent down AG we
have from (6) of Art. 144
Fi»7S
Bat
AC = yP sin d.
AC 'js AB sin u ;
aMiWNtaMliUHWUM' ^mmmmmimiammttm
iiUfM OF qmoKMar onacavt.
.'. AB»5fcf#»,
/2AB
which is constant, and shows that the time of falling down
any chord is the same as the time of falling down the
diameter.
Cor. — Similarly it may be shown that the times of
descent down all chords drawn to B, the lowest point,
are equal ; that is, the time down OB is eqaal to that
down AB.
146. The Straight Idna of QoickoBt Descent from
(1) a Qiven Point to a Qiven Straight lone (2) from
a Given Point to a Given Onrve.
(1) Let A be the given point and BO c
the given Uqq. Through A draw the
horizontal line AO, meeting OB in 0;
bisect the angle ACB by 00 which inter-
sects in the vortical line drawn through
A ; from draw OP perpendicular to BO;
Join AP ; AP is the required line of quick-
est descent
For OP is evidently equal to OA, and therefore the
circle described with as centre and with OP (= OA) for
radius, will touch the line BO at P, and since the time of
falling down all chords of this circle from A is the same,
AP must be the line of quickest descent.
(2) To find the straight line of quickest descent to a
given ewrvt, all that is required is to draw a circle having
the given point as the upper extremity of its vertical
diameter, and tangent to the curve. Hence if DE (Fig.
76) be the curve, A the point, draw AH vertical ; and, with
centre in AH, describe a circle p^wsing through A, and
IHI
m
1 falling down
ing down the
the times of
I lowest point,
equal to that
Descent from
U»9 (2) from
n«.7« ^
therefore the
}P {- OA) for
I the time of
ia tiie same,
descent to a
circle having
jf its Tertical
if DE (Fig.
Ell ; and, with
>agh A, «ad
XXAMPLSa.
249
touching DE at P, then AP is the required line. For, if we
take any other point, Q, in DE, and draw AQ cutting the
circle in q, then the time down AP = time down A.q<i
time down AQ. Hence AP is the line of quickesl descent.
Thu problem of 6ndiDg the line of quickest descent from a point to
a line or curve is thus found to resolve itself into the purely geometric
problem of drawing a circle, the highest point of which shall be the
given point and which shall touch the given line or curva
EXAMPLES.*
1. If the earth travels in its orbit 600 million miles in
365jt days, with uniform motion, what is its velocity in
miles per second ? Ana. 19 -01 miles.
2. A train of cars moving with a velocity of 20 miles an
hour, had been gone 3 hours when a locomotive was
dispatched in purstiit, with a velocity of 25 miles an hour;
in what time did the latter overtake the former ?
Ana. 12 hours.
3. A body moving from rest with a uniform acceleration
describes 90 ft. in the 5th second of its motion ; find the
acceleration,/, and velocity, v, after 10 seconds.
Ans. / = 20 ; v = 200.
4. Find the velocity of a particle which, moving with an
acceleration of 20 ft. per sec. has traversed 1000 ft.
Ans. 200 ft. per sec.
5. A body is observed to move over 45 ft. and 55 ft. in
two eaccessive seconds ; find the space it would describe in
the 20th second. Ana. 195 ft.
6. The velocity of a body increases every hour at the rate
of 360 yards per hour. What is the acceleration, /, in feet
per second, and what is the space, a, described from rest ?
Ans. f= 0-3; v = 60 ft.
* In the«c examplci Uke g--9an.
MM
WO
MXAMPLKM.
7. A body is moting, at a given instant, at the rate ^
8 ft per sec.; at the end of 5 aecs. its >tl(K!ity is 19 ft. per
sec. Aeeuming its acceleration to be uniform, what was its
velocity at the end of 4 sees., and what will be its velocity
at the end of 10 sees. ? Ana. 16- 8 ; 30.
8. A body is moving at a given instant with a velocity of
30 miles an hoar, and comes to rest in II sees.; if the
retardation is uniform what was its velocity 5 sees, before it
stopped ? Ans. 20 ft. per sec.
9. A body moves at the rate of 12 fi a sec with a
nniform acceleration of 4; (1) state exactly what is meant
the number 4 ; (2) suppose the aci deration to go on for
i 6 .sees., and then to cease, what distance will the body
1; describe between the ends of the 6th and 12th sees.?
Ans. 224 ft.
10. A body, whose velocity undergoes a uniform retarda-
tion of 8, describes in 2 sees, a distance of 30 ft.; (1) what
was its initial velocity ? (2) How much longer than the
2 sees, would it move before coming to rest ?
Am. (1)23; (2) | sec.
11. A body whose motion is uniformly retarded, changes
its velocity from 24 to 6 while describing a distance of 12
ft.; in what time does it describe the 12 ft.?
Ans. 0-8 sec.
12. The velocity of a body, which is at first 6 ft. a sec,
nndergoos a uniform acceleration of 3 ; at the end of 4 sec&
the acceleration ceases ; how far does the body move in 10
sees, from the beginning of the motion ? Ans. 156 ft
13. A body moves for a quarter of an hour with a uni-
form acceleration ; in the first 5 minutes it describes 350
yards; in the second 5 minutes 420 yards; what is the
whole distance described in a quarter of an hour ?
Ans. 1260 yds.
» at the rate of
3ity is 19 ft. per
m, what was its
I be its velocity
w. 16-8; 30.
ith a velocity of
I I sees.; if the
5 sees, before it
20 ft. per sec.
^ a sea with a
what is meaut
ion to go on for
will tho body
ith sees.?
Ans. 284 ft.
oiform retarda-
50 ft.; (1) what
onger than the
S3; (2) I sec.
ardcd, changes
distuuce of 12
ms. 0-8 sec.
frst 6 ft. a sec,
end of 4 sees;
move in 10
ins. 156 ft
|ur with a nni-
describes 3.50
what is the
jur i
1260 Yds.
MXAMPLSS.
Ml
14. Two sees, after a body is let fall another body is
projected yertioally downwards with a velocity of 100 ft
per sec. ; when will it overtake the former i*
Ana. 1| sees.
15. A body is projected upwards with a velocity of 100
ft per sec.; find the whole time of flight Ans. li^ sees.
IG. A balloon is rising uniformly with a velocity of 10 ft.
per sec, when a man drops from it a stone which reaches
the ground in 3 sees.; find tho height of the balloon, (1)
when the stone was dropped ; and (2) when it reached the
ground. Ans. (1) 114 ft; (2) 144 ft
17. A man is standing on a platform which descends
with a uniform acceleration of 6 ft per sec. ; after having
descended for 2 sees, he drops a bull ; what will be the
velocity of the ball after 2 more seconds ? Ans. 74 ft.
18. A balloon has been ascending vertically at a uniform
rate for 4.6 sees., and a stone let fall from it reaches the
ground in 7 sees.; find the velocity, v, of the balloon and
the height, «, fh)m which the stone is let fall.
Ans. V = 174| ft per sec.; s = 784 ft If the balloon
is still ascending when the stone is let fall v = 68-17 ft.
per sec; s = 306-76 ft?
19. With what velocity must a particle bo projected
downwards, that it may in t sees, overtake another particle
which has already fallen through a f t ?
Ans. V -. T + VZc^.
20. A person while ascending in a balloon with a vertical
velocity of V ft per sec, lets fall a stone when he is k ft.
above the ground ; required the time in which the ntony
will reach the ground. ^ y + ^yt — 2gh
Ans.
20S
XXAMPLS&
21. A body, A, is projected vertically downwards from
the top of a tower with the velocity V, and one sec. after-
wards another body, B, is let fall from a window a ft. from
the top of the tower ; in what time, t, will A overtake B ?
2a + g
Ana. t =
^(y + g)
32. A stone let fall into a well, is heard to strike the
bottom in t seconds ; required the depth of the well, sup-
posing the velocity of sound to be a ft. per sec.
^ a_~|»
'iff V¥gJ '
Ans. \/ at
+
23. A stone is dropped into a well, and after 3 sees, the
sound of the splash is heard. Find the depth to the
surface of the water, the velocity of sound being 1127 ft.
per sec.
24. A body is simultaneously impressed with three
uniform velocities, one of which would cause it to move
10 ft. North in 2 sees. ; another 12 ft. in one sec. in the
same direction ; and a third 21 ft. South in 3 sees. Where
will the body be in 5 sees. ? Am. 50 ft. North.
25. A boat is rowed across a river 1^ miles wide, in a
directioi^ making an angle of 87° with the bank. The
bviat travels at the rate of 5 miles an hour, and the river
nu\s at the rate of 2.3 miles an hour. Find at what point
of tlie opposite bank the boat will land, if the angle of 87°
be made against the stream.
Ans. 898 yards down the stream from the opposite
point.
26. A body moves with a velocity of 10 ft. yter sec. in a
given direction ; find the velocity in a direction inclined at
an angle of 30° to the original direction.
Am. 6 V3 ft. per sec.
i(^^!^p»i^S!W««!^ta^'''i
EXAMPLES.
963
wrnwards from
one sec. af ter-
dow a ft. from
overtake B ?
1 to strike the
the well, aup-
c.
ter 3 sees, the
depth to the
being 1127 ft.
I with three
se it to move
le sec. in the
sees. Where
ft. North.
C8 wide, in a
)ank. The
ind the river
what point
angle of 87°
the opposite
t)er sec. m a
u inclined at
rt. per sec.
27. A smooth piano is inclined at an angle of 30° to the
horizon ; a body is started up the plane with the velocity
Hg; find when it is distant 9^ from the starting point
Arts. 2, or 18 sees.
28. The angle of a plane is 30° ; find the velocity with
which a body mnst be projected up it to reach the top,
the length of the plane being 20 ft.
Ana. 8 VlO ft. per sec.
29. A body is projected down a plane, the inclination of
which is 45°, with a velocity of 10 ft.; find the space
described in 2^ sees. Ant. 95.7 ft. nearly.
30. A steam-engine atarts on a downward incline of
1 in 200* with a velocity of 7|^ miles an hour neglecting
friction ; find the space traversed in two minutes.
Ana. 824 yards.
31. A body projected up an incline of 1 in 100 with a
velocity of 15 miles an hour just reaches the summit ; find
the time occupied. Ana. 68.75 sees.
32. From a point in an inclined plane a body is made to
slide up the plane with a velocity of 16.1 ft. per sec. (1)
How far will it go before it comes to rest, the inclination
of the plane to the horizon being 30° ? (2) Also how far
will the body be from the starting point after 5 sees, from
the beginning of motion ?
Ana. (1) 8.05 ft. ; (2) 120.75 ft. lower down.
33. The inclination of a plane is 3 vertical to 4 hori-
zontal ; a body is made to slide up the incline with an
initial velocity of 36 ft. a sec. ; (1) how far will it go before
beginning to return, and (2) after how many seconds will
it return to its starting point?
Ana. (l)33f ft.; (2) 3| sees. ,
* An Incline 6r 1 In SOO mMns here 1 fnot vcrttcally to • length of iOO ft., Uumgh
U to nMd b; Bnglneen to mean 1 foot vertically to nO ft. AoH«mta^.
MMHMiMi
Ui
MXAMPLM$,
34. There is so inclined plane of 6 vertical to 12 hori-
zontal, a body «lidea down 52 ft of its length, and tLo
passes without b« of velocity on to the horizoittl pluuT
after how long from the beginning of the motion will it hj
at a distauce of 100 f L from the foot of the incline ?
Ans. 6.7 sees.
i/iu^ I^^ '' projected up an inclined plane, whose
length IS 10 times its height, with a velocity of si tiZ
«ec. ; m what time will its velocity be destroyed ? '
Am. 9| sees., if ^r = 32.
30. A body falls from rest down a given inclined plane;
compare the tiwes of describing the first and last halves
-4ns, As 1 : ^2 -f 1.
37. Two bodies, projected down two pianos in-^lmed to
he honzon at anglee of 45» and 60°. describe in the same
time sp-oces re8{.ectively as ^/i : ^3; find the ratio of the
initial vokKjities of the projected bodies.
Ana. V2 : 1/3.
38. Thmugh what chord of a circle must a body fall to
Zeter^ ''' "'"'^-^ '''"'' '' '*"*"^ "^-V th^
Am. The chord which is inclined at 60' to the vertical
il\ ^^^ ^^^ ""'^."^''^^ ""'^^ ^^'"'^ * ^y «^««5<1 be pro.
jocted down an inclined plane, /. «o that the time of
running down the plane shall be equal to the time of
falling down the height, h.
. A * «in «\
nf *r;. ^'f!''^^ inclination of this plane, when a velocity
of fths that duo to the height is sufficient to render the
timoH of running down the piano, and of falling down the
height, equal to eacli othor. ^„,, 30°.
tical to 12 hori-
eugth, and tlieu
lorizoQtal piano ;
aotion will it be
incline?
Ans. 6.7 sees.
d plaue, whose
ity of 30 ft. per
iyed?
38., if ^ ~ 33.
inclined plane;
and last halves
li Vi -h 1.
nes in'ilined to
be in the eanio
:he ratio of the
f. V2 : Vs.
a body fall to
through the
the rerticaL
ihould be pro-
the tiiuo of
the time of
o
h Bin n\
leu a velooitj
to render the
ng down the
Ans. 80°.
SXAMPLJB8.
41. Through what chord of a circle, drawn from tlie
bottom of the yertical diameter mu^t a body descend; so sm
to acquire a velocity equal to -th part of the velocity
acquired in falling down the vertical diameter ?
A»s. If denoi' aie angle between tho required chord
and the vertical diameter cos = - •
42. Find the incUnation, d, of the radius of a circle to
the vertical, such that a body running down will describe
the radius in the same time that another bodji n>quir&a to
full down the vertical diameter. Ans. 6 = 60°.
43. Find the inclination, 0, to the vertical of tho diam-
eter down which a body falling will describti the last half
in the same time as the vertical diamekvr.
8a/|j-4
/8 ""
Ah$. co« 6 =
44. Show that the times of descent down all the radii of
curvature of tho cycloid (Fig. 40, Calculus) ore equal; that
8r
is, the time down PQ is equal to the time down O'A =
y
46. Find the Inclination, ^, to the horizon of an inclined
plane, so that the time of descent of a particle down the
length may be » tiroes that down tho height of tho plane.
Ans. = sin
.1
46. Find the line of quickest descent from tho focus to
a parabola whose axis is verticul and vertex upwards, and
show that its length is equal to that of the latus rectum.
47. Find the lino of quickest descent from the foous of a
parabola to the curve when the axis is horisMirital.
"IMII
256
SXAMPLSa,
48. Find geometrically the line of quickest descent (1)
tfom a point within a circle to the circle : (2) from t\ circle
to a point without it.
49. Find gpometrically the straight line of longest
descent from a circle to a point without it, and which
lies below the circle.
60. A man six feet high walks in a straight line at the
rate of four miles an hour away from a street lump, the
height of which lis 10 foot : eupposing the man to starh
from the L*mp-i)ost, finu the rate at which the end of his
shadow travels, and also the rate at which the end of bis
shadow Be])arate8 from himself.
Ana. ;j! ,dow travels 10 miles an hour, and gains on
himseii (3 miles an hour.
'•i. Two bodies fall in the same time from two given
pointB in space in the same vertical down two straight
J-n. « drawn to any point of a surface ; show that the sur-
laco is an equilateral hyperboloid of revolution, having the
given jwintfl as vertices.
62. Find the form of a curve in a vertical plane, such
that if ht'iivy pa/tioles be simultaneously let full from ca^h
point Mt it 80 as to nlide freoly along the normal at that
point, they may all reach a given horizontal straight Hue at
the same instant.
53. Show that the time of quickest descent down a focal
chord of a parabola whoso axis is vertical is
m
where I is the latus rectum.
54. Particles slide from rest at the highest point of a
vertical circle down chords, and ore then allowed to moye
«<'■«'
t deacent (1)
from n circle
i of longest
t, and which
it line at the
•eet lump, the
man to Btarti
le end of his
le end of bis
md gains on
om two given
two straight
lit the sur-
huving the
! I;ine, such
1 from each
rmul at that
'uighk Hue at
lown a focal
fioint of a
to move
EXAMPLES.
freely ; show that the locns of the foci of their paths is a
circle of half the radius, and that all the paths bisect the
vertical radius.
55. If the particles slide down chords to the lowest point,
and be then suffered to move freely, the locus of the foci is
a cardioid.
66. Particles fall down diameters of a vertical circle ; the
locus of the foci of their subsequent paths is the circle.
ii:
CHAPTER II.
CURVILINEAR MOTION.
147 Remarlw on CnrTillnear Motion.— The mo-
tion, vrliioh waa coneidered in the laat chapter, wa8 that of
a partiola describing a rectilinear path. In this chapter V.\e
circnmstances of motion in which the path is ctirvilinear
will he considered. The conception and the definition of
Telocity and of acceleration which were given in Artfl. 134,
135, are evidently ad applicable to a particle describing a
curvilinear v" ha to one moving along a straight line;
and coDsequ i. Jie fo>-mulaB for velocity in Arts. 143, 143,
are ap{>licablc eiiher to rectilinear or to curvilinear motion.
In the last chapter the effects of the corapoeition and the
resolutiou of velocities were considered, when the path
taken by the particle in consequence of them was stmight ;
we have now to investigate the effects of velocities and of
accelerations in a more general way.
148. Composition cf Unifonn Velocit7 and Ac-
o«ler«tlon.—SupiwBO a body tends to move in one direc-
tion with a uniform velocity which would carry it from A
to B in one second, and also suV>ject to an
acceleration thai would carry it from A
to in one second ; then at the end of
the second the body will bo at D, the
opposite end of the diagonal of the par-
allelogram ABDC, just as if it had moved
from A to B and then from B to D in the second, bnt the
body will move in th^ nivve and not along the diagonal.
For, the body in its motion is making progress uniformly
iu the direction AB, at the same rate iw if it had no other
motion; and at the Raniw time it is being accelenUed in the
OOMPOBtTION OF AOCMLMMATIOKa,
N.
ion. — The mo-
■jN, waH that of
this chapter the
h is curvilinear
he definition of
m in Artfl. 134,
ilo describing a
I straight line ;
J Arts. 142, 143,
rilinear motion.
»itiou and the
rhen the path
was sti-aight ;
ocitit'B and of
ity and Ac-
in one direc-
rry it from A
cond, bnt the
the diagonal.
■em uniformly
had no othor
united in the
9M
direction AC, aa fast as if it had no other motion. Hence
the body will reach D as far from the tine AO as if it bad
moved over AB, and as far fi'om AB as if it had moved
over AC ; but since the velocity along AC is not uniform,
the spaces described in equal intervals of times will not be
equal along AC while they are equal along AB, and there-
fore the points a^, a,, o,, will not be in a straight line. lu
this case, therefore, the path is a curve.
149. Composition and Resolution of Accelera-
tions. — If a body i« Bnbjeot to two different accelerations
in different directions the sides of a parallelogram may bo
taken to represent the ComponeiU Accelerations, and
the diagonal will represent the Resultant Acceleration,
although the path of the body may be along some other
line.
Rem. — These results with those of Arts. 142, 148, may be
summed up in one general law: When a body tmds to
move with several different velocities ^'n different directions,
the body will be, at the end of any given time, at the same
point, as if it had moved with each velocity separately.
This is the fundamental law of the composition of veloci-
ties, and it shows that all problems which involve tenden-
cies to motion in different direotionH simultaneously, may
be treated as if those tendcDcios were successive.*
(jPg
If -js be the acceleration along the curve, and («, y, t)
be the place of the moving particle at the time, ij it is
evident that th(> component accelerations parallel to the
llf*' ^' 3^' ^^^°**'''°» ^^^^ ^y "» **" ««i
ares are
W8
have
df
a*'*
dP
«»;
di*
— «•;
and V<«(^ -I- o/ + «i^ i« the rumUani acceleration.
• flwi H^irairkii oo Newlon'pi M Ur, Art. I6E>.
IKHBHI
360
coMPoamoif or AccELSRATwna.
Also if a, P, y, be the angles which the direction of
motioQ makes with the axes, we have
^ = ^-,0080 = 0,;
d»2
The acceleration is not generally the complete resultant of the
threr component acceleratious, bat is eo only when the path ig a
9trai}i;'>^ line or the velocity ia lero. It is, however, the only part
(P*
of thew iHwultant which has any effect on the velocity, -r-^ is the
sum of the njaolved partB of the component accelerations in the direc-
tion of motion, as the following identical equation showa:
which follows immediately from (1) of Art. 187 by diil^rentlation.
Accelerations are therefore Bubject to the same laws of composition
and resoluti'm as velocities ; and ronseqaently the acceleration of the
particle along any lino is the sum uf the resolved |)art8 of the axial
accelerations along that line. Thus to find .^, the acceleration along •,
^ (' dj?
\,^ has to be multiplied by v-, which is the direction cosine of the
dC '^ ' d$
small arc ds. 'Hie other part of the resultant is at right biihUk to
this, and Its only effect iii to change the dircetion of the motion of the
l>oint. (S<><> Tait and Steele's Dynamics of a Partlcl.', also Thomson
and Tait's Nat Phil.)
The following are examples in wluch the preceding ex-
priHsio! > ar>' applii-d to onaei in which the laws of velocity
and of ac<^ler»tion ore given.
ONB.
the direction of
9 resultant of the
ten the path ia a
fer, the only part
locSty. ^ ia the
tiona in the direc
tiowa:
y diflerenttation.
of composition
cel«ratiun of the
•arts of the axial
eleration along »,
on coalne of the
right ati(^li« to
le motion of the
t), also Thomaon
)roceding cx-
vrs of velocity
EXAMPLES.
1. A particle moves so that the axial compononts of its
velocity vary as the corres wnding co-ordinates ; it is
required to find the equation of its path ; and the accel-
erations along the axes.
Here
dx
di
.— kx
dy
dt
= h>
.'. ^ = ^ = kdt;
' y
if (a, b) is the initial »faice of tb« particle,
is the equation of the path.
And the axial accelerations are
H-m
di»
dfl
2. A wheel rolls along a straight line with a nniform
velocity ; compare the velocity of a given point in the ilr-
cumference with that of the centre of the wheel.
Let the line along which the wheel rolls be the axis of x,
and let v be the velocity of its centre; then a point in its
circumference describes a cycloid, of which, the origin
being taken at its starting {mint, the equation is
a — a vera
a
(2ay - y)*;
iiiiii
It
!i-
i ,1
EXAMPLMa.
^ _ d^ __ _ <fe
y* (3a -"y)* ~ (2a)**
Bat
'^^ «' (iJay-S'*)* *
, da _ds dy /2y\i
which is the velocity of the point in the circumference of
the wheel. Thus the velocity of the highest point of the
wheel is twice as great as that of the centre, while the
point that is in contact with the straight line has no
velocity. (See Price's Aaal. Mech's., Vol, I, p. 416.)
3- ^^ ^ — ^'y> ^ — *•»> show that the path is an equi-
lateral hyperbola and that the axial components arc
4. A particle describes an eDiiwe so that the a^component
of its velocity is a constant, a ; find the y-com|)onent of \U
vt'l(Kjity and acceleration, and the time of describing the
ellipse.
Ijet the fif/iia«i/(n of the ellipse be
^i
\\
and lot {x, y) bo the position of the parHfiht at the time / j
then ^--. "-' ^V ^
dx . dy hh;
W^"' and J=-;^;
^ ^dy (h _
dt dx' dt ~
«» y'
which is the y-component of the velocity.
dy
jircumference of
lest point of the
entre, while the
?ht line has no
[,p.416.)
path is ao oqui-
lenta are
bo ^-Component
)m])onent of itti
doBcribiug the
at the timo t
wmmm
mmmm^.
MXAMPLB.
363
Also
9y _
dP ~
dx fly
hence the acceleration parallel to the axis of y varies
inversel; as the cube of the ordinate of the ellipse, and acts
towards the axis of x, as is shown by the negative sign.
The time of passing from the extremity of the minor
axis to that of the major axis i& fonnd by dividing a by a,
the constant velocity parallel to the axis of x, giving
, and the time of describing the whole ellipse is — •
If the orbit is a circle b = a, and the acceleration par-
allol to the axis of w is 3—
If the v'olooity parallel to the ^-axis is constant and equal
to P, then
di *» • V '
dh:
and the periodica time s=
4ft
a?
1 ; find
5. A particle descrihes the hyperbola -, — ^
(1) the acceleration parallel to the aiis of a; if the velocity
parallel to the axis of ^ is a constant, 0, and (3) find the
acceleration parallel to the y-axis if the velocity parallel to
the X-axis is a constuut a.
(1) Here we have
di -^'
and
<ii
^ = - . -•
ii!(iii««>t>m<V'iVtiiiiMii?i«« Mmifmimmn%wniSt>'>mi-iriysr,«isim>fMi'' •■ iUM
u
264
EXAMFhBS.
dx
dt
dx dy
dy ^
/3rt» y
which is the velocity parallel to the a;-axi&
Also
% dx
_ ^
hence the acceleration parallel to the ar-axis varies inversely
as the cube of the abscissa, and *he »-component of the
velocity is increasing.
(2) Here we have ,
dx
dt
= «;
dy_
dt ~
Z
dh, _
dt» ~
: —
ay
and
hence the acceleration parallel to the y-axis is negative and
the y-component of the velocity is decreasing.
6. A particle describes the parabola, «» + y* = a*, with
a constant velocity, c ; find the accelerations parallel to the
axes of X and y.
Here we have
da
and
d'X
dt
■A-
= c;
ds
y
* {x + y)^*
m
varies inversely
mponent of the
is negative and
'g-
- y^ = a*, with
B parallel to the
'^MM
m 0mmw
EXAMPLES.
.♦.
X
X + y X
i*X
y'
and
d^
~ dfi '
y _
X -\- y ~^ X
+ y'
differentiating
we get
d^
2{x + yr
^ _
d^ ~
(?{ax)^
■ 2 (a; + vY
865
7. A particle describes a parabola with such a varying
velocity that its projection on a line peipendicular to the
axis is a constant, v. Find the velocity and the accelera-
tion parallel to the axis.
Let the ec^uation of the parabola be
then
and
y» = %px;
dy
~dt
= v,
dx _ dx dy
dt ^ dy dt
vy.
which is the velocity parallel to x
Also
dfi
P
which shows that the particle is moving away from the
tangent to the curve at the vertex with a constant accelera-
tion.
,YS'
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PROJKCTIhM IN VACUO.
Hence ae the earth acts on partiolee near its surface with
a constant acceleration in vertical lines, if a particle is
projected with a velocity, r, in a hoiiiontal line it wU move
in a parabolic path.
160. Motiun of Protjeetiles in Vacua— If a particle
be projected in a direction obliqne to the bozixon it i«
called a Projectile, and the path which it describes is called
it« Trajectory. The ca»e which we ghali here considt-r is
that of a particle moving in vacuo under the action of
gravity; so that the problem is that of the motion of a
firojeciile in vacuo; and hence, as gravity does not »ii\ct
its horizontal velocity, it resoivea itself into the purely
kinematic problem of t particle moving so that its hori-
zontal acceleration i? and its vertical acceleration is the
constant, g, (Art 140).
151. TlM Patji of •
Projectile in Vacuo is a
Parabola.— Let the plane
in which the particle is pro-
jected be the piano of zy ;
let the axis of x bo horizon-
tal And the axis of y vertical
and positive upwards, tb3
origin being at the point of
projection; let the velocity
of projection =: v, and let the line of projection be inclined
at an angle a to the axis of x, so that t; co« «, and v bin «
are the rcbolved parts of the velocity of projection along the
axes of r and y. It is evident that the particle will oon-
tijue to move in the plane of xy, as it is projected in i*,
and is subject to no toroo which would tend to withdraw
it from that plane.
Lot {x, y) be the plaoe, P, of the partide at the time / ;
then the equations of motioa arj
'
its snrface with
if a particle is
line it vnll move
O. — If a particle
le bomon it ic
lescribes is called
here consider is
it the action of
the motion of a
' does not aij:>>ct
into the purely
30 that its hon-
ioeleration is the
^
c
3tion be inclined
M a, and V bin a
lection along the
)artiole will oon-
1 projected in i*,
tnd to withdraw
e at tb0 time / ;
PMOJBOTILg IN VACUO. 867
the aeceleration being negative since the y-component of
the velocity is decreasing.
The first and second integrals of these equations will
then be, taking the limits corresponding to < = / uid
< = 0,
_^„cos«;f =
g 5= vQMoti y
vmaa — gt\ (1)
vmuut-- yfl. (it)
Eqoationa (1) and ^2) give the coordinates of the particle
and its velocity parallel to either axis at any time, t.
Eliminating t between equations (3) we obtain
9)
y = *to»«-2jfe-
which is the equation of the traject<>ry, and shows that the
particle will move in a parabola.
153. TheParaawfeer; th« Range JB; tbaGhrMtaak
Height H; Height of the Directrix— Equation (3) of
Art. 1^1 may be written
. 2v* sin a cos a 2t/* coi^ «e
a^ _ x=s r !/,
or
ti* sin « COS o\' 2t)* cos* a i
9
o\' 2t)*co8'a/ »"8ln»«\ .,.
9
^
By comparing this with the equation of a parabola
referred to its vertex as origin, we find tor
., . , ... . V* sin a cos a ,a\
the absouM of the ^ «)rtex = —— } (2)
ft'
^^ I -Jill !
\
268 PROJSCTTLS IN VACUO.
the ordioate of the vertex = "° " ; (3i
the parameter (latus rectum) = ^^ — ^- (4)
And by transferring the origin to the vertex (1) becomes
. 2w» co8» a
«•= :; — y
(5)
i
which is the equation of a parabola with its axis vertical
and the vertex the highest point of the curve.
The distance, OB, between the point of projection and
the point where the projectile strilies the horizontal plane
is called the Range on the horizontal • plane, and is the
value of X when y = 0. Putting y = in (3) of Art. 161
and solving for x, we get
tlui horizontal range /2 =: OB = ^ "° ^" ; (C)
which is evident, also, geometrically, as OB = 200 ; that
is, the range is eqiml to twice the abscissa of the vertex.
It follows from (6) that the range is the greatest, for a
given velocity of projection, when a = 45°, in which case
the range = — •
Also it appears from (6) that the range is the same when
o is replaced by its complement ; that is, for the same
velocity of projection the range is the same for two differ-
ent angles that are complements of each other. If « = 46°
the two angles become identical, and the range is a
maximum.
OA is calltsd the greatett height, H, of the projectile, and
ia given by (3) which, when a = 45° becomee 7-. (7)
f
X
I* sin' g
''iff '
2 »» cos» g
(3)
(*)
tex (1) becomoB
(5)
h its axis vertical
irve.
of projection and
e horizontal plane
plane, and ia the
in (3) of Art. 151
w» sin 2a
(C)
OB = 200; that
of the vertex,
the greatest, for a
5°, in which case
is the same when
is, for the same
ne for two differ-
(ther. If « = 46°
the range ia a
;he projectile, and
4^' (y
Daes
vsLOcirr or tbs psojjbottlk.
I%e height of the directrix = CD
''iff'
— —^*Jt A. X ^'^ C OS* «
- 2g -^i ~g
269
(8)
Hence when « = 45° the focns of the parabola lies in
tlie horizontal line through the point of projection.
153. Tlie Velocity of the Partiole at any Point of
its Path. — Let V be the velocity at any point of its path,
then V^ = (|)V (I)*, or by (1) of Ait 151
= w* cos' « + (v* sin* a — 2v sin agt + g*P)
s=v> — 2gi/.
To acquire this veWity in falling from rest, the particle
y»
must have fallen through a height -^, (6) of Art. 140, or
itsoqual "
~-ff = M8-MPhy(8)
= PS.
Hence, the velocity at any point, P, on the curve is that
which the particle would acquire in falling freely in vacuo
down the vertical height 8P; that is, in falling from (ho
directrix to the curve; and the velocity of projection at O
is that which the particle would acquire in falling freely
through the height CD. The directrix of the parabola is
therefore determined by the velocity of projection, and is ai
u vortical distance above the point of projection equal to
that down which a particle flailing would have the velocity
of projection.
154. The Time of Flight, T, along a Horizontal
Plane.— Put y = in (a) of Art. 151, and solve for a-, the
i: ':.
•mta or ruosr op pnojsoi'tMs.
%ifi sin K COB CE
870
values of which are and
9
' But the horizon-
%v sin a
tal velocity ia v cos «. Hence the time of flight =
which varies as the sine of the inclination tx) the axis of z.
155. To FinA the Point at n^ch a PrOjoctlla will
Striko a Oiveii IseliiMd FbuM paacing tbrongh the
Point of Projeotion, and the Time of Flight— Let the
inclined plane make an angle /3 with the horizon ; it is
evident that we have only to eliminate y between y =: a: tan
/3 and (3) of Art. 151, which gives for the abscissa of the ''
point where the projectile meets the plane
_ 2t>' cos a sin (« — jB) ,
^coe/3 *
and the ordinate i«
x^ z=
(1)
_ a»» COB CT t a n /3 sin (« -- <3)
'* ~" g cos /3
Hence the time of flight
T =
Xf _ 2v sin (g — /3)
V cos « g cos /3
(2)
156. The Direotlon of Protjeotloii whiek giTea the
Oreateat Range on a GUven Plane. — The range on the
horizontal plane is
t^ sin 2a
which for a given value of v is greatest when « = 7 {^^
152).
The range on the inclined plane = a^j sec 0'
— 8*^ COB « «'P ( « "" <^) .
(1)
• But the horizon-
f flight = —^
D \a the axis of x.
a Prdjdctlle wlU
icing tisirongh the
ifriight— Letthe
the horizon ; it is
between y — x tan
' the abscissa of the
le
(1)
"&)
-J)
8
(2)
k ^hiek giTtts th«
—The range on the
when « = T (Art.
, Bee fi
(1)
ANOLX or MLBVATtOH Of PBOjaOTtLg.
271
To find the value of a which makes this a maximum, we
must equate to zero ita derivative with respect to «, which
gives
ooe (2a — /3) = 0;
and henoe
(9)
(8)
which is the angle which the direction of projection makes
wi*h the inclined plane when the range is a maximum ;
that is, the projection bisects the angle between the
inclined plane and the vertical.
In this case by subttitating in (1) the values of « and
(« — (3) as given in (2) and (3) and reducing, we get
the g«ate.t range =:^-j^j.
(*)
1S7. The «agl« fti Bltvatloii uo th«t tiui Partlcio
nwy pUM through a Given Point— From Art. 153,
there are two directions in which a particle may be pro^
jected so as to reach a given point; and they are equally
inclined to the direction of projection (« = -).
Let the given point lie in the plane which makes an
angle ft with the horizon, and Btijipoge its abscissa to be A ;
then we must have from (1) of Art. 155
2w»
^-^^cos«gin(a-/J)=A.
-f «' and «" be the two values of a which satisfy this
equation, we must have
ooi «' sin (•' - /I) = 008 «" sin («" ~ 0) ;
■"■•P|*|.
IIL
iT
27a EQUATION OF TBAJBCTOST, SECOND METHOD.
and therefore a" — ^ = 5 — «',
I
i
or
»"-ig + P)=i(i + /')-«'
(1)
But each member of (1) is the angle between one of the
directions of projection and the direction for the greatest
range [Art. 156, (2)]. Hence, as in Art 152, the two
directions of projection which enable the particle to pass
through a point in a given plane through the point of pro-
jection, are equally inclined to the direction of projection
for the greatest range along that plane. (See Tait and
Steele's Dynamics of a Particle, p. 89.)
158. Second Method of Finding the Equation of
the Trujectory.— By a somewhat simpler method than
that of Art. 151, we may find the equation of the path of
the projectile as the resultant of a uniform velocity and an
acceleration (Art 148).
Take the direction of projection (Fig. 78) as the axis of
X, and the vertical downwards from the point of projection
as the axis of y. Then (Art. 149, Rem.) the velocity, v,
due to the projection, will carry the particle, with uniform
motion, parallel to the axis of x, while at the same time, it
is carried with constant acceleration, g, parallel to the axis
of y. Hence at any time, t, the equations of motion along
the axes of x and y respectively are
X — vt,
That is, if the particle were moving with the velocity v,
alone, it would in the time t, arrive at Q , and if Jt were
then to move with the vertical acceleration g alone it would
in the same time arrive at P; therefore if the velocity v
f'
W MBTHOP.
— a.
(1)
etween one of the
u for the greatest
Lrt 152, the two
18 particle to pass
the point of pro-
ition of projection
I. (See Tait and
the Equation of
pier method than
tion of the path of
m velocity and an
78) as the axis of
point of projection
n.) the velocity, v,
;icle, with uniform
it the same time, it
parallel to the axis
18 of motion along
rith the velocity v,
Q , and if It were
on g alone it would
re if the velocity v
EXAMPLES.
S73
and the acceleration g are simuUaneotis, the particle will in
the time t arrive at P (Art. 149, Rem).
Eliminating t we have
g "
which is the equation of a parabola referred to a diameter
and the tangent at its vertex. The distance of the origin
from the directrix, being J^th of the coefficient of y, is
V*
H-, asm Art. 163, (8).
BXAMFLES.
1. From the top of a tower two particles are projected at
angles a and /3 to the horizon with the same velocity, v, and
both strike the horizontal plane passing tlirough the bot-
tom of the tower at the same point; find the height of
the tower.
Let A = the height of the tower; v = the velocity of
projection ; then if the particles are projected from the
edge of the top of the tower, and x is the distance from the
bottom of the tower to the point where they strike the
horizontal plane we have from (3) of Art. 151
- A = a; tan b - 1^ (1 + tan» a),
A = X tau /3 - g (1 + tan» /3),
(1)
(2)
by subtraction
X =
2t»»
Sw* cos « cos i9 ^
g (tan a + tanT^ "" ^sin (a + /3) '
which in (1) or (2) gives
k — ^^ ^'^^ " "*^" ^ ^ ^ ^^ "*" ^^
■■i
AlWi.SWl.i'Sirl
^■?su^i«< eiis^^T^ss^s^Sie^E^aSj^^Si^^
wmm
H
i
rf^
874
vuLooirr of mscuARoE of skslls.
2. Piurticles are projected with a giTen veiocitj in all
lines in a Tertioal plane £rom the point 0; it ia required to
find the loona of their highest points.
Let {x, y) be the highest point ; then from (2) and (3) of
Art 162, we have
«* sin o cos rt
g- ;
»» Bin» a
therefore sin* a = -^, and coa* « = —--•
1^ 2v^y
g '
ff
Adding 4^ + a:*
which is the equation of an ellipse, wliose major axis =
and t,he minor axis = „-; and the origin is at the extremity
of the minor axis.
8. Find the angle of projection, a, so that the area con-
tained between the path of the projectile and the hori-
zontal line may be a muximnm, and find the value of the
maximum area.
Ans. tt = 60° and Max. Area
(3)*.
4. Find the ratio of the areas A, and A, of the two
parabolas described by projectiles whose horizontal ranges
are the same, and the angles of projection are therefore
complements of each other. ^ A, , .
Ana. -jT^ = tan' a.
At
159. Velocity of Dischaxge of Balls and Shells
from the Mouth of a GNul— As the result of numerous
SHSLLS.
n velocity in all
it ia required to
■om (2) and (3) of
major axis = - ;
8 at the extremity
hat the area con-
ic and the hori-
i the value of the
rea
»ff
. (3)*-
1 Aj of the two
horizontal ranges
ion are therefore
-r^ = tan' a.
A,
ills and Shells
lult of numerous
m
ANGULAR VELOCITY.
2T5
experiments made at Wcolwich, the following formula was
regarded as a correct expression fov the velocity of balls and
shells, on quitting the gun, and fired with moderate
charges of powder, from the pieces of ordnance commonly
used for military purposes :
t; = 1600>^^,
where v is the velocity in feet per second, P the weight of
the charge of powder, and W the weight of the ball.
For the investigation of the path of a projectile in the
atmosphere, see Chap. I of Kinetics.
160. Angular Velocity, and Angnlar Aocelera<
tion. — Hitherto the method of resolving velocities and
wcelerations along two rectangular axes has been employed.
It remains for us to investigate the kinematics of a particle
describing a curvilinear path, from another point of view
and in relation to another system of reference. Before we
consider velocities and accelerations in reference to a
system of polar co-ordinates, it is necessary to enquire into
a mode of measuring the angular velocity of a particle.
Angular Velocity way be defined as the rate of angular
motion. Thus let (r, 0) be the position of the point P, and
suppose that the radius vector has revolved uniformly
through the angle 9 in the time t, then denoting the
angular velocity by w, we shall have, ae in linear velocity
(Art. 8)
6
a =
t
If however the radius vector doe-a not
revolve uniformly through the angle 6
we may always regard it as revolving
uniformly through the angle dd in the
infinitesimal of time dt ; hence we shall
have as the proper value of w,
•"aWBWaffihffMl
276
SXAMPLSa.
i !i
dB
dt
(1)
Hence, whether the angular velocity bo uniform or
variable, it is the ratio of the anglp described by the mdius
vector in a given time to the time in which it is described ;
thus the increase of the angle, in angular velocity, takes
the place of the increase of the distance frcm a fixed point,
in linear velocity, (Art 8).
Angular Acceleration is the rat- of tncretse of angular
velocity : it is a velocity increment, and is measnred in the
same way as linear accelerafion (Art. 10). Thus, whether
the angular acceleration is uniform or variable, it may
always be regarded as uniform during the infinitesimal of
time dt in which time the increment of the velocity will be
du. Hence denoting the angular acceleration at any time,
t, by (p, we have
. dot d /de\, .,.
d£*'
(2)
and thus, whether the increase of angular velocity is
uniform or variable, the angular acceleration is the increase
of angular velocity in a unit of time.
The following examples are illustrations of the preceding
mode of estimating velocities and accelerations.
EXAMPLES.
1. If a particle is placed on the icvolving line at the
distance r from the origin, and the line revolves with a
uniform angular velocity, w, the relation between the linear
velocity of the particle a&d the angular velocity may thus
be found.
•^mmm"-
BgWW i WgllJliJ
EXAMPLES.
m
(1)
bo uniform or
ed by the rudius
h it is described ;
IT velocity, takes
;m a fixed point,
rease of angular
measured in the
Thus, whether
variable, it may
I infinitesimal of
! velocity will be
ion at any time,
(2)
:nlar velocity is
•n is the increase
of the preceding
ons.
ring line at the
revolves with a
itween the linear
locity may thus
(
Let dd be the angle through which the radius revolves in
the time dt, and let ds be the path described by the particle,
so that ds = rdO ;
then
<l8
de
dt = ''di = "'"'
so that the linear velocity varies as the angular velocity and
the length of the radius jointly.
2. If the angular acceleration is a constant, as ^ ; then
from (2) we have
d*e ^
di»='^'
••• Ji = 'pi ^ <->of
and
e = i<t>p + ..,# + e„
where «„ and 6^ are the initial values of w and 9.
Hence if a line revolves from rest with a constant angular
acceleration, >. e have
and the angle described by it varies as the square of the
time.
3. If a particle revolves in a circle uniformly, and its
place is continually projected on a given diameter, the
linear acceleration along that diameter varies directly as
the distance of the projected place from the centre.
Let 6) be the constant angular velocity, 6 the angle
between the fixed diameter and the radius drawn from the
centre to its place at the time t, x the distance of this
projected place from the centre. Then, calling a the i
radius of the circle, we have
it = « cos d.
St^w"-
s t.
S78
HnKEnRRSMMSNfT^iB!!
SXAMPLB8.
TT = —asm 6^ = — ousinO;
dt
dh: M
which proved the theorem.
4. If the angular acceleration varies afi the angb
generated from a given fixed lipe, and is negative, find the
angle.
Here the equation which expresses the motion is of the
form
1*0.
Calling a the initial value of we find for the result
= a COB kt.
6. If a particle revolves in a circle with a uniform
velocity, ehow tliat its angular velocity about any point in
the circumference is also uniform, and equal to one-half of
what it is about the centre.
At present this is sufficient for the general explanation
of angular velocity and angular accoieration. We shall
return to the subject in Chap. 7, Fart IIL, when we treat
of the motion of i-igid bodies.
161. The Component Aooelerationa, at any Jlnstant,
Jbiaajg, and Perpendioolar to the RadiiM Voctor.—
Let {r, e) (Fig. 79) be the place of the moving particle, i\
at the time t, (x, y) being its place referred to a system of
rectangular axes having the same origin, and the a;«xis
coiucident with the initial line. Then
a; s r CO8 0; y = r sin 0;
11
WM
mmm
sin 0;
60 M the aiigb
negative, find the
motion is of the
for the reault
with a uniform
tbout any point in
ual to one-half of
metal explanation
ation. We shall
[L, when we treat
I, at any iinstant,
adiua Vector.—
oving particle, P,
red to a syetem of
Q, and the x-axia
(1)
SAUICAL AND TRANSVERSAL ACCBL8RATI0N8. 279
m
therefore tj- rs ^j co8 ff — ." ?tn tf tt ;
and
_ rd^r _
'd^~\M ^\di
d^i
-»-[»l5 + 'S]-»<')
cPx rtfir
Similarly
which are the accelerations i arallel to the axes of x and y.
Resolving these along the r«diu8 vector by multiplying (3)
and (4) by cos 6 and sin 8 respectively, since accelen-ationa
may lie resolved and compounded along any line the same
as velocities (Art. 149), and adding, we have
^cosfl^r J8ma = -^-^
m^
(«)
which is the acceleration along the radius vector. *
Multijil>iug (3) and (4) by sir. 6 and cos 6 respectively,
and subtracting the former from the latter, we get
OOS0
d*x .. ^dr dd m
-dfi''''^ = ^di'dt-^*^Jfi
_1 rf/.d<?\.
-idiVJtr
(6)
which is the acceleration perpendicular to the radius vector.\
163. The Component Aceelerationa, at any in-
stant, Along, and Perpendicular to the TanA^nu -
liCt (jj, y) (Pig. 79) he the place of the moving particle, P,
at the time t, and « the length of the arc described during
• Bomettntw called tht roMal ac»eltrtUiim.
t BomatfaBM oalkid fft* t r m memal 4UMkfalHm.
mi'
'•■i
hi-
^1
' , «■
»0
rAifOJSj(rr/aji acceleratiqii.
that time. Then the accelerations along the axes of a? and y
(fix , <Pv da: du
•"■^ wfrt **"" j^ ; and the direction cosines* are ~r- and -?•
"* B" as as
To find the acceleration along the tangent we must multi-
ply these axial accelerations by t- and ™, respectively, and
t^d. Thus the tangential acceleration, T, is
cPx dx
diP'ds
T = ^.~ +
d»y dy
d'fl ' ds'
(1)
Sinco rfs» = (ie» -{ rfys, therefore, by differentiation we
have
ds d^s = dx dh) -^ djf dhf ;
and dividing by ds dP we get
d's d*x dx
dfl
which in (1) gives
d»«
T =
dt*'
(2)
for the acceleration along the tangent.
Similarly we have for the normal acceleration, N,
_^ ^_dh: dy
dfl ' ds ~ di'i ' ds
1 <i8*
~ P ■ d/» ' ^'^^ ^** *' P* ^**' ^*lcnlu8),
where p is the radius of curvature ;
* CosineaofUwMiglw wUoliUuiUnKWitniakMwItbUMUMofaudy.
•| II • im miwi
TOS.
he axes of x and y
^ dx _, dti
!«* are ~j- and -#•
as ds
t we must multi-
, respectively, and
', is
(1)
differentiation we
(2)
ration, N,
it*
. 144, dalcalns).
Um um of at wad y.
mmmmma
NORMAL ACOBLBRATtON.
V*
" = ?•
881
(8)
if V is the velocity of the particle at the point («, y).
Hence at any point, P, of the trajectory, if the accelera-
tion is resolved along the tangent to the curve at P and
along the normal, the accelerations along the two lines arc
respectively
d^
and -•
9
163. When the Acceleration Z''erpendionlar to
the Rad5,ii8 Vector is sero. — Then from (6) of Art. 161
we have
r* T- = constant = h suppose ;
dd
h
•'• dt -
r*
>
dr
dr dd
h
dr
dt
~ d)' dt
•—
r>
de*
and
'•* dt* ~ r*' de» VW'
which in (6) of Art. 161 gives
the acceleration along the radius vector
'-~t*dm ^r»\dsi} f«'
an expression which is independent of t.
(l)
This may be put into a more convenient form as follows:
I«t f = - ; then
dr
de
du
de'
«mm.
28!^
CONSTANT ANeULAB VBLOCITT.
tPr
— 1 ^ J. ^ l^^
which in {1) and reducing, gives
the acceleration along the radias vector
(2)
From these two forranlse the 'iaw of acceleration along
the radias vector may be deduced when the curve is given,
and the curve may be deduced when the law of accelera-
tion along the radius vector is given. Examples of these
processes will be given in Chap. (2), Part III.
164. When th« Angular Velocity is Constant-
Let the angular velocity be constant = « suppose. Then
= w;
therefore from (5) of Art 161
the acceleration along the radius vector
d*r ,
The acceleration perpendicular to the radius vector
dr
= 2u
dr
(1)
(2)
and both of these are independent of ft
The following example is an illustration of these
formulte :
A particle describes a path r7ith a constant angular
velocity, and without acceleration along the radius vector ;
find (1) the equation of the path, and (2) the acceleration
perpendicular to the radius vector.
"^*»-,
mmt
acirr.
a
(2)
acceleration along
;he curve is given,
law of accelera-
ilxamples of these
III.
is Constant. —
► suppose. Then
adins vector
(1)
(2)
itration of these
constant angular
the radius vector ;
I the acceleration
j( Ba W tg ! iKWi'tn i |»JIWI | i|IW i W IL IWIIMI I II»^
COMITAA-T ANOULAR VELOCITY,
(1) From (1) we have, from the conditions of the
question.
dp
— w»r = 0.
Integrating ve have
dr'
if r = a when jj = 0.
at
Therefore
dr
— udt;
if r = o when t = 0,
r = ^ie^ + <r-0.
(8)
d6
Also, as -^f r= u, therefore d = W, if fl = when / = 0.
Substituting this value of ut, we have,
which is the path described by the particle.
(*y
(i) Let Q be the required acceleration perpendicular to
the radius veotor;, then from (2) wo have
I
== a«i («-« _ «--*), from (8)
Bte—
984
KXAMPLES.
aw* (e — «-•)
2a*»(r» — a»)*;
(6)
which is the acceleration perpendicnlar to the radias
vector.
The preceding discussion of Kinematics is safBcicnt for
this work. There are various other problems which might
bo studied as Kinematic questions, and inserted here ; but
we prefer to treat them from a Kinetic point of view.
For the investigation of the kinematics of a particle
describing a curvilinear path in space, see Price's Anal.
Mech's, Vol. I, p. 430, also Tait and Steele's Dynamics of
a Particle, p. 12.
EXAMPLES.
1. A particle describes the hyperbola, a;y = if ; find (1)
the acceleration parallel to the axis of z if the velocity
parallel to the axis of ^ is a constivnt, /3, and (2) find the
acceleration parallel to the axis of y if the velocity parallel
to the axis of x is a constant, a.
Ans. (l)^a*;(2)^V.
2. A particle describes the parabola, y* = ^ax ; find
the acceleration parallel to the axis of y if the velocity
parallel to the axis of a; is a constant, o. ^ . . 4a*a'
Ans.
f
3. A particle describes the logarithmic curve, y = of;
find (1) the z-component of the acceleration if the y-com-
ponent of the velocity is a constant, /3, and (2) find the
y-component of the acceleration if the x-component of the
velocity is a constant, «.
<^>-^^ga'<*^)«*^^''8''>'y-
Ans.
JBXAMPLBA
386
)
^; (5)
*r to the radins
M is Bnfficient for
lema which might
[iserted here; but
>int of view,
.ties of a particle
see Price's Anal,
ele's Dynamics of
y = /fc»; find (1)
X if the velocity
and (2) find the
} velocity parallel
'*•; (2) ^V.
f = iax; find
f if the velocity
Ans. --
carve, y :=: of;
3n if the y-com-
»nd (2) find the
imponent of the
!)a» (log a)«y.
4. A particle describes the cycloid, the starting point
being the origin; find (1) the a^component of the accel-
eration if the y-component of the velocity is (i, and (2) find
the y-component of the acceleration if the a;-componont of
the velocity is o. ^^ . ^^^ ffl«y . ^jj) _ ^.
Ata, (1)
(2«y-y)*' ^
5. A particle describes a catenary, y = s(«*' + e "1;
find (1) the a;-component of the acceleration if the y-com-
ponent of the velocity is /3, and (2) find the y-component
of the acceleration if the aMsomponent of the velocity is «.
Am. (1)
6. Determine how long a particle takes in moving from
the point of projection to the farther end of the latus
rectum. ^ ** / • , \
Ans. - (sm a + cos a).
9
7. A gun was fired at an elevation of 60°; the ball
struck the ground at the distance of 2449 ft. ; find (1) the
velocity with which it left the gun and (2) the time of
flight. (y = 32i).
Ana. (1) 200 ft. per sec; (2) 19.05 sees.
8. A ball fired with velocity u at an inclination a to the
horizon, just clears a vertical wall which subtends an angle,
(i, at the point of projection; determine the instant at
which the ball just clears the wall.
« sin a — \gt _
Ana.
u cos a
im(3.
9. In the preceding example determine the horizontal
distance between the foot of the wall and the point where
the ball strikes the ground.
Ana. — co8» « tan /3.
8
vT
386
BXAMPLSa,
10. At the diatanco of a quarter of a mile from the bot-
tom of a oUff, which ia 120 ft. high, a shot ia to be fired
which ahall just clear the clifiF, and paaa over it horizon-
tally ; find the angle, a, and velocity of projection, v.
Ant. a = 10° 18'; t> = 490 ft. per aec.
11. When the angle of elevation ia 40° the range is
2449 ft. ; find the range when the elevation is 29|°.
Ana. 2131.6 ft.
12. A body is projected horizontally with a velocity of
4 ft. per see. ; find the latua rectum of the parabola de-
Bcribed, (g = 32). Ana. 1 foot.
13. A body projected from the top of a tower at an angle
of 45° above the horizontal direction, fell in 5 seca. at a
distance from the bottom of the tower equal to its altitude ;
find the altitude in feet, (g = 32). Ana. 200 feet
14. A ball ia fired up a hill whoae inclination ia 15°;
the inclination of the piece is 45°, and the velocity of pro-
jection is 500 ft. per sec. ; find the time of flight before
it strikes the hill, and the distance of the place where it
falls from the point of projection.*
Ana. T = 16.17 aecs.; R = 1.121 miles.
15. On a deacending plane whose inclination is 12°, a
ball fired from the top hits the plane at a distance of two
miles and a half, the elevation of the piece ia 42° ; find the
velocity of projection. Ana. v = 579.74 ft. per aec.
16. A body ia projected at an inclination a to the hori-
zon; determine when the motion is perpendicular to a
plane which is inclined at an angle /3 to the horizon.
u sin a
Atu.
u COB a
^ = ±oot|9.
* The nnge on the Inclined plane.
mile from the bot-
Bbot is to be fired
)8 over it horiaon-
>rojectiun, v.
490 ft. per sec.
40" the range is
m is 29|°.
Ans. 2131.5 ft,
ffith a velocity of
'. the parabola de-
Ans. 1 foot.
tower at an angle
ill in 5 sees, at a
lal to its altitude ;
Ans. 800 feet.
iclination is 15°;
3 velocity of pro-
5 of flight before
e place where it
= 1.121 miles.
ination is 12°, a
. distance of two
is 42° ; find the
L74 ft. per sec.
1 a to the hori-
rpendicutar to a
e horizon.
^' = ± oot /3.
nHMmii
MXAMPLS&
287
17. Calculate the maximum range, and time of flight,
on a descending plane, the angle of depression of which w
15°, the velocity of \ ejection being 1000 ft, per sec.
Ana. Max. range = 7.98 miles ; T = 51.34 sec.
18. With what velocity doe^ the ball strike the plane in
the last example ? Ans. V = 1303 feet.
19. If a ship is moving hori«on tally with a velocity
= 3</, and a body is let fall from the top of the mast, find
its velocity, V, and direction, d, after 4 sees.
Ans. Y — 5g; e = tan-» f .
20. A body is projected horizontally from the top of a
tower, with the velocity gained in falling down a space
eqnal to the height of the tower ; at what distance from
the base of the tower will it strike the ground ?
Ana. R = twice the height of the tower.
21. Find the velocity and cime of flight of a body pro-
jected from one extremity of the base of an equilateral
triangle, and in the direction of the side adjacent to that
extremity, to pass through the other extremity of the base.
Ans.
» = \/f>
T = '^?'*
3 ' - 9
2'4. Given the velocity of sound, V; find the horizontal
range, when a ball, at a given angle of elevation, «, is so
projected towards a person that the ball and sound of the
discharge reach him at the same instant
Ans. — tan «.
ff
23. A body is projected horizontally with a velocity of
4g from a point whose height above the ground is 16jr ; find
the direction of moHon, d, (1) when it has fallen half-way
to the ground, and (2) when half the whole time of falling
has elapsed. ^^
(l)e = 46°; (2) e = tan-» -^-=.
288
EXAMPLBS.
I,-.
I; :•
24. Particles are projected with a given velocity, v, in
all lines in a vertical plane from the point ; find the locus
of them at a given time, /.
Ans. a:* + (y + io^Y — ^^> which is the equation of a
circle whose radius is vt and whoso centre is on the axis of
y at a distance ^P below the origin.
25. How much powder will throw a 13-inch shell*
40G0 ft. on an inclined plane whose angle of elevation is
10° 40' ; the elevation of the mortar being 35".
Ans. Charge = 4.67 lbs.
26. A projectile is discharged in a horizontal direction,
with a velocity of 450 ft. per sec, from the summit of a
conical hill, the vertical angle of which is 120° ; at what
distance down the hillside will the projectile fall, and what
will be the time of flight?
Ans. Distance = 2812.5 yards; Time = 16.23 sees.
27. A gun is placed at a distance of 500 ft. from the base
of a cliff which is 290 ft. high ; on the edge of the cliff
there is built the wall of a castle 60 ft. high ; And the
elevation, «, of the gun, and the velocity of discharge, v,
in order that the ball may graze the top of the castle wall,
and fall 120 ft. inside of it
Ans. « = 63° 19' ; v = 165 ft. per sec.
28. A piece of ordnance burst when 50 yards from ^-^
wall 14 ft. high, and a fragment of it, originally in con-
tact with the groand, after grazing the wall, fell 6 ft.
beyond it on the opposite side ; find how high it rose in
the air. Ans. 94 ft.
* The weight of • la-inch ihell U IM Ibe.
Ire :
m
mi
iven velocity, v, in
ifcO; find the locus
! the equation of a
e is on the tuis of
a 13-inch shell*
gle of elevation is
g35".
irge = 4.67 lbs.
rizontal direction,
the summit of a
is 120° ; at what
tile fall, and what
le = 16.23 sees.
[) ft from the base
edge of the cliff
t high; find the
7 of discharge, v,
)f the castle wall,
165 ft. per sec.
50 yards from ^
originally in con-
B wall, fell 6 ft.
»w high it rose in
An8. 94 ft.
Iba.
PART III.
KINETICS (MOTION AND FORCE).
CHAPTER I.
LAWS OF MOTION— MOTION UNDER THE ACTION OF
A VARIABLE FORCE— MOTION IN A RESISTING
MEDIUM.
165. Deflnitioiui. — Kinetics is that branch of Dynamics
which treats of the motion of bodies under tfie action of
forces.
In Part I, forces were considered with reference to the
pressures which they produced upon bodies at rest (Art.
16), i. e., bodies under the action of two or more forces
in equilibrium (Art 26). In Part II we considered the
purely geometric properties of the motion of a point or
particle without any reference to the causes producing it,
or the properties of the thing moved. We are now to
consider motion with reference to the causes which produce
it, and the things in which it is produced.
The student must here review (Chapter I, Part I, and obtain clear
conceptions of MomeiUutn, Aeeelsrattini tf Mimuintum, and the KinHie
meamre of Fane (Arts. 18. 14. 19, and 90), as this is necessary to a fall
understanding of the fundamental laws of motion, on the trutli of
which all our succeeding investigations are founded.
166. Newton's Laws of Motioa— The fundamental
13
1'
\
■1.
t.
mm
290
A^rrojvs LAWS or motion.
:• I
:
principles in aooordancfi with wliich motion takes pliicc are
embodied in three statements, generally known as New/on^
Laws of Motion. These laws must be considered as resting
on convictions drawn from observation and experiment,
and not on intuitive perception.* The laws are the fol-
lowing:
Law \.— Every body continues in its state of rest
or of uniform motion in a straight line, except in
so far as it is compelled hy force to change that
state.
Law II.— Change of motion is proportional to the
force applied, and takes place in the direction of
the straight line in which the force acts.
Law IIL— 2b every action there is always an
equal and contrary reaction; or, the mutual ac-
tions of any two bodies are always equal and oppo-
sitely directed.
167. Remarks on Law l— Uw i snppUeB m with »
definition of force. It indicates that force is that which tends to
change a body's state of rest or of uniform motion in a straight line ;
for if a body does not continue in its state of rest or of uniform mo-
tion in a straight line it mupt bo under the action of force.
A body has no power to change its own state as to rest or motion ;
when it is at rest, it has no power of putting itself in motion ; when
in motion it hr« no power of increasing or diminishing its velocity.
Matter is inert (Art. 8). If it is at rert, it will mnain at rest ; if it is
moving with a given velocity along a rectilinear path, it will continue
to move with that velocity along that path. It Is alike natural to
mat(*r to be at rest or in motion. Whenever, therefore^ a body's
state is changed either from rest to motion, or fr<mi motion to rest,
or when its velocity is Increased or diminished, that change is due to
some external cause. This cause is called force (Art. 16) ; and the
word /WW is used in Kinetics in this meaning only.
* Thonuoa and Tait'i Nat. FhU., p. Stt.
riON.
ion takes place are
known as Newfon'f
nsidcred as resting
I and exporiment,
laws are the fol-
its state of rest
' line, except in
to change that
portional to the
•he direction of
; acta.
is always an
he mutual ac-
'■qual and oppo-
snpplies as with k
bat which tends to
I in a straight line ;
It ot of uniform rao-
I of force.
a to rest or motion ;
)lf in motion ; when
tishing its velocity.
oain at rest ; if it is
tath, it will continue
is alike natural to
therefore, a body's
imi motion to rest,
at cban(re is due to
(Art. 16) ; and the
J.
a.
itnMAiiKa oy law it.
291
168. Remark* on Law IL — Law ll asserts that it any
force generates motion, a double force will generate douljle motion,
nnd so on, whether applied simultaneously or successively, instan
tiineously or gradually. And this motion, if the body was movin|<
iK'i'orehand, is either added to the previous motion if directly consj ir
ing with it, or is -^ubtnwrted if directly opposed ; or is geometrit-ally
compounded with it according to the principles already explained.
(Art. 29), if the line of previous motion and the direction of tho forco
are inclined to each other at an angle. The term motio;, here means
quantity of motion, and the phrase dutnge of motion here means mtc
of change of quantUy of motion (Art. 14). If tho force be finite It will
require a finite time to produce a iiensible change ot motion, and the
cliange of momentum produced by it will depend upon the time dur-
ing which it acts. The change of motion must then be understood to
be the change of momentum produced per unit of time, or the rate
of change of momentum, or acceleration of momentum, which agrees
wHh the principles already explained (Arts. 14 and 20). In this law
uotlilng is said about tho actual motion of the body before it was
acted on by tho force ; it is only the change of motion that concerns
us. The same force will produce precisely the same change of mo-
tion in a body ; whether the body be at rest, or in motion with any
velocity whatever.
Since, when several forces act at once or. ^. particle either
at rest or in motion, the second law of motion is true for
every one of these forces, it follows that each must have the
same effect, in so far as the change of motion produced by
it is concerned, as if ti were the only force in action.
Hence the assertion of the second law may be put in the
following form :
When any number of forces act simultaneously on a body,
whether at rest or in motion in any direction, each force pro-
duces in the body the same change of motion as if it alone
had acted oh the body at rest.
It follows from this view of the law that all problems
vhich involve forces acting simultaneously may bo treated
as if the forces acted successively.
The operations of thia law have already been considered in Kine-
IM
292
HEM ARKS ON LAW U.
I : •
iv
TDatics (Art. 14ft) ; hut motion there was underotood to mean wHocUy
only, since the mass of the body was not considered. Thia law in-
cludes, therefore, the law of the composition of velocities already
referred to (Art. 20 j. Another consequence of the law is the follow-
ing : Since forces are measured by the changes of motion they i ro-
duce, and thirT directions assigned by the directions in which these
chnnges are produced, ani since the changes of motion of one and the
same body are in the directions of, and proportional to, the changes
of velocity, therefore a single force, measured by the resultant change
cf velocity, and in its direction, will bo the equivalent of any uuml)er
of simultaneously acting forcus.
Hence,
Hie remltant of any number of concurring forces is to be
found by the same geometric protxds as the reauUant of any
number of simultaneous velocities, and conversely.
From this follows at once the Polygon of Velocities anrl
thp Parallelopiped of Velocities from the Poly»joii and
Parallelopipcd of Forees, as was described in Art. 142.
This law also ^ves us the meniw of measuring force, and also of
measuring tho mats of a body : for the actions of different forces upon
the bame body for oijual times, evidently produce changes of velocity
which ore proporliuitai to the forces. Also, if equal forces act on dif
f.jrent Ixxlios for e<iual times, the rhanjres of velocity ])rofluced must
Iw invfi'tfly as the mat»exo( the bodies. Again, if different Ijodies,
each acted on by a force, acquire In the same time the same changes
of velocity, the forces must be proportional to the musses of the
bodies. This means of measuring force is practically the same us
tliot already deduced by abstract reasoning (Arts. 19 and 20).
It appears from thie law, tliat every theorem of Kine-
matics ctmnceted with uccelerat'on has its counterpart in
Kinetics. Thi's, the meauure of acceleration or velocity
increment, (Art. 9), which wua discussed in Chap. I (Arts.
9 and 10), and in Kinematics (A: 135), and which is
(Ps
denoted by / or its equal ^-j, is also the effect and the
measure of force ; therefore all the results of the equation
itood to mean ndocUy
lidered, Thia law in-
of velocities already
the law is the follow -
s of motion tliey jro-
Btions in which tlicso
motion of one and tlif;
ional to, the changes
' the resultant change
vaieat of any number
ring forces is to bo
he resultant of any
nversely.
n of Velocities anfl
the Poly»joii and
I in Art. 142.
ng fiirce, and also of
' difibrent forces upon
3 clmngeR of velocity
iual forces act on dif-
iocity ])ro(luced mast
in, if different IwxlirB,
me the same fhangos
n the musses of the
Helically the same us
. 19 and 20).
tlieorom of Kine-
it« counter})art in
ration or velocity
in Chap. I (Arts.
6), and which is
lie effect and the
J of the equation
H
RBMARKS ON LAW IT.
293
d*8
(1)
'■f
1
its various forms, and the remarks which have been made
on it, are applicable to it when / is the accelerating force.
Thus, (Art. 162), we see that the force, under which a
particle describes any curve, may bo resolved into two
components, one in the tangent to the curve, the other
towards the centre of curvature ; their magnitudes being
tlie acceleration of momentum, tiud the product of the
momentum into the angular velocity about the centre of
curvature, respectively. In the case of uniform motion,
the first of these vanishes, or the whole force is perpen-
dicular to the direction of motion. When there is no force
perpendicular to the direction of motion, there is no curva-
ture, or the path is a straight line.
Henco if we suppose the particle of mass wi to be at the
iwint {x, y, z), and resolve the forces acting on it into the
three rectangular components, X, Y, Z, wo have
m
dt*
m
^l = Y:m^ = Z.
dh
dC^
dl^
(2)
In several of the chapters those equations will be sim-
plified by assuming unity as the maas of the moving
particle. When this Cannot be done, it is sometimes con-
venient to assume X, Y, Z, as the component forces on the
unit mass, and (8) becomea
m
d*.t
dfi
mX, etc.
It will be ob-
from which m may of course be omitted,
served that an equation such as
^, = X
may be interpreted either as Kino.ical or Kinematical ; if
.;•!« ■
...ritfM
IMM
2H
aSMASKS ON LAW W.
%/
the former, the uuit of mass must be understood as a fac-
tor on the left-hand ide, in which case X is the a;-eom-
ponent, for the unit oi mass, of the whole force exerted on
the moving body.
The fl.«t two lawB, h»ve, therefore, furniahed us with r definition
and a meaturtof forte; and they aleo tsho* i ow to oompoaud, aod
therefore bow to rwjolve, forcee; and also how to investigate the
conditions of equilibrium or motion of a single particle subjected to
given forces.
169. RemarkB on Law IH— According to Law III. if one
body presses or draws another, it is pressed or drawn by this other
with an wjiial force In the opposite direction (Art. 10). A horse
Vjwing a boat on a canal, is pulled backwards by a force equal to that
which he impresses ou tho towiogro[je fcrwarda. If one body strikes
another body and changes tho rootiou of the other body, its own
motion will bo changed in an equal quantity and in the opposite
direclioa ; for at each instant during the impact the bodies exort un
each other cqur 1 and opposite pressures, and ihe momentum that one
body loses is jMjual to that which the other gains.
Tlie earth atiracts a ikUing pebble with a certain foiw, while the
pebble attracu the earth with an equal force. The result is thot
while the pebble moves towards tho earth on «•< ..iint of ito attrac-
tion, th«? earth aleo moves towards the pebble under the influence o'
tho nttruction of the latter ; but tlie mass of the earth being enor-
mously greater than that of the pebble whUe the forces on the two
arising from their mutual attractions are equal, tho motion produced
thereby in the earth is almost incomparably leas than that pnxluced
in the pebble, and is consoquintly insensible.
It follows that the sum of the qusni.Mes of motion parallel to any
fixed direction of the particles of any syrtem influencing one another
in any po.«ible way, remains unchanged by their mutual action.
Therefore if the centre of gravity of any system of mutually
In'luenciug iiarticles is in motion, it continues movii.g unlfomi'v In a
«truiglit lino, unless In s > far as the direction or velocity of i» motion
is chonged by forces bctwet-n the psivicles and some other matter not
Moiiging to the gyHem : also the centre of gn-vity of any system of
particles moves Just as all the matter of the system, if .».»centrated in
a point, would move under tho lntiuenc(> of forces (>*]Ual ami parallel
to ihe foreve ivsHy miiag on its difll>rent parts. (For furthitr
-(._^,
wm
TWO LA Wa OP MOTTOIf.
295
tideratood as a fac-
e X is the a'-com-
i force exerted on
as with A definition
>w to oompouud, aod
r to investigate the
particle subjected to
n^ to Law III, if one
drawn hy tliis other
(Art. 10). A horse
a force equal to that
If one body strikes
uther body, its own
and in the opposite
i the bodle8 exert un
momenttim that one
tain foix^e, while the
The result is that
W( oiint of its attrac-
der the influoncii o*
e earth being enor-
le forces on the two
he motion produced
than that produced
tinn parallel to any
ucncing one another
leir mutual action.
yRt«>m of mntually
Hiiir untform'v in a
alooity of i's motion
Tie other matter not
y of any systom of
a, it ..tincontrated in
(Mjual ami parallel
rts. (For fortliMr
remarkr on tbese laws see Tait and Steele's Dynunics of a Particle,
Thomson and Tait's Nat. Phil., Pratt's Mechanics, etc.)
170. Two Laws of Motion in the Fronch Troa-
tises. — Newton's Laws of motion are not adopted in the
principal French treatises ; bet we fiud in them two prin-
ciples only as borrowed from experience, viz.:
First. — ^The Law of Inertta, that a body, not acted
upon by any force, would go on for ever with a uniform
velocity. This coincides With Newton's First Iaw.
Second. — That the velocity communicated is proportional
to the force. The second and third Laws of Motion are
thus reduced to this second priticiple by the French writers,
especially Poisson aud Laplace.*
171. Motion of a Particle under the Action of an
Attractive Force. — A particle moves under a force of
attraction which is in its line of motion, and varies directly
as the distance of the particle from the centre of force; it is
required, to determine the motion.
The point whence the it fluenco of a force emanates is
called the centre of force ; and the force is called an attrac-
tive or a repulsive force according as it attracts or repels.
jjet be the centre of force, P the
position of the particle at any time, t, v ~
its velocity at that time, and let OP = x,
and OA = a, where A is the position of the particle when
/ = ; let ^ = the absolute force ; that is, the force of
attraction on a unit of mass at a unit's distance from 0,
which is supposed to be known, and is sometimes called
the strength of the attraction. At preaeo* we shall suppose
* ParkiDaou'i MMbaaiei, p. 187. See paper by Dr. Whewell on ttip principle*
of DTiuunks. iMitUcnlarty as alalsd by FrMieb wiiten, Is tbo Idinboiili jonrsal gf
Hci*w». Vol. VaL
.-^
Fi|.M M»
^m
39i
aSMAMKS ON LAW Ut.
the former, the unit of mass must be imilerstood aa a fac-
tor on the left-hand side, in which case X ia the x-com-
ponent, for the unit of 'Tiass, of the whole force exerted on
the moving body.
The flnt two Uwb, h»ve, therefore, furniahed aa with a definition
and a meature of force ; aud they alau ahow how to oompouud, and
lherefoi« how to reeolve. forcea; and also how to inveetigata the
conditiona of equilibrium or motion of a aingle particle aubjectod to
given forces.
169. Ztomarka on Law ni.— According to Law III, if one
bodjr presaea or drawa another, it ia prnsaed or drawn by thia other
with an equal force in tlie oppoeite directior (Art. 10). A horse
towing a boat on a canal, is pulled backwards by a force equal to that
whic>i he impreasee ou the towing-rope forwarda. If one l)ody strikes
another Ixhdy and changes the motion of the other body, ita own
motion will l>o changed in an equal quantity and in the opposite
dirucUun; fur at each instant during the impact the Ixidiea exert <>n
each othi>r oqnal and opposite pressures, and the momentum that one
liody luBcs is equal to that which the other gains.
The eorth attracts a falling pebble with a certain force, while the
pebble nttracta the eurth with an equal force. The reault is that
while the pebble moves towards the earth on account of its attrac-
tion, the <>arth also moves towards the pebble under the infauenoe of
the nttractitm of the latter ; but the mass of the earth being enor-
WDUiily greater than that of the pebble while the forces on the two
arising from their mutual attn ^tions are equal, the motion produced
thereby in the earth is almost incomparably leas than thai produced
iu the pebble, and is consequently insensible.
It follows thai the sura of the quantities of mai.ion parallel to any
fixed direction of the partlclro of any system influondng one another
in any poHsible Avsy, remains unchanged by their mntual action.
Therefhro if the c-entro of gravity of any system of mntnally
influencing particles is in motion, it oontinuea movir^|r uniformly in a
straigRt lino, unless in bd fur us the direction or velocity of i's motion
is changed by forcea between the particles and some other matter not
hflonging to the tyntem : also the centre of gravity of any system of
]iartirles moves Just as all the matter of the system, if concentrated in
11 |M>lnt, would move under the influenc«< of .orres •■<)ual and parallel
to ihe forcva leaHy acttHg on its ditlcrr>at parts. (For furtliMT
' ^ '' fl B Sfc^ i^ ' flt P i lESB fflll
TWO LA wa OP Morrox;
395
derstood as a fac-
X ia the ar-com-
force exerted on
38 with a tkfinition
V to eompouud, »ad
to investigats the
article subjected to
Of to Law III, if one
Irawn by tliia other
Art. 16). A horse
I force equal to that
If one body atrikea
ther body, its own
id in the opposite
the IxHiiea exert on
lomentum that one
lin force, while the
The result is that
count of its attrac-
er the infauencti of
earth being enor-
I forces on the two
e motion produced
than thai, produced
nn parallel to any
pncing on<» another
>ir mutual action.
Rtem of routnally
rfr uniformly in a
ority of i's motion
o other matter not
of any systom of
. if concentrated in
Hfual and parallel
ts. (Por furtliMf
remarks on these laws see Tait and Steele's Dynamics of a Particle,
Tikomaon and Tait's Nat Phil., Pratt's Mechanics, etc.)
170. Two Laws of Motion in the French Troa-
tiaoa — Newton's Laws of motion are not adopted in the
principal French treatises ; bnt we find in them two prin-
ciples only as borrowed from experience, viz.:
First. — ^The Law of Inertia, that a body, not acted
upon by any force, would go on for ever with a uniform
velocity. ThiS coincides with Newton's First T^iw.
Second. — That the velocity communicated ia proportional
to the force. The second and third Laws of Motion ai'e,
thup reduced to this second principle by the French writers,
especially Poisson and Laplace.*
171. Motion of a Pazticto nnder tho Action of an
Attr a ct i ve Force. — A particle moves under a force of
attraction whtck is in its line of motion, and varies directltf
as the distance of the particle from the centre nf force; it is
required, to determine the motum.
Tho point whence the influence of a force emanates is
called the centre of force ; and the force is called an attrac-
tive or a repulsive force according as it at/racla or repels.
Ijet be the centre of forco, P the ^,
position of the particle at any time, /. v
its velocity at that time, and let OP = x,
and OA = a, where A is tho position of the particle when
t — 0; let /i = the absolute force ; that is, the force of
attraction on a unit of mass at a unit's distance from O,
which is supposed t,o be known, and is sometimes called
the strength of the attracti .i. At present we shall suppose
4-^
Ft|.N M»
Ail*
1 n
9.1 !
* ParWnion'* Xachanles, p. Wl. See paper bj Dr. Wkcwell on ibe principle*
of Djmamiea, parttcalarly •■ ■tated by Viench wrlten, In the Bdlnborgh joomai of
•, Vol. VUL
S96
A VARIABLE ATTRACnVB tORCU.
the maas of the particle to be unity, aa it aimplifles the
eqaatiouB. Then fix is the magnitude of the force at the
distance x on the particle of unit mass, or it is the accelerr.-
tion at P ; and the equation of motion is
(1)
the negative sign being taken because the tendency of the
force 18 to diminish x\
idzcPx
= — %\ix dx.
Integrating, we get
d^
dp
= fi (a» - a^),
(2)
if the particle be at rest when x = a and t — 0,
.'. -^^===V^dt,
Va* - a?
the negative sign being taken, because x decreases as t
increases. Integrating again between the limits correspond-
ing to ^ = / and t = 0,
COS"
i? = H*/.
1 -1*
< = -j- COS ' -•
(3)
Prom (2) it appears that the velocity of the particle is
zero when a; = a and - o ; and is a maximum, viz.: «^*,
when a; = 0. Hence the particle moves ftom rest at A : its
velocity increases until it reaches where it becomes a
r rOHCll.
88 it simplifies the
of the force at the
or it is the accelei^-
is
the tendency of the
(2)
i< = 0,
e X decreases as t
e limits correspond-
(3)
ty of the particle is
laximaro, viz.: mju*,
from rest at A ; its
here it becomes a
-.
-«•
'jMiaiim'""* •^I'liiin 'MmxxmsiattKnx^itri/msmmsimieKiiiienis^j
A VAMlABm ATTRACTIVM Wn^BGJl,
S97
masimnm, and where the force is zero ; the putiole passes
through that point, and its velocity decreases, and at A', at
a distance = — a, becomes zero. From this point it will
return, under the action of the force, to its original posi-
tion, and continually oscillate over the space 2a, of which
O is the middle point.
From (3) we find when a; == a, < = and when x = 0,
t = — T ; 80 that the time of passing irom A to = — r .
and the time from to A' is the same, so that the time of
oscillation from A to A' is -r* Tbis result is remarkable,
as it shows that the time of oscillation is independent of
the velocity and distance of projection, and depends solely
on the strength of the attraction, and is greater as that is
less.
This problem includes the motion of a particle within a
homogeneous sphere of ordinary matter in a straight shaft
through the centre. For the attraction of such a sphere on
a particle within its bounding surface varies directly as the
distance from the centre of the sphere (Art. 1338). If the
earth were such a homogeueoua sphere, and if AOA' (Pig.
80) represented a shaft running straight through its centre
from surfitce to surface, then, if a particle were free at one
end. A, it would move to the centre of the earth, 0, where
its velocity would be a maximum, and thence on to the
opposite side of the earth. A', where it would come to rest ;
then it would return through the centre, O, to the j- ie. A,
from where it suirted ; and its motion would continue to bo
oscillatory, and thus it would move backwards and forwards
from one side of the earth's surface to the other, and the
time of the oscillation would bo independent of the earth's
radius; that is, at whatever point within the earth's sur&ce
the particle be placed it would reach the centre in the
same time.
Sit
■ t '
298
A VARtABhM BSPITLBIVK t^jXOX.
in
jjifr
i
OoB. — To find this time. Since /i is the attraction at a
unit of distance and g the attraction at the distance B^ we
hare fi = -^, which in t =. — j gives
for the time it wonld take a body to moTe from any point
within the earth's surface to the centre.
If we put g = Z^ feet and R = 3963 miles we get
^ = 21 m. 6 s. about,
which would be the time occupied in passing to the earth's
centre, however near to it the body might be placed, or
however far, so long as it is within th<) surface.
172. Motton of a Particle under the Action of a
Variable RspnlaiTe Force. — Let the force be one of
repulsion and vary as the distance, then the equation of
motion is
dhi ~
dt»
= fix.
Let us suppose the particle to be projected from the cen-
tre of force with the velocity v^ ; then we have
dt*
F«^ + »,» ;
(1)
As t increases x also increases, and the particle recedes
further and further from the centre of force; and the
velocity also increases, and ultimately equals ao when x =r
t = cc. Thus in this case the motion is not oscillatory.
ii
ffjRCB.
9 the attraction at a
b the distance R^ we
OTe from any point
3 miles we get
wiing to the earth'g
aight be placed, or
surface.
r the Aetioii of a
he force be one of
en the equation of
ected from the oen-
ire have
(1)
*'>
the particle recedes
of force; and the
equals 00 when x =r
is not oscillatory.
'■imm
ivmtim mmmm^j i i '-'Mi i i i ii iii i i i i||
•mm
A VASTABIX ATTHACnva POSOX.
390
X73. Motton of « Particle under the Aotioa of an
Attractive Force which ia in the line of motion, and
which variea Invervely aa the Square of the Uatance
from the Centre of Force.
Let O (Fig. 80) be the centre of force, P the position of
the particle at the time t; and A the position at rest when
< = 0, 80 that the particle starts from A and moves to-
wards 0. Let OP = X, Ok = a, and /« = the absolute
force as before or the acceleration at unit distance firom 0.
Then the equation of motion is
ft
dx
Multiplying by 2 tj and integrating, we get
dt*
=^e-^.
(1)
which gives the velocity of the particle at any distance, x,
from the origin.
From (1) we have
2/u\/<
ax
dt
the negative sign being taken because in the motion to-
wards 0, X diminishes as / increases. This gives
^J^dt^
— xd x
= u
['
a — 2x
Vox — afl 2 y/ax _ ^^ J
\dx.
■'* #
800
VMLOCtrr IN FALLING.
Integrating and taking the lunits corresponding to ^ s ^
and / = 0, we have
(2)
which gives the valne of t
When the particle arrives at 0, a; =: 0, therefore the
time of felling to the centre from A is
From (1) we see that the velocity = when a; = a ; and
= « when a; = ; hence the velocity increases as the
particle approaches the <»ntre of forco, and ultimately,
when it arrives at the centre., becomes infinite. And
although at any point very near to there is a very great
attraction tending towards 0, at the point itself there is
no attraction at all; therefore the particle, approaching
the centre with an indefinitely great velocit;y, must pass
through it. Also, everything being the same at equal
distances on either side of the centre, we see that the
motion must be retarded as rapidly as it was accelerated,
and therefore the particle will proceed to a point A' at a
distance on the other side of equal to that from which it
started ; and the motion will continue oscillatory.
174. Vdloeiiy acquired in Falling throng a Oraat
Height above the Earth.— The preceding case of motion
includes that of a body falUng iVom a great height above
the earth's surface towards its centre, the distance through
which it falls being so great that the variations of the earth's
attraction due to the distance must be taken into account
For a sphere attracts an external particle with a force which
varies inversely as the square of the distance of the particle
a.
■espondrng tot x t
1+™] (')
= 0, therefore the
) when a; = a ; and
Y increases as the
0, and ultimately,
kes infinite. And
lere is a very great
nt O itself there is
rticle, approaching
relocit;y, must pass
;he same at equal
, we see that the
1 it was accelerated,
to a point A' at a
that from which it
scillatory.
; tliroii^ a Or««t
ling case of motion
great height above
B distance through
ations of the earth's
taken into account
) with a force which
iuce of the particle
fit il.lWIHH
ivapii
vtLocirr m fallino.
301
from the centre of the sphere (Ait 133a); therefore if i? is
the earth's radius^ g the kinetic measure of gravity on a
unit of mass at the earth's surface (Arts. 20, 23), and z the
distance of a body from the centre of the earth at the time
t, then the equation of motion is •
if ^'
which is the same as the equation in Art 173 by writing n
for gR" , therefore the results of the last Art. will apply to
this case. Substitu'ang gIP for /* in (1) of Art 173 we
have
/« -- x\ ^y
«" = ^*(~>
When the body reaches the earth's surfoce, x = R and
(1) becomes
^^%gR(^^). P)
H o is infinite (3) becomes
so that the velocity can never be so great as this, however
fiir the body may fall; and hence if it were possible to
project a body vertically upwards with this velocity it would
go on to infinity and never stop, supposing, of course, that
there is no resisting medium nor other disturbing force.
If in (2) we put g = 32^ feet and R = 3963 miles we
get ,
V = [2- 321-3963. 6280]* feet = 6-95 miles;
BO that the greatest possible velocity which a body (»n
acquire in falling to the earth is less than 7 miles per
second, and if a body were projected upwards with that
"■^s^^as^agyAjEFgCi^'S
mm^mmmm
303
MOTION IN A RXaiSTlNQ MSDIUJl.
I*;
veloci V, and were to meet with no resistance except
gravity, it would never return to the earth.
Cob.— To find the velocity which a body would acquire
in falling to the earth's surface from a height h above the
surface, we have from (1) by putting z = Rm^a = h + R,
If A be small com^)ared with Ji, this may be written
which agrees with (6) of Art 140.
The laws of force, enumerated in Arts. 171, 173, are the
only laws that are known to exist in the universe (Pratt's
Mecbs., p. 212).
175. Motion in a Resiatisg Modiun.— In the pre-
ceding discussion no account is taken of the atmospheric
resistance. We shall now consider the motion of a body
near the surfoce of the earth, taking into account the
resistance of the air, which we may assume varies as the
square of the velocity.
A particle under the action of gravity, as a constant force,
moves in the air supposed to be a resisting medium of
uniform density, of which the resistant^ varies as the square
of the velocity required to determine the motion.
Suppose the particle to descend towards the earth from
rest Take the origin at the starting point, let the line of
its motion be the axis of x ; and let x be the distance of
the particle from the origiii at the time t,' and for con-
venience let gifi be the resistance of the air on the particle
for a unit of velocity; gJi^ is called the eoeffioient of resist-
ance. Then the resistance of the air at the distance x fh>m
rtte
MBDIUM.
resistance except
rth.
body would acquire
height ft above the
= iff and a = A + i?,
fiffRh
R + h'
Y be written
». 171, 173, are the
le UDiTerse (Pratt's
iiun. — ^In the pre-
)f the atmospheric
motion of a body
into account the
lume Taries as the
M a constant force,
fisting medium of
'aries as the square
otion.
is the earth from
nt, let the line of
be the distance of
e t,' and for con-
^ron the particle
^ffii^ient of resist-
te distance x fh>m
>>«;W«V«<bwr^tMT«i4«fri«m9a^ian:4aaet*vMi^ •
MOTION Ilf A RJHSISmrO MJSDtUM.
the origin is gi^ (^) , which acts upwards, and the force of
gravity is g acting downwards, the mass l)eing a unit.
Hence the equation of motion is
(1)
d
.♦. gdt =
m
-Q'
Integrating, remembering that when ^ = 0, t> = 0, we
get
1 + h-
t = ^log ^, (Calculus, p. 259, Ex. 5).
2k
^ "dt
Passing to exponentials we have
dt
ke^g* + e-^*'
(«)
which gives the velocity in terms of the time. To fina it in
terms of the space, we hav« from (1)
««(§)'
-^=:2gi^dx;
observing the proper limits;
m
if
L
fit
■■-'WfiiiifflWifflgitl^^
3C4
MOTION or ASCENT IN TETM AlB,
which gives the velocity in terms of the distance.
(4)
Also, integrating (2) taking the same limits as before,
we get
gli^x = log (e*»* + C"***) — log 2 ;
2«ff»'« = «*»« + e-*Ot,
(5)
which gives the relation between the distance and the time
of falling through it.
As the time increases the term e-*v diminishes and from
(5) the space increases, becoming infinite when the time is
infinite; bat from (2), as the time increases the velocity
becomes more nearly uniform, and when < = oo, the
velocity = r ; and although this state is never reached, yet
it is that to which the motiuu approaches.
176. Motion of a Particle Aaeending in the Air
agsdnst tbe Action of Oravity.— Let ns suppose the
particle to be projected upwards, that is, in a direction
contrary to that of the action of gravity, with a given
velocity, v, it is required to detennine the motion.
Let UB suppose the particle to be of the same form and
siso as before, and the same coefficient of resistance.
Then, taking x positive upwards, both gravity and the
resistance of the air tend to diminish the velocity as t
increases; so tliat the equation of motion is
tPx
-,<)•;
(1)
(4)
istance.
limits as before,
log 2;
(5)
mce and the time
linishes and from
when the time is
aases the velocity
len ^ = 00, the
never reached, yet
ling in the Air
as suppose the
8, in a direction
ty, with a given
motion.
le same form and
it of resistance,
gravity and the
the velocity as t
is
(1)
MOTION or ASCMXT IX TBM .AIM,
dx
dh
dl
i + KI)
.'. tan-> 1e^ — *»«"* (H - O^d »
(Calculus, p. 244, Ex. 3), since the initial velocity is v.
Taking the tangent of both members and solving for
dx .
^ _ 1 v l — tan kgt .
di ~ k' 1 + vk toD kgt*
(8)
which gives the velocity in terms of the time. To find it
in terms of the difltanoe, w» 1mv(> from (1)
-(f)'
-, s= -«jfifc»<toi
m
... log ^ ^^ =:^%gm',
.. . (^)' = ^-*^'^ - ^ (1 - ^**'''>'
which gives the velocity in terms of the distance.
Also, integrating («) irfter sabstituting sine and cosine
for tangent, and taking the same limits w before, we get
^/L*» =3 log (»* sin /is'' +• oo* M) ; (*)
which gives the space desori'jed by the particle in terms of
tiie tim«.
i
UBI
mozaon or AsoMifT isr tsjb air.
OoB. 1.— To find the greatest height to which the par-
tide will ascend put the velocity, -^ = 0, in (3) and get
which is the distance of the highest point
da
Putting ^ = v> in .^s Te ge*;
(«)
kg
m
which is the time required for the particle to reach the
highest point. TIaving reached the greatest height, the
particle will begin to foil, and the circnmfitances of the
fall will be giyen by the equations of Art 175.
OoB. 2.— Since h is the same in this and Art 176, we
may compare the velocity of projection, t', with that which
the parcicle would acquire in descending to the point
wh< nee it was projected. Denote by v^ the velocity of
the particle when it reaches the '^nint of starting. From
(3) of Art 176 we have
and placing this value of x equal to that given in (6),
we get,
r-w:*^^-^^'
.'. », =
(l + *»l^)*'
which is less than t>; hence the velocity »^aired in the
•Ai
f.W"WtW»K5?2?l
m
S AIR,
t to which the par-
0, in (3) and get
(6)
t
(7)
irtiole to reach the
latest height, the
cnm^tances of the
;. 175.
and Art 176, we
t>, with that which
ling to the point
t'o the velocity of
)f starting. From
that given in (6),
ty »^aired in the
wmmmm
UOTIOH OF A PSOJBCTtLB,
307
descent is less than that lost in the ascent, as might have
been inferred.
Cob. 3.— Substituting (6) in (5) of Art 176, we get for
the time of the descent.
/
^log(Vl + iV + *»),
which is difFennt from the time of the ascent as given in
(7). (See Pri je's Anal. Mech's, Vol. I, p. 406 ; Venturoli's
Mech's, p. 81) ; Tait and Steele's Dynamics of a Particle,
■AHi. )
177. Motion of a Protjectile in • Resisting M^-
dimn. — The theory of the motion of projectiles in vacuo,
which waf; examined under the head of Kinematics, affords
results which differ greatly from those obtained by direct
experiment in the atmosphere. When projectiles move
with but small velocity, the discrepancy between the para-
bolic theory, and what is found to occur in practice, is
small ; but with increasing velocities, as those with which
bftils and shells traverse their paths, the air's resistance
increases in a higher ratio than the velocity, so that the
discrepancy becomes very great.
The most important application of the theory of projec-
tiles> is that of Gunnery, in which the motion takes place
in the air. If it were allowable to negWct the resistance of
the air the investigations in Part II would explain the
theory of gunnery ; but when the velocity is considerable,
the atmospheric resistance changes the nature of the tra-
jectory so much as to render the conclusions drawn from
the theory of projectiles in vacuo almost entirely inap-
plicable in practice.
The problem of gunnery may be stated as follow^:
Given a projectile of known weight and dimensions,
starting with a known velocity a^ a known angle of eleva-
308
XOTtON or A PKOJBCTTLE.
tion in a calm atmosphere of approximately known density ;
to find its range, time of flight, velocity, direction, and
position, at any moment ; or, in otb'jr words, to construct
its trajectory. This problem is not yet, however, suscepti-
ble of rigorous treatment ; mathematics has hitherto proved
unable to furnish complete formulae satisfying the condi-
tions. The resistance of the air to slow movements, say of
10 feet per second, seems to vary with the first power of
the velocity. Above this the ratio increases, and as in the
case of the wind, is usually reckoned to vary as the square
of the velocity ; beyOnd this it increases still further, till at
1200 feet per second the resistance is found to vary as the
cube of the velocity. The ratio of increase after this point
is passed is supposed to diminish again ; but thoroughly
satisfactory data for its determination do not exist.
From experiments* made to determine the motion of
cannon-balls, it appears that when the initial velocity is
considerable, the resistance of the air is more than 20 times
as great as the weight of the ball, and the horizontal range
is often a small fraction of that which the theory of pro-
jectiles in vacuo gives, so that the form of the tngectory is
very different firom that of a parabolic path. Such experi-
ments have been made with great care, and show how little
the parabolic theory is to be depended upon in determining
the motion of military projectiles.
178. Motion of « Prc^otile in the Atmosphere
Snppoeing its Resistance to wary ss the Square of
the Telocity. — A particle under the action of gravity ia
projected from a given point in a given direction loith a
given velocity, and moves in the atmosphere whose resistance
is assumed to vary as the square of the velocity ; to deter-
mine the motion.
* B«e BDcrolo)Midto BrllMiiiiea, Art ChuuMrr ;
Hnttoa'f Tncu.
•iw Bobln'i Qnaasrj, and
IJP.
3ly known density;
ity, direction, and
ords, to construct
however, suscepti-
las hitherto proved
isfying the condi-
movements, say of
the first power of
ses, and as in the
rary as the square
till farther, till at
nd to vary as the
se after this point
; but thoroughly
not exist,
le the motion of
initial velocity ia
lore than 20 times
; horizontal range
the theory of pro-
I the trajectory is
th. Such experi-
d show how little
m in determining
M Atmoaphera
■ the Square of
fion of gravity ia
direction with a
a whoae reaiatanca
'eiocitjf ; to deter-
Bobln'i OnniierT, aaA
mwaOH OP A PROJEOfiLA
Take the given point as origin, the axis of x horizontal,
the axis of y vertical and positive upwards, so that the
direction of projection may be in the plane of xy. Let v
be the velocity of projection, g the acceleration of gravity,
a the angle between the axis of x and the ^'le of projection,
and let the resistance of the air on the particle be i for a
nnit of velocity ; then the resistance, at any time, t, in the
line of motion, is h ijA ; and the x- and y-components of
thie resistance art, respectively,
, da dx J
* jj • TV, and
dt dt
da dy
dt dt
Then the equationfl of motion are, resolving horizontally
and vertically,
(Pa; _ ,da dx .^v
dfl- ~^dt rr ^ '
dP ~
9
Prom (1) we have
tdx\
dx
dt
= -h-la;
dx
log
^dt dt
dm
dt
voosa
(»)
— ha\
since when < = 0, -^ =:^ v c(x a',
dx
di
= v cos « «-■
(8)
Maltildying (1) and (2) by dy and dx, respectively, and
aabtraoting the former from the latter we have
d^dx — iPxdy _ ^
dfi
gdx.
(4)
ii
310 MonoN or a projbctilm.
Substituting in (4) for dP ita value from (3) we get
Substitu ting in t he second member of (6) for dx its value
dxV ^ da«/ i;» co8» - '^'"*
(6)
Pat ^ = P' and (6) becomes
"+'■>** = pi?-. •**•
Integrating, and remembering that when « = 0,^ = tan <^
we get
P (1 + J»«)* + log [p + (l+ JB»)*]
-^,-e«
(7)
Ar* Cf /8» a
where e is the constant of integration whose value
= tan o sec « + log (tan o + sec «) + , -^ . . fg^
' «t)»COS»o ^ '
From (5) we have
t>»oos»a dxXdxf'
which in (7) gives
i'a +i^)* + log [i» + (1 +i,»)*] - c = ^|,
, dp
" /'(l+jJ^+l0g[iB + (l+^)*]_c^**'' ^*^
ttHH
tM.
(3)
we get
9
%
(6)
i) for dx its value
«»»»(fo.
(6)
»rf».
« = 0,jp = tan
«>
/^)*]
Mseralne
ifct)»C08»tt
(7)
(8)
— = kdx, (9)
.uu l.i'.JCJiW gWWWff^^^^Wff!— W^gl'^ff'wW
and
MOTtON or A PROJMCTILa.
pdp
311
=:Hy. (10)
/» (1 +i^* + log O + (1 + pO*] - c
From (4) ..e have
dX'dp = — gdfi.
Sabgtituting this value of dx in (9) and solving for dt we
get
|c - p (1 + p»)* ~ log [/» + (1+P»)*] f*
the negative sign of dp being taken because j» is a decreas-
ing function of ^.
Bephicing the value of i> = ^, (9), (10), and (U) become
(fo = T
4^
dx
*i(i^^Wwl^(^*]-«'
, 1
dx dx
'l('+g)*-'-[|+(' + g)*]-«'
-d
dt=:
dx
♦^{-K'+gr-Hi^^OT"
(A)
(B)
, (0)-
from which equations, were it possible to integrate them,
X, y, and t might be found in terms o* ^ > *»^ *' ^ ""^^
eliminated from the two intends, of (A) and (B), the re-
sulting equation in terms of x and y would be that of the
I
1
1 1
312
Monoir or a pxwjwmjk
required trajectory. But these eqnatiors cannot be inte-
grated in 6nit6 terms; only approximate .eolations of them
can be made ; and by means of these the path of the pro-
jectile may be constructed approximately. (See VentoroM's
Mechs., p. 92.)
Squaring (A) and (B), and dividing their suit by the
square of (0) we get
di*
9
k
1+^
-n'^%r-^<^^\^
+ V
^m
(D)
which gives the velocity in terms of ^.
179. Motion of a Projectile in the Atmosphere
under a small Angle of Elevation.— The case fre-
quently occurs in phwtice where the angle of projection }*
very small, and where the projectile rises but a very little
above the horizontal line. In this case the equation of the
part of the trajectory that lies above the horizontal line
may easily bo found; for, the angle of projection being
dy
very small, ^ will be very small, and fterefore, throughout
the path on the upper side of the axis of x, powers of
^ higher than the first may be neglected. In this case
then
diszix', .•. » = «}
which in (6) of Art. 178, becomes
rfy
d^ =
«»co«*«
«**<fe;
»r8 cannot be inte-
« solutions of them
he path of the prp-
j. (Sm Ventiuroli's
; their sun? by the
(D)
/' _. ^\*
\*
the Atmosphere
•n. — The case fre-
{le of projection }#
ses bat a very little
he equation of the
he horizontal line
f projection being
irefore, throughoat
cis of X, powers of
ited. In this case
i i i Tjj i . i iiimfjw i i.il
BXAMFLBa.
913
Integrating, we get
tana = —
dx
2kir cos' a
(«*»-!);
dv
since when x = 0, -^ =: iaa a.
Integrating again we get
y =2;tan « -f
« 4ifct;» cos* o ^ / \ /
2;{;t;» COB*
Expanding e**^ in a series, (1) becomes
_ ga^ gkafi
y _ a: tan « - c^^^^ " 3«» cos* « ~
(2)
the first two terms of wHich represent the i ijectory in
vacno. [See (3) of Art. 151.]
From (3) of Art. 178, we have
dt =
dx
t -
vcosa
flto-1
V cos a
m
which gives the time of flight in terms of the absdraa.
The most complete and valuable series of experiments
on the motion of projectiles in the atmosphere that has yet
been made, is that of Prof. F. Bashforth at Woolwich.
EXAMPLES.
1. Find how far a force equal to the weight of n Ibe.,
would move a weight of m lbs. in / seconds ; and find the
velocity acqaiied.
Here P = n, and W=zm; therefore from (1) of Art
"^ we have
** == — /;
• • / — — >
which in (4) and (5) respectively of (Art. 10), gives
tn
; and « = 4 -^ /*.
* m
2. A body weighing n lbs. is moved by a constant foroe
which generates in the body in one second a velocity of a
feet per second ; find the force in pounds.
Ana.
5? lbs.
9
3. Find in what time a force of 4 lbs. wonld move a
weight of 9 lbs. through 49 ft along a smooth horizontal
plane ; and find the velocity acquired.
21
4. Find the number of inches through which a force of
one ounce, constantly exerted, wiU move a mass weighing
one lb. in half a second. ^„,, 3^ /ijs,
6. Two weights, P and Q, are connected by a string
which passes over a smooth peg or pulley ; required to
determine the motion.
Since the peg or pulley is perfectly
smooth the tension of the string is the
same throughout; hence the foroe which
causes the motion is the difference between
the weights, P and Q, the weight of the
string being neglected. The moving force
therefore is P — 0; but the weight of the
mass moved is P + ^. Hence substituting
in (1) of Art 25, we get
P+ Q
9
X
Q*
■p-e
'f.
Fig. SOo.
•re from (1) of Art
m
i (Art. 10), gives
by a constant force
Jond a velocity of a
is.
Ana. — lbs.
lbs. would move a
emooth horizontal
21
V^'
V = ^t.
gh which a force of
e a mass weighing
Am. 3ffH)\
leoted by a string
alley ; required to
Btly
the
lich
een
the
>rce
the
ing
X
Q*
Fig. SOo.
•mtt 'm i't w mv ta gmm mw' i m i mi P i f smr miUKa
XXAMPLSa.
.-. /=
P-Q.
which is the acceleration.
Substituting this in (4) and (5) of Art 10, we have
815
(1)
V = -i^.— j<//,
P— o
(8)
(8)
which giTCS the velocity and space at the time t, the initial
velocity v, being 0.
6. A body whose weight is Q, rests on a smooth hori-
zontal table and is drawn along by a weight P attached to
it by a string passing over a pulley at the edge of the table ;
find the motion of the bodies.
Since the weight Q is entirely supported by the resistance
of the table, the moving force is the weight P, hanging
vertically downwards, and the weight of the mass moved is
P + Q; therefore from (1) we have
/ =
P + Q
(I)
and this in (4) and (5) of Art. 10 gives the velocity and
space.
7. Required the tension, T, of the string in the pre-
ceding example.
Here the tension is evidently that force which, acting
along the string on the body whose weight is Q, produces
p
in it the acceleration, - „ , ^ g, and therefore is measui^
P + Y
by the mass of Q into its acceleration. Henoe
m
316
JfXAMPLSa.
'8. Find the tension, T, of the string in Ex. 5.
Here the tension equals the weight Q, pins the force
which, acting along the string on Q, produces in it the
acceleration
P+Q^'
•• ^-^^-g'PTQ^'
- ^^g
~ P + Q'
or it equals P the accelerating force which, of course,
gives the same result,
9. Two weights of 9 lbs. and 7 lbs. hang over a pulley, as
in Ex. 5 ; motion continues for 5 sees., when the string
breaks ; find the height to which the lighter weight will
rise after the breakage.
Substituting in (2) of Ex. 5 we have
v = V^32.5 = 20;
therefore each weight has a velocity of 20 feet, when the
string breaks. Hence from (6) of Art. 10, we have (calling
^32ft.)
that is, the lighter weight will rise 6J feet before it begins
to descend.
10. A steam engine is moving on a horizontal plane at
the rate of 30 miles an hour when the steam is turned ofE ;
supposing the resistance of friction to be ^ of the weight,
find how long and how far the engine will run before it
stops.
PQ
: in Ex. 5.
; Q, pins the force
produces in it the
■^9,
irce which, of course,
uig over a pulley, as
58., when the string
lighter weight will
>f 20 feet, when the
10, we have (calling
feet before it begins
horizontal plane at
team is turned off ;
e jfy of the weight,
le will run before it
BXAMPhK3.
»17
Let ff be the weight of the engine; then the rosistance
W
of friction is -r^t and tl»8 is directly opposed to motion,
400
w
m
9^'
400
30x1760x8
=:44
The velocity, v, is 30 miles an hour = — ^ ^ g^
feet per second. Substituting these Vfiues of / and v in the
equation v = //, we get
44 = ^«;
,'. t = 550 sees.,
which is the time it will take to bring the engine to rest if
the velocity be retarded ^ feet per second.
Also r» = Ufa, therefore
g — « «yy ^x*oo — 12100 feet.
11. A man whose weight is W, stands on the platform
of an elevator, as it descends a vertical shaft with a uniform
acceleration of |jf ; find the pressure of the man upon the
platform.
Let P be the pressure of the man on the platform when
it is moving with an acceleration of \g ; then the moving
force is W — P; and the weight moved is W', therefore
W
W^P = ^\9;
tr.
J2. A plane supporting a weight of 12 ozs. is descending
with a uniform acceleration of 10 ft. per second ; find ,the
pressure that the weight exerts on the plane.
Am. S\ ozs.
318
SXAMPLSa.
13. A weight of 24 lbs. hanging over the edge of a
smooth table drags a weight of 12 lbs. along the table;
find (1) the acceleration, and (2) the tension of the string.
Aw. (1) 5^ ft. per sec. ; (2) 20 lbs.
||; !*• A weight of 8 lbs. rests on a platform; find
its pressure on the platform (1) jf the latter is de-
scending with an acceleration of \g, and (2) if it is
ascending with the same acceleration.
Ans. (1) 7 lbs.; (2) 9 lbs.
15. Two weights of 80 and 70 lbs. hang over a smooth
pulley as in Ex. 5 ; find the space through which they will
move from rest in 3 sees. A7i8. 9| ft.
16. Two weights of 16 and 17 ounces respectively hang
over a smooth pulley as in Ex. 5 ; find the space de-
scribed and the velocity acquired in five seconds from rest
Ana. » = 26, V = 10.
17. Two weights of 5 lbs. and 4 lbs. togethei pull one
of 7 lbs. over a smooth fixed pulley, by means of a con-
necting string; and after descending through a given
space the 4 lbs. weight is detached and taken away without
interrupting the motion ; find through what space the
remaining 6 lbs. weight will descend.
Ans. Through J of the given space.
18. Two weighta are attached to the extremities of a
string which is hung over a smooth pulley, and the weights
are observed to move through 6.4 feet in one second ; the
motion is then stopped, and a weight of 5 lbs. is added
t.) the smaller weight, which then descends through the
same space i\» it ascended before in the same time ; deter-
mine the original weights. Ans. | 11)8. ; V lbs.
19. Find what weight must Iw added to the Bnmller
wcigltt in Ex. 5, so that tlie acceleration of the systen. may
over the edge of a
t)e. along the table;
jnaion of the string,
jr sec. ; (2) 20 lbs.
a platform ; find
the latter is de-
, and (2) if it ia
I 7 lbs.; (2) 9 lbs.
hang over a smooth
igh which they will
Ans. 9| ft.
es respectively hang
find the space de-
! scoouds from rest
» = 26, V = 10.
. togethei pull one
by means of a cou-
; through a given
taken away without
B[h wi)at space the
f the given space.
10 extremities of a
ey, and the weights
in one second ; the
of 5 lbs. is added
icends through the
same time ; deter-
s. I 11)8. ; V lbs.
I'd to the Bnjaller
of the syateiL may
XX'
nBS.
810
have the same nnmerioal value aa before, but may l)e in
the opposite directisn.
Ana.
20. A body \s. projected np a rough inclined plane with
the velocity which would be acquired in faUing freely
through 12 foet, and just reaches the top of the plane ;
the inclination of the plane to the horizon is 60°, and the
coeflScient of friction is equal to tan 30°; find the height of
the plane. Ana. 9 feet.
21. A body is projected up a rough inclined plane with
the velocity ^g ; the inclination of the plane to the horizon
is 30°, and the coeflHcient of friction is equal to tan 15° ;
find the distance along the plane which the body will
describe. ^na. g (v^S + 1).
22. A body is projected up a rough inclined plane ; the
inclination of the plane to the horizon is «, and the coef-
ficient of friction ie tan e ; if m be the time of ascending,
and n the time of descending, show that
\« / ~ sin (a + e)'
23. A weight P is drawn up a smooth plane inclined at
an angle of 30° to the horizon, by means of a weight Q
which descends vortictjUy, the weights being connected by
a string passing over a small pulley at the top of the plane ;
if the acceleration be one^fourth of that of a body fa]lh.j
freely, find the ratio of ^ to P. Ant. Q = P.
24. Two weights P and Q are connected by a string,
and Q hanging over the top of a smooth plane inclined at
30° to the horizon, can draw P up the length of the plane
in just half the time that P would take to draw up Q ;
show that Q U, half m heavy again as P.
830
MXAMPhaa.
25. A particle moyes in a straigfat line undo* the action
of an attraction varying inversely as the (|)th power of
the distance; sho\» that the velocity acquired hy falling
from an infinite distance to a distance a from the centre is
equal to the velocity which would be acquired in moving
from rest at a distance a to a distance ^•
m
te nndear the action
the (|)th power of
acquired by falling
[ from the centre is
acquired in nioTing
CHAPTER II.
CENTRAL FORCES.*
180. Definitlona. — A central force is one which acts
directly towards or from a fixed point, and is called an
attractive or a repulsive force according as its action on
any particle is attraction or repulsion. The flxed point is
called the Centre. The intensity of the force on any par-
ticle is some function of its distance from the centre.
Since the case of attraction is the most important af>plica-
tion of the subject, we shall take that as our standard case ;
but it will be seen that a simple change of sign will adapt
our general formnlsa to repulsion. If the centre be itself
in motion, we may treat it as flxed, in which case the term
"actual motion "of any particle means its motion "rela-
tive" to the ct itre, taken as fixed.
The line firom the centre to the particle, is called a
Radius Vector. The path of the particle under the action
of an attraction or repulsion directed to the centre is
called its Orbit.^ All the forces of nature with which we
are acquainted, are central forces ; for this reason, and be-
cause the motion of bodies under the action of central
forces is a branch of the general theory of Astronomy, we
shall devote this chapter to the consideration of their
action.
181. A PuHele under the Aotion of a Central
Attnotion; Reqnired the Polar Equation of the
Path. — The motion will clearly take place in the plane
passing through the centre, and the line along which the
• TbiM chapter conUln* the Ant prinolplei of lUtbWDaticai Attronomy. It
nuy, howaver, be omlltad bj the •tudent of KngtMering.
t OkU«d Omtnl Urbiu.
:il
M
CKlfTSAL ATTRACnO'cf.
particle is initially projected, as there is nothing to with-
draw the particle from it. Let the centre of attraction, 0,
be the origin, and OX, OY, any
two lines iirough ftt right angles
to each other, be the axes of co-
ordinates. Let (x, y) be the
position of the particle M at the
time t, and (r, 9) its position
referred to polar co-ordinatos,
OX being the initial line. Then,
calling P the central attractive
force, we have for the components parallel to the axes of x
and y,* respectively, -- P-, — P^, the forces being nega-
tive, since they tend to diminish the co-ordinates. There-
fore the equations of motion are
(1)
Multiplying the former by y, and the latter by x, and
iubtracting, wo have
; — 2 _ « —
a;^-y,^ =0.
(3)
Integrating we have
.%
dt "dt
where A is .a undetermined constant
Since x =5 r oos 0, and y = r sin 0, we hav«
dx = COS B dr — r Bin do,
dy =z nine dr + rcoaO d9,
which in (3) gives
(8)
(*)
o:y.
is nothing to with-
itie of attraction, 0,
Fli.M
llel to tlie axes of x
e forooB being nega<
D-ordinates. There-
■ Pf. (1)
he latter by x, and
(3)
(8)
ire iuiye
ide,
)d$,
<*)
CMlfTBAL ATTSACnOJf,
d( ^ '*•
898
(5)
Again, multiplying the first and second of (1) by 2tbe
and 2dy respectively, and adding, we get
2dxdh: + %dy dhf _ 2P{xdx + ydy)
d0^ ~ r *
d
da*
,5? "•■<««
-2/Wn
m
Substituting in (6) the values of dafi and Jy* from (4), we
have
(7)
Put r ss - ; and . •. dr as — — | ; and (7) becomes
2P
./dvF .A 2P ,
performing the differentiation of the fint member> and
dividing by idu, and transposing, we get
dht
+ «♦-
]?u*
m
which i$ th* differential equation of the orbit described ;
and as, in any particular instance, the force P will be given
in terms of r, and therefore in terms of u, the integral of
this equation will be the polar equation of the requijr^
path.
6<dving (8) for P w« have
324
CENTRAL ATTRACTtON.
^ = »^(^" + «)'
(»)
which is the same resolt that was found by a different pro-
cess in Art 163 for the acceleration along the radius
vector.
OoB. 1. — The general integ.als of (1) will contain four
arbitrary constants. Ona, A, that was introduced in (6),
and two more will be introduced by the integration of (8).
If the value of r ia terms of 0, deduced from the integral
of (8), be substituted in (5), and that equation be then
integrated, the fourth constant will be introduced, and the
path of the particle and its position at any time will be
obtained. The four constants must be determined from
the initial circumstances of motion; viz., the initial
position of the particle, depending on two independent
co-ordinates, its initial velocity, and its direction of pro-
jection.
Cor. a.— By means of (9) we may ascertain the law of
the fcAse which must act upon a particle to cause it to
describe a given curve. To eflfect this we must determine
the relation between v, and B from the polar equation of the
orbit referred ^ > the required centre as pole ; we must then
differentiate . twice with respect to 0, uid substitute the
result in the expression for P, eliminating 9, if it occurs,
by means of the relation between « and 9. In this way we
shall obtain P in terms of u alone, and therefore of r
alone.
OoB. 8.— When we know the relation between r and fl
from (9), we may by (6) determine the time of describing
a given portion of the orbit ; or, conversely, find the posi-
tion of the particle in its orbit at any time.*
* IM lUt ud StMle'i DyuuniM of > I^wtkla, p. IM; alw Pm(t'« Ifack't,
V.
; (9)
by a different pro-
along the radius
I will contain four
introduced in (5),
integration of (8).
from the integral
equation be then
itrodnced, and the
; any time will be
I determined from
viz., the initial
two independent
I direction of pro-
certain the law of
;icle to canse it to
re must determine
liar equation of the
ole ; we must then
and aubstitnte the
ng 6, if it occurs,
>. In this way we
ind therefore of r
I between r and 6
time of describing
sly, find the posi-
ic.*
lU; atao Pmtt'* Ifach't,
THB BBCTIONAL AREA.
OoB. 4, — If p is the perpendicular from the origin to
the tangent we have from Oalcnlus, p. 176,
xdy — ydx =: pd9 i
which in (3) gives
and this in (6) gives
ds
dt
P'
(10)
,h*
Differentiating, and solving for P, we have
A»rfr
f dp'
(11)
which is the equation of the orUt between the radius vector
and the perpendicular on the tangent at any point.
182. The Etoctorial Area Swept over by tiie
Raditis Vector of the Particle In any time is Pro-
portional to the Time. — Let A denote this area ; then we
have £rom Calculus, p. 364,
if A and t be both measured from the commencement of
the motion. Therefore the areas swept over by the radius
vector in different times are proportional to the times, and
equal areas will be described in equal times.
Cob. — If t = 1, we have A = |A. Hence h = twice
the sectorial arja described in one unit of time.
1B3. The Velocity of the Particle at any Point
of ita OrMt— We have for the velooity,
A = i/r»rfd
= ^fhdt,hj (5) of Art. 181,
i-
= iAf,
1
886 vsLocmr at ant poi^rr or rmi obbit.
= j^ by (10) of Art 181. (1)
Hence, the velocity of the patticle at each point of its
path is inversely proportional to the perpendicular from the
centre ott the tangent at that point.
OoB. 1.— We have, by Calculus, p. 180,
2- _ 1 1 rff*
= «» + ^', since f = - (Art. 181),
which in (1) gives
•'=1 = »•(«•+ a.
another important expression for the velocity.
Cob. 2.— From (6) of Art. 181, we have
(2)
(8)
I*t V be the velocity at the point of projection, at
which let r =: B, and since P is some function of r, let
P = /(r), then integrating (3) we get
which is another expression for the velocity ; and since this
is a ftinotion only of the corresponding distance*, R and r,
it follows that th$ velocity at any point qf tho orbit is
rjfV OBBJT.
of Art 181.
(1)
t each point of its
pendicular from the
30,
36 f = - (Art 181),
).
(2)
locity.
IT©
Fdr.
(8)
t of projection, at
e fanotion of r, let
(r)],
{*)
sity ; and since this
diatanoM, E and r,
nnt qf the orbU %$
vBLOcrtr AT Atrr potsr of wr« oMBtt. 88?
independent of the path described, and dtpmdt iokljf on the
magnitude of the atttaetion, the dietanoe of th$ point from
the centre, and the velocitif and distance of projection.
From (4) it appears that the velocity is the same at all
points of the same orbit which are equally distant from the
centre ; if r =: i?, the velocity = V; and thus if the orbit
is a reentering cunrc, the particle always, in its successive
revolutions, passes through the same point with the same
velocity.
If the velocity vanishes at a distance a from the centre
(4) becomes
t;» = 3[A(a)-A(r)] (6)
and a is called the radius of the circle of zero Velocity.
OoB. 3.-
-From (3) we have
rf(r») = -
2Pdr;
.*. vdv = —
Pdr.
Taking the
logarithm of (1) we
have
log w = log A
-log/.
Differentiating we get
dv
P
(«)
(*J
Dividing (6) by (7), we get
= 2P X J chord of curvature* through the centre ; (8)
• To prove that f ~ U one-f onrth the cbord of OWmttM.
8 dp
Let MD (Fig. 81), ba tbe Ungent to the orbit, end C the MDtre of cnrratara ; let
OD =|), CM = p, tbe radliM of curvature ; end the angle HSS - ^. Then MS, the
338
VSLOCJTT AT Afrr POINT OP THE OBBIT.
and, comparing this with (6) of Art. 140, it appears that
the particle at finy point has the same velocity which it
would have if it mov^ from rest at that point towards the
centre of force, under the action of the force continuing
constant, through one-fourth of the chord of the circle of
curvature.
Hence, the velocity of a particle at any point of a central
orbit is the same as that which would be acquired by a
particle moving freely from rest through one-fourth of the
ciiord of curvature at that point, through the centre, under
the action of a constant force whose magnitude is equal to
ihat of the central attraction at the point.
Cob. 4. — If the orbit is a circle having the centre of force
part of the radlns vector CM, which la intercepted by the circle of cnrratON is
Called the chord tf curvature. Ita valae is determined as follows ;
We have (Fig. 81)
* = fl + OMD
= » + sln-i
rVr'—tf
Viom Calcoluf, p. 180, (10), we have
pdr
a$ = —
r Vr" — p"
and
Subatitiitlng CI) in (1) we get
<$, = '■
rOr
<l*
P
_ dp
yfr' — p*
But Oalcalns, p. 381, we bare
dr
Now MS (Tig. 81) = 8M0 sin OMD,
= ^? = *|.hy(5)
= the chord of cnrratiue ; therefore
tap
? :j- = one-fourth the chord of carvatniv.
«
m
(«>
m
HE OBBIT.
10, it appears that
velocity which it
point towards the
I force coutiDuing
rd of the circle of
point of a central
' be acquired by a
i one-fourth of the
h the centre, under
titude is equal to
the centre of force
tbe circle of cnmtara ia
FoUowb;
0)
W
tare.
TBB ORBIT mrDER VARtABLB ATTRACTION.
320
in the centre, and R, V, P, are reapeotiyely the radius,
velocity and central force, we have
r» = PR.
OOR. 5.— From (5) of Art 181, we have
de
dt
(9)
The first number, being the actual velocity of a point
on the radius vector at the unit's distance from the centre,
is the angular velocity of the particle (Art. 160). Hence
the angular velocity of a particle varies inversely as the
square of the radius vector.
ScH.— A point in a central orbit at which the radius
vector is a maximum or minimum is called an Apse ; the
radius vr^tor at an apse is called an Apsidal Distance ; and
the angle between two consecutive apsidal distances is called
an Apsidal Angle of the orbit The analytical conditions
for an apse are, of course, that ^ = 0, ai.d that the first
derivatiye which does not vanish should be of ati even
order. The first condition ensures that the radius vector
at an apse is perpendicular to the tangent.
184. The Orbit when fhe Attraction Vaxiee In-
v-ersely as the Square of the Diatanca— J particle is
projected from a given point in a given direction with a given
velocity, and moves under the action of a central attraction
varying inversely as the square of the distance ; to determit.e
the orbit.
Let the centre of force be the origin; V = the velocity
of projection ; i? =r the distance of the point of projection
from the origin ; /3 = the angle between R and the line of
Hi
880 TBlt OBBIT Umxn VABIABhW ATTBACTIOB.
projeotioii ; and let ^ = the absolate force and < s=
when the particle is projected. Then since the relooity s
- (Art 183), and at the point of projection /> = ^ sin /3,
wc have
K Sin p
As the force yaries inversely as the square of the distance,
wehaye
p SB ^ SB fiv^, ^since »" = z)- (2)
which in (9) of Aft 181 gives
Multiplying by ddu and integrating, we get
S + «* = 2?.« + C;
(3)
d0»
¥
1 1 du* F*
when « =: 0, tt =± - i= -g, and ^ + «» sa -p-, (Art 183,
Cor.l); therefore
' ~ A» A»i2
A»i?
Substituting thl« valtte for c We get
Therefore (Art. 183, Cor. 1) we have
(velocity)' = F» + 2^ (^ - ^
(4)
(5)
' ATTBACTIOir.
te force aod < r=
since the Telocity =z
jection /> = i2 sin /3,
f2 sin /3.
(1)
qnare of the distance,
(8)
re get
.tt»s=-^, (Art.188,
-2p
»i2
8p«
-i)
w
0)
TJ» OSBtT tniJ>MB VABtABLM ATfttAOTtOlf. 381
which showa that the velocity it th« grmtest whm ritth*
least, and the leaet when r is the greatest.
Changing the fonn of (4) we Uave
d0*
- h»R ■^h*-\h*-'V'
(6)
To express this in b bimpler form, let
g = *, and ^^^~ + ^ = <^.; a^^ (6) becomes
_=c!-.(«-J)a;
d9»
— du
[c» -. (» - J)«]*
= dO,
the negatiTe sign of- the radical being taken. Integrating
we have.
COS"
where c' is an arbitrary constant;
.'. tt =a 4 + C DOS (« — C*).
(7)
Replacing in (7) the values of h and c, and the value of h,
from (1), and dividing both terms of the second number by
Hf we have for the equation of the path.
1 + ri ( F»iJ - 8^) /? r» Bin» /3 + 1 J 008 (»-«')
ti = 1 _____^___- ,
^r»8in»/J
(8)
which is the equation of a conio section, the pole being at
the focus, and the angle (9 — c') being measared from the
THE ORBIT UNDER VARIABLE AVtRACTION.
shorter length of the axis major. For if is the eccentricity
of a conic section, r the focal radias vector, and ^ the
angle between r and that point of a conic section which is
nearest the focus, we hare, '
1 1 + c cos A
— H* ta^m ' ■ ■ ■ '■ - •
Comparing (8) aud (9), we see that
fl» = -(FJ72-2^)i?r«3in»/3 + 1;
4» = d
e\
(»)
(10)
(11)
Now the couis section is an ellipse, parabola, or hyper-
bola, according as e is less than, equal to, or greater than
unity ; and from (10) e is leas than, equal to, or greater
than, unity according as V*B — 2fi is negative, zero, or
positive ; therefore we see that if
2n
F* < -^, e < 1, and the orbit is an ellipse,
2u
F» = -p, e = 1, and the orbit is a parabola.
(12)
(13)
2u
F» > ~, « > 1, and the orbit is a hyperbola. (14)
CoR. 1.— By (1) of ;^rt. 173, we see that the square of
the velocity of a particle falling from infinity to a distance
Ji from Ihe centre of force, for the la^ of atti-action we
, , . . 2«
arc considering, is -^. Hence the above conditions may
be expressed more concisely by saying that the orbit,
d^mihed about this centre 0/ force, wiil be an eUipse, a
Mfe
tirRACfioir.
6 is the eccentricity
vector, and ^ the
aic section which is
(»)
*/3 + l;
(10)
(11)
parabola, or hyper-
to, or greater than
iqnal to, or greater
1 uogative, zero, or
an ellipse, (12)
a parabola, (13)
a hyperbola. (14)
that the square of
finity to a distance
h^ of atti'action we
ve conditions may
ng that the orbit,
ill be an ellipse, a
TBS ORBIT AN BLLIPSJB.
883
parabola, or a hyperbola, according as the velocity is ksa
than, equal to, or greater than, the velocity from infinity.
The species of conic section, therefore, does not depend
on the position of the line in which the particle is pro-
jected, but on the velocity of projection in reference to the
distance of the point of projection fh)m the centre of
force.
Oor. 2.— From (11), we see that 6 — c' is the angle
between the focal radius vector,
r, and that part of the principal
axis which is between the focus
and the point of the orbit which
is nearest to the focus ; i. e., it
is the augle PFA (Fig. 82) ; and
therefore if the principal a^Js is the initial line c' = 0.
185. Suppose the Orbit to be an EUipse.— Here
p-s ^ ^ . eo that from (10) we have
c» = 1 - i (2?* - F»/Z) i? r» sin» /3.
(1)
Now the equation of an ellipse, where r is the focal
radius vector, 9 the angle Iwtween r and the shorter seg-
ment of the miyor axis, 2o the major axis, e the eccon-
'■"" 1 -f- eome*
1 fl Cos
•■* »* = o (1 _ «») + ^JT^ e») '
comparing (2) with (8) of Art, 184, we have
1 ± .
(»)
4
834
TBB ORBIT Air SLUPSK.
substituting f or 1 — «» its yaliw from (1), and aolTingfor
a, we have
" - 2^ _ yJR'
(8)
which shows that ths major axis is independent of the direc-
tion of projection.
We may explain the several quantities which we have
used, by Fig. 82.
B is the point of projection; FB = 5; DB is the line
along which the particle is projected with the velocity V;
FBD = p, the angle of projection; FP = r; PFA = d;
FD = /2 «n /3; if (9 = 90°, the particle is projected from
an apse, i. «., from A or A'.
Cob. 1.~To determine the apsidal diitances, FA and
FA', we must put ^ = 0, (Art 183, 8oh.), and (4) of
Art 184 give us the quadratic equation
(*>
the two roots of which are the reciprocals qf the two apsidal
diitances, a{l —e) and a (1 -f e).
Cor. a.— Since the coefficient of the second term of ( )
is the sum of the roots with their signs changed, we have
a(l-e) ^ a^l-e)
«(l-^=^'l
which gives the latus rtetum of the «rUi.
(»)
rj&
(1), and lolTing for
(3)
mdeni of the direc-
es whioh we have
2; DB is the line
ith the velocity V;
= r; PFA = 6;
is projected from
ittanccB, FA and
Soh.), and (4) of
: 0. (4)
jff the two apaidal
econd term of \ >
langed, we have
in.
(»)
MBPLgR'S LAWS.
CJOB. 3.— From Art 183 we have, calling Tthe time,
T=:
aA
where A is the area swept over by the radius vector in the
time T. Therefore for the time of describing an ellipse,
we have
_ a area of ellipse
2na* Vl — e«
, from (5),
Sff
Vf'
which is the time oceupiad bif the particle in passing from
any point of the ellipse around to the same point cgain.*
l^'i. KflplftX** Xiaws. — By laborious calculation from
an immense series of observations of the planets, and of
Mars in ptsrticular, Kepler enanoiattd the following as the
Ifiws of the planetary motions aboat the Sun.
/, 'A%e orbits of ihe planed are ellipses, of whioh
the Sun occupies a focus.
II. The radius vector of each planet describes
equal areas in equal times.
III. The squares of the periodic times of the
planets are a<i Ine cubes of the ,naJor axes of their
orbits.
187. To Detannin« the Mature of the force which
Acta apon the Planetary Byatem.— (1) ^^"^ ^''"
• oidlad AHmHo nmt.
^^
336
PLAJfBTASr srsTxir.
second of these laws it follows that the planets arc retained
in their orbits hy fm attraction tending to the Snn.
Let {x, y) he the position of a planet at the time t
referred to two co-ordinate axes drawn through the Sun in
the plane of motion of the planet ; X, Y, the component
accelerations due to the attraction acting on it, resolved
parallel to the axes ; then the equations of motions are
^ _
Y;
yX
(1)
Bnt, by Kepler's second law, if A be the area described
by the radius vector, -^ is constant,
rfA ^r*d6
" dt '' ^ di
^♦(''f -yfl^*^"***"*-
DifliBrentiating, we have
d»y dh! __
''dfi ^dfl-^'
.'. xY — ifX = 0, from
(1).
which shows that the axial components of the acceleration,
due to the attraction acting on the planet, are proportional
to the co-ordinates of the planet; and therefore, by the
parallelogram of forces (Art. 30), the resultant of X and Y
puMes through the origin.
W^W V*-^ *K»5SB?*;
planets arc retained
to the Snn.
[anet at the time t
hrough the Sun in
V, the component
ing on it, resolved
of motions are
yx (1)
the area described
Qstant
a(l),
3f the acceleration,
it, are proportional
I therefore, by the
lultant of Jr and F
PLANBTART araTJBM.
887
Hence the forces acting on the planets aU pass throiigh
the Sun's centre.
(2) From the first of these laws it follows that the
central attraction varies inversely as the square of the
distance.
The polar equation of an ellipse, referred to its focus, is
__ o(l- e')
** "^ 1 +• « cos ©'
or
Hence
u =
+
1 + e cos
a(%-^'
1
^ ,
rfe» ■*■ ** ~ a (1 - fl») '
and therefore, if P is the attraction to the focus, we have
[Art 181, (9)],
h»
~ a (1 - e») r»
Hence, if the orbit he an ellipse, described about a centre
of attraction at the focus, the law of intensity is that of the
inverse square of the distance.
(3) From the third law it follows that the attraction of
the Sun (supposed fixed) which acts on a unit of mass of
each of the planets, is the same for each planet at the same
distance.
By Art. 186, Cor. 3, we have
in*
15
S?8
KXAMPLMB.
But by the third kWj T» « o«, and therefore )« tottst be
constant ; i. e., the strength of attraction of the Sun must
be the same for all thi planets. Hence, not only is the law
of force the same for all the planets, but the abaotuU force
is the same.
This very brief discussion of central forces is all that we
II' have space for. To pursue these enquiries farther would
compel us to omit matters that are more especially entitled
to a place in this book. The student who wishes to pursue
the study further is referred to Tait and Steele's Dynamics
of a Particle, or Price's Anal. Mech's, Vol I, or to any
work on Mathematical Astronomy. We shall concludD
with the following examples.
EXAMPLES.
1. A particle describes an ellipse under an ftttraction
always directed to the centre ; it is required to find the law
of the attraction, the velocity at any point of the orbit, and
the periodic time.
(1) The polar equation of the ellipse, the pole at the
centre, is
cos»0 sin'O
a* "•■ i» '
«f>
(1)
• • ""dd = iji - aJ sine cos », (»)
But [Art. 181, (9)] we have
MXAMPLWa,
lerefore ju tiiiist be
a of the Snn must
not only is the law
t the abaotule force
)rce8 is all that we
iries farther would
3 especially entitled
bo wishes to pursue
Steele's Dynamics
, VoL I, or to any
We shall conclude
nder an attraction
ired to find the law
at of the orbit, and
•
e, the pole at the
(1)
J 9, (2)
- Bin' 6). (3)
by (3),
(COS* e — sin* d)], by (2),
by factoring,
A* 1 1 u /iv ** -
m
and therefore the attraction varies directly as the distance.
If ^ = the absolute force we have, by (4),
A» = ^a»ft». (6)
(2) If t; = the velocity, we have, by Art. 188,
t»» = ^ = ^J (Anal. Geom., p. 133)
= tib\ by (6),
where V is the semi^diameter conjugate to t.
.'. V tsh' V/ii.
(3) If 2* = the periodic time, we have, by Art. 182,
_ 2iTfl& %n
V^
, by (6),
and hence the periodic time is independent of the magni-
tude of the ellipse, and depends only on the absolute
central attraction. (;See Tait and Steele's Dynamics of a
840
BXAMPLSa.
Particle, p. 144, also Prioe'fl AnaL Mech's, Vol. I,
p. 516.)
%. A particle describes an ellipse ander an attraction
always directed to one of the foci ; it is required to find the
law of attraction, the velocity, and the periodic time.
(1) Here we have
1 + 6 cos
0(1 -c)» '
u =
and
tPu
dp
du _
dd - '
ecos 9
e sin
(1)
a(l-e»)'
which in (9) of Art. 181 gives
a(l -fl») ~ o(l-c») r»'
(2)
hence the attraction varies inversely as the square of the
distance. Ufi = the absolute force, we have by (2)
(2) By Art 183, Cor. 1, we have
1 - , dM» 20W-1 . ...
j^ = *^' + ^ = 5Mrr^'*'y<^>'
. h* ^(2o« — 1) . ,_. , ...
• *• *^ = p = ~~S ' y ^^^ ^^^ ^*^'
(3) If r = the periodic time we have (Art. 182)
y^ 2rroMl-<^)*
- g^g* (1 -- <**)* _ ^ i
"[^(l-e»)]*~ v^"'
(3)
(4)
(6)
(6)
. Mech's, Vol. I,
ander an attraction
required to find the
periodic time.
e Bin
(1)
_ i
(8)
,8 the square of the
e have by (2)
(3)
ry^JWi (4)
3) and (4). (6)
e (Art. 182)
3ff 1
= — a*,
(8)
SXAMPLSA
841
and hence the periodic time varies aa the square root of
the cube of the major axis.
3. Find the attraction by which a particle may describe
a circle, and also the velocity, and the periodic time, (1)
when the centre of attraction is in the centre of the circle,
and (2) when the centre of attraction is in the circum-
ference.
(1) Let a = the radiiu; then the polar equation, the
pole at the centre, is
Also
1 rfu <i»M _
r = a; .-. u = -; ^ = ^ ^- 0;
. A2 , _ 27ra»
v» = -i, and T= -j-»
(1)
(8)
From (1) and (2) we have
"=1'
and hence the central attraction is equal to the square of
the velocity divided by the radius of the circle.*
and
(2) The equation, is
r = 2a cos 6 ;
t* +
P := 8a»A V =
. • . 2au = sec d,
= 8a V;
8a»A«
and hence the attraction varies inversely as the fifth
• Called tbe OUUrVugal 9bre$. Sm Art. IflS.
343
BXAMPLSa.
powOT of the distance ; and if ju = the absolute force, we
have fi = 8aW;
h> =
and «• =
If T
the periodic time, we have
fir*
2«
na'
(See Priced Anal. Mech., Vol. Ill, p. 518.)
4. Find the attraction by which a particle may deseribe
the lemnisoate of Bernouilli and also the velocity, and the
time of describing one looj^i, the centre of attraction being
in the centre of the lemn ideate, and the equation being
r» = c» cos 3ff.
Ans. P =
«2 —' T
5. Find tlie attraction by which a particle may describe
the cardioid and also the velocity, and the periodic time,
the equation being r = a (1 + cos 0).
(>. Find the attraction by which a particle may describe
a i)arabola, and also the velocity, the centre of attraction
being at the focus, and the equation being r = = i«
Compare (13) of Art. 184.
7. Find the attraction by which a particle may describe
a hyperbola, and the velocity, the centre of attraction being
at the focus, and the equation being r = z — ^•
' ** 1 + e cos ©
Ans. P =
_ JL_ \. ^- M(2a« + 1)
a ( 1 - e») fS ' *^ ~ a
solute force, we
III, p. 518.)
le may describe
jlocity, aud the
.ttraction being
equation being
e may describe
periodic time,
= (Sfia')K.
le may describe
9 of attraction
*" ~" 1 + cos e'
) of Art. 184.
e may describe
ittraction being
(g*-!)
■f fl cos 6
o
MAAMPl^MS.
343
8. K the centre of attraction is at the centre of the
hyperbola, find the attraction, and velocity, the equation
. . eos^ 9 sin* d
Deing — ^ ^
= «».
AtU. P =z 55*' =
■fir', v» =r ^ (f» -> a» + ¥).
9. Find the attraction to the pole under which a particle
will describe (1) the curve whose equation is r = 2a cos nd,
and (2) the curve whose equation is r = -"- - -.
1 — e cos hS
(1 - n») A»
n»)A».
That is, the attraction in the first curve varies
partly as the inverse fifth power, and partly as the inverse
cube, of the distance ; and in the second it varies partly as
the inverse square, aud partly as the inverse cube, of the
distance.
10. A planet revolved round the sun in an orbit with a
major axis four times that of the earth's orbit ; determine
the periodic time of the planet. Ans. 8 years.
11. If a satellite revolved round the earth close to its
surface, determioe the periodic time of the satellite.
Ana. ——J of the moon's period.
13. A body deecrihefl an 'ellipse under the action of a
force in a focus : compare the velocity when it is nearest
the focus with its velocity when it is furthest from the
focus.
Am. As 1 + c : 1 — c, where e is the eccentricity.
13. A body describes an ellipse under the action of a
force to the focus 8', if if be the other focus show that the
844
MXAMPLES.
velocity at any point P may be resolved into two velocities,
respectively at right angles to 8P and HP, and each vary-
ing as HP.
14. A body describes an ellipse under the action of a
force in the centre : if the greatest velocity is three times
the least, find the eccentricity of the ellipse. Ans. | \/2.
15. A body describes an ellipse under the action of a
force in the centre : if the major axis is 20 feet and the
greatest velocity 20 feet per second, find the periodic time.
Ans. It seconds.
16. Find the attraction to the polo under which a par-
ticle may describe an equiangular spiral. ^^^^^
17. M P = ^ (5r» + 8c»), and a particle be projected
from an apse at a distance c with the velocity firom infinity ;
prove that the equation of the orbit ia
r = |(c«'-e-«').
18. M P = 2fi iX — ^, and the particle be projected
from an apse at a distance a with velocity — , prove that
it will be at a distance r after a time
+ y/i» — a*
-f r
Vh"^^)'
!s«)SW!?wse?»«s»wiSMP«w»«*«^
i into two velocities,
HP, and each vary-
tder the action of a
locity is tliree times
llipse. Ana. | ■>/2.
nder the action of a
i is 20 feet and the
id the periodic time.
Ans. It seconds.
under which a par-
al. . » 1
Ana. P<^^'
particle be projected
relooity firom infinity ;
particle be projected
ocity —, prove that
r Vr» — en'
CHAPTER III.
CONSTRAINED MOTION,
188. Definitioiis. — A particle is constrained in its mo-
tion when it is compelled to move along a given fixed curve
or surface. Thus far the subjects of motion have been
particles not constrained by any geometric conditions, but
free to move in such paths as are due to the action of tho
impressed forces. We come now to the ctvse of tho motion
of a particle which is constrained ; that is, in which the
motion is subject, not only to given forces, but to unrieter-
mined reactions. Such cases occur when the particle is in
ft small tube, either smooth or rough, the bore of which is
supposed to be of the same size as the particle ; or when a
small ring slides on a curved wire, with or without friction ;
or when a particle is fastened to a string, or moves on a
given surface. If we substitute for tho curve or surface a
force whoso intensitj'^ and direction are exactly equal to
those of the reaction of the curve, the particle will describe
the same path as before, and we may treat the problem as
if the particle were free to move under the action of this
system of forces, and therefore apply to it the general equa-
tions of motion of a free particle.
189. Kinetic Energy or Vis Viva (Living Force),
and Work. — A particle is constrained to move on a given
smooth plane curve, under given forces in the plane of the
curve, to determine the motion.
Let APC be the curve along which the particle is com-
l)elled to move when acted upon by any given forces. Let
Ox and Oy be the rectangular axes in the plane of the
^i
III!
346
KTNSTIC ENERGY.
curve, ilie axis y positive up-
wards, and {x, y) the place of
the particle, P, at the time t ;
let X, Y, parallel respectively to
the axes of x and y, be the axial
components of the forces, the
mass of the particle being m ;
let R bo the pressure between
the cuvve and particle, which
acta in the norma.1 to the curve, since it is smooth. Then
the equations of motion are
FI«.83
(IP as
(1)
(3)
Multiplying (1) and (2) respectively by dx and dy, and
atlAiag, we have
„d^^y^fl = Xdx + Ydy.
lutograting between the limits t and i^, and calling v^ the
initial velocity, we have
!J ,^ - 'I r,« = CiXlx + Ydy
m
The t«rm ^ ^ »« called the vis vivo/*, or Kinetic Energy
of the mass w; that is, vis viva or kinetic energy is a
quantity which varies as the product of the mass of the
particle and the square of its velocity. There is particular
advantage in def.ning vis viva, or kinotic ener gy, as half
t IB smooth. Then
; (1)
(3)
y by dx and dy, and
J, and calling v^ the
+ rrfy
(3)
, or Kinetic EniTgy
kinetic energy is a
of the mass of Ihe
There is purlicnlar
notic energy, a« half
til, p. MB.
KINETIC SNSR&r>
847
the product of the mass and the »quaro of its velocity.*
The lirst member, therefore, of (3) is the via viva or kinetic
energy of m acquired in its motion from (x,, y^) to {x, y)
under the action of the given foi-ces.
The terms Xdx and Ydy are the products of the axial
compoL-nts of the forces by the axial displacements of the
mass in the time dt, and are tlierefore, tlie elements of work
done by the accelerating forces X and Y in the time dt,
according to the definition of work given in Art 101, Eeni.;
80 that the second member of (3) expresses the work done
by these forces through thfe spaces over which they moved
the mass in the timo between t^ and /. This equation is
called ihe equation of kinetic energy and of work; it shows
that the work done by a force exerting action through a
given distance, is equal to tl»o increase of kinetic energy
which has accrued to the mas« in its motion through that
distance.
If in the motion, kinetic energy is lost, negative work is
done by the force ; i. e., the work i;- -.orcd up as potential
work in the mriss on which the foicp has acted. Thus, if
work is spent on winding up a wulch, that work is stored,
in the coiletl spring, and is thus potential and ready to be
'•estored under adajjted circumstances. Also, if a weight is
raised through a vertical distance, work is Bi)ent ui raising
it, and that work may l)e recovered by lowering the weight
through tlie same voitieal distance.
Tliis theorem, in its most general form, is the modern
principle of nmservation of eneryy ; and is made the funda-
mental theorem of abstract dyiiamics as applied to natural
philosophy.
In this case we have an instance of spaee-intecirah, which,
as wo have seen, gives uw kinetic energy and work ; tlio
Holution (»f problems of kinetic energy and work will 1)6
oxplaiuod in Chap. V.
• Si)ni(> wnUTH iltifliiK viR Ttvk a* llii> whiil<i pcndiict uf the UUM and tb« iquar*
<A Ute vviueltjr. So* MovUi'h II%M Uywuak*, |>. KM.
'jTtyiiffli'wwrontU Bfffl w
SB ACTION OF TBS COMiTRATNTNO CUB VS.
Now if X and Y are functions of the co-ordinates x and
y the second member of (3) can be integrated ; let it be the
differential of some function of x and y, as ^ (ar, y). Inte-
grating (8) on this hypothesis, and supposing w and v^ to
bb the velocities of the particle at the points (x, y) and
(*«> Vit) corresponding to t and t^, "ve have
"c
v«») = <^ (^, y) - ^ («o. yo)
(4)
»^'i
i-i'i
which shows that the kinetic energy gained by the particle
constrained to move, under the forces X, F, along
any path whatever, from the point (a-^, y,) to the point
{x, y), is entirely independent of the path pursued, and
depends only upon the co-ordiniites of the points left and
unived at; the reaction R does not appear, which is clearly
as it should be, since it does no work, because it acts in a
line perpendicular to the direction of motion.
190. To Find the Reaction of the Constraining
Curve. — For conveniens, the muss oi the particle iniiy bo
taken as unity. Multiplying (1) and (2) of Art. 189 by
^ and -V-; subtracting the former from the latter, and
solving for R, we have,
„ _ (Pydx — rf»x d y -^dy ydx
""- " ilPda "^ ds ' ds
«^-. + ^% - Y%^'^^n^)<^^ ^rLm (1)
in which p is the radius of curvature at the point P. The
lu«t two terms of (1) are the nornml conipoiionts of the
impressed fonrs; and (liorefore, if the purticlf were at rej^t,
tlioy would donule I be whole prcsaure on the curve ; but
Mto
fO CUB vs. ,
co-ordiuates x and
•ated ; let it be tho
as ^ (x, y). lutc-
posing V and v, to
poiuts (x, y) and
ve
(«o. yo)
(4)
led by the particle
jes X, F, along
j»,) to tho point
)ath pursued, and
the points left and
ir, which is clearly
icause it acts in a
bion.
;he Constraining
he particle nuiy bo
2) of Art 189 by
in the latter, and
at
:;i) of Art. 102 (1)
the point P. The
•onipononts of the
iirtich' Were at rest,
on the curve ; but
REACTION OF THE CONSTRAWirfO CURVE.
349
the particle being in motion, there ;.> an additional pressure
on the curve expressed by -•
In the above reasoning we have considered the particle to
be on the concave side of the curve, and the resultant of X
and Y to act towards the convex side along some line as PF
so as to produce pressure against the curve. If ou the
contrary, this resultant acts towards tho concave side, along
PF' for example, then, whether the jiartiole be on tho
concave or convex side, the pressure against the curve will
.3
bo the diflEerence between - and tho normal resultant of X
9
and Y.
191. To Find the Point where the Particle WiU
Leave the Constraining Curve. — It is evident that at
that point. R = 0, as there will be no pressure against the
curve. Therefore (1) of Art. 190 becomes
- = Y~- X^
p ds ds
= F' cos F'PR
if F' be tho resultant of X and Y.
.-. «)»= F'p con F'PR
= 2F'-i chord of curvature in the direction PF'.
Comparing this with (6) of Art. 140, we see that the
particle will have the curve at the point where its velocity is
such as timtild be produced by the resultant force then acting
on it. if continued constant during its fall from rest through
a space equal to J of the cnord of curvature parallel to that
resultant. (See Tait and Steele's Dynamics of a Particle,
p. 170.)
m
■/miii i>u a»!aiM .i »t v i «
350
COySTRAlNSD UOTtOK.
192. Constrained Motion Undor the Action
Gravity. — When gravity is the only force acting on thi
particle, the formnlsB are simplified. Taking the axis of y
vertical and positive downwards, the forces become
X=0, and r= -{-^;
and for the velocity we have, by (3) of Art 189,
where y, is the initial space corresponding to the time t^.
"^r the pressure on the curve we have, by (1) of Art. 190,
p ^ da
m
If the origin be where the motion of the particle begins,
the initial velocity and space are zero, and (1) becomes
i«^==i?y.
(8)
This shows that the velocity of the particle at any time
is entirely independent of the form of the cnrve on which
it moves ; and depends fjolely on the perpendicular distance
through which it falls.
193. Motion on a Circnlar Aro in a Vertical
Plane. — Take the vertical dianiete' a? axis "of y, and its
lower extremity m origin ; then the equation of the circle is
3f = 2ay~f;
dT
a-y X
dy __ ds
j4
^■' "(PP ■ -^f^.
a)
' the AotLon
irce acting on thi
;ing the axis of y
B8 become
r*
rt 189,
I (1)
g to the time t^.
by (1) of Art. 100,
(3)
le particle begins,
I (1) becomes
(3)
tide at any time
le cnrve on which
endicalar distance
In a Vertical
nxis'of f/y and its
ion of the oirclo is
(1)
1
MOTTON ON A CIXCULAR ABC
Let (ife, A) be the point K where
the particle starts from rest, and (x, y)
the point P where it is at the time /.
Then the particle will have fallen
through the height HM — h — y,
and hence from (3) of Art. 492 we
have
351
(8)
Hence the velocity is a minimum when y = h, and a
maximum when y = 0; and this maximum velocity will
carry the particle through to iT' at the distance h above
the horizontal line through 0.
To find the time occupied by the particle in its descent
from K to the lowest point, 0, we have from (2)
dt
da
Vajr (* - y)
— ady
Vfi9{h-y)(iay-f)
by(i) (3)
the negative sign being taken rinoe / is a decreasing func-
tion of 8.
This expression does not admit of integration ; it may be
reduced to an elliptic integral of the first kind, and tables
are given of the approximate values of the integral for
given values of y.*
If, however, the radius of the circle is large, and the
greatest distance KO, over which the particle moves, is
Biurtll, wo may develoiw (3) into a series of teruis in ascend-
ing powers of | , and thus find the integral approximately.
• See Leuondre i Twit* des, PoncU<in« KUIplUiaei.
nwimMiiWi
•
'isSuSistsosiiLWit
363
TBS St MPLS PENDULUM.
Let ST be the time of moti'm of the particle from K\o K',
i. e., from y = h, thiougli y =: 0, to y = h again, then (3)
becon\08
r = - \ /? r~~J^-- (i - ^v^
~ V^'A. L
i + *^ + H
(ir--]
</y
V/iy — y* '
integrating eacli t«rm eeparately we have
^-^«4m
W
which is the complete expression for the time of moving
from the extreme position K on one side of the vertical to
the extreme position K' on the other; this is called an
OBcillatiori. (See Price'e Anal. Mcchs., Vol. III., p. 588).
If the arc le very small, h is very small in comparison
with a, and all the terms containing — will be very small,
and by neglecting them (4) becomes
'■=Vi-
(«)
194. The Simple Pendulnm.— Tri'^toad of 8uppoBin/>:
the j article to move on a curve, wo may imagino it r,aa-
pended by a string of invariable length, or a thin rod
considered of no weight, and moving iu a vertical plane
ftlmnt Mil' point 0: for, whether the force acting on the
jmrticle be the reaction af the curve or the tension of the
string, its intensity is the Hame, while its direction, in
either caae is along the uormai Lu the curve.
MM
if.
icle from K to K',
: h again, then (3)
•]
dy
^hy — y» '
•4/ W
]
(4)
le time of moving
of the verticiil to
this is called an
"ol. in., p. 588).
nail in comparison
will be very small,
(5)
toad of supposing
ly imagine it ^ias-
th, or a thin rod
1 a vertical piano
rce acting on the
the tension of the
e its dtrection, in
•ve.
S^^^m^^^^^E^B^Sn
RSLATION OF TIME, LBNOTH, STC.
353
When the particle is supposed to be snspended by a
thread without weight, it bocomes what is termed a simple
pendulum ; and although such an instrument can never be
jjerfectly attained, but exists only in theory, yet approxima-
tions may be made to it sufficiently near for practical pur-
poses, and by means of Dynamics we may reduce the
calculation of the motion of such a pendulum to that of
the simple pendulum.
If I is the length of the rod, the time of an oscillation is
approximately given by the formula
= ''v/l
0)
when site angle <rf oscillation is very small, i. «., not ex-
ceeding »dout 4°;* and therefore, for all angles between
this and aero, the tones of oscillation of the same pen-
dulum will not perceptibly differ ; i. e., in very mall arcs
the os<;illatio7iH may be regarded as isochronal, or as all
performed in the same time.
195. Relation ol Time, Length, end Force of
aravity.— From (I) of Art. 191, we have Tx y/lif yis
constant ; Tec -— ii I is conBtpnt; gcc l\t Tn^ constant,
yg
that is
(1) For the same place the times of osdUaiion are as the
square roots of the lengths cf the pendulums.
{%) For the same pendulum the times of ountfntion are
inversely as the square roofs of the force of gravity a4
different places.
• If the InliUl lnriln»iloo Is 6', the seeoad tmna of (4) U only O.WOm ;»»•*•
•MOB* Krm la only O.UOOOitt.
md
354
HSrOET or MOUNTAIN BSTSBMINSD.
(3) For tbs same time the lengths of pendulums vary as
the force of gravity.
Hence by means of the pendulum the force of gravity at
different places of the earth's surface may be determined.
Let L be the length of a pendulum which vibrates seconds
at the place where the value of jr is to be found ; then ftom
(1) of Art. 194 we have
1 = •^Y -; •'• 9 ~ "*^;
(1)
and from this formula g has been calculated at many places
on the eai-tb. Tho method of determining L accurately
will bo investigated in Chap. VII.
Cor. — If n bo the numl>er of vibratioacs performed dur-
ing ^seconds, and iTthe time of one vibration,
then
= ^by<l)ofArt. 194. = fy^f-
(2)
Since gravity decreases according to a known law, as we
ascend above the earth's surface, the comparison of flbe
times of vibration of the same pendulum on the top of a
monnfain and at its base, would give approximately its
height.
106. The Height of a Mountain Determined with
the Pendulum. — A hcciiuiIh jnniiuUtm is carried to the top
of a mountain ; required to find ^h$ huighl nf the mountain
by observing the chat^e in the lime of oscillation.
Tjet r be the radius of the «f( Ih onnsidered spherical , h
the height of the mountain above tho surface ; I the length
of the pendulum ; g and g' the values of gravity ou the
earth's surface, and at the top of the mountain respectively.
Then (Art. 1T4) we have
\XINMD.
fndulmns vary as
force of gravity at
ay be determined.
I vibrates seconds
found ; then from
*L;
(1)
;ed at many places
niiig L accurately
IS performed dur-
iratioD,
(2)
known law, as we
omparison of the
I on the top u( a
approximately its
>etermiiiecl with
s carried to the top
hi of the mountain
illation.
dt red spherical , h
rface ; I the length
of gravity ou the
intiiiu respectively.
SB^HssMSssw^rfj&a^ss^caB
mmm
BEI6BT or MOUXTAIlf DMTKRMINED.
9
g'~\ r /'
9
a' - ^ •
y - (r -f hy*
355
(1)
which is the force of gravity at the top of the mountain.
Let n — the number of oscillations which the seconds
pendulum at the top of the mountain makes in 24 houi-s ;
24 X 60 X 60
then the time of osciUatioo
(1) of Ai-t. 195, we have
24 X 60 X 60
Hence from
h
r
24 X 60 X 60
n
-1, (since Tr'Y/- = l), (2)
which gives the height of the mountain in terms of the
radius of the earth. For the sake of an example, suppose
the pendulum to lose 5 seconds 1n a day ; that is, to make
'< oscillations less than it would make on the surface of the
earth.
Then
» = 24 X 60 X 60 — 5 ;
ii in (2) gives
A
= ('-
24 X 00 X
_U X 60 X 60
24 X 60 X 00 - I
-1
h =
4000
24 X 60 X 12
= J mile, nearly,
nearly ;
»4 X 6») X 12
r being 4000 miles (approximately).
197. The Depth of a Mine Oetenniued by Ob-
serving the Change of Oscillation in a Seconds
Pendulum.— [jf I r ho the radius of the earth ae in the
356
CENTRIPETAL FORCE.
last case ; h the depth of the mine ; g and g' the values of
' gravity on the earth's surface and at the bottom of tlio
mine. Then (Art. 171) we have
g' r~h
(1)
Let n = the number of oscillations which the S'econds
pendulum at the bottom of the mine makes in 24 hours.
Then
24 X 60 X 60
n
_ / Ir
r — h'
r ~ V24 X 60 X 60/ '
from which h can be found. If, as before, the pendulum
loses 5 seconds a day, we have
* = '-(■
24
__j y
X 60 X 12/
nearly.
12 X 60 X 12
. • . h = ^ mile nearly.
(See Price's Anal. Mech's, Vol. I, p. 590, also Pratt's
Mech's, p. 376.)
198. Centripetal and Centrifugal Forces.
;iuce
the pressure — , at any point, depends entirely upon tlie
velocity at that point and the radius of curvature, it would
remain the same if the forces X and Y wer^ *K)th zero, in
which case it would be the whole norma pressure, B,
I g' the values of
lie bottom of tlu;
(1)
hich the seconds
es in 34 hours.
-h)
re, the penduhim
3 _,
rJ
590, also Pratt's
Forces. — Face
jntirely upon tlie
rvfjture, it would
yer« fwth zero, in
mil; pressure, /?.
,,«,, r.'»>mmimmmKgmmtmmmKt&iKKmm^
CBN'^ rrVQAL FORCS.
867
against the curve. It is easily seen, therefore, that this
pressure arises entirely from the inertia of the moving
particle, i. e., from it- tendency at any point, to move in
the direction of a tangent ; and this tendency to motion
along the tangent uecessurily causo-- it to exert a pre iire
against the deflecting curve, and wluJi requires the curve
4
to oppose the resistance -• Hence, since tl- particle if
left to itself, or if left to the action of a force uong the tan-
gent, would, by the law of inertia, continue to niovi' along
that tangent, - is the effect of the force which deflects the
particle from its otherwise rectilinear path, and draws it
towards the centre of curvature. This force is called the
Centripetal Force, which, therefore, may be defined to be
the force which deflects a particle from its otherwise recti-
linear path. The efjual and opposite reaction exerted away
from the centre is called thi' Centrifugal Force, which may
be defined to be the resistance which the inertia of a particle
in motion opposes to whatever deflects it from its rectilinear
path. Centri[)etal and centrifugal are tlei-efore the same
quantity under different aspects. The action of the former
is towards the centre of curvature, while that of the latter
wfron the centre of curvature. The two are called central
forces. They determine the direction of motion of the par-
ticle but do not affect the velocity, since they aet continu-
ally at right angles to its puth. If a particle, attached to a
string, be whirled about a centre, the intensity of these
central forces is measured by the tension of the string. If
the string be cut, the particle will move along a tangent to
the curvo with unchanged velocity.
Cob. 1.— If m be the mass moving with velocity v, its
centrifugal force is »i -• If w bo the angular velocity
^f^i^ifv^" ,'WrTr^
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■■ *T"^fif i rftVTMi'fr'^lnfimitTWr
898
cMNimaveAh forcm,
described by the mdms of corvature, thea (Art. 160, Ex. 1),
V =. pu, and CQQfleqaently
the centrifugal force of wt = mw^.
(1)
Cob. 8. — Let m move in a cinjle with a constant velocity,
v\ lot a = the radius of the circle, and T the time of a
complete revolution ; then drra := vT'y
4rr*fl
the oentriftagal force of m = «• -^ ;
w
and thus the centrifugal force i% a circle varies directly as
the radius of tlie circle, and inversely as the square uf the
periodic time.
Cob. 3.— 11' fa irsoves in the ci:K!le with a constant
angular velocity, w, then (Art. 160, Ex. 1), t* = ou ;
the centrifugal force of «i = tn «- *a ;
(8)
and therefore varies directly as the radius of the circle.
Thus if a particle of mass m is fastened by a string of
length a to a point in a horizontal plane, and describeb a
circle in the plane about the given point as centre, the cen-
trifugal forco produces a tension of the string, and if u is
the constant angular velocity, the tension = tn u^
199. Th* OMtriftigia Foree Kt tii« Bqnator.— Let
R demote the equatorial radius of the earth = 209:88802*
feet, T the time of revolution upon its axis = 86164
seconds, and -n = 3.1418926. Substituting these values in
(2) of Art. 198, and denoting the centrifugal fbroe at the
equator by/, and the mass by unity, we have
4ffS/Z
/= ^^ = 0.11126 feet.
d)
* Bnejp. Brit., Art. Qweimf.
1 (Art. 160, Ex.1),
muV«
(1)
k constant velocity,
T the time of a
(2)
( varies directly as
the square uf the
with a constant
m itra \
(3)
8 of the circle.
sned by a string of
le, and describes a
as centre, the cen-
string, and if u is
1 = m <.<*a,
!• Bqwitor.— Let
arth = 2093680S*
its axis = 88164
ing these valaes in
Ifugal fbroe at the
have
ct.
(1)
Hi
nn
OMiTTBiruaAL roBom,
86d
The force «rf gravity at the equator has been found to be
32.09023 ; if this force were Eot dinainidbed by the cen-
tiifogal force; t. «., if the earth d.l not revolve on its
axis the force of gravity at the equator would be
G = 32.09022 + 0.11126 = 32.20148 feet. (2)
To determine the relation between the oentriftigal force
and the force of gravity, we divide (1) by (2) which gives
l_ 0.11126 _ 1^ J
(8)
N
C /(?
that IS, tU centrifugal force at the equator is gijf of that
which the force of gravity at the equator would be if the
earth did not rotate.
200. CeiitiiltagidFonMatZMffir-
ent LatitadM on tlM Barth.— I at
P be any particle on the earch's surface
describing a circumference about the
axis, *.V5, with tlie radius PL. Let
<l> z= AOP = the latitude of P; J?
the radius, AC, of the earth ; and I^
the radius PD of the parallel of lati-
tude passing through P. Th«m we have
i2' :^ i2 cos <t>.
Let the centrifagal force at tha point P, whidi is exerted
in tho direction of the radius DP, be repwier.ted by the
lino PB. Resolve this into the two components Pf, act-
ing along the tangent, and PE, acting along the normal.
Then by (2) ol Art 198 we have
s
n|.«
(1)
, by (1).
w
360
CSITTRtmjftAL PORCJS.
Hence, the centrifugal force at any point on (he earth's
surface varies directly as the cosine of the latitude of the
place.
For the normal component we have
PB = PB cos ^
i-n*R co8» . ...
= jiT— ^7 (2)
= / co8» ^, by (1) of Art 197. (3)
Honce^ the component of the centrifugal force which directly
opposes the force of gravity, at any point on the earth's sur-
face, is equal to the centrifugal force at the equator, mul-
tiplied by the square of the cosine of the latitude of the
place.
Also
PF = PB Bin
— ^^^ **'" cos ^
. by (2)
.kl .
= J sin 20. by (1) of Art. 197 ; (4)
that is, the component of the centrifugal force which tends
to draw particles from any parallel of laltfnde, P, towards
the equator, and to cause the earth to assume the figure
of an oblate spheroid, varies as the sine of twice the
latitude.
The preceding calculation is made on the hypothesis that
the earth is a perfect sphere, whereas it is an oblate
spheroid ; and the attraction of the eaith on particles at
its surface decreases as we pass from the poles to the
equator. The t>endalum furnishes the most accurate
m
m'nt on (he earth's
' the latitude of the
by (2)
(1) of Alt 197. (3)
force which directly
! on the earth's sur-
it the equator, mul-
the latitude of the
(l)of Art. 197;(4)
I force which tends
lattinde, P, towards
} assume the figure
sine of twice ike
the hypothesis that
as it is an oblate
arth on particles at
1 the poles to the
the most accnrate
TBJl CONICAL PENDULUX.
861
na-N
method of determining the force of gravity at different
places on the earth's surface.
201. The Conical Pendn-
lum.— The Gtovemor. — Suppose
a particle, P, of mass m, to be at-
tached to one end of a string of
length /, the other end of which is
fixed at A. The particle is made
to describe a horizontal circle of
radius PO, with uniform velocity
rouml the vertical axis A0,&o that
it makes n revolutions per second.
It is required to find the inclina-
tion, 0, of the string to the vertical,
and the tension of *he string.
The velocity of P in feet per second = 2rrn • OP = 2ir« I
sin e. The forces acting upon it are the tension, 7, of the
string, the weight m, of the particle, and the centrifugal
force, m — ^j— ^ — (Art 198). Hence resolving, we have
for horizontal forces, T* sin (9 = »j. 4n»»»» / sin e ; (1)
for vertical forces, iTcos© = mg. (%)
From (1) 7'=m.4r.V?, (3)
which in (2) givea
whore 7* and 9 ere completely determined.
If the string be replaced by u rigid rod, which can turn
about ^ in a ball and socket joint, the instrument is called
a conical pendulum, and occurs in the governor of the
steam-engine.
16
d6d
MXAMrXtMB.
EXAMPLES.
1. If the length of the seconds pendalam be 39.1393
inches in London, find the valne of jr to three places of
decimals. Ana. 82.191 feet.
2. In what time wiU a pendnlum nbratc whose length is
15 inches ? Ans. 0.63 sec. nearly.
3. In T.hat time will a pendnlnm vibrate, whose length is
double that of a seconds pendulum P Ang. 1.41 sees.
4. How many vibrations will a pendalam 3 feet long
make in a minute ? Ant. 62.65.
5. A pendulum which beats seconds, is taken to the top
of a mountain one mile high ; it is required to find the
number of seconds which it wiP lose in 12 hours, allowing
the radius of the earth to be 4000 miles. Ans. 10.8 sees.
6. What is the length of a pendulum to beat seconds at
the place where a body falls 16^ ft in The first second 7
Ann. 39.11 ins. r early.
7. If 39.11 ins. be taken as the length of the seconds
pendulum, how long must a pendulum be to beat 10 times
in a minute ? Ans. 117^ ft
8. A particle slides down the arc of a circle to the
lowest point ; find the velocity at the lowest point, if the
angle described round the centre is 60°. Ana. Vgr.
9. A pendulum which oscillates in a second at one place,
is carried to another place where it makes 120 more oscil-
lations in a day; compare the force of gravity at the latter
place with that at the former. Ana, (|fH)**
10. Find the number of vibrations, n, which a pendulum
will gain in Jf seconds by shortening the length of the
peudulum.
m
ilam 1)6 39.1393
) three places of
1. 32.191 feeL
whose length is
63 sec. nearly.
, whoso length is
[ng. 1.41 sees.
urn 3 feet long
Ans. 62.66.
aken to the top
ired to find the
honre, allowing
Ins. 10.8 sees.
beat seconds at
first Hecond ?
II ins, r«arly.
I of the seconds
bo beat 10 times
Ans. m^it
a circle to the
est point, if the
Ana. Vffr.
•nd at one place,
120 more oscil-
ity at the latter
^»*. (IIH)'. _
ioh a pendulnm
le length of the
Let the length, i, be decreased by a small qnantity,
1 1, and let n bo increased by »( ; then from (2) of Art 196
we get
which, divided by (2) of Art 195, gives
Ilenoe «, = — 1.
11. If a pendulnm be 45 inches long, ho«r many vibra-
tions will it gain in one day if the bob* be screwed up one
turn, the screw having 32 threads to the inch P
Ans. 30.
13. If a clock loses two minutes a day, how many turns
to the right hand must we give the nut in order to correct
its error, supposing the screw to have 50 threads to the
inch? • Ans, 5*4 tarns.
13. A mean solar day contains 24 hours, 3 minates,
56 '5 seconds, sidereal time; calculat«d the length of the
pendulum of a clock beating sidereal seconds in London.
See Ex. 1. Ans. 38-925 inches.
14. A heavy ball, scspended by a fine wire, vibrates in a
small arc ; 48 vibrations are counted in 3 minutes. Cal-
culate the length of the wire. Ans. 45 -87 feet
15. The height of the cupola of St Paul's, above the
6oor, is 340 ft; calculate the number of vibmtions a heavy
body would make in half an hour, if suspended from the
dome by a fine wire which reaches to within 6 inches of
the floor, Ans. 176-4.
• The kwer extremity ot the pendnhua.
'I :!,
.■I :,
wm
16. A seconds pendulnm is carried to the top of a
mountain m milea high ; aasuming that the force of
gravity varies inversely as the square of the distance from
the centre of the earth, find the time of an oscillation.
/4000 + m\
17. Prove that the lengths of pendulums vibrating dar-
ing the same time at the same place are inversely as the
squares of the number of oscillations.
18. In a series of experiments made at Harton coal-pit, a
pendulum which beat seconds at the surface, gained 2|
beats in a day at a depth of 12C0 ft. ; if g and g' be tho
force of gravity at the surface and at the depth mentioned,
show that
9'-9 _ 1
• "- itto'o-
19. A pendulum is found to make 640 vibrations at the
equator in the same time that it makes 641 at Greenwich;
if a string hanging vertically can just sustain 80 lbs. at
Greenwich, how many lbs. can the same string sustain at
tho equator ? Am. 80J lbs. about.
20. Find the time of descent of a particle down the arc
of a cycloid, the axis of the cycloid being vertical and vertex
downward ; and show that the time of descent to the lowest
point is the same whatever point of the curve the particle
starts from. /^
\ 9
21. If in Ex. 20 the particle begins to move from the
extremity of the baae of the cycloid find the prtssure at the
lowest point of the curve.
Ans. 2^; i. e., the pressure ia twice the weight of tho
particle.
to the top of a
bat the force of
the distance from
m oscillation.
tOOQ + m\
4000 )
Bees.
ms vibrating dnr-
ft inversely as the
Harton coal-pit, a
iirface, gained 2|
if g and g' be the
depth mentioned,
Tibrations at the
541 at Greenwich ;
sustain 80 lbs. at
string snetain at
80^ lbs. about.
tide down the arc
irertical and vertex
Micnt to the lowest
curve the particle
Am.
Wy
to move from the
he pressure at the
he weight
mm m m smmmmmmm
HXAMPLSa.
d65
%%. Find the prepare on the lowest point of the carve
in Art 193, (1) when the particle starts from rest at the
highest point. A, (Fig. 84), (ii) when it starts from rest at
the point B.
Ana. (1) 5^; (2) 3^; i.e., (1) the pressure is five times
the weight of the particle and (2) it is three times the
weight of the particle.
23. In the simple pendulum find the point at which the
tension on the string is the same as when the particle
bangs at rest
Am. y = f.'., where h is the height from which the
pendulum has fallen.
24. If a particle be compelled to move in a circle with a
velocity of 300 yards per minute, the radius of the circle
being 16 ft, find the centrifugal forco.
Ana. 14' 06 ft. per sec.
25. If a body, weighing 17 tons, move on the circum-
ference of a circlo, whose radius is 1110 ft., with a velocity
of 16 ft per sec., find the centrifugal force in tons (take
g = 32-1948). Am. 0-1217 ton.
26. If a l)ody, weighing 1000 lbs., be constrained to move
in a circle, whose radius is 100 ft, by means of a string
capable of sustaining a strain not exceeding 450 lbs., find
the velocity at the moment the string breaks.
Ana. 38-06 ft per sec.
27. If a railway carriage, weighing 7-21 tons, moving at
the rate of 30 miles per hour, describe a portion of a circle
whose radius is 460 yards, find its centrifugal force in tons.
Am. 0-314 ton.
28. If the centrifugal force, in a circle of 100 ft radius,
be 146 ft per sec, find the periodic time.
Ana. 6-2 sees.
4^1 ;
366
XXAMPnSA
29. If the oentrifa^l force be 131 oss., and the radias
of the circle 100 ft., the periodic time being one hour, find
the weight of the body. Ans. 386- 309 tons.
30. Find the force towards the centre required to make
a body move uniformly in a circle whose mdius is 5 ft.,
with Huch a velocity as to complete a revolution in 6 sees.
5 ■
Ans.
31. A stone of one lb. weight is whirled round horizon-
tally by a string two yards long having one end fixed ; find
the time of revolution when the tension of the string is 3 lbs.
Ana. 2rT
VI
sees.
38. A weight, w, is placed on a horieontal bar, OA,
which is made to revolve round a vertical axis at 0, with
the angular velocity «; it is required to determine the
position, A, of the weight, when it is upon the point of
sliding, the coefficient of friction being /.
Am. OA
33, Find the diminution of gravity at the Sun's equator
caused by the centrifugal force, the radius of the Sun being
441000 miles, and the time of levolation on his axis being
607 h. 48 m. Ans. 0- 0192 ft. per 8e<;.
84. Find the centrifugal force at the equator of Mercury,
the radius being 1670 miles, and the time of revolution
24 h. 6 m. Ans. 0- 0435 ft. per sec
35. Find the centrifugaJ force at the equator, (1) of
Venus, radius being 3900 miles and time of revolution
23 h. 21 m., (2) of Mars, radius being 2050 miles and
iwriodic time 34 h. 37 m., (3) of Jupiter, radius being
43500 miles and periodic time 9 h. 56 m., and (4) of Saturn,
radius being 39580 miles and periodic time 10 h. 29 m.
(8., and tbe radias
;ing one hour, find
s. 386-309 tons.
required to make
086 rndiuB is 5 ft.,
olution in 6 sees.
Ana. -=-'
ed round horizon-
ne end fixed ; find
! the string is 3 lbs.
s. 2n
VI
sees.
rieontal bar, OA,
cal axis at 0, with
to determine the
ipou tbe point of
Ans. OA
the Sun's equator
8 of the Sun being
1 on his axis being
Qin ft. per 8e<;.
jnator of Mercury,
ime of revolution
0435 ft. per sec
le equator, (1) of
mo of revolution
r 2050 miles and
iter, radius being
and (4) of Saturn,
le 10 h. 29 m.
mmmimmmmmmsmmmim
MXAMPLXa.
867
Ana. (1) 0> 11504 ft. per sec.; (2) 0>0544 ft per sec.;
(3) 7-0907 ft per 8oa; (4) 6- 7924 ft per sea
36. Find the effect of centrifngal force in diminishing
gravity in the latitude of 60°. [See (3) of Art. 200).
Ana. 0-028 ft. per sec.
37. Find (1) the diminution of gravity caused by cen-
trifugal force, and (2) the component which urges particles
towards the equator, at the latitude uf 23°.
Ans. (1) 0-09 ft per sec; (2) 0-04 ft per sec.
38. A railway carriage, weighing 12 tons, is moving
along a circle of radius 720 yards, at the rate of 32 miles
an hour; find the horizontal pressure on the nils.
Ana. 0-39 ton, nearly.
39. A railway train is going ouoothly along a curve of
500 yards radius at the rate of 30 miles an hour ; find at
what angle a plumb-line hanging in one of the carriages
will be inclined to the vertical Am. 2° 14' nearly.
40. The attractive force of a mountain horizontally is/,
and the force of gravity is ^; show that the time of vibra-
tion of a pendulum will be ww j - y^ ;
of the pendulum.
41 1 In motion of a particle down a cycloid prove that the
vertical Telocity is greatest when it has completed half its
vertical descent
42. When a partide falls from the highest to the lowest
point of a cycloid show that it describes half the path in
two-thirds of the time.
43. A railway train is moving smoothly along a curve at
the rate of 60 miles an hoar, and in one of the carriages a
pendulum, which would ordinarily oscillate seconds, is
observed to oscillate 121 times in two minutes. Show that
the radiOB of tiia ourve is very nearly a qiiarter of a mile.
a being the length
368
EXAMPLES.
44. One end of a string is fixed ; to the other end a
particle is attached which describes a horissontal circle with
uniform velocity so that the string is always inclined at an
angle of 60" to the vertical ; show that the velocity of the
particle is that which would be acquired in falling freely
from rest through a space equal to three-fourths of the
length of the string.
■15. The horizontal attraction of a mountain on a particle
at a cciiain place is such as would produce in it an accelera-
tion denotfld by "• Show that a seconds pendulum at that
, .,, . 21600. , . , ,
place will gam — j— beats ma da>, very nearly.
46. In Art 201, suppose I equal to 2 ft. and m to be 20
lbs., and that the system makes 10 revolutions per sec., and
^ = 32; find e and T.
Ans. e — co8-» — ^; T = SOOtt* pounds.
47. A tube, bent into the form of a plane curve, revolves
with a given angular velocity, about its vertical axis ; it is
required to determine the form of the tube, when a heavy
particle placed in it remains at rest in all parts of the
tube.
(Take the vertical axis for the axis of y, and the axis of x
horizontal, and let <o = the constant angular velocity).
Ans. aj^uS = 'igy, if a; = when y = 0, t. c, the curve
is a parabola whose axis is vertical and vertex downwards.
48. A particle moves in a smooth straight tube which
revolves with constant angular velocity round a vertical
axis to which it is perpendicular, to determine the curve
traced by the particle.
Let <i) = the constant angular velocity ; and (r, 0) the
position of the particle at -the time /, and lot r =: a when
the other end a
lontal circle with
^s inclined at an
le velocity of the
in falling freely
:e-fourtha of the
tain on a particle
in it an accelera-
lendulnm at that
early.
and nt to be 20
ions per sec., and
SOOttS pounds.
le curve, revolves
ertical axis ; it is
}, when a heavy
all parts of the
and the axis of x
Qgular velocity).
t. c, the curve
bex downwards.
ight tube which
round a vertical
rmine the curve
; and (r, 0) the
. lot r =s a when
M''t(WJSlflfiWltfl!rtfll|JWlWffe:J'v
SXAMPLSa.
369
/ = 0. Then since the motion of the ^mrticle ia due
entirely to the centrifhgal force, we have
dr
if ^ = 0, when r = o. Hence we have
CHAPTER IV. .
IMPACT.
202. An Xsppnlsive Force.— Hitherto wc have con-
sidered force only as continuous, i. e., as acting through a
definite and finite portion of time, and producing a finite
chango of velocity in that time. Guch a force is measured
at any instant by the mass on which it acts multiplied by
the acceleration which it causca If a particle of mass m be
moving with a velocity v, and be reta;ded by a constant
ibrce which brings it to rest in the time /, then the measure
of this force is -r- (Art. 20). Now suppose the lime t dur-
ing which the particli is brought to rest to be made very
ginall then tl»o force required to bring it to rest must bo
▼cry large ; rod if wo suppose t so smal! that we are unable
to measure it, then the force becomes so great that we are
nnablo to obtain its measure. A typical cnse is the blow of
a hammer. Here the time during which there ir contact is
apparently iinlnitcsimal, certainly too small to be measured
by any ordinary methwle ; yet the effect produced is con-
Bidcrable. Similarly wiien a cricket ball ia driven back by
a blow from a but, the original velocity of the ball is
destroyed and a new velocity generated. Also \,hen a bul-
let is discharged from a gun, a large veloc'ty is generattid
in an extremely brief time. Forcee acting in this way are
called impulsive forces. An impulsive force may thertfore
h' defined to he a forc^. which protluces a finite change cf
motion in an indefinitely brief time An Impulse is the
effect of o hhw.
In duch cases as ^hese it is inipossiblo accurately to
determine the force and time ; out wo can dotormino
■i
^
imi
tmm--
to wo have con-
octiog thiough a
i-oducing a finite
brcc is measured
icta multiplied by
^!cle of mass m be
ed by a conataut
then the measure
ie the time t dyr-
to be mmle very
to rest must bo
lat wo are unable
great that we are
ise is the blow of
here ir contact is
11 to be measured
prodticed is con-
la driven back by
ty of the ball is
Also \,hen a bul-
c'ty is generattid
[ in this way are
'ce may thert/ore
I finite change cf
n Impulse is the
»lo accurately to
can dotormino
IMPACT on COLLISrOlf.
371
their product, or P^, since thia is merely the change
in velocity caused by the blow (Art. 20). Hence, in
the case of blows, or impulsive forces, we do not attempt
to measure the force and the time of action separately, but
simply take the whole momentum producefl or destroyed, as
the mcasnre of the impulse. Because impulsive forces pro-
duce their effects in an indefinitely short time they are
sometimes called instantaneous forces, i. c., forces requinng
no time for thdr ection. But no such force exists in
nature; every force requires time for its action. There is
no case in nature in which a finit« change of motion ie
produced in an infinitesimal .-f time ; for, whenever a
finite velocity is generated or destroyed, a finite time is
occupied in the process, though we may be unable to
measure it, even approximately.
203. Impact or CoUisioa— When two bodies in rela-
tive motion come into contact with each jther, an impact
or collision is said to take place, and pressure begins to act
between them to prevent any of their parts from jointly
occupying the same space. This force increases from zero,
when the collision begins, np to a very large magnitude at
the instant of greatest compression. If, as is always the
case In nature, each body possesses some degree of eksticity,
and if they are not kept together aft«r the impact by
cohe, ion or by some artificial means, the mutual pressure
l)ctwe^n them, after reaching a madmum, will g.-adually
<limin.8h to zero. The whole process would occupy not
greatly more or less than an hour if the bo'iios were of such
dimensions as the earth, and such degrees of rigidity m
copper, steel, or glass. In the case, however, of globes of
tliese substances not exceeding a yard in diameter, t>e
whole process is probably finished within a thousandth of
11 socond.*
The impulsive forces are bo much more intense than the
• TbomMD tud TtliV Nkt. Pbtl., p. fti.
SdiP
9n
DIBBCT AND CJSNTRAL IMPACT.
ordinary forces, that daring the brief time in which the
former act^ an ordiuar}' force does not produce an effect
comparable in amount with that produced by an impulsive
force. For example, an impulsive force might generate a
velocity of 1000 in less time than one-U^nth of a second,
while gravity in one-tenth of a second would generate a
- elocity of about three. Hence, in dealing with the effects
of impulses. Unite forces need not be conaidei'ed.
204. Direct and Central Impact— When two bodies
impinge on oach other, so that their centres before impact
are movic^? in the same straight line, and the common tan-
gent at the point of contact is perpendicular to the line of
motion, the impact is said to be direct and central. When
these conditions are not fulfilled, the impact is said to be
obligue.
When two bodies impinge directly, one upon the other,
the mutual action between them, at any instant, must be
in the line joining tbeir centres ; and by the third law
(A?*t. 166), it must be equal in amount on the two bodies.
Hence, by Law II, they must experience equal changes of
motion iu contrary directions.
We may consider the impact as consisting of two parts ;
during the first part the bodies are coming into closer con-
tact with each other, mntvally displacing the particles in
the vicinity of the point of contact, producing a compres-
sion and distortion about that noint, which increases till it
reaches a maximum, when the molecular reactions, thns
called into play, are sntfioient to resist farther compression
and distortion. At this iusiant it is evident that the
points in contact are moving with the same velocity. No
body in nature is perfectly itietantic ; and henoe, at the
instant of greatest compression, the elastic force* of resti-
tution are brought into action ; and during -the second part
of the impact the mutual pressure, produced by the elastic
forcoB, which were brought into action by the compression
«■«
ACT.
nc in which the
produce an cfFect
by an impnlsive
might generate a
nth of a second,
vonid generate a
% with the effects
dei-ed.
-When two bodies
es before impact
the common tan-
lar to the Hne of
I central. When
uct is said to be
e upon the other,
instant, must be
by the third law
I tlie two bodies,
equal changes of
ng of two i»rt« ;
f into closer con-
the particles in
icing a compres-
b increases till it
T reactions, thus
■ther compression
evident th&t the
me velocity. No
nd henoe, at the
c fortm of resti-
gthe second part
led by the elastic
the compression
SLASTICJTy OF BODIMS.
mmmsmamn mmi j.
373
during the first part of the impact, tend to separate the
two bodies, and to restore them to their original form.
205. Elasticity of Bodiea.-^o«Aci«at of Resti-
tution. — It appears from experiment that bodies may be
compressed in various degrees, and recover more or less
their original forms after the compreMing force has ceased ;
this property is termed ekiHticity. The force urging the
approach of bodies is called the force of compression ; the
force causing the bodies to separate again is called the
force of reditution. Elastic bodies are such as regain a
part or all of their original form when the compressing
force is removed. The ratio of the force of restitution to
that of compression is called the Coefficimt of JiestituHon.*
It has been found that this ratio, in the, same bodies, is
constant whatever may be their velocities.
When this ratio is unity the two forces ore equal, and the
body is said to lie perfectly elaeticj when the ratio is aero,
or the force of restitution is nothing, the body is said to bo
non-elastic; when the ratio is greater than zero and less
than unity, the body is said to be imperfectly elastic. There
are no bodies either perfectly elastic or perfectly non-elas-
tic, all being more or less elastic.
In the cases discussed the bodies will be supposed ppher-
ical, and in the case of direct impact of smooth spheres it
is evident that they may be considered as particles, since
they are symmetrical with respect to the line joining their
centres.
The theory of the impact of bodiea ia chiefly due to
Newton, who found, in his experiments, that, provided the
impact is not so violent as to make any seuHiblo iidenttition
in eitiier body, the relative velocity of separation after the
impact bears a ratio to the relative velocity of approach
before the impact, which is constant for the same two
• BoneUmea called CmIIIcIsd( of IhMtlcitr. To<bmlar'» Heck. . p. tm.
374
DIRECT IMPACT OF I.VELASTIC BODIES.
bodies. In Newton's experiments, however, the two bodies
seem always to have been formed of the same sub-
stance. He found that the value of this ratio (the coeffi-
cient of rettitution), for balls of compressed wool was about
f, steel about the same, cork a little less, ivory {, glass \\.
The results of more i-ecent experiments, made by Mr.
Hodgkinson, and recorded in the Report of the British
Association for 1834, show that the theory may be received
as satisfactory, with the exception that the value of the
ratio, instead of being quite constant, diminishes when the
velocities are very large.
206. Direct Impact of XneUuitic Bodies. — A sphere
of mass M, moving with a velocity v, overtakes and impinges
directly on another sphere of mass M', moving in the same
direction with velocity v', and at the instant of greatest
mutual eomj'Tession the spheres are moving with a com)non
velocity V. Determine the motion after impact, and the
impulse during the compression.
Jjst R denote the impulse during the compression, which
acts en each body in opposite directions ; and let us sup-
pose the bodies to be moving fh)m left to right. Then,
since the impulse is measured by the amount of momentum
gained by one of the impinging bodies or lost by the other
(Art 202), we have
Momentum lost hy M = M{v — V) - R, (1)
« gained by M' = M' (V - v') = R, (2)
.-. M{v- V) = M'iV-v'). (3)
Solving (3) for V wo get
F =
which in (1) or (2) givea
Mv + M'v'
M+sr*
(*)
BODtSS,
r, the two bodies
the same sub-
J ratio (the coeffi-
l wool was about
'^orj I. glass +f
I, made by Mr,
t of the British
may be received
he value of the
iiishes when the
dies. — A sphere
tea and impingea
ing in the same
iant of greatest
' with a conAnon
impact, and the
ipresBion, which
md let us sup-
o right. Then,
It of momentum
)8t by the other
ft i ^M H i w W ius^a iLi i lw g ' MW
DIBBCT IMPACT OF mSLASTiC BODIES,
„ _ MM'j V - V)
Hence the common velocities of the two bodies after impact
is equal to the algebraic sum of their momenta, divided by
the sum of their masses, and also, from (3), t/is whole
momentum after impact is equal to the sum of the momenta
before.
Cob. 1. — Had the balls been moving in opposite direc-
tions, for example had M' been moving from right to left,
v' would httve been negative, in which case we would have
^ Mv - M'v' - _ MM' (r + t;')
(6)
) = Ii,
(1)
v') = B,
(2)
V').
(8)
(♦)
1
1
■
From the first of these it follows that both balls will be
reduced to rest if
Mv = M'v';
that is, if before impact they have equal and opposite
momenta.
Cob. 2. — If M' is at rest before impact, v' = 0, and (4)
becomes
If the masses are equal we have from (4) and (6)
V
* ~ 2 '
or
— V
(8)
according as they move in the same or in opposite direc-
tions.
207. Direct Impact of Ela«1ac Bodies.— When thei
balls are elastic the problem is the same, up to the instant
of greatest compi-ession, as if they were inelastic ; but at
; i '
•■ I
mm
376
VIBMCT IMPACT OF tHELASTtC BODUta.
tbw initant, the force of restitation, or th«t tendency which
elMtio bodies have to regain their ranginal form, begins to
throw <me ball forward with the same momentum that it
throws the other back, and this matnal pressure is propor-
tional to R (Art. 205).
I«t s be the coefficient of restitation ; then daring the
second part of the impact, an impulse, eR, acts on each
ball in the same direction respectirely as R acted daring
the compression. Let v, and v,' be the velocities of the
balls JTand M' when they are finally separated. Then we
have, as before.
Momentum lost hy M r= M(V —v^) = eR, (l)
* gained by M' = M' (r/ - F) = eR. (2)
From (1) we have
eR •
V, = V-
M
i.j
Mvj- M'v
M+M' ^^ - "^
by (4) and (6) of Art. 306,
*' - mTjt <^ + '> <" -"')•
p Similarly from (8) wo have
M
*'•'=•»'' ^ft:f'(i + ')(*'-»^)5
(3)
(*)
which are the velocities of the balls when finally eeparated
These results may be more easily obtained by the oon-
sidorution that the whole impulse is (1 + e) R; for this
gives at once the whole momentum lost by IT or gained by
M' dnring oomprMnon and restitution as follows :
M(v - V,) = (1 -f- «) J?,
(6)
■\?ia»»-'ijMWW«Bi*w«wi»»»ww» 'w iiii i i w'» i M n w ^
tt tendency which
il fonn, begins to
lomentum that it
reflsure is propor-
; then daring the
eR, acta on each
18 B acted daring
i velocities of the
irated. Then we
■ v.) = en, (1)
- F) = eB. (2)
-V)
d (5) of Art 306,
"'). (3)
t^h
(*)
uiUy eeparakd.
ined by the oon-
+• «) -S ; for this
Mot gained by
Pollows :
and
DISKCT IMPJLOT Of IttBhASTIO BODTKS. 377
M'{v,'^v')=z{l + e)Jt. (6)
Substituting in (5) and (6) the value of It from (5) of Art.
206, we have the values of v and v,' immediately.
Cob. 1. — If the balls are moving in opposite directions*
«' becomes negative. If the balls are non-elastic, e = 0,
and (3) rad (4) reduce to (4) of Art 206, as they should.
Cob. 2.— If the balls are perfectly elaatie, « = 1, and (3)
and (4) become
^^' = V+-J^,(v-v').
(8)
M+ M
Cob. 3,— Subtracting (4) from (3) and reducing, we get
», — »,' = r ~ t>— (1 + e) {v r- v'),
Hence, tks relative velocity after impact «« — e times the
relative velocity before impact.
Cob. 4.— Multiplying (8) and (4) by M and M', respeet-
ively, and adding, we get
Mv, + M'v,' =r ift? ^. M'v'.
(10)
Hence, as in Art. 206, the algebraic turn of the mometUa
after impact ie the same at before; i. e., there i» no mo-
mentum lost, which of conne is a direet oonseqnence of the
third law of motion (Art 169).
Cob. 6. — Suppose «* = 0, so ^t the body ot mass M,
moving witii velocity v, impinges on a.tody of mass M' at'
rest, th$n (3) and (4) become
878
LOSB OP KINSTW BNXROT.
M-eM'
. , i/ (1 + «)
V, and t», = jy^jyrt'.
(11)
Hence the body wJiich is struck goes onwards ; and the
striking body goes onwai-ds, or stops, or goes backwards,
according as Jif is gfreater than, equal to, or less than eM'.
If M' = eM, then (11) becomes
V, = (1 — e) V, and »,' = v.
(12)
OOB. 6.— If M = M and e = 1 ; that is, if the balls
are of equal mass, and perfectly elastic,* then (7) and (8)
become, respectively,
Vx = »', and < = t>;
(13)
that is, the balls interchange their velocities, and the
motion is the same as if they had passed through one
another without exerting any mutual action whatever.
Cob. 7.— If M' be inflnite, and v' = 0, we have the case
of a ball impinging directly upon a fixed surface ; substi-
tnting these values in (3) it becomes
Vt= —w,
(14)
that is, the baU rebounds from the fixed surface unth a veloc-
ity e times that with which it impinged.
208. Lou of Kinetic Bnexgyf in the Zmpaet of
Bodies.— Squaring (9) of Art 207, and multiplying it by
MM', we have
MM' (r, - »,')» = MM' ^(v-i/y
= MM'{v- v'y -(!-«») MM' (V - »')».
(1)
• ThU it the nioal phnMoiogy, bat iiiialaadiiiK, Incy. Brit, Vid. XV, Art.
Xceh'f.
t 8m Art. 189.
kmf
<r.
r^jgrv. (11)
nwards; and the
' goes backwards,
ar less thau eJf'.
= V.
(12)
at is, if the balls
then (7) and (8)
(13)
locities, and the
ssed through one
m whatevef.
we have the case
I surface; snbsti-
(U)
•face with a vdoe-
the Impact of
multiplying it by
iV-vJ' (1)
By, Brft, Vd. XV, Art.
mMimHiiyi
Bm
LOSS OF KINSTtC ENSROT. 379
Squaring (10) of Art 207, we have
(J/», + M' y = {Mv + M'v'f. (2)
Adding (1) and (2), we get
(if + M") (ifv,» + Jf'»,'») = ( Jr -I- M') (Mv* + M'v'*)
- (1 - e») MM' {V - v'y ;
.-. iJfr,» + ii/"r,'» = ^Mv»+ iJfV*
the last term of which is the loss of kinetic energy by
impact, since e can never be greater than unity. Hence,
there is always a loss of kinetic energy by impact, except
when e = 1, in which case the Iris is zero; i. «., when the
coeflBcient of restitution is unity, no kinetic energy is lost
When « = the loss is the greatest, and equal to
. ^STTM'^''-''^''
From (3) we see that during compression kinetic energy
MM'
to the amount of \ m ^ -u' (^ — *')* *^ ^<^* ' *°^ ***®"
during restitution, e* times this amount is regained.
RiBM.— From the theory of kinetic energy it appears
that, in every case in which energy is lost by resistance,
heat is generated; and from Joule's* investigations we
learn that the quantity of heat so generated is a perfectly
definite equivalent for the energy lost ; and also that, in
• Bee "The ConelatioB eDd Coiuemtlon of Force*," by Heimholta, FuwUtTt
LieMg, etc. ; elM " Hett M • Mode of Motion,'' by Prof. TyndaU. Also a Stewart's
" OoneerrMbw of Boefgjr."
380
OBLiqtrS IMPACT.
any nataral action, there is never a development of energy
which cannot be accounted for by the disappearance of an
equal amount elsewhere by means of some known physical
agency. Hence, the kinetic energy which appears to be
lost in the above cases of impact, is only transformed,
partly into heating the bodies and the surrounding air, and
partly into sonorous vibrations, as in the impact of a ham-
mer on a bell.
209. ObUqne Impact of Bodiea.— The only other
case which we shall treat of is that of oblique impact when
the bodies are spherical and perfectly smooth.
A particle impinges with a given velocity, and in a given
direction, on a smooth plane ; required to determine ike
motion after impact.
Let AC represent the direo*
tion of the velocity before im-
pact, me^ng the plane at 0,
and CB the direction after
impact. Draw CD perpen-
dicular to the plane ; then
since the plane is smooth its impulsive reaction will be
along CD.
Let V and v, denote the velocities before and after
impact, respectively ; and let o and /3 denote the angles
ACD and BCD.
Resolve v along the plane and perpendicular to it. The
former will not be altered, since the impulsive force acts
perpendicular to the plane ; the latter may be treated as in
the case of direct impact, md will therefore, after impact,
be e times wL.,t it was before (Art. 207, Cor. 7). Hence,
resolving Vi
have
Ftg.87
and perpendicular to the plane, we
Vt sin j3 = V sin a, (1)
V|Cosj9=: — evooso.
(»)
[)ment of energy
ppearance of an
known physical
\\ appears to be
ily transformed,
Duuding air, and
npact of a bam-
The only other
ue impact when
th.
and in a given
to determine the
Ftg.a7
reaction will be
efore and after
inote the angles
Blar to it. The
iilsire force acts
be treated as in
re, after impact,
3or. 7). Hence,
I the plane, we
(1)
m
OBLIQVM ntPACT.
881
Dividing (2) by (1), we get
cot /3 = — e cot «.
(8)
Squaring (1) and {%), and adding, we get
«,« = t)» (sin* a 4- fl» cos* «).
(4)
Thus (3) determines the direction, and (4) the magnitude
of the velocity after impact.
The angle ACD is called the angle of incidence, and the
angle BCD the angle of reflexion.
Cor. 1.— If the elasticity be perfect, or e = 1, we have
from (3) and (4),
and
oot /3 = — cot «, or i3 = ~ « ;
t),' = v", or r, = V.
(6)
(6)
Hence, in perfectly elastic balls the angles of incidence
and reflexion are numerically equal, and the velocities before
and after impact are equal. This is the ordinary rule in
the case of a billiard ball striking the cushion.
Cor. 2.— Suppose e = 0; then from (8), — 90".
Thus, if there is no elasticity, the body after impact moves
along the plane with the velocity v sin a.
If o = 0, so that the impact is direct, we have from (4),
vt = ev; i. e., after the impact the body rebounded along
its former course with e times its former velocity.
If a = 0, and « = 0, then from (4), «, = 0, and the
body is brought to rest by the impact.
SoH.— Of course the results of this article are applicable
to oases of impact on any smooth surface, by substituting
for the plane on which the impact has been supposed to
383 OBLiqUS IMPAC'i OF TWO SMOOTH SPBSREa,
\'m
vm
m
take place the plane which is tangent to the surface at the
point uf impact.
210. Oblique Impact of Two Smooth Spheres.—
Two smooth spkfres, moving in given directions and with
given vehcitiea, impinge; to determine the impulse and the
subsequent motion.
Let the masses
of the spheres be
M, M' ; their cen-
tres C, C; their
velocities before
impact V and -d,
and after impact
V, and w,'. Let ED be the line which joins their centres at
the instant of impact (called the line of impact): CA and
CB the directions of motion of the impinging sphere, M,
before and after impact ; and O'A' and C'B' those of the
other sphere; let «, a, be the angles, ACD and A'C'D,
which the original directions of motion make with the line
of impact; /?, /J, the angles, BCD and B'C'D, which their
directions make after the impact
It is evident that, since the spheres are smooth, the
entire mutual impulsive pressure takes place in the line
joining the centres at the instant of impact. Let R be the
impulse, and e the coefficient of restitution. Resolve all
the velocities along the line of impact and at right angles
to it ; the latter will not be affected by the impact, and the
former will be affected exactly in the same way as if the
impact had been direct. Hence, since the velocities in the
line of impact are v cos «, v' cos a', t?, cos j3, «,' cos /S", we
have, by substituting in (3) and (4) of Art 207,
p, cos /3 = i; cos « - jjqT^ (1 + e) (» cos a— if cos «'), (1)
r 8PHSRES.
he surface at the
>oth Spheres.—
•ections and with
impulse and the
i their centres at
mpact): GA and
iging sphere, M,
I'B' those of the
CD and A'C'D,
ike with the line
D'D, which their
are smooth, the
ilace in the line
t. Let R be the
ion. Resolve all
at right angles
impact, and the
ne way as if the
Telocities in the
I j3, «,' cos P", we
207,
i—i/ COB a'), (1)
SXAMPLSa.
M
383
v,'co8/}' = v'oosa'+^-^y, (l-i-e) («oo8«-r'co8 «'), (a)
which are the final velocities c/the two spheres along the line
of impact ED.
Also, from (6) of Art. 306, we obtain by substitution,
~ W+l^' '" °°'' ** — *' ''OS a'), (8)
(See Tait and Steele's Dynamics of a Particle, p. 323.)
Cob. 1.— Multiplyin/j (1) by M, and (8) by M', and add-
ing we get
Mv, cos ^ + M'vt' cos i3' = Mv cos a + M'v^' cos «', (4)
which shows that the momentum of the systtm resolved
along the Urn of impact is the same after impact as before.
Cor. 2.— Subtracting (2) from (1) we obtain,
t>, COS ^ — r/ cos /3' = — e (» cos a — v' cos «'). (6)
That is, the relative velocity, resolved along the line of
impact, after impact is — e times its value before.
EXAMPLES.
1. A body* weighting 3- lbs. moving with a velocity of
10 ft. per second, impinges on a body weighing 2 lbs., and
moving with a velocity of 8 ft per second ; find the com-
mon velocity after impact Ans. 7\ ft per second.
2. A body weighing 7 lbs. moving ll ft per second,
impinges on another at rest weighing 15 lbs.; find the com-
mon velocity after impact Ans. 3^ ft per second.
• The liodiM are inelaaUc aalen otherwiM itatcd. The flnt W ezamplee are in
Mnet Impact.
384
EXAMPLES,
hiu} \
3. A body weighing 4 lbs. moving 9 ft per second,
impinges on another body weighing 2 lbs- and moving in
the opposite direction with a velocity of 5 .ft. per second ;
find the common velocity after impact.
Ana. 4^ ft. per second.
4. A body, M', weighing 6 lbs. moving 7 ft per second,
is impinged upon by a body, M, weighing 6 lbs. and mov-
ing in the same direction; after impact the velocity of M'
is doubled : fi ad the velocity of M before impact
An$. 19f ft. per second.
6. Two bodiea, weighing 2 ll»., and 4 lbs., and moving in
the same direction with the velocities of 6 and 9 ft. respec-
tively, impinge upon each other; find their common
velocity after impact Ans. S ft per second.
6. A weight of 2 lbs., moving with a velocity of 20 ft
per second, overtakes one of 6 lbs., moving with a velocity
of 5 ft per second ; find the common velocity after impact.
Ana. 94 ft per second.
7. If the same bodies met with the same velocities find
the common velocity after impact
Ans. 24 ft per second iu the direction of the first
8. Two bodies of different mtiSRcs, are moving towards
each other, itnth velocities of 10 ft and 12 ft. per second
respectively, and continue to move after impact with a
velocity of !• 2 ft per second in the direction of the greater;
compare their masses. Arts. As 3 to 2.
9. A body impinges on another of twice its mass at rest:
■how that the impinging body lo, -s two-thirds of its
velocity by the impact
10. Two bodies of unequal masses 1 viag in opposite
directions with momenta numerically equal meet; show
that the momenta are numerically equal after impact.
Mai
ii
J ft per second,
3. aud moving in
5 .ft. per second;
ft per second.
7 ft. per second,
g 6 lbs. and mov-
lie velocity of M'
mpact
ft. per second.
s., and moving in
and 9 ft. respec-
their common
ft per second.
velocity of 20 ft
: with a velocity
nty after impact
ft per second.
te velocities find
on of the first.
moving towards
2 ft. per second
' impact with a
n of the greater;
\n3. As 3 to 2.
its mass at rest:
ro-tbirds of its
ring in opposite
lal meet; show
her impaot
EXAMPLS&
11. A bodr, M, weighing 10 lbs. moving 8 £1. per second,
impivjges on M', weighing 6 lbs. and moving in the same
direction 5 ft. per se^wnd ; find tbeir velocities after impact,
supposing 6 = 1.
Am. Velocity of if = 6| ; velocity otM' = 8f.
12. A body, M, weighing 4 lbs. moving 6 ft per neoond,
meets M' weighing 8 lbs. and moving 4 ft per second;
find their velocities after impact, « = 1.
Ans. Each body is reflected back, M with a velocity of
7| andif ' with a velocity of 2|.
13. Two balls, of 4 and 6 lbs. weight, impinge on each
other when moving in the same direction with velocities of
9 and 10 ft respectively ; find their velocities after impact,
supposing e = ^. Am. 10-08 and 9-28.
14. Find the kinetic energy lost by impact in example 5.
Ans. f^.
IL Two bcJies weighing 40 and 60 lbs. and moving in
the same direction with velocities of 16 and 26 ft respec-
tively, impinge on each other: find the loss of kinetic
energy by impact. Ana. 37 -3.
10. An arrow shot from a bow starts off with a velocity
of 120 ft per second; with what velocity will an arrow
twice as heavy leave the bow, if sent off with three times
the force ? Ana. 180 ft per second.
17. Two balls, weighing 8 ozs. and 6 ozs. respectively,
arc feimultanoously projected upwards, the former rises to a
height of 324 ft. and the latter to 266 ft ; compare the
forces of projection. Ana. As 3 to 2.
18. A freight train, weighing 200 tons, and traveling 20
miles per hr. runs into a passenger train of 60 tons, stand-
ing on the same track; find the velocity at which the
remains of the passenger train will be prcpelleO along the
tnwk, snpposing « = f Aiu. 19- 3 miles per hr.
386
^^J[AMFL£8.
19. Thejie is a row of ten perfectly elastic bodies whose
masses increase geometrically by the constant ratio 3, and
the first impinges on t second with the velocity of
6 ft per second ; find the velocity of the last body.
Ana. -yf , ft per second.
20. A body weighing 5 lbs. moving with a velocity of 14
ft. per second, impinges on a body weighing 3 lbs., and
moving with a velocity of 8 ft. per second; find the veloci-
ties after impact supposing e = \. Ana. 11 and 13.
21. Two bodies are moving in the same direction with
the velocities 7 and 6 : and after impact their velocities
are 6 and 6; find c, and the ratio of their masses.
Ans. e = 4; J/' = 2M.
22. A body weighing two lbs. impinges on a body weigh-
ing one lb.; e is ^, show that v, = r ^- v', and that v,' = v.
23. Two bodies moving with numerically equal velocities
in opposite dirv'KJtious, impinge on each oth r; the«result is
that one of them turns hack with its original velciity, and
the othgr follows it with half that velocity; show that one
body is four times as heavy as the other, and tluit e = J.
24. A strikes B, which is at reist, and after impact the
velocities are Dumerically equal ; if r be the ratio of B'g
2
ma9s to A'b mass, show that e is 7, and that B's mass
r — 1
is at least three times A's mass.
25. A body impinges on an equal body at rest ; show
that the kinetic energy before impact caunot be greater
than twice the kinetic energy after im^Mct
26. A beries of perfectly elastic balls are arranged in the
same straight line ; one of them impinges on the next,
then this on the noxt and so on ; show that if their masses
form a geometric progression of which the oommou ratio
atic bodies whose
tant ratio 3, and
the velocity of
wt body,
ft per second.
1 a velocity of 14
;hiiig 3 lbs., and
; find the velocl-
[ns. 11 and 13.
le direction with
!t their velocities
masses.
4; J/' = 2M.
on a body weigh-
and that v/ = v.
y equal velocities
n:v; the<resnlt ia
inal velciiiy, and
; show that one
id tliitt e = |.
after impact the
the ratio of B'g
ind that B's mass
ly at rest ; show
Eiunot be greater
I arranged in the
![eH on the next,
t if their masses
he oommou ratio
KXAMPLS8,
m
is r, their velocities after inajiact form a geometric progres-
sion of whicli the common ratio is
r 4- 1
27. A ball falls from rest at a height of 20 ft. above ft
fixed horizontal plane; find the height to which it will
rebound, e being J, and g being 32. Arts. 1\\ feet.
28. A ball impinges on an equal ball at rest, the elas-
ticity being perfect; if «he original direction of the strik-
ing ball is inclined at an angle of 45" to the straight line
joining the centres, determine the augle between the
directions of motion of the striking ball before and after
impmt. Ans. 45°.
29. A ball falls from a height h on a horizontal plane,
and then rebounds; find the height to v/hicb it rises in its
iiscent. Ana. e*//.
30. A ball of mass M, impinges on a ball of mass M', at
i"est ; show that, the tangent of tlie augle between the old
and new directions of the motion of the impinging body is
1 4- iTjin^o
% ~M + M' (sin* a — cos* a)
31. A ball of mass M impinges on a ball of mass M' at
rest ; find the condition in order that the directions of
motion '^f the impinging ball before and after impact may
be at right angles.
Ana. tan" n =
Me
M' H Af'
83. A ball impinges on an equal ball at rest, the angle
between the old and new directions of motion of the
impinging ball ia C0° ; find the velocity after impact, »
being 1. Ang. v sin 30°.
33. A ball impinges on an equal ball at rebt, e being 1 ;
liu'i the conditiou under which the velocities will ht; vi\uaX
uftev impact. Ant. u ^^ 45°
n
388
MXAMPLM8.
34. A ball is v.<ojected from the middle point of one Bide
of a biUiard table, eo as to strike first an adjacent side, and
then the middle point of the side opposite to that from
M'hich it started ; find wk iro the ball must hit the adjacent
side, its length being b.
Am. At the distance -^ — : — from the eud nearest the
opposite aide.
l + «
W'
e point of one side
adjacent side, aud
osite to that from
3t hit the adjacent
) eud nearest the
m
CHAPTER V.
WORK AND ENERGY.
211. Deflnitioii and BfA^sora of Work.— Work i»
the production of motion against rmstanee. A force is said
to do work, if it moves the hody to which it is af plied ;
and the work done by it is moasared by the product of the
force into the B{>aoe through which it moves the body
(Art 101, BciD.).
Thus, the work done in lifting a weight throagh a ver-
tical distance is proportional to the weight lifted and
t)ie vertiukl distance through which is is lifted. The unit
of work used in England imd in this country «'« thtU which
is required to overcome the weight of a pound through the
vertical height of a foot, and is called a foot-pound. For
instance, if a weight of 10 lbs. is raised to a height of
5 ft., or 5 lbs. raised to a height of 10 ft, 50 foot-ponnds of
work must have been expended in overcoming the resist-
ance of gravity. Similarly, if it requires a force of 50 lbs.
to move a load on a horizontal plane over a distance of
100 ft., 5000 foot-pounds of work must have been done.
If a carpenter urges forward a plane through 3 ft. with a
force of 13 Ibe., he does 36 foot-pounds of work ; or, if a
weight of 1 lbs. descends through 10 ft., gravity does
70 foot-pounds of work on it,
Her^e, the number of units of work, or foot-pounds,
ne <essary to overcome a constant resistance of P pounds
through a distance of S feet is equal to the product PS.
From this it appears that, if the point of application
move always perpendicular to cne direction in which the
force acts, such a force does no work. Thus, no work ui
done by gravity in the caf:e of a particle moving ou %
390
roRK DOXE itr A Foncs.
liurizontal plane, and when a particle moves on any smooth
curface no work is done by tlio force which the surface
e.ierts upon it.
Neither /ore? nor motion alone is sufficient to constitute
work; so that a man who merely Hupports a load without
niiiving it, does no work, in the sense in which that term is
used mechanically, any more than a column does which
sustains a heavy weight npon its summit.
If a body is moved in the direction opposite to that in
which its weight acts, the agent raising it does work ufKin
it, while the work done by the earth's attraction is neffa-
tii'e. When the work done by a force is negative, i. e.,
when the point of application moves in the direction oppo-
cite to that in which the force acts, this is frequently
expressed by saying that work is done against the force.
In the above case work is done by the force ? if ting the
body, Kud against the earth's attraction.
212. General Caae of Work done by a Force.—
When either the magnitude or direction of a force varies, or
if l)oth of tliom vary, the work done by the force during any
finite displacement cannot be defined as in Art. Sill. In
this case the work done during any indefinitely small dis-
placement may be found by supposing the magnitude and
direction of the force constant during the displacement, and
finding the work done as in Art. 211 ; then taking the sum
of all 8uoh elements of work done during the consecutive
HinuU displacements, which t^>gethcr make up the finite
(lispliicemeut, we obtain the whole work done by the force
during such finite displacement.
TliuB let a force, P, tot a( a p<^t, O, In the direction OP (Fig. ffO),
and let us Buppose tlie point, O, to move into any other poaitiun, A,
very near 0. If Iw tiie angl'^ betwwn the direction, OP, of tlie
I'orce and tlie direction, OA, of the diiplaoement of tlie point of appli-
cation, then the product, POA ocmH,\B called the work done by the
forco. If we drop a porp«>ndlcalar, AN, on OP, the work done l>y the
force i8 alio equal to the pruduet P ■ ON, where ON ii to lie eati-
:jiia"^:iPMMiiJiifly4&Myi:
iCS.
MEABURE OF WORK.
391
oves on any smooth
which the Burfaoe
icient to constitute
orts a load without
I which that term is
jolumn does which
t.
opposite to that in
it does work u[)on
attraction is nega-
a is negative, /. c,
tiie direction oppo-
, this is frequently
) agaim,. the force,
he force Jifting the
M by a Force.—
of a force varies, or
he force during any
8 in Art. ail. In
lefinitely small dis-
the magnitude and
B displacfment, and
len taking the sum
ag the consecutive
[lake up the finite
done by the force
direction OP (Fig. SO).
vay other poritiun, A,
direction, OP, of tlie
1 "f tJie point of appli-
the work done by the
the work done liy the
lere ON is to l>e wti-
mated as positive when in the direction of the lorce. If several forces
act, the work done by each can be found in the same wiiy ; and the
sam of all tlieoe is the work done by the whole gyatem of force».
It appears from this that tlie work done by any force during an
infinitesimal displacement of the point of application, is the product
of the resolved part of the force in the direction of the displacement
into the displacement ; and this is the same as the virtual momtnt of
tlie force, which has been described in Art 101. In Statics we are
concerned only with the small hjfputJutkai displacement which we
give the point of application of the force in appIyiMg the principle of
virtual velocities. But in Kinetics tlie liodies are in motion ; the
force aetitaUff disphues its point of application in such a manner that
the displacement has a projection along the direction of the force. If
ds denote the projection of any elementary arc of a curve along the
direction of P, the work done by P in this displacement is Pd». The
sum of all these elements of work done by P in its motion over a
finite space is the whole work found by taking the integral of Pd$
between proper limits.
Hence generally, if « be an arc of the path of a particle, P the
tangential component of the for«»c •-aich act on it, the work done on
the particle between any two points of its path is
/Pd», (1)
the integral being taken between limits corresponding to the initial
and final positions of the particle.
213. Work on an Inclined Plane.— Let a be the
inclination of the plane to the horizon, W the weight
moved, « the distance along the plane through which the
weight is moved. Resolve W into two components, one
along the plane and the other perpendicular to it; the
former, W sin a, is the component which resists motion
along the plane. Hence the amount of work required 'o
draw the weight up the plane = W sin « • s = Wxtho
vertical height of the plane ; t. e., the amount of work
required is unchanged by the substitution of the oblique path
for the vertical. Hence the work in moving a body up an
incUntnl plane, without friction, is equal to the product of
the weight of the body by the vertical height through which'
it ia raised.
4
\'i
393
WORK ON AN INCLINHD PLANW.
I
Cob. 1. — If the plane be rough, let ft = the coefficient
of friction ; then since the normal component of the weight
is FF COB a, the resistance of friction is /i fF cos a (Art 92).
The work required consists of two parts, (1) raising the
weight along the plane, and (2) overcoming the resistance
of friction along the plane, the former = (F sin « • «, and
the latter is ^ H^ cos « • «. Hence the whoh work necessary
to move the weight up the plane is
FT (sin a 4- |t* cos cc) «.
(1)
Since » sin a represents the vertical height through
which the weight is raised, and » cos a the korwmtal space
through which it is drawn, tiiis result may be stated thus :
The work expended it the same as that which would be
required to raise the weight through the vertioal height of
the plane., together with that which would be required to
draw the body aUmg the base qf the plam horitojUaUif
against friction.
GoR. 3. — If a body be dragged through a space, s, down
an inclined plane, which is too rough for the body to slide
down by itself, the work done is
fT (^ cos a — sin «c) s.
(»)
Cor. 3. — If h — the height of the inclined plane, and
d = its horizontal base, then the work done against gravity
to move the body up the plane = Wh ; and the work done
against friction to move the body along the plane, suppos*
ing it to be horixontal, = nb W. Hence (Cor. 1) the total
work done is
Wh + yhW. (8)
If the body be drawn down the phne, the total work
expended (Cor. 2) is
- Wh + ftbW. (4)
~
: the coefficient
at of the weight
C08 a (Art. 92).
(1) raising the
[ the resiBtance
tF sin « . 9, and
teork necessary
(1)
leight through
\or%»ontai spuoe
e stated thas :
ohi<^ would be
HmI height of
be required to
M horitoiUaUif
^Mce, 8, down
' body to slide
(2)
led plane, and
igainst gravity
the work done
plane, euppoB-
r. 1) the total
(3)
le total work
JSJCAMl'LBS.
If in (4) the former term la greater than the latter,
gravity does more work than what is expended on friction,
and the body elides down the plane with accelerated
velocity.
ScH. 1.— If the inclination of the plane is small, as it ia
in most cases which occur iu pmctice, as in common roads
and railroads, cos a may without any important error be
taken as equal to unity, and the expression for the work
becomes (Coi-s. 1 and 2)
W^ ()">'' ± « Bin a),
(«)
the upper or lower sign being taken according as the body
is dragged up or down the plane.
ScH. 2.— If the inclination of the piano is small, as in
the case of railway gradients, the pressure upon the plane
will be very nearly equal to the weight of the body ; and
the total work in moving a body atong an inclined plane
wiU be from (3) and (4),
1^1 W± Wh,
(«)
where nlW k the work due to friction along the plane
of length I, and Wh is the work due to gravity, the proper
sign being taken as in (5).
EXAMPLES.
1. How much Work is done in lifting 150 and 200 lbs.
through the heights of 80 and 120 fl. respectively.
The work done = 150 x 80 -f 200 x 120
= 3G000 foot-pounds, Ans.
2. A body weighing 500 lbs. elides on a rough horizontal
plane, the coefficient of friction being 0.1 ; how much work
must be doue against friction to move the body over
100 ft. ?
P
■i
894
sxAitPLsa.
Here the friction is a force of 60 lbs. acting directly
o|)iK>site to the motion ; hence the work done againgt firio-
tion to move thd body ovef 100 ft it
60 X 100 = 6000 foot-pounds. Am,
t. A tniin Weigtll lOO totisi the total resistance is 8 lbs.
tier ton ; how mnoh mot\ mart be expended in l«ising it
to the top of aa inclined plane a mile long, the inclination
of the plane behig 1 rertical to 10 horizontal.
Here the work done against friction (Sch. 2)
= 800 X 6380 =r 4224000 foot-pounds,
and the work done against gravity
= 8a4000* X 6280 x V» = 16896000 foot-ponnds,
10 that the whole work = 21120000 foot-pouncis.
4. A train weighing 100 tons moves 30 miles an hout
along a horizontal rood ; the resistances are 8 lbs. per ton ;
find the quantity of work expended each hour.
Aha. 126720000 foot-pounds.
6. If 26 cubic feet of water are pumped every 6 minutes
from a mine l40 fathoms deep, required the amount of
work expended per minute, a cubic foot of water weighing
62}^ lbs. Am. 262500 foot-iK>undB.
6. How much work is done when an engine weighing
10 tons moves half b mile on a horizontal road, if the
total resistance is 8 lbs. per ton.
Atu. 211200 foot-pounds.
7. If a weight of 1120 lbs. be lifted up by 20 men, 20 ft.
high, twice in a minute, huw much work does each man
do per hour P Am. 1344000 foot-ponnds.
* Uue lou being 3MB Wtti. uoIudh otht-rwlM: stated.
1^
BOXSJt POWXM.
890
aotihg directly
oe againtt firio-
poonds, Am.
Istanoe is 8 11^
3d in inising it
the inclination
il.
2)
rands,
bot-poanda,
uncis.
miles an houf
8 lbs. per ton ;
ur.
I foot-pounds.
ivery 5 minutes
the amount of
water weighing
foot-|H>undB»
ngine weighing
ai road, if the
foot-pounds.
30 men, 20 ft.
does each man
foot-pounds.
•ted.
8. A body falls down the whole length of an inclined
plane on which tiie coefficient of friction is 0.3. The
height of the plane is 10 ft and the base 30 ft. On reach-
ing the bottom it rolls horisonti^y on a pluie, having the
same coefficieDt of friction. Find how far it will roll.
Ans. 20 ft.
9. How much work will be requiisd to pump 8000 cubic
feet of water from a mine whose depth is 600 fathoms.
Atu. 1600000000 horse-power.
10. A hv.se draws 150 lbs. out of a well, by means of a
rope going over a fixed pulley, moving at the rate of
2^ miles an hour; how many units of work does this horse
perform a minute, neglecting friction.
Ana. 83000 units of work.
214. Hone Power. — It would be inconvenient to
express the power of an engine in foot-pounds, since Lhis
unit is so small ; the term Horse Potoer is therefore used
in measuring the performance of steam v<)ngines. From
experiments made by Boulton and Watt it was estimated
that a horse could raise 33000 lbs. vertically through one
foot in one minute. This estimate is probably too high on
the average, but it is still retained. Whether it is greater
or lets than the power of s horse it matters little, while it
is a power so well defined. A Morse Power there/ore means
a power which can perform SSOOO foot-pounds of work in a
minute. Thus, when we say that the actual horse power
of an engine is ten, we mean that the engine u able to per-
form 330000 foct-poantl^ of work per minute.
It has been eatimated that | of the 88000 foot-pounds would be
about the work of a hone of avenge Btreng^h. A mule will perform
f the work of a horse. An aaa will perform about \ the work of a
horse. A man will do about ^ the work of a horse, or about 8800
units of work per mluiute. See Even' Applied Mech's; also Byrne's
Practical Meoh's,
390 WORK OF RA1SISO A SVSTSM OF W BIO UTS.
21S. Work of Rntoing a System of Weights.—
Let P, Q, R, bfc any three weights at the distances, p, q,
r, rospeotivoly above a fixed horizoutal plane. Then [Art.
50 (3)] or (Art 73, Oor. 3), the distances of the ceutie of
gravity of P, Q, B, above this fixed horizoutal plane is
Pp + Qq + Jir
P+Q-{ Ji '
(1)
Now suppose that the weights are raised vertically
tlirongh the heights o, b, c, respectively. Then the dis-
tance of the centre of gravity of the three weights, in the
new position, above the same fixed hcrizontal plane is
P(p + a) + Q{q + b) + R{r + c)
P+Q+Ji
Subtract! ug (1) from (2), we have
Pa + Q b + lie
P+ Q + Ji '
(2)
(3)
for the vertical distance between the two positions of the
centre of gravity of the three bodies.
Now the work of raising vertically a weight equal to the
sum of P, Q, Ji, through the space deuoted by (3) is the
product of the sum of the weights into the space, which is
Pa+ Qb + Re,
(4)
but (4) is the work of raising the three weights P, Q, R,
through the heights a, b, c, respectively. In the same way
this may be shown for any number of weights.
Herux when several toeighta are raised vertically through
different heights, the whole work done is the same as that of
raising a weight equal to the sum of the weights vertically
from the first position of their centre of gravity to the last
position. (See Todhuutcr's Mech'a, p. 338.)
Weights.—
istunces, p, q,
. Then [Art.
' the ceiiti'o of
\\ plaue is
(1)
led vertically
rhcn the dis-
eigbts, in the
plane is
£).
(2)
(3)
jitioDs of the
equal to the
by (3) is the
yfx, which is
(4)
hts P, Q, R,
the same way
cally through
me as that of
hts vertically
■y to the last
' ■iMmmmk^-mmmm^Mo^KmMm^
SXAMPLSa.
S97
EXAM PLES.
1. How many horse-power would it take to raise 3 c«fc
of coal a minute from a pit whose deptii is 110 fathoms?
Dopt!i = 110 X 6 = C60 feet.
3 cwt. = 112 X 3 = 336 lbs.
Hence the work to be done in a minute
= 600 X 336 = 221760 foot-pounds.
Therefore the horse-power
= 221760 -^ 33000 = 6.72, Ans.
2. Find how many cable feet of water an engine of
40 horse-power' will raise in an hour from a mine 80
fathoms deep, supposing a cubic foot of water to weigh
1000 ozs.
Work of the engine per hour = 40 x 33000 x 60 foot-
pounds.
Work expended in raising one cubic foot of water
through 80 fathoms = i^f^ x 80 x 6 = 30000 foot-
pounds.
Hence the number of cubic feet raised in an hour
= 40 X 33000 X 60 -!- 30000 = 2640, Ana,
3. Find the liorse-power of an engine which is to move
at the rate of 20 miles an hour up an incline which rises
1 foot in 100, the weight of the engine and load being
60 tons, and the resistance from friction 12 lbs. per ton.
The horizontal space passed over in a minute = 1760 ft. ;
the vertical space is one-hundredth of this = 17.60 ft.
Hence from (6) of Art. 213, we have
12 X 1760 X 60-1-60 x 2240 x 17.6=1760 x 2064 foot-pounds.
Therefore the horse-power
= 1760 X 2064 + 83000 = 110.08, Am.
4 A well ig to be dng 20 ft. deep, and 4 ft. in dmmetor ;
find the work in raising the material, snppw'.ng that a
cubic foot of it weighs 140 lbs. » «^ e>
Here the weight of the material to bo raised
= 47r X 20 X 140 = 140 X 80rr lbs.
The work done is eqaiyalcKt to raising this throngh the
height of 10 ft (Art 316). Hanoe tU whole work
= 140 X 80rr X 10 = lUJOOOrr foot-pounds, Ane.
5. Find the horse-power of an. engine QMt would nuso
7 tons of coal per hour from a pit whose depth is a
fathoms.
Work per minute = ZLiH^J^liL? .== ^ZiaT;
.-. the horse-power = ^^, Atu.
6 Required the work in raising water ftom thi«o different
levels whose depths are a, *. c fathoms respoctirely ; from
the first A, from the second £, 'rom the third C, cubic
feet of water are to be raised per minute.
Work in raiting water from the first kfvel
= 62.6 J X c X 6 = 376 A-a;
and 80 on for the work in the other lev»li ;
. •• work per min. =r ^76 (^ .a-|. 5.*+ C-c) foot-pounds.
7. Find the horso-powor of an engine which draws a
load of r tons along a level road at the rate of m miles
>, Ant.
ft in dicmetor ;
ppocing that a
Bd
lis throQgh the
b work
mds, Ane.
\9t would nieo
om depth is a
8.
three different
5ctiTely; from
bird C, cubic
foot-pounds.
lioh draws a
e of M miltHi
fa[A.WPM,lbi,
an how, lOie Wcdon being p pooadi per to», all other
resistances being ne^ciotdL
Work of tbe wgine per minute
. . U.-F. ~ gj^QQQ - 3000 » ^"«
8. Bequired the nnmber of horse-power to nuse 2260
onWc ft. of water an hour, feenii a mine whose deptii a 63
fathoms. J«*. 26i,
9. What weight of coal will an engine of 4 horse-power
raise in one hour from a pit wnose depth is 200 ft. P
Am. 89600 lbs.
10. la what time will an engine of 10 ho?»^ppwer raise
5 tons of material from the depth of 182 ft.?
Ans. 4>48 minutes.
11. How many cubic feet of water will an engine of 86
horse-power raise in an hour from a mine whose depth is 40
fathoms P ^»»*- 4762 cubic feet
12. The piston of a steam engine Ut U ins. in (Hametw 5
its stroke is 2^ ft. longi it makes 40 strokes per minute ;
the mean pressure of the ateam on it is 16 lbs. pw sqnara
inch; -^vhat number of foot-pouada is done by the steam
per minute, and what is the hone-power of the engine ?
Ans. 866072 foot-po«nds ; 8-08 H.-P.
18. A w<vight of li tons is to be nuwd from a d^th cl
00 fathoms in one minute; datermine the hone-power of
the eni^na oaipthh of doing the work.
4n$, 80A H.-P,
4(K)
MODTTL'S OF A MArniNE.
14. The resistance to the motion of a certain body is
440 lbs.; how many foot-pounds must be expended in
making tins body move over 30 miles in one hour? What
must be the horse-power of an engine tliat does tlic siime
number of f oot-^wunds in the same time ?
Am. C9G9C00U foot-pounds ; 35^ H.-P.
15. An engine draws a loud of CO tons at tlie rate of 20
miles im hour; the resistances aro at the rate of 8 lbs. ixsr
ton; liud the horse-poV' rol i ■^nginp. Ans. :J5»C.
1 6. How many cubic feet of water will an engine of 250
horse-power raise \m: minute from a depth of 200 fathoms?
Ann. 110 cubic ft.
17. There is a mine with three shafts which are respeo-
tlvoly 300. 450, and 500 't. deep; it is required to raise
from tlio first 80, f.om the second 60, from the tiiird 40
cubic ft. of water per minute; tlnd the horse-power of the
engine. An*. li34{|.
216. Modulus'^ of a Machine.~The whole work \)or-
formed by a nia'jhine consists of two parts, the useful work
and the I-'-at work. The useful work is that which tiio
tliat which is
^.r•Q lost work
'if unavoidably
machine is detiiirned to produce, or «'
employed in overcoming vaefiil rcsiritav
is that M'bich is not wanted, Imt w' ';
II produced or it is tliat which is sjwni iu ' ' < i^ ning waste-
fill resistances. For instance in drawing a •. . of cars, the
useful work is perf>>rmod in m«»ving the train, but the lost
work is that which is xlone in overcoming the friction of
the train, the resistance of gravity on up grades, the resist-
ance of the nir. etc. The work appliinl to a machine is
equal to the whole work done by tl.e machine, both useful
ond lost, therefore the useful work is always less than the
work Hp|>lied to the machine.
* Domotinet oailad Hffidojcj. <An. lOB.)
5ertmn body is
expended in
honr? What
does the same
j; 35^ H.-P.
the rate of 20
of 8 lbs. jier
Ans. 26' C.
engine of 250
200 fathoms?
10 oiibic ft.
ch are reapeo-
lired to raise
1 the tiiird 40
-power of the
Ins. 134|J.
[)le work j)or-
' useful work
lit which the
Imt which is
I'e lost work
unavoidably
•sing waste'
of cars, the
1)11 1 the lost
iio friction of
8. the rosist-
t machine is
, both nseful
\e8H than the
•
BXAMPLBS.
The Modulus of a machine is tfte ratio of the useful ioork
done to the work applied. ThiiB, if the work applied to mi
engine be 40 horse-power, and the engine delivers only 30
horse-power, the modulus is |, t. ft, one-quarter of the work
applied to the machine is lost by friction, etc.
Lot W be the work applied to the machine, W, the use-
ful work, and »» the modulus. Then wo have from the
above definition
W,
m
W
U)
Ii a machine were perfect, i. e., if there t ere no lost work,
the modulus would be unity; but in every machine, some
of the work Ib lost in overcoming wasteful resistances,
80 that the modulus is always less than unity ; and it is of
course the . bject of inventors and improvers to biing this
fraction as neer to unity aa possible.
EXAMPLES.
1. An engine, of N effective horse-power, is found to
pump A cubic ft of water per min., from a mine a fathomi
deep*; find the modulus of the pumps.
Work of the engine per min. = 88000 N H.-P.
*rhe ufl«ftil work, or work expended in pumping water,
= 62.6 A X 6a = 375 A^o;
hence from (1) we have
M =
876 Ao __ A^ .
8800O- "" 88 JV" ^'"'
2. There wore A cubic ft of water in a mine whose depth
is a fethoms, when an engine of N horse-power bogap to
work the pump; the water continued to flow iato the mine
at the rate of B cubic ft yot minute i required the time
^ BXAMPLMB.
in which the mine wouM be cleared of wftter, the modolus
of the pump being nu
Let a; = the number ot minntea to dear the mine of
W«Am. Then
weight of waier to be pumped = 62- 6 (A + Bar) ;
work in pnmpiiig water =r 376a (A -|- Bar) foot-poundo;
effective work of the engine = m- JV.33000ar ;
.'. 33000 mNxss S76a (A + Bar) ;
•"• * = 88mJ^-A.«' ^**
8. An engine has a 6 foot ojlinder; the shaft make§ 80
revolutions per minute; the average steam preasure ig 25
lbs. per aquMe inch ; reqnisred the iMmw-power when the
area of the piaton ia 1800 aqnare inches, the modnlas of
the engine being \^.
Work done in one minute = 1800 x 25 x 6 x 2 x 30
foot-pounda. Wo multiply by twice the length of the
atroke, becanae the piaton is driven both up and down in
one revolution of the ahaft.
The efleotive horse-power = mmmniMJ x \\
= 450, Ana.
i. The diameter of the piaton of a steam engine ia 60
ins. ; it makea 11 strokes per minute ; the length of each
atroke ia 8 ft. ; the mean presanre per square in. is 16 lbs.*
reqnireil the number of cubic ft. of water it will raise per
hour from a depth of 60 f^honu, tho modulus of the
engine being 0' 65.
The number of foot-poanda of naefnl work done in one hoar and
•pent In railing water = ir X 80« X 8 X 15 X 11 y 80 X • 66. therefore, etc.
Am. 7763 cubic ft.
nil"!-
ife
the modulus
tibe mino of
+ Rr);
KJt-poundB ;
OOOic;
ft makes 30
"esBurs is 25
ar when the
modnlas of
6 X 2 X 30
igth of the
uuidownm
mgine is 60
igth of each
. is 16 lbs.;
rill raise per
ulns of the
one hoar and
therefore, etc.
cubic ft.
MXAMFLMS.
409
5. An engine i» required to pump lOQQOOO fgallons of
water every 12 hours, from a mine 132 fSktboms deep ; ftod
the horse-power if the modolns be {^, faid a gallon of
water weighs 10 lbs. An*. 363jV H.-P.
6. What must be the horre-power of an engine working
e hours per day, to supply n families with g gallons of
water each per day, supposing the water to be raised to the
mean height of h feet, and that a gallon of water weighs 10
lbs., the modulus being m. ^^^ ngh „ p
Ana.
198000 «m
,
7. Water is to be raised from a mine at two different
levels, viz., 50 and 80 &tiiomB, from the former 30 cubic ft,
and firMU the latter 16 «nbic ft per minute ; find the horse-
power of the machinery that wiU be required, assaming
the modulos to be 0* 6. Am. 61 • 12 H.-P.
8. The dianieter of the piston of an engine is 80 ins., the
meMx pressure of tiie steam is 12 lbs. per square inch, the
length of the stzoke is 10 ft, the nitmber of strokes made
per minute is 11 ; how many oiibio ft. of water will it rside
per minute from a depth <^ 250 fatbomii, its modulus being
0*6? Ans. 42*46 cubic i't
9. If the engine in the last example had raised 55 cubic
ft of water per minute from a depth of 250 fathoms, what
would have been its modulus ? Am. 0- 7771.
10. How many strokes per minute must the ungine in
Sx. 8 make in order to raise 16 onbio ft otf water psr
minute from the given depth ? Am. 4.
11. What must be the length of the stroke of an engine
whose modulus is 0' 66, and whose other dimensions and
conditions of working are the same as in Ex. 8, if they both
do the same quantity of useful work ? .^»«. 9- 23 ft
404
KIXETIC AXD POTKyriAL ENRROT.
217. Kinetic and Potential Energy. Stored
Work.— 77ie energy of a body means its power of doing
work J and the total amount of enenjy posscsaed by the body
is inmsnred by the total amount of work which it is capable
of doing in passing from its present condition to some
standard condition.
Every moving body possessea energy, for it can bo mado
to do work by parting with its velocity. The velocity of
the body nmy Iw used for causing it to ascend r^rtically
against the attraction of tho earth, i. e., to do work against
the TCsistunce of gravity. A cannon Ijall in motion can
penetrate a resisting body ; water flowing against a water-
wheel will turn tho wheel ; tho moving air drives the ship
through the water. Wherever we find matter in motion
■Wo have a certain amount of energy.
Energy, as known to us, belongs to one or the other of
two classcM, to which tho pames kinetic* energy and
potential energy are given.
Kinetic energy is energy that a body possmet ir virtue of
its being in motion. It is energy acturlly in use, energy
ihat is constantly being spent The en< rgy of a ballet in
motion, or of a fiy-wheel revolving rapidly., or of a pile-
driver just before it strikes the pile, are examples of kinetic
energy. The work done by a force on a body free to move,
oxortod through a given distance, is always equal to the
corresiwnding increase of kinetic energy [Art 189 (3)]. If
ii mass, m, is moving with a velocity, v, its kinetic energy
is iwi-* [(3) of Art 189]. If this velocity be generated by a
oonstjant force, P, acting through a space, «, wo have,
(Art 211)
Ps = •«!»», (1)
that is, tho work done on the body is the exact equivalent
of the kinetic energy, and the kinetic energy is recon-
* OnIM alao aeltuU iMrgy, or *im^ */ meUon.
m
r.
r. Stored
wer of doing
■ by the body
it is capable
■ion to some
can bo made
le velocity of
id r^rtically
rork against
motion cau
nst a water-
ves the ship
r in motion
the other of
energy and
iV virtue of
1 use, energy
a ballet in
or of A pile-
OS of kinetic
ree to move,
$qaal Co the
189(3)]. If
lotio energy
Derated by a
», wo have,
(1)
b equivalent
gy is reoon-
KINETIC AND POIfSNTIAL SNBROT.
406
vertible into the work; and the exact amount of work
whiub the masH m, with a velocity v, can dc against resist-
ance befoie its motion is completely destroyed is \m\fi.
This is called stored work,* and is the amount of work that
any opposing force, P, will have to do on the body before
bringing it to rest. Thus, when a heavy fly-wheel is in
rupid motion, a considerable portion of the work of tlie
engine must have gone to produce this motion ; and before
the engine can come to a state of rest all the work stored
up in the fly-wheel, as well as in the other parts of the
machine, must be destroyed. In this way a fly-wheel acta
as a reservoir of work.
If a body of mass m, moving through a space s, change
its velocity from v to v^ tlie work done on the body as it
moves through that space, (Art 189), is
im(v*
••o»).
(«)
If the body is not perfectly free, ». e., if there is one force
urging the body on, and another force resisting the body,
the kinetic energy, \mr^, gives the excess of the work done
by the former force over that done by the latter force.
Thus, when the resistance of friction is overcome, the
moving forces do work in overcoming this resistance, and
all the work done, in excess of that, is stored in the moving
mass.
Potential energy is energy that a body possesses in virtue
of its position. The energy of a bent watch-spring, which
does work in uncoiling ; the energy of a weight raised
above the earth, as the weight of a clock which does work
in falling ; the energy of compressed air, as in the air-gun,
or in an air-brake on a locomotive, which does work in
expanding ; the energy of water stored in a mill-dam,' and
of steam in a boiler, are all examples of potential energy.
* C*n«d ibo mnemmiiabd vol*, See Todlw iter'* Moclw., alio Btorad mmgj mmI
not work. Btowmi'd Mechulca, p. 178.
«D6
SXAMPLBS.
«
Such energy may or may aot be called ioto ection, it may
be dcHrmant for years ; Uie power exiflts, but tbo aotion will
begin only when the weight, or the water, or the steam is
released. Henoe the word potential, is aignifipant, as
expressing that the energy is in existence, and that a now
power has been conferred upon it by the act of raisin;): or
oonfiDing it
For example suppose a weight of 1 lb. be projected
Yfflrtioally upwards with a velocity of 32 '2 ft {ler second.
The energy imparted to the body will carry it to a height
of 16. 1 ft., when it will cease to have any vwocity. The
whole of its kinetic energy will have been expended ; but
the body will have acquired potential energy instead ; i. e.,
aie kinetic enwgy of ttie budy will all have been converted
into potential energy, which, if the weight be lodged for
any time, is stored up and ready to be freed whenevc the
body shall be permitted to fall, imd bring it back to its
starting point with the velocity of 32' 2 ft per second ; and
tiins the body will feaoquire the kinetic eneigy which it
originally received. Benoe kinetic energy imd potential
energy are mutually oonvoftible.
Let A be the height through which a body must lUl to
acquire the velocity v, m and W the mass and weight,
respectively. Then since •» as 2j*, we have, for the ii<WBd
work,
W
2ffA= Wh.
m
Hence we may say that the work stored in a moving body
is measured by the product of the weight of the body into the
height through which it muaifall to acqmre the vtlociiy.
BXAMP1.B8.
1. Let a bullet leave the barrel of a gan ^i^ the velocity
of 1000 ft. per second, and suppose it to weigh 2 oas.; find
■k
Dto action, it may
ut tho action will
r, or the steam is
is significant, as
, and that a now
e act of rai^inc: or
lb. be projected
2 ft per second,
ry it to a height
my Velocity. The
en expended ; but
jrgy instead ; i. e.,
re been converted
;bt be lodged {or
•oed whenevc the
ing it back to its
L per second ; and
3 enei^gy which it
rgy and potential
body most fall to
mass and weight,
ive, lor the iU»«d
Wh.
(8)
n a moving body
f the body into the
r« th0 velociiff.
a ^th the velocity
weigh 2 086. ; find
BXAMPLS&
the w<»rk stored up in ^e bullet, uid the height from which
it must fall to acquire that velocity.
Here we have from (3) for the stored work
2
(1000)» = Wh
2x1^
= 1941 foot-pounds.
.-. A = 16628 feet.
2. A ball weighing to lbs. is projected along a horizontal
plane with the velocity of « ft. p?>- second ; what space, «,
will the ball move over before it comes to a state of rest,
the coefBcient of friction being/?
Here the resistance of friction is fw, which acts directly
opposite to the motion, therefoK the work done by friction
while the body mov% over 8 feet = fws ; the work t ored
up in the ball = ^t^ = ^ ; therefore from (1) we have
/W8 =
8 =
3. A railway train, weighing 7* tons, has a velocity of v
ft. per second when the steam is turned off ; what distance,
s, will the train have moved on a level rail, whose friction
iap lbs. per ton, when the velocity is v^ ft. per second ?
Here the work done by friction = pT8', hence from (2)
we have
_ . 2240 7*.. „
1120<t>»-V)
~ 9P
8 =
4. A train of T tons descends an incline of « ft in ,
length, having a total rise of A ft.; what will be the velocity,
V, ocquued by the train, the friction being/* lbs. per ton P
408
KINETIC BNBROY OF A RIGID BODY.
Here we have (Art. 213, Sell. 2), the work done on the
train = the work of gravity — the work of friction
= Z%iQ Th — pTs',
which is equal to the work stored up in the train. Hence
2240 7V»
^
= %UQTh—pT8\
• •• V = y/'igh — rh^P'-
5. If the velocity of the tnun in the last example be
t\ ft. per second when the steam is turned off, what will be
its velocity, v, when it reaches the bottom of the incline ?
Ans. V = Vn'o^ + ^h — rhvgP''
6. A body weighing 40 lbs. is projected along a rough
horizontal plane with a velocity of 150 ft. per sec. ; the
cneflicient of friction is |; find the work done against
friction in five seconds. Ans. 3500 foot-pounds.
7. Find the work accumulated in a body which weighs
300 lbs. and has a velocity of 64 ft. per second.
Ans. 19200 foot-pounds.
2ia Kinetie Bnergy of a Rigid Body revolving
round an Axis. — L«t m be the mass of any iMtrticle of
the body at the distance r firom the axis, and lev u bo the
angular velocity, which wil; be the same for every particle,
since the body is rigid; then the kinetic energy of m =
\m (rw)«. The kinetic energy of the whole body will bo
found by taking the sum of these expressions for every
particle of the body. Hence it may be written
£ Jf?ir«w» = •a'S, mi*.
«
(1)
•'-^"^w^iffimiwwiii^^
M
) BODY.
vork done on the
of friction
be train. Hence
; last example be
i off, what will be
» of the incline ?
ed along a rough
I ft. per sec. ; the
ork done against
>00 foot-pounds.
>ody which weighs
jcond.
too foot-pounds.
Body revolving
of any i)article of
and leu u bo the
for every particle,
c energy of f?» =
hole body will bo
ressious for every
ritteu
(1)
EX A MP LBS.
409
S wr» is called the moment of inertia of the body about the
axis, and will be explained in the next chapter.
Hence the kinetic energy of any rotating body = f I<o»,
where I is (he moment of itiertia round the axis, and w the
angular velocity.
in the case of a fly-tcheel, it is sufficient in practice to
tieiit the whole weight as distributed uniformly along the
circumference of the circle dtccribcd by the mean radius
<»f the rim. Let r be this radius ; then the moment of
inertia of any particle of the wheel = mr», and the moment
of inertia of the whole wheel = Mr^, where M is the total
mass. Hence, substituting in (1) wo have 'i Mi^, which
id the kinetic energy of the fly-wheel.
EXAMPLES.
1. Two equal particles are made to revolve on a vertical
axis at the distances of a and b feet from it ; required tlie
point where the two particles must be collected so that the
work may not be altei-ed.
Let m = the mass of each particle, h = the distance of
tlio required point from the axis, and w = the angular voloc-
ity ; then we have
Work stored in both particles = \m («w)» + |m (Sw)* ;
Work stored in both particles collected at point = m (I'w)*; •
.-. m (iw)» = \m (aw)» -f \m (iw)«;
.-. Jc = ViW+~^).
This point is called the centre of gyration. (See next
chapter.)
2. The weight of a fly-wheel is w lbs., the wheel makes
n revolutions per minute, the diameter is 3r feet, diameter
410
SXAMPLSa,
of axle a inches, and the ooeffioieot of friction on the axle
/; how many rovolutioni, «, will tho wheel make before it
stops ?
Work stored in the wheel = ^ (~^f*,
3^ V 60 /
~ 2g'90O '
Work done by friction in z rovolations'
and when the whe< is, we have
J, na to it*n*t*
J -19. — 9 •
12
z =
000
3. Beqnired the nnmber of strokes, x, which the fly-wheel
in the last example, will give to a fozge hammer whose
weight is W lbs. and lift A feet, supposing the hammer to
make one lift for every revolution of the wheel
Here the work dae to raising hammer = Whx. . ■ . Ac.
Ans. X =
«>n*nM
150^(12IFA + na/w)
4. The weight of a fly-wheel is 8000 lbs., the diameter
20 feet, diameter of axle 14 inches, coefBcient of fHction
0.2 ; if the wheel is separated from the engine when mak-
ing 27 revolutions per minute, find how many revolutions
it will make before it sti^s {g taken = 381.2).
Ana. 16.9 revocations.
'riction on the axle
heel make before it
f".
[8
irhich the fl^-wheel
<rg& hammer whoae
ng the hammer to
I wheel
iee. .• .3k.
wn*nh*
i2Wb + ira/w)'
) lbs., the diameter
tfBcient of fHction
engine when mak-
many revolutions
16.9 revocations.
KXAMPLBS.
411
210. Foro* of • Blow.— In order to express the
amount of foroo between the face of a hunmer, for in-
Btancc, and the head of a nail, we mnst consider what
weight mast be lidd upon tho head of the nail to force it
into th« wood. A nul requires a largo force to pull it out,
when friction done is retaining it, and to force it in must
of course require a still larger force.
Now the head of the hammer, when it delivers a blow
upon the head of the nail, must be capable of developing a
force equal for a short time to the oonttnued pv^ssure that
would be produced by a very heavy load. Henee, the effect
of tho hammer may be cxplcmed by the principles of ottergy.
When the hammer is in lotion it has a quantity of kinetic
energy stored up in it, und when it comes in contact with
tlic nail this energy is insttmtly oonvertod into work which
forces the nail into the wood.
VXAIiPLBS.
1, iSappose that a faammefr weighs 1 lb., and that it is
ijnpciled downwards by the arm with considerable force, so
that, at the instant the bead of ^e hamm^jr reaches the
Tiail, it is moving witii a vel6dty of 80 ft pe^||||ond ; find
tiie force which the hammer exerts on t^e nail if it is
driven into the wood one-tmth of an inch.
Let P be the force which the hammer exerts on the nail,
then the work done in forcing the. nail into the wo»>d =
P X ffir, and the energy stored up in the hammer
=: ^„^ = »' = 6.2.
o4
Since the work done in forcing the nail into the wood
mnst be equal to all the work stored in the hammer, (Art.
ai7), wehave
— = 6.2; .'. F=7441b*
412
EXAMPLES,
Hence the force which the hammer exerts on the head of
the nail is at least 744 lbs.
2. If the hammer in the last example forces the nail into
the wood only 0.01 of an inch, find the force exerted on
the nail. Ans. 7440 lbs.
Heace, we see that, nccoiding as the wood m hanlor, i. e., accord-
ing e.s the noil enters I088 at each stroke, the force of the blow
beconios greater. So that when we speak of the " force of a blow,"'
vnc. must B|)ccify how aeon the body givinf; the blow will come t)
r^^t, otherwise the term is meaningless. Thus, suppose n ball of
100 11/6. weight have a velocity that will cause it to ascend 1000 ft. ;
ir' t^io UII is to be stopped at the end of 1000 ft., a force of 100 ll«.
will do it ; but If it is to be stopped at the end of one foot, it will
u((h1 a force of 100000 lbs. to do it ; and to 9top it at the cud ot one
inch will require t/ force of 1200000 Iba, and h) on.
220. Work of a Water-FalL—When water or any
body falls from a given height, it is plain that the work
which is stored up in it, and which it is capable of doing, is
otjnal to that winch would Ikj required to raise it to the
height from which it has fallen ; i. e., if 1 lb. o1^ water
descend tl^|pfgh 1 foot it must accumulate as much work
m would be required to raise it through 1 foot. Hence
when a fall of wz^m is employed to drive a wator-whc"l, or
any other hydraulic niichiue, whose modulus is given, the
work done upon the maciiino is equal to the weight of the
water in pounds x its fall in feet x the modulus of the
machine.
BX AMPLBS.
1. The breadth of a stpoam in h feet, depth n feet, mean
Telocity r feci per minute, and the height of the fall A feet ;
find (1) the horse-power, N, of the water-wooel whose
modulus is m, and (2) find the number of cubic feet. A,
which the wheel will pump per minute from the Iwttom of
the fall to the height of At feoU
erts on the head of
9 forces the nail into
the force exerted on
Ans. 7440 lbs.
iB harder, i. e., acconi-
, the force of the blow
f the " force of a Wow, '
the blow win come t^^
hu8, suppose n ball of
le it to ascend 1000 ft.:
ft., a force of 100 Ite.
! end of one fo»>t, it will
top it at the end ol one
con.
VTien water or any
plain that the work
8 capable of doing, is
id to raise it to the
<•., if 1 lb. o1^ water
lulate as much work
ugh 1 foot. Hence
ive a water-whc"l, or
odulus is given, tlie
to the weight of the
the modulus of the
depth a feet, mean
fht of tlie fall A feet ;
watcr-Wiioel whose
tcr of cubic feet, A,
from the Iwttom of
SXAMPLSS.
413
62.6 Aht
.-. A -
(8)
Weight of water going over the i*t;l per min. = 62.6 abv.
.'. Work of wheel per min. = 62.5 abvhm. (1)
•*' ~33000 ' ' '
Work in pumping water per min. = 62.5 J A, ;
which must = the work of the wheel per min.; hence
from (1) we have
62.5 abvhm ;
abvhm
2. The mean section of a stream is 6 ft. by 2 ft. ; its
mean velocity is 35 ft jjer muiate ; there is a fall of 13 ft.
on this stream, at which is erected n water-wheel whoso
modulus is 0.65 ; find the horse-power of the wheel.
Ans. 6.6 H.-P.
3. In how miiny hours would the wheel in Ex. 2 grind
8000 bushels of wheat, supposing each horse-power to grind
1 bushel per hour ? Ans. 1428i^ hours.
4. How many cubic feet of water must be made to
descend the full per minute in Ex. 2, that the wheel may
grind at the rate of 28 bushels jwr hour ?
Am. 1749.5 cu. ft,
5. Given the stream in Ex. 2, what must be the height
of the fall to grind 10 bushels per hour, if the modi'lus of
the wheel is 0.4 ? Ana. 87.7 feet.
C. Find the useful horsc-jwwer of a wat«r-wheel, lujp-
l)08ing the stream to bo 5 ft. broad and 2 ft. deep, and tx>
ilow with a velocity of 30 ft. per minute ; the 'uight of tho
full being 14 ft.., and tliu modulus of tho machine 0.05,
Ans. 5.2 nearly.
I
4U
MXAMPLES.
221. The IDntjr of an 'Ba^aam.'-Tkt duty of an engine
is the number of units of toork which it is capable cf doing
by burning a given quantity of fud. — It has been found by
experiment that, wha^<)Ter may be the pressure at v, hich
the steam is formed, the quantity of fuel necessary to
evaporate a given volume of water is always nearly the
same ; hence it is most advantageous to employ steam of a
high presavre.*
In good ordinary engines the dnty variea between SOOOGO and
600000 units of work for a lb. of eoal. The extent to which tbe
eoonumy of fool may be carried is well illuBtrated by the engines eni>
ployed to drain the mines in Cornwall, England. In 1816, the
average duty of then engines was 20 million units of work for a
boshelf of coal : in 1848, by reason of snocnsive improvements, tlie
averagv duty had beoome 60 millions, eflSecting a saving of £80000
per aunun. It Is stated ^at in ths ease of oa» engine, the duty waa
raised to 196 millions. The duty of the engine depends largely on
tlie oonatraotloa of the boiler ; 1 lb. of eoal in the Coniish boHer
evaporates 11} lbs. of water, while la a diflferently-shaped boiler 8.7
IB the maximum.)
EXAMPLES.
1. An engine burns 2 lbs. of coal for each horse-^iower
per hour ; find the duty of the engine for a lb. of coal.
Here the work done in one hour
= 60 X 830C0 foot-po ands ;
therefore the duty of the engine = 30 x 33000 foot-pounds,
= 990000 foot-iwunda.
% How many bushels of coal must be expended in a
day of 24 hours in raising 160 cubic ft. of wttt«r \vir minute
-^^ tm TaM In EeshanlM' Macaslna, in the .tmut IMl.
t On« bstbol of coal s 64 or M tbi., depcndliif upon wbura It la. a<i«d«<'e,
p. lao.
t Doarna on Ibe BMkm Bni^nc p. IT1, and PalrbBli.i. ITnofol Infor.iMtlon,
V. m
MP
'^« duty of an mgitie
I is capable of doing
[i has been found by
e pressure at \,hich
)f fuel necessary to
3 always nearly the
► employ steam of a
B between 3000GO and
je extent to which the
kted by the engines eni'
agland. In 181S. the
)n unita of work for a
sive improvement«, tlie
ig ■ MiTlnflr of £80000
# engine, the dutjr waa
Ine depends largeljr on
In the Cornish boiler
eotlj-shaped boiler &7
T each horse^iower
)r a lb. of coal.
ids;
33000 foot-pounds,
00 foot-jwunds.
be expended in •
f wat«r j)or minnte
on wbur* it In. Goodc'e,
tli.<. Uiipful tnror.iMlloo,
WOSK OF A VARtABLX PORCS.
m
-U^
from ft depth of 100 fathoms; the doty of the engine
being 60 millions fof a bniifael of coal ?
Ana. 185 bnsbela.
3. A steam engine is reqaired to raise 70 cubic ft. of
water per minute from a depth of 800 ft. ; find how many
tons of coal will be required per day of 24 hours, supposing
the duty of the engine to be 260000 for a lb. of ooal.
Ana. 9 tona.
222. Work of a Vavtebte Potc«.--When the force
which performB Work through a pTen epaee varies, the
work done may be determined by maltiplying the given
space by the mean of all the variable forces.
Le* AG reprraent " e spfioe in units
of feet through which a variable
force is exerted. Divide AG into
six equal parts, and 8n|^)ose i*„ /»„
P,, etc., to be the forces in pounds
applied at the points A, B, 0, etc.,
respectively. At thep ■ points draw the ordinatcs y,, y,, y„
etc., to represent the lorces Trhioh act at the points A, B,
C, etc. Then the work done from A to B will be equal to
the space, AB, multiplied by the mean of the forces /*,
and P^, i. «., the work will be represented by the area of
the snr&oe AabB. In like manner the work done from
D to will be represented by the area BicO, and so
on ; so that the work done through the whole space, AG,
by a force which varies continuously, will be represented by
the area Aa^O. This area may bo found approximately by
IJio ordinary rule of MenaurtUion for the area of a curved
surface with equidistant ordinates, or more accurately by
A B c E F a
n|.it
Simpson's* rule, the proof of which wo shall now give.
223. Simpson'* Rnla.—Let y,, y„ y„ etc., be the
* Althoanti II WM not luveutud by Hituimin. Sv« 'PDdtaonter.
416
SlitPSOS'B rCLS.
equidistant ordinatcs (Fig. 89) and / the distance between
any two consecutive ordiaates; then by taking the sum of
the trapezoids, AabB, BicC, etc., we have for the area of
AagQ,
i^(yi + Vt) + Viye + yi) + iUy. + y*) + etc.
= \i (y. + 2^8 + 2^8 + 2^4 + ^i + 2ye + yi)-> (0
which id the ordinary formula of mensuration.
Now it is evident that when the curve is always concave
to the line AG (1) will give too small a result, and if con-
vex it will give too large a result
Let Fig. 90 represent the space between any two odd
consecutive ordinates, as Cc and Ee(Fig. 89); divide CE
into three equal parts, CK = KL = LE, , JlAJ
and erect the ordinates Bufc and U, dividing
the two trapezoids CcdD and Drf«E into the
three trapezoids CckK, KklL, and LfcE.
The sum of the areas of these three trapezoids
K OL
ng.M
= ICK (Cc + 2Kk + 2U + E«)
= ^l (Oc + 2Kt 4- 2LZ 4- Ee), (since ^CK = iCD = \l)
= \l (Cc + 4Do + Ec), (since 2K* + 2U = 4Do), (2)
which is a closer approximation for the area of CceE
than (1).
Now when the curve is concave towards CE, (2) is
smaller than the area between CE and the curve ckdle ; if
wo substitute foi Do, the ordinate Drf, which is a little
greater than Do and which is given, (2) becomes
\l (Cc + 4D(/ + E«),
whi'-' is u still closer api>roximatiuu than (2).
(8)
distanco between
iking tlie sum of
B for the area of
+ y*) + ete.
f-2y, +y,); (i)
tion.
18 always concave
38ult, and if oon-
?cn any two odd
;. 89) ; divide CE
[iE,
ing
[the
eE.
>ids
K OL E
d = 4Do), (2)
le area of CwE
rards CE. (2) is
c curve ckdle ; if
which is a little
icomca
(8)
(2).
Simil -y we have for the areas of AccC and "EetfO,
V (Aa + 4B6 + Cc), and \l (Ee + 4F/ + G^).
Adalag (3) and (4) together, we have for an approximate
value of the whole area,
V [yi + ^7 + 2 (f/s + y,) + Hy,+y, + ,/,)], (5)
which is SiinpaoH's Formula. Hence Sirapsou's rule for
finding the area approximately is the following : Divide the
abidssn, AG, into an even number of equal parts, and erect
ordinates at the points of division j then add toijethcr the
first and last ordinates, twice the sum of all the other odd
ordinates, and four times the sum of all the even ordinates ;
multiply the sum by otie-tfiird of the comuon distance
between any two adjacent ordinates. (See Todbuntcr's
Mensuration, also Tate's Geometry and Meusurution, also
Morin's Mech's, by Bouuett.)
EXAMPLES.
1. A variable force has acted through 3 ft. : the value of
the force taken at seven successive oqnidistaut points,
including the first and the last, is in lbs. 189, 151.2, 126,
108, 94.6, 84, 75.6 ; find tiie whole work done.
Ans. 346,4 foot-jwunds.
2. A variable force has acted through 6 ft. ; the Ytduo of
the force taken at seven successive equidistant pointij,
including the first and the lust, is in lbs. 3, 8, 15, 24, 35,
48, 63 ; find the whole work done.
Ans. 162 foot-j)onnds.
8. A variable force has a<'tod through 9 ft.; the value of
the force taken at se^en successive equidistant points,
including the first and the lost, is in lbs. 6.082, 6.164,
6.245, 6.403, 6.481, 6.557; find the whole work done.
Ans. 56.907 foot-puund«.
':m.
MMMnppMKMMT^cwu^ .-
416
aXAMPLMS.
Should any of the ordinates become zero, it will not pre-
vent the use of Simpeon's rale.
Simpson's rule is a^Uoable t:) other investigations as
well as to that of work done by a variable fOTce. For
example, if we want the velocity generated in a given time
in a particle by a variable force, let the straight line AG
represent the whole time during which the fcnrce actii, and
let the stTMght lines at right angles to AG reprei>6ui the
force at the corresponding instants; then the area will
represent the whole space described in the given time.
BXAMPL.BS.
1. The ram of a pile-driving engine weighs half a ton,*
jmd has a fall of 17 ft. ; how many units of work are per-
foi-med in raising this ram P Ana. 19040 foot-pounds.
a. How many units of woA are required to raise 7 cwt
of coal from a mine whose depth is 13 fathoms ?
Ans. 61162 foot-pounde.
3. A horse is used to lift the earth fi-om a trench, which
he does by moans of a cord going over a pulley. He pulls
up, twice v^ery 5 minutes, a uan weighing 130 lbs., and a
barrowf ul of earth weighing 860 lbs. Each time the horse
gooa forward 60 ft. ; find the units of woik done by the
horse per hour. Ana. 374000.
4. A railway i ain of T tons asoends an inclined plane
which has a rise of e ft. in 100 ft., with a uniform speed of
m miles per hour ; find the horse-power of the engine, the
friction being/) lbs. per ton.
Ans. ^y(P+/^-^ ) H..P.
6. A railway train of 80 tons ascrnds an incline which
rises one foot in 50 ft., with the uniform rate of 16 miles
* Ont lull -- tW ewt. - 9S40 lbs.
wmmmmtmtm
mi
rot it will not pre-
' investigations as
riable force. For
d in a given time
straight line AG
he force acts, and
A.G repreucui the
len the area will
e given time.
reighs half a ton,*
of work are per-
'40 foot-pounds.
d to raise 7 cwt
loms ?
52 foot-ponnde.
1 a trench, which
pulley. He pulls
g 130 lbs., and a
h time the horse
oi'k done by the
Am. 374000.
m inclined plane
miform speed of
' the engine, the
+ 22^) „ p
76 "'^^
an incline which
rate of 16 miles
SXAMPCES.
419
per hour ; find the horse-power of the engine, the friction
being 8 lbs. per ton. Ans. 169.96 H.-P.
6. If a horse exert a traction of t lbs., what weight, v>,
will he pull up or down a hill of small inclination which
has a rise of in 100, the coefficient being/?
. 100/
Ana. w = -— r-; •
100/±«
7. From what depth will an engine of 22 horse-power
raise 13 t«ins of coal in an hour ?
Am. 24.9 ft.
8. An engine is observed to raise 7 tons of material an
hour from a mine whose depth is 85 fothoms ; find the
horse-power of the engine, supposing | of its work to be
lost in transmission. Ans. 4.4829 H.-P.
9. Required the horse-power of an engine that would
supply a city with water, working 12 hours a day, the
water to bo raised to a height of 50 ft. ; the number of
inhabitants being 130000, and each person to use 5 gallons
of water a day, the gallon weighing 8| lbs. nearly.
Aug. 11.4 H.-P.
10. Prom what depth will an engine of 20 horse-power
raise 600 cubic feet of water per hour P Ans. 1056 feet
11. At what rate per hour will an engipe of 30 horse-
power draw a train weighing 90 tons gross, the resistance
being 8 lbs. per ton ? Ant. 16.628 miles.
12. What is the gross weight of a train iHliioh an engine
of 25 horse-power will draw at the rate of 26 miles an
hour, resistances being 8 lbs. per ton ?
Ans. 46.875 tons.
13. A train whose gross weight is 80 tons travels at the
rate of 20 miles an hour; if the resistance is 8 Uw.
per ton, what is the horse-power of the engine ? '
Am. Ui\ H.-P.
r
i'.
420
X^ AMPLSa.
14. What must be the length of the stroke of a piston
of an engine, the surfuco of wliich is 1500 square iiicliee;,
which makes 20 strokes per minute, so that with u weuii
pressure of 12 lbs. on each square aua of the piston, the
engine may be of 80 horse-power ? Ans. 7| ft.
15. The diameter of the piston of an engine is 80 ins.,
the length of the stroke is 10 ft, it makes 11 strokes per
minute, and the mean pressure of the steam on the piston
is 12 lbs. per square inch ; what is the horse-power ?
Ans. 201.0CH..P.
16. The cylinder of a steam engine has an internal
diameter of 3 ft, the length of the stroke is 6 ft., it miikcs
6 strokes per minute; under what efFective pressure ^ar
square inch would it have to work in order that T5 hoi'sc-
power may be done on the piston? Ans. 67' 54 lbs.
17. It is said that a horse, walking at the rate of 2^ miles
an hour, can do 1G50000 units of work in an hour ; what
force in pounds does ho continually exert ?
Ans. 126 lbs.
18. Find the horse-power of an engine which is to move
at the rate of 30 miles an hour, the weight of the engine
and load being 50 tons, and the resistance from friction
10 lbs. jier ton. Ans. 64 H.-P.
19. There were 0000 cubic ft. of water in a mine whoso
depth is 60 fathom^, when an engine of 50 horse-power
began to work the pump ; the engine continued to work 5
hours before tho mine was cleared of the water ; required
the number of cubic ft of water wliich had run into the
mine during the 5 hours, supposing \ of the work of the
engine to be lost by transmission. Ans. 10500 cubic ft
20. Find the horse-jiower of a steam engine which will
raise 30 cubic it. of water ^Ksr minute from a mine 440 ft
doop. Ann. 25 ll.-P.
"oke of a piston
square iiiclictj,
it with u lucuii
' the piston, tbo
Ans. 7|ft.
igiue is 80 ins.,
11 strokes per
1 on the piston
-power ?
J01.00H.-P.
as an internal
6 ft., ft miikcs
e pressure ^ ar
r that 75 hoi-se-
8. 67. 54 lbs.
^to of 2^ miles
an hour ; what
ns. 125 lbs.
ich is to move
of the engine
from friction
|w. 64H.-P.
a mine whoso
horse-powor
uod to work 5
iter ; required
run into the
ifi work of the
00 cubic ft.
36 which will
a mine 440 ft.
6. aoU.-P.
,jjji|i,|gi!a > ;iii | ^p i! ij^^ i iyffyy?tjMW
SXAMPLSS.
21. If a pit 10 ft. deep with an area of 4 square ft. be
excavated and the earth thrown up, how much work will
have been done, supposing a cubic foot of earth to weigh
90 lbs, Ans. 18000 ft-lbs.
22. A well-shaft 300 ft. deep and 5 ft. in diameter is full
of water ; how many units of work must be expended in
getting this water out the well ; (t. «., irrespectively of any
other water flowing in)? Ans. 165223850 ft.-lbs.
23. A shaft o ft. deep is full of water; find the depth of
the surface of the water when one-quarter of the work
required to empty the shaft has been done. . a „
i
24. The diameter of the cylinder of an engine is 80 ins.,
the piston makes per minute 8 strokes of lOJ ft. under a
mean pressure of 15 lbs. per square inch ; the modulus of
the engine is 0'65; how many cubic ft of water will it
raise from a depth of 112 ft. in one minute?
Ans. 485. 78 cub. ft.
25. If in the last example the engine raised a weight of
66433 lbs. through 90 ft. in one miuute, what must bo the
mean pressure jier square inch on the piston ?
Ans. 26.37 lbs.
26. If the diameter of the piston of the engine in Ex. 24
had been 85 ins., what addition in horse-power would that
make to the useful power of the engine ?
^»«. 13-28 H.-P.
27. If an engine of 60 horse-power raise 2860 cub. ft. of
water per hour from a mine 60 fathoms deep, find the
modulus of the engine. Atu. '65,
28. Find at what rate an engine of 30 horse-power could
draw a train weighing 50 tons up an incline of 1 in 280,
the resistance from friction being 7 lbs. per ton. ,
Ans, 1320 ft, per minute.
422
SXAUPLSS.
29. A forge hammer weighing 300 lbs. makes 100 lilts a
minate, the perpendicniar height of each lift being 2 fL;
what is the horse-power of the engine that gives motion to
20 such hammers? Ans. 36*30 H.-P.
30. An engine of 10 horse-power raises 4000 lbs. of coal
from a pit 1200 ft. deep in an hour, and also gives motion
to a hammer which makes 50 lifts in a minute, each lift
having a perpendicular height of 4 ft.; what is the weight
of the hammer? Ans. 1250 lbs.
31. Find the horse-powor of the engine to raise T tons of
coal per hour from a pit 'vhose depth is a fathoms, and \jA
the same time to give motion to a forge hammer weighing
to lbs., which makes n lifts per minute of h ft each.
224flr + nhw „ _
Ana, 7^-^t ■ xl.-r.
32. Find the useful work done by a fire engine per
second which difioharges every second 13 lbs. of water with
a velocity of 60 ft. per second. Am. 508 nearly.
33. A mil way truck weighs m tons ; a horse draws it
along horizontally, the resistance being « lbs. per ton ; in
passing over a sjiace a the velocity changes from « to f ;
find the work done by the horse in this space.
2240W , . J, ,
Ana. —X — («» — «') + mns.
34. The weight of a ram is 600 lbs., and ai the end of
the blow has a velocity of 32J ft.; what work has Ijcen
done in raising it P Am. 9G50.
35. Retiuired the work stored in a cannon ball whose
weight is 32^ lbs., and velocity 1500 ft. Am. 1126000.
86. A ball, weighing 20 lbs., is projected with a velocity
of 60 ft. a second, on a bowling-green ; what space will the
ball move over before it comes to rest, allowing the friction
to be T«5 the weight of the ball? Am. 1007.3 ft.
':'mimimimmmmmmf>mm
akes 1001iit« a
lift being 2 ft.;
jives motion to
36.36 H.-P.
OOO lbs. of coal
10 gives motion
linuto, each lift
t is the weight
ns. 1250 IbH.
raise T tons of
ithoms, and "ut
nmer weighing
t. each.
■h nhw
H.-P.
ire engine per
. of water with
508 nearly.
Iiorse draws it
per ton; in
from u to V ;
«') + mns,
at the end of
'ork has Ixjen
Ans. 9650.
ball ^.vhose
1125000.
ith a Telocity
space will the
the friction
1007-3 ft.
SXAMPLKS.
m
87. A train, weighing 198 tons, haa ft velocity of 80
miles an hour when the steam is turned off; how far will
the train move on a level nul before coming to rest, the
friction being 6^ Iba per ton !* Am. 12266 ft.
38. A train, weighing 60 tons, has a velocity of 40 miles
an hour, when the steam is turned off, how Ux will it
ascend an incline of 1 in 100, taking friction at 8 lbs. a ton ?
Jim. 3942i ft
39. A carriage of 1 ton moves on a level rail with the
speed of 8 ft a second; through what space must the
carriage move to have a velocity of 2 ft., supposing friction
to be 8 lbs. a ton? J >w. 348 ft
40. If the carriage in the last example moved over 400
feet before it comes to a state of rest, what is the resistance
of friction per ton ? Ans. 6.57 lbs.
41. A force, P, acts upon a body parallel to the plane;
find the space, % moved over when the body has attained a
given velocity, v, the coefficient of friction being/, and the
body weighing «? lbs. a*,, , ?£??__.
Ana. " - 2g{P -fw)
42. Suppose the body in the last example to bo moved
for t seconds ; reqair«)d (1) the velocity, v, acquired, and
(2) the work stored.
A.ta. (1)
■1
to
w
t9\ (2)
2w
43. A body, weighing 40 lbs., is projected along a rough
horizouta! plane with a velocity of 150 ft per second ; the
coefficient of friction is \ ; find the work done against fric-
tion in 5 seconds. Ans. 3500 foot-pounds.
44. A body weighing 60 lbs., is projected along a rough
horizontal plane with the velocity of 40 yards per second ;
find the work expended when the body comes to rest.
Ah8. 11260 ft-lbs.
■ ■y f ^ « »'» , ' ' «■*■ <
424
-.. - EXAJtPLKS.
'
45. If a train of cars weighing 100000 lbs. is moving on
ft horizontal truck with a velocity of 40 milos an hour when
the Bteam is turned off ; through what space will it move
before it ig brought to rest by friction, the friction being
8 lbs. i)er ton ? Am. 13374.8 ft.
40. What amount of energy is acquired by a body weigh-
ing 30 lbs. that falls through the whole length of a rough
inclined plane, the height of which is 30 ft., and the base
100 ft., the coefficient of friction being \ ?
Ans. 300ft.-lb8.
47. If a train of cars, weighing 7' tons, ascend an in-line
having a raise of e ft. in 100 ft., with the velocity v^ ft. per
second when the steam is turned off; through what space,
X, will it move before it comes to a state of rest, the friction
being j» lbs. per ton ? . liaOi--^
Ann. X = — TT-r-.— ^
g {•l-i.\e, + ;))
48. Suppose the train, in Ex. 4, Art. 217, to be attached
to a rope, passing round a wheel nt the top of the incline,
which has an empty train of T, tons attached to the other
extremity of the rope: what velocity, r, will the train
acquii-e iu descending s ft. of the incline ?
Ans, r
Ti
gps
1126"
49. An engine of 35 horse-power makes 20 revolutions
per minute, the weight of the Hy-wheel is 20 tons and the
diameter is 20 ft.; what is the accumulated energy in foot-
pounds? Ans. 307000.
50. If the fly-wheel ui the last example lifted a weight of
4000 lbs. through 3 ft., what part of its angular velocity
woitlu it lose ? Ans. ^.
51. If the axis of tlie above fly-wheel be 6 ins. in
diameter, the coefficient of friction 0-075, what fraction,
. IS moving on
3 an hour when
aco will it move
3 friction being
r. 13374.8 ft.
T a body weigh-
>gth of a rough
, and the base
?. 300ft.-lbs.
fend an in':!line
)eity j'p ft. per
jh what space,
)st, the friction
_l]20j;o2
{•i'ZAe+p)'
to be attached
nf the incline,
to the other
tvill the train
9P«
1126'
!0 revolutions
tons and the
lergy in foot-
n». 307000.
d a weight of
pilar velocity
Ans. ■^.
be 6 ins. in
'hat fraction,
Hi
EXAMPLES.
approximately, of the 35 horse-power is expended in turn-
ing the lly-wheel ? Ans. ^.
52. In pile driving, 38 men raised a ram 12 times in an
hour ; the weight of the ram was 12 cwt., and the height
through which it wan raised 140 ft.; find the work done by
one man in a minute. Ans. 990 ft-lbs.
53. A battering-ram, weighing 2000 lbs., strikes the
head of a pile with a velocity of 20 ft per second ; how far
will it drive the pile if the constant resistance is 10000 lbs.?
Ans. 1.25 ft.
54. A nail 2 ins. long was driven into a block by suc-
cessive blows from a monkey weighing 5.01 lbs.; after one
blow it was found that the head of the nail projected 0.8
of an inch above the surface of the block ; the monkey was
then raised to a height of 1.5 ft, and allowed to fall uiwn
the head of the nail ; after this blow the head of the nail
was 0.46 of an inch above the surface ; find the force which
the monkey exerted upon the head of the nail at thia blow.
Ans. 265.66 lbs.
55. The monkey of a pile-driver, weighing 500 lbs. is
raised to a height of 20 ft, and then allowed to foil upon
the head of a pile, which is driven into the ground 1 inch
by the blow; find the force which the monkey exerted
upon the head of the pile. Ans. 120000 lbs.
56. A steam hammer, weighing 500 lbs., falls through a
height of 4 ft. under the action of its own weight and a
steam pressure of 1000 lbs.; find the amount of work
which it can do at the end of the fall.
Ans. 6000 ft-lbs.
57. The mean section of a stream is 8 square ft.; its
mean velocity is 40 ft per minute ; it has a fall of 17^ ft.;
it is required to raise water to a height of 300 ft by means
of a water-wheel whose modulus is 0.7 ; how many cubic ft
will it raise per minute ? Ans. 13.07 cub. ft.
-t'
426
XXAMPLBA
68. To what height would the wheel in the last example
raise 2^ cub. ft of water per minute ? Ans. 1742} ft
59. The meaa eectioo of a stretam is 1| ft by 11 ft; ita
ntean velocity is 2^ miles an hour ; there is on it a fall of
6 fk. on which is erected a wheel whose modulus is 0.7 ; this
wheel is employed to raise the hammers of a forge, each of
which weighs 2 tons, and hw a lift of l^ft.j how many
lifts of a hammer will the wheel ,^eld per minute P
Am. 142 nearly.
60. In the last example determine the mean depth of
the stream if the wheel yields 136 lifts per minute.
Ana. 1.43 ft
61. In Ex. 59, how many cubic ft of water must descend
the fall per minute to yield 97 lifts '>f the hammer per
minute ? Ans. 2483 cub. ft
62. A stream is a ft broad and b ft deep, and flows at
the rate of e ft per hour ; there is a fall of A ft ; t lie water
turns a machine of which the modulus is e ; And the num-
ber of bushels of corn which the machine can grind in an
hour, supposing that it requires m units of work per
minute for one hour to grind a bushel. ^ _ lOOOabche
Ans.
16 X 60m
63. Down a l4-ft fall 200 cnK ft. of water descend every
minute, and turn a wheel whose modulus is 0.6. The
wheel lifts water from the bottom of the fall to a height of
64 ft; (1) how many cubic ft will be thus raised jkt
minute? (2) " the water were raised from the top of the
fall to the same [loint, what would the numl)cr of cubic ft
then be? Ans. (1) 31.1 cub. ft.; (2) 34.7 cub. ft.
In tlut ispcond caim the numbnr of cub. ft. of water taken from the
top of tho fnll bi-lDg <r, tbe number of ft. tliAt will turn the wheel will
bcaOO-sr.
04. Find how many units of work are stored up in a
.•M
last example
19. 174iJfft.
by 11 ft. ; ita
>n it a tall of
las is 0.7 ; this
forge, each of
!t.j how many
lute?
142 nearly.
leao depth of
nute.
fu. 1.43 ft.
' mast descend
I hammer per
1483 cub. ft.
and fiowH at
ft; die water
Ind the num-
grind in an
of work per
lOOOabcJte
16 X 60m'
descend every
is 0.6. The
to a height of
us raised jwr
ic top of the
r of cubic fU
4.: cub. ft.
takeu from the
the wheel will
Drcd up iu a
sm
I HiiiJiMIIUL,
KXAMPhSa.
427
mill-pond which is 100 ft. long, 60 ft. br^, «nd 3 ft. deep,
and has a fall of 8 ft. Ans. 7600000.
66. There are three distinct levels to be pamped in a
mine, the flrst 100 fathoms deep, the second 120, the thiid
150 ; 30 cub. ft. of w&ter are to come from the first, 40 from
the second, and 60 from the third per minute ; the duty of
the engine is 70000000 for a bushel of coal Determine (1)
its working horse-power and (2) the consumption of coal
per hour. Ana. (1) 191 H.-P. ; (2) 5.4 bushels.
66. In the last example suppose there is another level of
160 fathoms to be pumped, that the engine docs as much
work as before for the other levels, and that the utmost
r»ower of the engine is 276 H.-P. ; find the greatest number
of cub. ft. of water that can be raised from the fourth level.
An$. 46^ cub ft.
67. A variable force has acted throngh 8 ft.; the value
of iho force taken at nine successive equidistant points,
including the first and the last, is in lbs. 10.204, 9.804,
9,434, 9.090, 8.771, 8.475, 8.197, 7.937, 7.692; find the
whole work done. Ana. 70.641 foot-pounds.
68. The value of a variable force, taken at nine succes-
sive equidistant points, including the first and the last
points, is in lbs. 2.4849, 2.6649, 2.fS391, 2.7081, 2.7726,
2.8332, 2.8904, 2.9444, 2.9967, the common distance between
the points is 1 ft ; find the whole work done.
An». 22.0967 foot-ponnda
69. A train whose weight is 100 tons (including the
engine) is drawn by an engine of 160 horse-power, the fric-
tion being 14 \h% per ton, and all other resistances neglect-
ed ; find the maximum speed whicli the engine is capable
of sustaining on a level rail. An». 40^ miles per hour.
7d. If the train described in the hist example bo movipg,
tit a particular instant with a velocity of 16 ^ailes per hour.
EXAMPLES.
and the engin? workirig at full power, what ia the accelera-
tion at that instant ? (Call g = 3*^.) Ans. -^^V
71. Find the horse-power of an engine required to drjig a
train of 100 tone up an incline of 1 in 60 with a velocity of
30 miles an hour, the friction being 1400 lbs. .
Ans. The engine must be of not lees than 470f horse-
IK)wer. This is somewhat above the power of most locomo-
tive engines.
72. A train, of 200 tons weight, is ascending an incline
of 1 in 100 at the rate of 30 miles per hour, the friction
being 8 lbs. per ton. The steam boiug shut off and the
break applied, the train is stopped in a quarter of a mile.
Find the weight of the break-van, the coefficient of fric-
tion of iron on iron being |. Atis. 11-^ tons.
is the acceltra-
Ans. ^^.
[uired to drag u
th a velocitj' of
1. .
lan 470| horse-
>f most locomo-
ling an inclino
lur, the friction
mt oflf and the
rter of a mile,
flicient of fric-
s. 11^ tona.
CHAPTER VI.
MOMENT OF INERTIA*
224. Moments of In6rtia.'-The quantity Lmr' in
which m is the mem of. an element of a body, and r its
ilistance from an axis, occurs frequently in problems of
rotiition, so that it becomes necessary to consider it in
detail ; it is called the moment of inertia of the body about
the axis (Art, 218). Hence, "moment of inertia" may be
defined as follows : If the mass of every particle of a body be
multipiied by the square of its distance from a straight line,
the sum of the products so formed is called the Moment of
Inertia of the body about that line.
If the mass of every particle of a bjdy be multiplied by
the square of its distance from a ^iven plane or from a
given pointy tiie sum of the products so formed is called the
moment of inertia of the body with reference to that plane
or that point.
If the body bo referred to the axes of x and y, and if the
mam of each parttcio be multiplied by its two co-ordinates,
.r, y, i\> sum of the products so formed is called the
product of inerd-a of Ihe body about those two axes.
If dm denote 'he mass of an element, p its distance from
the axis, and / tl>e moment of inertia, we have
• / = l.jMm. (I)
If the body Ix? leforred to rectangular axes, and x, y, «,
l)e the co-ordinatob of any ''lomont, then, according to the
dcfiuitioDs, the moments of inertia alM)ut the axes of x, y,
z, respectively, will be
* Thit term wm Introduced b; Bulor, knd bav now g<A into general umc wbeii-
eror Kigtd DjmMnlw U iIimUoiI.
^mmmmmmmmmsmsmm
mmmmmmmammmmT
I
480 EXAMPLES.
The momentB of inertia with respect to the planes yz, zx,
xy respectively, are.
S vNhn, S yWfW, S Mm.
(3)
The prodacts of inertia with respect to the axes y and t,
K and X, X and y, are
l,ytdm, Izxdm, Xxydm. (4)
The momont of inertia with respect to the origin is
S (a^ + y» + 1^) rfw = 1 r*dtH, (S)
where r is the distanoe of the particle from the origin.
The moment of inertia of a lamina, when the axis lies in
it, is called a rectangular moment of inertia, and when it is
perpendicular to the lamina it is called a polar motnent of
inertia, and the corresponding axis is callod the rectangular
or the polar axis.
The process of finding momenta and produete of inertia
is merely that of integration ; but after this has been accom-
plished for the simplest axes possible, they can be found
without integration for any other axes.
EXAMPLES.
1. Find the moment of inertia of a uniform rod, of mass
m, and length I, about an axis through its centre at right
angles to it.
Lpt X he tne distanoe of any element of the rod from the
centre, and ft the mass of a unit of length ; then dm = ftdx,
which in (1) gives for the moment of inertia 1 itsMx, or
-A
/*:r»<fo,
JtXAMPLS.
431
[a^ + f^dm, (2)
I the planes yc, zx,
I. (3)
the axes y and t,
«. (4)
the origin is
dm, (ft)
n the origin,
en the axis lies in
ia, and when it is
polar motnetU of
)d the recianffular
roduett of initio
is has been aooom-
ley can be foond
orm rod, of mass
ts centre at right
the rod from the
then dni = ftdx,
;ia £ lisMz, or
remembering that the symbol of summation, £, includes
integration in the cases wherein the body is a continuous
muss.
Hence / — ^j^ = itmP.
If the axis be drawn throngh one end of the rod and
perpendicular to its length we shall have for the moment
ol" inertia
/ = \mP.
2. Find the moment of inertia of a rectangular lamina*
iihout an axis through it? centre, parallel to one of its sides.
Let b and d denote the breadth and depth rospoctively of
the rectangle, the former being parallel to the axis. Im-
agine the lamina composed of elementary strips of length b
piuullel to the axis. Let the distance of one of them fi-om
ilio axis be y, and its breadth dy ; then, denoting the moss
of a unit of area by ft, we have dm = ftbdy, which in (1)
gives
If the axis be drawn through one end of the rectangle, we
shall liave for the moment of inertia
/ = twd».
3. Find the moment of inertia of a circular lamina with
re8)>ect to an axis through its centre and perpendicular to
il8 surface.
Ijet the radius = a, and fi the mass of a unit of area as
before, then wo have
t/o ''0
• In all MM« mv shall anamo the thlcknens of tho laminin nr plateit lo he
liiaultcalmal.
..■ mamm
is*
433
PARALLSC AXSS,
4. Find tho moment of inertia of a oircnlar plate (1)
about u diameter as an axis, and (2) about a tangent.
Ans. (1) imtfi; (2) Jma*.
5. Find the moment of inertia of a square plate, (1)
about an axis through its centre and perpendicular to itH
plane, (2) about an axis which joins the middle points of
two opposite sides, und (3) about an axis passing tlirough
an angular point of the plate, and perpendicular to its
plane. Let a = the side of the plate and ^ the miiss of a
unit of area.
(2) T»,»irt3; (3) |wfl«.
6. Find the moment of inertia of an isosceles triangular
plate, (1) about an axis through its vertex and perpen-
dicular to its plane, and (2) about an sxis which passes
through its vertex and bisects the base.
Let 2b = the base and a = the altitude, then
^0 ''0
m
fi (a^ + y») rfy dx = ^ (3«» + b>) ; (2) ^»»i».
225. Moments of Inertia relative to Parallel
Axes, or Planes. — The moment ofimrtia qfa body about
any axis is equal foils moment of inertia abopt a parallel
axis through the centre of gravity of the body, plus the
product of the mass of the body into the square of the dis-
tance between the axes.
Let the piano of the paper pass
tlirough tho centre of gravity of tho
body, and be perpendicular to tho two
parallel axes, meeting them in and
O, and let P be the projection of any
element on the plane of the pa])er.
a oircalar plate (1)
oat a tangs^nt.
L) im^; (2) |»ia«.
a square plate, (1)
perpendicular to itn
ihe middle points of
sis passing tlirough
)orpeudicuIar to its
and n the mass of a
6 ~ c" ■
I isosceles triangular
vertex and pcrpen-
sxis which passes
ido, then
k» + «») ; (2) ^mb*.
tive to Parallel
tia qfa body about
rtia nbo^t a parallel
the body, plus the
he square of the die-
EH
BXAMPLES.
433
Take the centre of gravity, O, as origin, the fixed axis
through it perpendicular to the plane of thepapw a» tlie
axis of z, and the plane through this and the pai-allcl axis
for that of 2*; and let /, be the moment of inertia abont
tlio axis through G, /that for the parallel axis (hrougn 0,
a the distance, OG, between the axes, and (.r, //) any point,
/'. Then we shall have
/, = 2 (.i;» + yi) dm ; / = X [(a; + a)^ + f] dm.
Hence I — I^ = 2« Zxdm + a'lrfm = a'hti,
aineo I.xdin ~ 0, as the centre of gravity is at the origin.
.-. /--= 7, +rt»m, (1)
wliich is called the formula of reduction.
llonco the moment of inertia of a body relative to anv
axis can be found when tiiat for the parallel axis through
its centre of gravity is known.
Cor. 1. — The moments of inertia of a body are iho same
for all pai-allel axes situated at the same distance from its
centre of gravity. Also, ot all jjarallel axes, that which
passes through the centre of gravity of a body has the least
moment of inertia.
Cor. 2.— It is evident that the same theorem holds if the
moments of inertia bo taken with respect to i)arallel planen,
instead of parallel axes.
A similar proi)erty also connects the moment of iiupiiii
relative to any point with that relative to the centre of
gravity of the body.
EXAMPLES.
1. The moment of inertia of a rectangle* in reference
to an axis through its centi-o and parallel to one end is
* Sec Note to R«. t, Art. *M ; strictly fpt^WiiR, an area bat s moment of inertta
no more tUau it bw weight.
■ mmxiini i Jnm
434
SADIU8 or' aVBATtOff.
■^^mtP ; find tho motoent of iaertia in rdercnee to a parallel
axis throngh one end.
From (1) we have
/, = ^m<P + -J- w = Jtnrf*. ^^B
3. The moment of inertia of an isosceles triangle about
an axis throngh its vertex and perpendicular to its plane
is \m (3efl + i»), (Art 224, Ex. 6) ; find its moment about
a i^irallel axis through the centre.
From (1) we have
J = ^(3a» + i») - ♦a»»» = im(V»» + J»).
3. Find the moment of inertia of a circle about an axis
through its circumference and perpendicular to its p^ine
(See Ex. 3, Art. 224). Ans. fnufi.
4. Find the moment of inertia of a square about an axis
through the middle point of one of its sides and perpen-
dicular to its plane (Ex. 5, Art 224). Am. ^fgtnaK
226. Radiiui of Qyratic^— Let k be such a quantity
that the moment of inertia = mk?, then we shall have
/ = I,r*dm = mk^.
(1)
The distance * is called the radius of gyraUon of the
body with respect to the fixed axis, and it denotes the
distance from the axis to that point into which if the whole
mass were concentrated the moment of inertia would not be
altered. The point into which the body might be concen-
trated, without altering its moment of inertia, is called the
centra of gyration. When the fixed axis passes through tho
centre of gravity, tho length k and the point of oonoentra-
tion are called pritmjHtl radius and principal centre of
gyration.
DQC8 to a parallel
P.
B triangle about
ular to its plane
s moment about
e about an axis
liar to its p^'*ne
Ans. |«M^.
ire about an axis
rides and perpen^
Ant. ^^ftna*.
such a quantity
e shall have
(1)
'" gyration of the
I it denotes the
rhioh if the whole
irtia would not be
might be concen-
rtia, is called the
lases through the
>int of oonoentra-
ineipal centre of
BADWa OF aniATIO/f.
Let A:, = the {Nrinoipal radius of gyration and r, the
distance of an element fivm the axis through the centre of
f^ravity; then from (1) we have
= £ Ti^dm + m<fi, [by (I) of Art. 225]
= mk^^ + mo*;
.-. A* = *j» + ^», (2)
from which it appears that the principal radius of gyration
is Ifie hast radius for parallel axes, which is also evident
from Cor, 1, Art 226.
Son. — In homogeneous bodies, since the mass of any part
Miiies directly as its volume, (1) may be written
J.f*dV=z V1^,
(8)
where rfT denotes the element of volume, and V the entire
volume of the body.
Hence, in homogeneous bodies, the value of A is inde-
pendent of the density of the body, and depends only on its
form ; and in determining the moment of inertia, we may
tiike the element of volume or weight for the element of
mass, and the total volume or weight of the body instead
of its mass.
Also in finding the moment of inertia of a lamina, since
k is independent of the thickness of the lamina, we may
hike the element of area instead of the element of mass,
iind the tot^ area of the lamina instead of its m f w.
From (1) we have
*« = ^. (4)
m
Similarly,
*.'
= i.
m
(6)
430
POLAR MOMENT OF lySHT/A.
henco, the square of the radius of gyration with reaped to
any axis equals the moment of inertia with respect to tha
same axis divided by the mass.
EXAMPLES.
1. Find the princij)al imlins of gyration of a straight
liuc.
From Ex. 1, Art. 224. wc have
therefore from (5) we have k\^ = ^P.
2. Find the principal radiu.s of gyration of a circle (1)
witli respect to a polar axis, and (2) with respect to u
rectangular axis.
Ans. (])|««; (2)K-
3. Find the principal radius of gyration of a rectangli!
with respect to a rectangular axis. Ans. ^d^.
4. Find the principal radius of g}'ration (1) of a square
with resiwct to a pohir axis, and (2) of an isosceles triangle
with rcsjtect to a polar axis.
Ans. (1)K; (2) ^(^^3 + ^2).
227. Polar Moment of Inertia— If any thin plate, or
lamina, l)e referred to two rectangular axes and x, y he the?
co-ordinates of any elemont, then (Art. 224) the momentw
of inertia about the axes of x and y respectively, are S yHm
and £ x^dm ; and therefore the moment of inertia witli
respect to the axis drawn perpendicular to the plane at the
intersection of the axes of x and y is
S (^ + y^) dm.
Hence the polar moment of inertia of any lamina is equal
to the sum of the tnomrvts of inertia with respert to any two
rectangular axes, hjing in the plane of the lamina.
w
th
be
nu
re
III]
of
its
|.h
itti
774.
m with respect to
vith respect to the
ktiou of a straight
tion of a circle (1)
with rcspoot to n
tion of a rectangle
A lis. ^(P.
m (1) of a square
ti isosceles triauglo
(2) i(i«» + *»).
f any thin plate, or
:e9 and x, y he tho
334) the momenta
ctively, are X y^dni
»t of inertia with
» the plane at the
ny lamina is equal
respert to any two
1 lamina.
POI.Ali MOMENT OF TXtlRTIA.
4;j7
Coil.— For every two rectangular axes in the plane of
the lamina, at any jwint, we have
S xMm + 2; i^dm = const
that is, the sum of the moments of inertia with respect to a
pair of rectatigulur axes is constant. Heucu, il one be a
maximum, the other is a minimum, and vice versa.
EXAMPLES.
1. Find the moment of inertia of a rectangle with respect
to an axis tlirough its centre and perpendicular to its plane.
From Ex. 2, Art. 324, the rectangular momenta of
inertia are
-^^miJP and ^»iA» ;
tlierefore the polar moment of inertia = ^m {iP + 4«) ;
2. Find the moment of inertia of an isosceles triangle
, with respect to an axis through its centre parallel to its
1)086, a being the altitude and and 2b the base.
Ans. ^ina'^; P = -^ciK
228. Moment of Inertia of a Solid of Revolution,
with respect to its Geometric Axia— Let ihe axis be
that of x; and let tho equation of the generating curve
''0 y =f{x). Let the solid be divided into an inlinite
number of circular plates perpendicular to the axiii of
revolution ; let the density be uniform and ft the mass of a
unit of volume; and denote by x tho distance of the centre
of any circular plate from tho origin, y its radius, ami rfx
its thickness ; thou the moment of inertia of this circular
plate about an axis through its centre and j)er[K?ndicul!ir to
\ia i>lane, by (Ex. 3, Art. 334), is
S^Sa^^S
4E3SaaswTOiasar
438
XXAMPLXS.
therefore the moment of inertia of the whole solid is '
f/[/W^; (1)
the integration being takeji between proper limits.
EXAMPLES.
1. Find the moment of inertia of a right circular cone
about its axis.
let h =r. the height and * = the radius of the base ;
then the equation of the generating curve is y = * a-,
which in (1) gives for the moment of inertia,
10
m* Jo ^ ~
= A'"**, (since m = ^f^hlA'
I since m
Therefore *,« = ^.
2. Find the moment of inertia (1) of a solid cylinder
about its axis, b being its radius and h its height, and (2)
of a hollow cylinder, b and b' being the external and
internal nsdU. Am. (1) f«*» ; (2) ^m {W + b^
3. Find the moment of inertia of a paraboloid about its
axis, h being its altitude and b the radius of the base.
Am, -g-.
229. Moment of Inertia of a Solid of Revolntion,
with respect to an Axis Perpendicnlar to its Geo-
metric Aada.— Take the origin at the iiitcrscctiou of the
rholo solid is
por limita.
right oircalar coiic
dias of the base;
b
curve 18 tf = T X,
irtia,
>^y
t a solid cylinder
ts height, and (2)
the external and
2) *»« (i» + b'^.
raboloid about its
I of the base.
TTfihb*
6
Ant.
1 of ReTOlntion,
Dlar to its Ooo-
iitcrsc'Ctiou of the
EXAMPLSS.
489
axis of revolution with the axis about which the moment
of inertia is required ; and denoting by x the distance of
the centre of any droular plate from the origin, y its
radius and dx its thickness, we have for the moment of
inertia of this circular plate, about a diameter, by Ex. 4,
Art 224,
therefore (Art 225) the moment of inertia of this plate
about the pandlel axis at the dutanoe x from it is
therefore the moment of inertia of the whole solid is
''t'f(l+!f^)^^> (1)
the integration being taken between proper limita.
EXAMPLES.
1. Find the moment of inertia of a right circular cone
about an axis through its vertex and perpendicular to its
own axis.
Let A = the height and b = the radius of the base, then
the moment of inertia from (1)
2. Find the moment of inertia of a cone, whose altitude
= h, and the radius of whose base = b, about an axis
through its ceuire of gravity and perpendicular to its own
axis. Ans. ^m{h* + 46»).
*fci«
i>
440
EXAMPLES.
3. Find the moment of inertia of a pamboloid of revolu-
tion about an axis through its vertex and periHjndicular to
its own axis, the altitude being A and the radius of tho
b
230. Moment of Inertia of Varioua Solid Bodies.
EXAMPLES.
1. Find tho moment of inertia of a rectangular parallel-
epiped about an axis through its centre of gravity and par-
allel to an eilge.
Let the edges be a, b, c; !»inee a parallelopiped may bo
conceived as consisting of an infinite number of rectangular
laminje, Oaeii of which has tiie same mdius of gyration
relative to an uxi« ixfrpendicilar to its plane, it foUowa
that the radius of gynition of the parallelopiped is tho
same as that of tho lamina!. Hence, the moments of
inertia roltUivo to three a.\e8 through the centre and par-
allel to the edges a, h, c. n^spectively, are by Ex. 1, Art.
237, ^m (ft» + c^), ^m («» -f r^), ^,tn (a* + 6»).
2. Find the moment of inertia of a rectangular parallel-
opiped about an edge.
This may be obtained immediately from the last oxam*
pie by using Art. 225, or otherwise iudefx-ndently afi
follows :
Take the thi-ee edges a, b, c for the axes of x, y, «,
ro8i)ectively ; let [t be the mass of a unit of volume, then
tho moment of inertia relative to the edge a is
= r I A (»»+«») rf« rfy ''«
•/Q •■ '0
'm-
iboloiil of revolu-
peri^ndicular to
the radius of tho
Sclid Bodies.
tanpular parallel-
gravity aud par-
lelopijwd may bo
xsr of retitangulur
idius of gyration
plauo, it follows
illelopiped is tlio
the moments of
contro and par-
e by Ex. 1, Art.
lingular parallel-
m the last oxam*
indofK-ndcntly as
axes of X, y, z,
of volume, then
a is
,'/ 'f^
c'U
MOTIOy OF IS'ERTIA OF A LAMIXA.
441
and similarly for tho moments of inertia about the edLn^s
/' and c. * '
The moment of inertia of a cube whose ed^jo is a with
respect to ono of its edges m \iia^ = ^i„u\
a. Find the u.jraent of inertia of a segment of a -phero
relative to a diameter parallel to tho pkue of flection ^ho
nidms of tho sphere being « and the distance of tho lilane
section from the ct-ntre b.
Am. i^TT (IGflS + \^b + 0)0^1^ ._ y^),
231. Moment of Inertia of a Lamina with respect
to any Aaris.— When the moment of inertia of a plane
fijruro about any axis is known, we easily find tho moment;
oL inertia about any {)arullel axis (Art. 225) ; ,il,wo. whoa
the moments of inertia about two rectangular axes in tin-
j.lano of the fignrc are known, the moment of inertia about
i li(} straight line at right angles to the plane tf these axes
it their intersection is known immotliately. (A^ Ui) ; wo
now proceed to find the moment of inertia about 'any
.straight lino- in the plane inclined to these axes at anv
JillglO.
Through any point, 0, as
■rigin, draw two rectangular
)>;es, OX, OY, in the ])lane of
Pig.w
!he lamina; and draw any
iraight line, OX, in the plane.
ii is re<piired to find tiie nio-
'n at of inertia about OX in
t. ima of iho uiomoutp of inertia about OX and OY.
\A^iP\^ any point of the lamina...r. v, its n-otimgular,
.Hid r, e, its polar co-ordinates./; =: I'M, and « fhe\u,glo
'OX. Then if f bo the moment of iuerllH of the laniiim
ivlativo to O.c, aand b Iho moments of inertia relative to
tlic axes of x and ?/ n^speetively, and h tho product of
mertitt relative to the same axes, we have
-■.i.-««ry.r>iH^i*^^.^-,-— -Hr^--..---..^:^^-^^^-.^^-.
^:;^A..:;^;i^Vft..-..-.t/ntYi"--->--'i-'-'i r-T\i.i "1
442
PRINCIPAL AXJSS OT' A BODY.
fil!
7 = X ;)«Jm = £ r» 8in« (0 - a) rfm
= 1 (y cos « — a; sin «)* rfm
= cos* a £ y*<i»t + sin* « £ a^ywi — 2 sin « cos a £ aryJw
= a cos* « -f ft sin' « — 2A sin « cos «. (1)
If wo choose the axes so that the tenn h or £ xydm — 0,
the expression for / becomes much simpler. The pair of
axes Bo chosen are called the principal axes at the point ;
und ^ho corresponding moments of inertia ar( called the
principal moments of inertia of the lamina, relative to the
point
If A and B represent these principal moments of inertia,
(1) becomes
7 = J co8» « + 5 8iii» «. (2)
Hence, tlie moment of inertia of n lamina with respect to
any axis through a point tnay be found when the principal
moments with respect to the point are determined.
232. Principal Axeb of a Body.— .1^ any point of a
rigid body and in any plane there is <i pair of principal
axes.
Lot OX, OY (Pig. 92), be any rcctiingular axes in tlii^
plane ; let Ox, Oy, Imj another set of rcctungulur axes in
tli« same plane, inclined to the former at an angle «; li't
a, ft, and h, as before, denote the moments ami product of
inertia .ibout OX, OY, and let {x', y') be any point, P,
referred to the axes Ox, Oy. Then, using the notation of
the last article, wo have
x' = r cos (0 — «) ; y' = r sin (9 — «) ;
£ x'y'dm = j^£ r» sin 2 {0 — «) dm
= cos 2« £ r* sin cos d dm
— I sin 2« £ r» (cos* — sin» 6) dm.
Putting this = 0, and solving for k. wo nbtain
toor.
gin a cos a £ xi/dm
(1)
h or X xydm = 0,
pier. The pair of
%xes at the poiut;
rtia ar( called Ibo
na, relative to the
aoments of inertia,
(2)
ina with respect to
when the principal
^rmined.
—At any point of a
t pair of principal
igular axes in tlie
cctaiigulur axes in
at an angle « ; Int
its ami product of
be any point, P,
iig the notation of
II (0 - ft) ;
- a) dm
' cos 9 dm
)b!i — sin" 0) dm.
ol'.taiu
wfm
mmmm.
THRBE PRINCIPAL AXXa.
22 r» sin e COS (? d:.n
tan 2« =
£ r» (cos« e — sin" 1^)
_ _^2Sa;y^H__ _ 2h
^(^- y') dm - i^T,
As the tangent of an angle may have any value, positive
or negative, from to co, it follows that (1) will always
give a real value for 2«, so there is always a set of princi-
pal axes ; that is, at every point in a body there exists one
pair of rectangular axes for which the quantity h or
:^ xy dm = 0.
CoK.— It may also be shown that at every point of a
rigid body there are three axes at right angles to ouo
another, for which the products of inertia vanish.*
♦ Let a, 6, «, bo the moments of Inertia about throo
«xo», ox, OT, OZ, at right aiigies to one Mother ; rf, «
/, the prodncts of Inertia (Smiw, Xmsr. \mxy, rc-
cixcilvely). L<tt Ox be any lino drawn ihronjfh tlio
.irlf-in, making ungles ,,, /?, ,, wl(|, the co onllualo
axon.
Let Ol,, IM, Mf, he the coordinate* a-, y, t, of any
point P i,r the body at wlilca an eicnuiit of uiaiw i» i»
Hlliiaiod. Draw I'N ixTpendlciiiar (o Ox.
l'rf)jo<ting tlie brolion lino, GLKP, on ON, (Art.
lOS), we have
ON = * cog >i ^ycotp\■^ ooa y
'Ih. 0P» = «• 4- »• + i •, and 1 =co»>- a + «»•/? + ooa* y.
The rnomert of inertU I atjont Ote = SmPN"
- ttn (OP» - 0K»)
= SW It' + V' * »' (* cos « f V com /f + • eo* y)«]
=• l>»»[(*'+|f' + »')(COB«afC0«»^KC0B« ^)-(irCOB« + jrCO«/J+fC<K|y).]
- £m (If* f #•) roB* a + Sm(f « + «•) con* fi i- tm (t« + y') coa' y
- ISmp* rati t) cm y — tSmmt: coa ^ co« a - ttm tmaiiot$
= • cofi" o + A co»« /)<-<• con" K - W coa /? coa )-
-9»co« > eosa - ycodocoii/?. (1)
To rapreaont lhi» Ruometrleally, take a polut Q on ON ; and let Ita dJatue*
from O be r, and Ifcs co-ordinate* be*,, y,, I,. Then
«, -recMa. jr, - rcuK^, «. ^rco*).
Fig. 814
■■T--^'it''-^iii'iii'wi».i
i'-'^
h'f:
444
TURSS PRtSCIPAL AXXS.
ScH. — In many cases the position of the principal
can bo seen at once. Suppose, for example, we wish
principal axis for a rectangle when the given point is the
centre. Dravr through the centre straight lines parallel to
the sides of the rectangle ; then these will be the principal
[Mil axes I
mh the I
f ia flw>
Tbaufor* (1) becomef
I -
<w,' t ay,' 4 <».' -1W».«, - at>m, -ytj.y.
But the equation
dpnoleg a^vflT^iXfJit' srivjgit tavbf- Is sU O; becanw a, ti, e are nccegsBrily posUlvo,
eiuee » mr- I at lovrtift U MWfitlally positivu, being the sain u( • bobiIwi' of
Miuarr 1 ({ U tt point on tbU uUlpcoid, (3) Ut'oomee
I c SmPn* <c
!i
or thn moment of in«rtift »bont tny line thruugh O, h mcMnred by the sqaut; of
tho roctiiruckl of thfl radlun Tector of tUi» elll[>M>id, which coloeidos wllh the
liuo.
'{Ilia li ckUed Uiit niommtat cS^/mld, and ww first aaed by Oauchy, Mrtrcimt d»
Hath., Vol, n. It tuti no phyxical oxlcteDoe, hat In an artiRoi to bring under Ute
rauthodH of (jeoinctry tho pro|)ert5en of inomcTiln of tiioKla. The mouienlal elllp.
gdid han a drtlulte form tor cvory polBt of a rigid body.
Now every elli|if>old hoa thvoe axoc^, to which If il \* ivlbrred, the coefficients of
y», cr, xy vanlidi, aud therefore (!)), when tnuiBfbriued to those asus takes the
form . _
A/,' ( By," \ 0,« = 1; («)
Biul hence (1) or (») wh'in raAnred to these axcB, becoroee
1 « A eoe* • + B to** ^ + COB" »-,
m
whoro A, B, C, are Uie niomentti of inertia of the body about then! a: m.
When three recxangalar axe;!, meeting In a t^veu point, arc chuRon w) that ihe
prodncM of Inertia all v«Hl»h, tlie> are c*lle<l the f)riiteip3l epsu at the given
piiliil.
Tht) Ihnw^ planen thtoogh any two priticlpal axM iir:> eallw' itio prindija/ }>!cme^
at lltt' ^Ivon point.
The monionts of inertia ahont lh« piincjial nxwi at o«iy point are railed the prin-
fi;in/ tncnumtJi tf i»*rHa at thai p«ilnt.
If the (hreo prlnii|)al moinenti> of Inertia ofa body are Mjniil to one another, the
oUl|iHoi'l '4) I«comeii a Hp'jere, ulni^c A =. B - V ; 3nJ therofotc the inumeut of
iMrlla about every other atlt it oqiuU to L-nae, tor (R) becoiuua
1 ■-'< A <oot' • ♦ cm* <$ 4^ ow< i) o" A ;
and every <uti» In a pru»lp«l Ml«. (See IbiUth'g UI>rii] Dynamic*, p. I«, Price's
AuftV S.cliv Vui, a. 1. IW, I-irWrJUglilDyuaiulcc, p TO, iie -
#
principal oxoa
1, we wish tlio
a point is tliu
11C8 parallel to
I tbo priuci(Ml
*.
= I.
m
(8)
9i'0a««riiy postllro,
un uf t DDiDlter of
td by the aqnsro of
coloeldM with the
^auchy, Mtereitet d»
I to bring under tlie
lie mouienlal I'lliji-
Ihc cocfllcleiit« of
boDe iLYuii takuD tlio
(*)
(»)
iP.ni a: :m.
churan no that the
aiM at the givvu
ilio prlneiiMtl planti
tarct-«nei1 IhopHn-
lo on« another, tbo
'ore (ho inomeut of
laniic*, It. 1«. IMco's
TBRSE PBTNCIPAI, AXES.
««8
axes; becaase for orery element, dm, on one side of the
axis of X at the point («, y), there is another elenient of
equal mass on the other side at the point {x, —y). Hence,
yLxydm consists of terms which may bo arranged in pairs,
so that the two terms in a pair arc numerically equal but
of opposite signs ; and therefore £ xy din — 0.
Agiiin, if in any uniform body a straight lino can Imj
drawn with respect to which the body is exactly symmetri-
cal, this must be a principal axis at every point in its
length. Any diameter of a uniform circle or sphere or the
axis of a parabola or ellipse or hyperbola is a principal axis
at any point in its line; but the diagonal of a rectangular
])lat« is not for this reason a principal axis at its middio
point, for every straight line drawn perpendicular to it is
not equally divided by it.
Let the body be symmetrical about the plane of xy, then
for every element dm^ on one side of the plane a*: the point
{x, y, «), tliero is another elem nt of equal mjusa on the
other side at the point {x, y, — *). Hence, for such a body
2 xz dm = and £ yz dm ■=. 0. If the body bo a lamina in
the plane of xy, then z of every element is zero, and wo
have again £ xz dm = 0, !• yt dm = 0.
Thufi, in tlic case of the ollipsoid, the three principal
sections are all planet of symmetry, and therefore the three
axes of (he ellipsoid are principal axes. Also, at every
jKiint in a lamina one principal axis ia the pcrpendioular to
tlie pltme of the lamina,
EXAMPLES.
1. Find the moment of inertia of a rectangular lamina
:il)uut a diagonal.
From Ex. 2, Art. 5S24, the moments of inertia about two
linos through the centre parallel to the sides (principal
moments of inertia) ore
fm
446
BXAMPLSa.
• »
I
^m<2* and VfWii*;
■whoro h ami rf aro the breadth and dopth respectively.
Also, if « be the angle which the diagonal raakee with
the tiide i, we have
«n» a =
I>> + (P'
00^ rt =
d2 + <^
SnVwtituting these values for A, B, sin* «, cos» «, in (2) of
Art. 231, wo have
= ^m
2. Find the moment of inertia of an isosceles trfangnlar
plat« about an axis through its centre and inclined at an
angle « to its axis of syrametry, a being its altitude and 2ft
its baae. Ans. ^ (^a* cos* « H- J» sin* «).
3, Find the moment of inertia of a square plate about a
diagonal, « being a sidi! of the square. Jns. ^maK
233. ProductB of Inertia.— The value of the product
of inertia at any iwint may be made to depend on the value
of the product of inertia for parallel axes through the cen-
tre of gravity. Let {x, y) lie the position of any element^
dm, referred to axes through any a88ign<'d point ; {x', y') the
jwsition of the element referred to parallel ixes through
the centre of gravity, and (A, k) the centre of gravity
referr(Hl to the firat pair of axea Then
ar = «' + A, y =. y' + h)
tlM»fore "£ xydm^I. {x' + h) {y' + h) dm
= Y,r!y' dm + hk^dm, (1)
since S wwf' = 0, and S my'
0.
J,#Sv
SXAMPLSA
U1
KJCfcively.
U makes with
a« «, in (2) of
elea triangular
inclined at an
Ititude and 2b
H- J» 8in» «).
plate about a
Ans. '^ftnaK
of the prodnct
id on the value
rough the cen-
f any element,
iut ; {x', tj') the
il ixes through
itre of gravity
Im
(1)
ScH.— By (1) we may often find the product of inartia
for an assigned origin and axes. Thus, suppose we roqnire
the product of inertia in the case of a rectangle, when the
origin is at the comer, and the axes are the edges which
meet at that ooruer. By Art 332, Sch. we have I.z'y'din
= 0; therefore from (1) we have
and as h and k are known, being half the longthn of the
edges of the rectangle to which they are rcsiiectivcly
parallel, the product of inertia is known.
BXAM PL.BS.
Find the expressions for the moments of in«tia in the
following, tiie hwiies beiag wipposed homogeoeous in all
cases.
1. The moment of inwtia of a rod of length a, with
respect to an axis perpendicular to tite rod and at a distance
d from ita middle point , /a" \
* Ans. m{^~ + dn-
2. Tlie moment of inertia of an arc of a circle whose
radius is a and which subtends an angle 2« at the centre, (1)
about an axis through its centro perpendicular to its plane,
(2) about an avis through its middle point perpendicular to
its plane, (3) about the diameter which bisects the arc.
Ans. (l)«,a- (2) 2„. (!_?!£_«)«.; (8) «, (l -.!l|^ «*.
3. The moment of inertia of the arc of a complete
cycloid whose length is a with respect to its base.
Ans. ^ffna',
4. The moment o( inertia of an equilateral triai.gle, of
aide a, relative to a line in its plane, parallel to a side, at
the disUutoo d twiu its oeuke of gravity. . '
Am. m (I + <p).
riM
di«..^».i■>..W.t.■l.,.L.-^:.ljiA^t>fc•^.^^|J.^,■^.,,, .y:...r..^^-....^-^- .
418
SXAMPLSS.
h
6. Oivcn a triangle whoBc- sides are a, b, e, and whoso
jwrpendiculars on these sides, from the opposite vortices
are p, q, r, respectively ; find the moment uf inertia of the
triangle about a line drawn through each vertex and
parallel respectively, (1) to the side a, (2) to the side b, {?,)
to the side e. Arts. (1) Impf^; (3) |tw7« ; (3) ^m,^,
C. Find the moiTient of inertia of the h'iiinglo in the last
example relative to the three lines drawn through the
centre of gravity of the triangle and parallil respectively
to the sides a, b, c. Ann. ^mf\ if,m(j*\ i-xiniK
T. Find the moment of inertia of the triangle iu Ex. 5,
relative to the three sides a, h. c, respectively.
Am. {nlf^'f ifUKf; \mr'K
8. The monioiit of inertia of a right angled triangle^ oi
hvputlienose c, relative to a perpendicular to its plane
J)a8>itig through the right augle. Av.^. \m^,
!t. Tiic moment of inertia of n ring vvhcisc outer atttf
inner radii aie a and b respectn ely, (1) with respect lo a
IM)lar axis tiiroiigh its centre, and (2) with respect to a
diameter. Am. (1) ^m («» -|- h^) ; {%) \m («» + S^).
10. The mof'ienf of inertia of an ellipse, (1) with respect
to ')\f> miijor axis, (i) with respect to its mhior axis, and (3)
with reaiiet to an tt>«iH through its centre and pc-riiendieular
U) it« plane.
4n». (\) ^110 \ |l) Imn'i; (.1) \m («» + A«).
11. The mmmui uf inertia of Hie surface of a rfphero of
radius a ahout it.s diameter. Aiis. ImaK
li. The moment of inertia of a ri(i)i( pri«m whose haae
is !i rigiit anglal triangle, witli l'OB[wot to au axis passing
{limutrh thr eentres of gravity of the eud% tUo tldoa cou-
laining the right angle of the triangular base being a and b
and the height of the prism c. Ans. ^m{a^+¥).
ifeite
MMMMMM
', c, and whoso
jposite vortices
L inertia of tho
3h vertex and
tlie side h, (3)
r« ; (3) \m,^.
iglo in the last
1 through the
It'l respectively
angle in Ex. 5,
r.
\ /«'/' ; f/Hr*.
led ihimgh, of
' to its phe
u.-ir outer s»j«t
h respect to a
I respect to a
[m (rt» + HI).
I) witli rosjwct
»r axis, and (3)
ptriiendicuUir
ni {(ii + *«).
I if (I riphore of
Alts. |i//«^,
(III whoge baao
lu avis pasHtng
tUo tidofl oou-
being a and b
XXAMPLJB&
M»
13. The m(»D(«t of ioertia of a right pmnt vhoae height
is c, about an axis passing throngh the centres of gravity of
the ends, tho base of the piistQ being an isosceles triangle
whose base is a and height b. /a> Nt\
■^-id + r)-
14. The moment of inertia of » sphere < f. radins a, (1)
relative to a diameter, and (3) relative to a tangent.
Ans. (l)fwo»; (3) |»ja».
15. Tho moment of inertia, about its axis of rotation, (1)
of a prolate spheroid, and (3) of an oblate spheroid.
Ans. (l)|«ift»; (3)|7»fl2.
16. Tho moment of inertia of a cylinder, relative to an
axis perpendicular to its own axis and intersecting it, (1) at
a distance c from its end, (3) at the end of the axis, and (3)
■.hi f ho middle point of the axis, the altitude of the cylinder
boujg fi and radius of its base a.
Ans. (1) ima» + ^m (A» + 8Ac + c») ;
(3) ^m (3«» + 4/t«) ; (3) ^m (7*3 + 3«»)
moment of inortirt of an ollipso about a central
radius vector r, making an angle « with the major-axlB.
Am. im-^.
18. The moment of inertia of the area of a parabola cut
off by any ordinate at a distance a, from the vertex, (1)
aliout the tangent at tho vertex, and (3) about the axis of
tho iMuiibola.
Ans. fwa? ; (3) ^my^ where y is the ordinate correspond-
ing to «.
l». The moment of inertia of the area of the lomniscatey
»■» =z «' COS m, about a lini* through the origin in its plane
uiid periRiudicular to it^ a^is. Am. ^ftn (3rr f 8) «».
wg,?i i i wwgig flFgy
*■
■■immii fin fiifrfrrtiii
460 MXAMP/.ES.
20, The moment of inertia of the ellipfsoid,
^ + ^ + ^-^-
about the axis a, ft, c, respectively.
Am. (1) im(*» + c») ; (2) i«i (c» + a») ;
(3) iw (a» + J»).
j^ s
1,
J«i (c» + a») ;
''s^rtliBSUBSSH
CHAPTER VII.
ROTATORY MOTION.
234. Impressed and Infective Forces.— All wes
acting on a body other than tho mutual actionft , tho
particles, are called the Impressed Forces that act on the
body.
ThuB, when a ball is thrown in vacuo, the impresst*!
force is gravity; if a ball is rotating about a vertical axis,
the impressed forces are gravity and the reaction of tho
axis.
The impnssod or external forces are the cause of the motion and of
all tlio other forces. Which are the impraosed forces depends npon
the particular system which is under consideration. Tho same force
may be external to one system and inn. rnal to another. Thus, tho
pressure between the foot of a man and tlie deck of a ship on which
he ifl, is external to tho ship and also to the man and is thfl cause of
his own forward motion and of a slight backward motion o, Jio ship ;
but if the man and ship are considered as parts of one system the
pressure is internal.
When a particle is moving as part of a rigid body, it is
acted on by the external impressed forces and also by the
molecular reactions of the other particles. Now if ".his
particle were considered as separated from the rest of the
body, and all the forces removed, there is some one force
which, singly, would move it in the same way as before.
This force is called the Effective Force on the particle; it is
evidently the rtsultant of the impressed and molecular
forces on the particle.
Thus, the effective force is that part of the impressed force which
is effective in causing actual motion. It is tho force which is required'
for prodncing tUo deviation from the straight line and the change of
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452
D 'A LEMnKR T 'S PRINCIPLE.
vplocitv. If a particle is revolving with constant velocity round a
fixed axis, the effective force is the ceatripetal forc« (Art. 198). If a
heavy Ixxly fulls wiihoiit rotation, the whole force of gravity is
effective ; but if it \» rotating about a liorizoutal axis the weight goeti
partly to kiJauce the prcHsure on the axis.
If we suppose tlie particle of mass m to bo at the point
(r, y. z) at niiy time, /, flnd resolve the forces acting on it
into the tiireo axial com])onent8, X, Y, Z, the motion may
be found [Art. 108 (2)] by solving the Bimultj.neou8 equa-
tions
(Pi/
= r
m ^, := Z.
a)
■^« ite-
If we regard a rigid body as one in which the particles
retain inviiriable po-sitions with respect lO one another, so
that no external force can alter thon (Art 43), we might
write down the equations of the several particles in accord-
ance with (1), ii" all the forces wore known. Such, how-
ever, is not the ease. We know nothing of ihe mutual
actions of the particles, and ccn80<pjently cannot determine
the motion of tlio bodv by calculating the motion of its
particles separately. ^\ hen there are several rigid bodies
which mtitually act and react on one another the problem
becomes still more comidicated.
235. D'Alombert'B Principle, •—By D'Alembert's
Principle, however, all the necessary equations may be
ol)tained without writing down the equations of motion of
llio several iwirHoles. and witnout any assumption as to the
nalureof the mutual actions except the following, which
may l»e i-egardxHi as a natural oonsoqueitco of the laws of
motion.
TfiK infernnl nrtiorm and rcaciions of any ayntnn of rigid
bodies in motion arfi in (fiuiUbrinin nmonff fhemsehvu.
* latrodaccd »>y y'Alenibcrt lii n«.
D 'ALEMBERT'S PRI^CtPLB.
458
velocity round a
(Art. 198). If a
CO of gravity is
the weigbt goes
I at the point
J acting on it
e motion may
ituneous cqvia-
\ the particlcd
)ne another, so
43), we might
cles in accord-
I. Such, hew-
)f the mutual
inot determine
! motion of its
111 rigid bodies
or tlie problem
D'Alombert's
fttions may be
18 of motion of
ption a8 to the
•llowing. which
of the laws of
system of rigid
■temselws.
\
The axial accelerations of the particle of mass m, whicli
cPx dt^v (Pz
is moving as part of a rigid body, ai-e ^, -^i^ -^^'
Let/ be their resultant, then the eflfeotive force is measured
by mf Let ^and R be the resultants of the impressed and
molecular forces, i-espectively, on the particle. Then mf
is the resultant of F and R. Hence if mf bo reversed, tlio
three foi-ces, F, R, and mf, are in equihbrium.
The same reasoning may be applied to every particle of
eiich body of the system, thus furnishing three groups of
forces, similar, respectively, to F, E, and mf: and these
three groups will form a systeir of forces in equilibrium.
Now by D*Alembert's principle the group 7^ will itself
form a system of forces in equilibrium. Whence it follows
that the group F will be in equilibrium with the group mf.
Hence,
If forces equal and exactly opposite to the effective ^forces
were applied at each parfick of the system, they would l>e
in equilibrium with the impressed forces.
That is, H'A'emhcrCs principle asserts that the lohoU
effective forces of a system are together eqnivaknc to the
impressed forces.
SoH. — By this principle the solution of a problem in
Kinetics is reduced to a problem in Statics ha follows : We
first choose ihe co-ordinates by means of which the position
of the system in space may be fixwi. Wo then express the
effecii^o forces on each element in terms of its cc-oi'dinates.
Those effective forces, reversed, will be in equilibrium
with the given impressed forces. Lastly, the equations of
motion for each body may b^ formed, as is usually done in
Statics, by resolving in three directions and taking mo-
ments abont- throe straight linos. (See llouth's Rigid
Dynamics, Pirio's Rigid DynamicB, Pratt's Mech's, Price's
Anal. Mech's, Vol. IL)
^.^•i^
^HMMWKMMMWMM
464
ROTATION OF A liWID BODY.
236. Rotation of a Rigid Body about a Fixed
ii^B under the Action of any Forces.— Let any
plaue pagRing through the axis of rotation und fixed in
space be taken as a piano of reference. Let m be the mass
of any element of the body, r its distance from the axis,
and the angle which a pkuo through the axis and the
element makes with the plane of rekronce.
Then the velocity of m in a direction yrpendicular to
the plane containing the element and the axis is r-^-.
The moment of the momenluni* of this particle about
the axis ia mr' --,-• Hence the moment of the momenta of
at
all the particles is
d6
at
0)
Since the particles of the body are rigidly connected,
it is clear that -=j is the same for every particle, and is the
angular velocity of the body. Hence the moment of the
momenta of all the particles of the body about the axis is the
moment of in^tia of the body about the axis multiplied by
the angular velocity.
The acceleration of m perpendicular to the direction in
whicii r is measured is r -^ , and therefore the moment of
the moving forces of m about the axis is 'wr*-,^- Hence,
the moment of fhe effective forces of all the particles of the
body about the axis is
Iwr»
(m
(2)
which is the moment of inertia of the body about tJie axis
mkltiplied by the angular acceleration.
* Called alHi) Angular Momentum. (Soo Itrlo'ii Rlcld Dy niunlce, p. 44.) .
>ut a Fixed
ses.— Let any
1 and fixed in
m be the mass
from the axis,
axis and the
^rpendicular to
. . do
LC axis \a ''j-.'
particle about
;he momenta of
(1)
idly connected,
icle, and is the
moment of the
t the axis is the
is multiplied by
the direction in
the moment of
fir^-jp- Hence,
1 particles of the
dy about the axis
DyniunlcB, p. 44.)
ROTATION OF A RIGID BODY.
455
(1) Let the forces be impulsive (Art. 203) ; let 6), w', be
the angular velocities just before and just after the acaon
of the forces, and N the moment of the impressed forces
about the axis of rotation, by which the motion is pro-
duced.
Then, since by D'Alembert's principle the effective
forces when reversed are m equilibrium with the impressed
forces, we have from (1)
w' S mi* — w X rnr* = JV";
«
(J :=
N
"Lmt*
' moment of imp ulse abo ut axis ,
~ momo^IofTncrtia about axis '
(8)
that is, the change in the ai^^^^r velocity of a body, pro-
duced byan impulse, is equal to the m^^ient of the impulse
divided by the moment of inertia of the body.
(2) Let the forces be finite. Then taking tiicr.'^^nts
about the axis as before, we have from (2)
(Pd
dt^
JL
^mi^
moment of fo r ces about axis .
~ moment of inertia about axis '
(4)
that is, tU angular acceleration of a body, produced by a
force, is equal to the mommt of the force divided by the
moment of inertia of the body.
By integrating (4) we shall know the angle through
which the body has revolved in a given time. Two arbi-
trary constants will appear in the integrations, whose
values are to be determined from the given initial value?
of and ~- Thus the whole motion can bo L\mA, and
456
MXAMPLE.
WO shall conseqaontly be able to determine the position of
the body at any instant
ScH. — It appears from (3) and (4) that the motion
of a rigid body round a fixed axis, under the action of any
forces, depends on (1) the moment of the forces about
that axis, and (2) the moment of inertia of the body about
the axis. If the whole mass of the body were concentrated
into its centre of gyration (Art. 226), and attached to the
fixed axis of rotation by a rod without mafls, whose length
is the radius of gyration, and if this system were acted on
by forcoe having the same moment as before, and were set
in motion with the same initial values of 6 and the angular
velocity, then the whole subsequent angular motion of the
rod would be the same as that of the body. Hence, we may
say br'ofly, that a body turning about a fixed axis is
Irinetically given when its maaa ana radius of gyration are
known.
EX ADA? L F.
A rough circular horizontal boi. i is capable o£ vixoh']ng
freely round a vertical axis through its centre. A man
walks on and round at the edge of the board; when he
has completed the circuit what will be his position in
space ?
Let a be the radius of the board, M and M' the maeses
of the board and man respectively ; B and 0' the angles
described by the board and man, and i^^the action between
the feet of the man and the board.
The equation of motion of the board by (4) is
Fa - ^*.»^'
( Since the action between the man and the board is con-
tinually tangent to the path described by the man, the
equation of motion of the man is, by (5) of Art. 20,
mmmm*
the position of
tat the motion
le action of any
be forces about
the body about
ere concentrated
attached to the
!8, whose length
oa weie acted on
re, and were set
and the angular
ar motion of the
Hence, we may
a fixed axis is
I of gyration are
able o£ Vixohing
centre. A man
board; when he
I his position in
id M' the masses
and 0' the angles
e action between
(4)i8
the board is con-
by the man, the
,f Art. «0,
"'^wnmmP'-
m!m»!»s^mmmmm>efmBii>i^>m^f>f»»!"
TBI! COMPOUND VENDULUM.
IP V ^
m
Eliminating i^and integrating twice, the constant being
zero in both cases, because the man and board start from
rest, wo get
Mk?0 - M'aW. (1)
When the man has completed the circuit we have + 0'
= %iT\ also ^'i* = -. Substituting these in (1) we get
V
* —
2nM
-iM' + M'
which giTOs the angle in space described by the man.
If M — M', this becomes
and
d' = \ir;
= f7r,
which is the angle in space described by the board. (See
Rye^b's Rigid Dynamics, p. G7.)
237. The Compound Pcc<3nlum.— ^1 body moves aiont
n fixed horizontal axis acted on by gravity (?vli/, to determine
the motion.
Lot ABO be a section of the body made by
the iilane of the paper passing through O,
tlie centre of gravity, and cutting the uxis
of rotation iierpondioularly at 0. Lot —
the angle which 00 makes with the vertical
OY; and let h = 00, t, = the principiil
radius of gyration, and M = the mass of
the body. Then hy (4) of Art. 336, we
have
20
•iMfl-TT-- TlWiai
''''"•""' I'i ii "r"'n'r i - rn' ' ■imrr
468
TH£ COirPOUND PENDULVM.
"~f
Mgh ain d
_ Mgh Bin 6
= ~ ^;nnra sin e [by (2) of Art. 226], (1)
the negative sign being taken because 9 is a decreasing
function of the time.
This etjuation cannot be integrated in finite terms, but
if the oscillations be small, we may develop sin f> and reject
all powers above the first, and (1) will become
(Id
gh
*,» + A«
e.
(2)
Multiplying by 2 -^ and integrating, and supposing that
the body began to move when 6 was equal to «, (2)
becomes
rffl* _ gh
dP
M(«»-n
*,^ + A«
Hence denoting the time of a complete oscillation by T,
we have
/h" + k{'
(8)
which gives the time in seconds, when h and Jfe, are meas-
ured in feet and g = 32.18.
When a heavy body vibrates about a horizontal axis, by
the force of gravity, it is called a compound pendulum.
CoE. 1. — If we suppose the whole mass of the compound
pendulum to be concentrated into a single point, and this
point connected with the axis by a medium without weight,
it becomes a simple pendulum (Art. 194). Denoting the
distance of the point of concentration from the axis by I,
wc luvvc for the time of an oscillation, by (1) of Art. 194,
afl
of Art. 326], (1)
is a decreasing
inite terms, but
sin and reject
me
(«)
i supposing that
equal to «, (2)
oscillation by T,
(3)
and yfci are meas-
orizontal axis, by
d pendulum.
1 of the compound
;le point, and this
m without weight,
t). Denoting the
•om the axis by I,
y (1) of Art. 194,
■ ' '^fSISWI^SWiSSW'
CBXTSES OF OSCILLATION AND SUSPENSION. 459
"V^-
If the point be so chosen that the simple pendu-
lum will perform an oscillation in the same time as the
comijound pendulum, these two expressions for the time of
an oscillation must be equal to each other, and wo shuU
liave
1 =
h
A + ^ = 00',
(4)
(0' being the point of concentration).
Cob. 2. — This length is called the length of the simple
equivalent pendulum ; the point is called the centre of
suspension ; the point 0', into which the mass of the com-
pound pendulum must be concentrated so that it will
oscillate in the same time as before, is called the centre of
oscillation; and a line through the centre of oscillation
and ps-i-allel to the axis of suspension is called an axis of
oscillation.
From (4) we have
{l-7i)7i = *,»;
or
GO'. GO = *,».
(5)
Now (5) would not be altered if the place of and 0'
were interchanged;* hence if O' be made the centre of
suspension, then will be the centre of oscillation. Thus
tlie centres of oscillation and of suspension are convertible,
and tfie time of oscillation about each is the same.
CoR. 3. — Putting the derivative of I with respect to h in
(4) equal to zero, and solving for h, wo got
h — k^,
460
EXAMPLES.
wliich makes I a minimum, and therefore makes / a mini-
mum. Hence, when the axis of suspension jjasses thmuijh
the principal centre of gyration the time of oseillaiion is a
minimum.
Rem. — The problem of dctnrmining the law under which a heavy
body swings about a liorizontftl nxia is one of tlie most iniiwrtnnt in
tlic lilgtory of science. A simple pendulum is a thing of theory ; our
accurate Imowlodge of the acceleration of gravity depends tberoforo
on our understanding the rigid or compound pendulum. This was
th<> first problem to which IVAlembert applied his principle.
The problem was called in the days of D'Alembert, the " centre of
oscillation." It was required to find if there were a point at which
the whole mass of the body might be concentrated, so as to form u
simple pendulum whose law of ot^cillation was the same.
The jxiRitlon of the centre of oscillation of a body was first correctly
determined by Huyghens and published at Wans in 1678. As
D'Alembert's principle was not known at that time, Huyghens had to
discover some principle for himself.*
EXAMPLES.
1. A material straight line oscillates about an axis jier-
pendicnlar to its length ; find the length of the equivalent
simple pendulum.
Let 'ia = the length of the line, and h the distance of its
centre of gravity from the point of suspension. Then since
ij« = ^-, we have from (4)
l = h +
U
(1)
Cor. 1. — If the point of suspension be at the extremity
of the line (1) becomes
* Rontb's Rigid DTDiimlcB, ]>, 69,
EXAMPLES.
461
makes t a mini-
fy 2Jasses through
f oscillation is a
ndor which a heavy
! most iniiKirtnnt in
liing of tlieory ; our
y depends therefore
oduluiu. This was
I principle,
ibert, the " centre of
e a point at which
ted, BO as to form a
9 same.
dy was first correctly
tans in 1678. As
le, Huyghens had to
xbout an axis -pev-
of the equivalent
the distance of ite
ision. Then sine©
(1)
le at the extremity
that is, the length of the equivalent simple pendulum is
two-thirds of the length of the rod.
Cob. 2.— Let /t — |a ; then (1) becomes
I = !«..
Hence, the time of an oscillation is the same, whether tho
line bo suspended from one extremity, or from a point one-
third of its length from the extremity. This also iilustnites
the convertibility of the centres of oEoillation and of sus-
pension (See Cor. 2).
Cob. 3.— If h = 10a, then (1) becomes
1 =
2. A circular arc oscillates about an axis through its
middle point i^erpendicular to the plane of the arc. Prove
that the length of the simple equivalent pendulum is
independent of the length of the arc, and is equal to twice
the radius.
From Ex. 2, Art. 333, we have
P
^ /^ sin <*\ o
From Ex. 1, Art. 79, we have
h — a — a
sin a
Therefore (4) becomes
« .. /^ sin «\ /i sin «\ „
I = 20" ^1 ^— J -7-a\i ~j = 2rt.
,finn
Iff
LENOTII OF TBTS SECONffS PEXDVLCM.
0jl
m'
Sll
3. A right cono oscillates about an axis poasiug through
its vertex and perpendicular to its own axis ; it is required
to find the length of the simple equivalent pendulum, (1)
when h is the altitude of the cone and b tlie radius of the
baflo, and (;j) when the altitude = the radius of the base = h.
Ans. (1) —gy- ; (3) h.
That is, in t!ie second cono, the centre of oscillation is in
the centre of the bawc ; so that the times of oscillation are
equal for axes through the vertex and the centre of the
base perpendicular to the axis of the cone.
4. A sphere, radius a. oscillates about an axis ; find the
length of the simple eciuivalent pendulum, (1) when the
axis is tangent to the sphere, (3) when it is distant 10«
from the centre of the sphere, and (3) when it is distant
- fi'om the centre of the sphere. ,
Ans. {I) la; (2) W«; O^) V«.
238. The Length of the Second's Penduluir.
Determined Experimentally. — The time of oscillulion
k *
of a compound pendulum depends on h + -,- by (4) of
Art. 337. But there are difficulties in the way of determin-
ing /» and ky The centre, G, can not be got at, and, as
every body is more or less irregular and v&riablo in density,
h^ cannot bo calculated with sufficient accuracy. These
(piantities must therefore be determined from experiments.
Bessel observed the times of oscillation about different
axes, the distances between which wore very accurately
known. Captain Kator employed the property of the
convertibility of the centres of suspension and oscillation
(Art. 237, Cor. 2), as follows :
' Let the pendnlum consiHt of an ordinary stralglit bar, CO, and a
small weight, m, wliich may be clamped to it l>y means of a screw,
uud shifted fruiu ouu positiuu to auuiher cm the peuduluiu. At tho
(2) h.
DVLOM.
paasiug through
8 ; it is required
it pendulum, (1)
Llie radius ai the
8 of the base = h.
f oscillation ia in
of oscillation are
he centre of the
m axis ; find the
m, (1) when the
it is distant lOrt
vhen it is distant
W«;
(:i)
id's Penduluir.
nie of oscillation
I + ^^ by (4) of
way of detcrmin-
be got at, and, as
iriablo in density,
iiccnraoy. These
Prom experiments.
I about different
9 very accurately
property of the
m and oscillation
aigbt bar, CO, and a
ly means of a screw,
B penduluiu. At llio
"ipwn' ."r^TO tKV^^r^pKV
LENGTH OF THE SECOND'S PENDULtm.
403
[Tin*
o I
>ifl
l1]»»
Fls.94
points C and O in tvjo triangular aper-
tures, at the distance I apart, let two knife
edges of hard steel bo placed parallel to
each other, and at right angles to the
pendulum, so that it mny vibrate on either
of them, OS in Fig. 04. Let m be shifted
till it is found that the times of oscillatiuu
about and O are exactly the same. It
remains only to measure CO, and observe
the time of oscHlation. The distance be-
tween the two points C and O is the length
of the simple equivalent pendulum. This distance between the knife
edges was measured by Captain Kater with the greatest care. The
mean of three measurements differed by less than a ten-thoustindth
of an inch from each of the separate measurements.
The time of a single vibration cannot be observed directly, because
this would require the fraction of a second of time as shown by tho
clock, to be estimated either by the eye or ear. The difflouity may
be overcome by observitg the time, say of a thousand vibrations, and
thus the error of the time of a single vibration is divided by a
thousand. The labor of so much counting may however bo avoided
by the use of " the method of coincidences." The pendulum is placed
iu f.-ont of a clock pcudulum whob- time of vibration ia slightly
different. Certain marks made on the two pendulums are observud
by a telescope at the lowest point of their arcs of vibratior The field
of view is limited by a diaphragm to a narrow aperture "cross which
the marks are seen to pass. At each succeeding vibration one
pendulum follows the other more closely, and at Inst its mark is
completely covered by the other during their passage acrosr the field
of view of the telescope. After a few vibrations it appears again
preceding the other. In the interval from one disappearance to the
next, ono pendulum has made, as nearly as possible, one complete
oscillation more than the other. In this manner 530 half -vibrations of
a clock pendulum, each equal to a second, were found to correspond to
533 of Captain Eater's pendulum. The ratio of the times of vibra-
tion of the pendulum and the clock pendulum may thus be calculated
with extreme accuracy. The rate of going of the clock must then bo
found by astronomical means.
The time of vibration thus found will require several corrections
which are called "reductions." For instance, if the oscillation be
not so small that we can put sind — in Art. 237, wo must make a
reduction to infinitely small arcs. Another reduction is necessary if
fill
m
r<,l
M'
j%t*«3»iri<os»iii»>oi«maa4K»S;»iK<»»a^
JaSJ)^
4G4 MOTION OF A BODY WHEN UNCO NSTBAI NED.
IIP
we wish to reduce the rpsult to what It would liAve been at the level
of the sea. Tha altfaction uf the interveDing land Toay be allowed
for by Dr. Young's rule, (Phil. Trans., 1819). We laay thuu obtain
the force of gravity at the level of the sea, annpoeing all the laud
above this level were cut ott' and the sea cuaBtraincd to keep its
present level. As the level of the sea is altered by the attraction of
the land, further correcticas au) still necessary if we wish to reduce
the result to the snrfaco of that spheroid which most nearly repre-
sents the earth. See Routh's lligid Uynamios, p. 77. For the details
of this experiment the Htudent \a referred to the Plill. Trans, for 1818,
and to Vol. X.
< 239. Modon of a Body when Unconstrained.— If
an impulf'o be commuuicated to any point of a freo body
ill a direction not pfwsing through the centre of gravity, it
will ])roditce both translation and rotation.
Let P be the impulse imparted to
the body at A. At B, ou the opposite
eide of the centre G, a distance GB ^
r= AG, let two opposite impulses be > j
applied, each equal to \P ; they will
not alter the effect. Now if \P
applied at A is combined with the ^7'
at B which acts in the same diraotion, their resultant is P,
acting at G and in the «.ame direction, and this produces
translation only. The remaining ^P at A combined with
the remaining \P at B, which acts in the opposite direc-
tion, form a couple wuich produces rotation about the
centre G.
Hence, when a body receives an impulse in a 'lirection
which doer not pnifS through the centre of gravity, thxt centre
ml' assume n motton of trnndalion as ihongh the ImpuUe
were applied immediately to it ; and the body will have a
motion of rotation about the centre of gravity, as though
Uiat jwint were fixed.
240. Centre of Fercusaion. -Axis of Spontaneous
Rotation. — Lot Jli> rcprcsuut the impulse imprensod upon
IP
FI0.9S
BAILED.
3een at the level
I tnay be allowed
!uay thiu obtain
sing all the laud
aiued to keep Uh
- the attraction of
we wisb to reduce
noBt nearly repre-
, For the dutoila
I. Trana. for 1818,
aBtrained.— II
of a free body
re of gravity, it
rihjmmmm'mm^ ■
'fm
If
Ftg.9S
rc&ultant is P,
id this produces
L combined with
I opposite direfi-
xion about the
J tu a 'lirecHon
ivity, th tt centre
ugh the mptihe
'jody will have a
vily, as though
f BpontaneouB
imprt'ttsiod ui>ou
CENTRE OP PERCUSSION.
4G5
the body (Fig. 95) whose mass is M, and
h the perpendicular distance, 00, from
the centre of gravity, G, to the line of
action, OP, of the impnlse. The centre
of gravity will assume a motion of trans-
lation with the velocity v, in a direction
parallel to that of the impulsive force.
Then from (3) of Art 236, we havo for the angular
velocity
Mvh _ vh
Fl0:M
w =
The absolute velocity of each point of the body will bo
compounded of the two velocitieft of translfition and rota-
tion. TL " point 0, for example, to which the impulse is
applied, haa a velocity of translation, Oa, equaUto that of
the cen*^^re of gravity, and a velocity of rotation, ab, about
the centre of gravity ; so that the velocity of any point at
tf distance a from the centre, 0, will bo expressed by
I? i: aw ; the upper x - lower sign being taken according as
the point is, or is uot, on the same side of the centra of
gravity as the point 0. Thus, if we consider the motion of
the body for a vevy short interval of time, the line 000
will assume the position bOC, the point C remaining at
rest during this interval ; that is, while the point would
ho carried forward over the line Cc by the motion of trans-
lation, it would bo carried backward through the same
distance by the motion of rotation. Ueuce, since the abso-
lute velocity of O is zero, wo have
ow .-= ;
« = - = ^ ;
w A '
(1)
lud hence denoting 00 Ity I we havo
^uM#
466
AXIS OF SPOyTAffXOUS fiOTATIOff.
1 =
(3)
Now if there had been a fiied axis throngh C perpen-
dicular to the plane of motion^ the initial motion would
liave been precisely the same, and this fixed axis evidently
would p.ot have received any pressure from the impulse.
"When a rigid body rotates about a fixed axis, and the
bod^ can be so struck that there is no pressure on the axis,
any point in the line of action of the force is called a centre
of percussion.
When the line of action of the blow is given a^ d the
body is free from all constraint, so that it is capable of
translation as well as of rotation, the axis about which the
body begins to turn is called the axis of spontamous rota-
tion. It obviously coinoidcs with the position of the fixed
axis iu tbtv first case.
Cou. 1. — From (1) we have *
ah = GC'OQ = *,»;
hence the points and C are convertible, that is, if the
(ms of rotation be supposed to pass throvgh the point 0,
the centre of spontaneous rotation will coincide with the sen-
tre of percussion.
Cob. 3, — From (2) it follows, by comparison with (4) of
Art. 237, that if the axis of spontaneous rotation coincides
wi*h the aids of suspension, the centre of percussion coin-
cides with the centre of oscillation.
Son. — It is evident that if there be a fixed obstacle at 0,
and '<- be struck by the body OC rotating about a fixed
axis through C, the obstacle will recei"e the whole force
of the moving body, and the axis will not receive any.
Hence the centre of percuss Icn also dct?rrainos the position
' in which a fixed obBtuclo niunt bo ])laccd, on which if the
rotating body impinges nud iu brought to rest, the axis of
rotation will suflior no pretwure.
TIOH.
(3)
rough C perpcn-
ial motion would
jd axis evidently
1 the impulse,
ced axis, and the
jsure on the axis,
I is called a centre
is given a' d the
it it is capable of
about which the
spontamous rota-
litiou of the lixed
EXAMPLES.
467
lie, that is, if the
ovgh the point O,
icide with the cen-
arison with (4) of
rotation coincides
\f percussion coin-
ixed obstacle at 0,
,ing about a fixed
the whole force
1 not receive any.
mines the position
1, on which if the
,0 rest, the axis of
An axis through the centre of gravity, parallel to the
axis of spontaneous rotation, is called the axis of instantane-
ous rotation. A free body rotateo about this axis (Art, 239).
*
EXAMPLF. S.
1. Find the centre of percussion of a circular plate of
radius a capable of rotating about an axis which touches it.
Here /;," = j, and h = a. Hence from (3) we have
I = a + ■T = {a.
2. A cylinder is capable of rotating about the diameter
of one of its circular ends ; find the centre of percussion.
Let a = its length, a.nd b = the raf'ius of ltd base.
3i» + 4a^
Ans. I =z
Ga
Henca if 2&' = 2a\ the centre of iwrcussion wiE be at
the end of the cylinder. If b is very small compared with
a, I = ^i; thus if a straight rod of small transverse section
is held by one end in the hand, I gives the point at which
it may be struck so that the hand will receive no jar.
2^1. The Principfil Radius of Gyratior Deter-
mined Practioally. — Mount the body upon an axis not
passing through the centre of gravity, and cause it to
oscillate ; from the number of oscillations performed in a
given time, say an hour, the time of one oscillation ia
known. Then to find h, which is the distance from the
axis to the centre of gravity, attach a spring balance to the
lower onil, and bring the centre of gravity to a horizontal
])lane through the axis, which position will be indicated by
the maximum reading of (he balance. Knowing the maxi
mum reading, R, of the balance, the weight, H", of the
body, and the distance, a, from the axis of snKiHjusion to
IP"
i
468
TUB BALLISTIC PENDTTLVM.
tho point of attachment, we have from the principle of
moments, Ra = Wh, from which h is found. Substitut-
ing in (3) of Art 237, this value of h, and for T the time
of an oscillation, kt becomes known.
242. The Ballistic Fendnlnm. — An interesting aj)-
plication of the principles of the compound pendulum is
the old way of determining the velocity of a bullet or can-
non-ball. It is a matter of considerable importance in tho
Theory of Gunnery to determine the velocity of a bulk ' as
it issues from the mouth of a gun. It was to determine
this initial velocity that Mr. Robins about 1743 invented
the Ballistic Pendulum. This consists of a large thick
heavy mass of wood, suspended from a horizontal axis in
tiie shape of a knife-edge, after the manner of a compound
pendulum. The gun is so placed that a ball projected
from it horizontally strikes this pendulum at rest at a cer-
tain point, and gives it a certain angular velocity about its
axis. The velocity of the ball is itself too great to be
measured directly, but the angular velocity communicated
to tho pendulum may be made as small as we please by
increasing its bulk. The arc of oscillation being meas-
ured, the velocity of the bullet can be found by calcu-
lation.
Tho time, which the bullet takes to penetrate, is so short
that we may suppose it completed before the pendulum has
sensibly moved from its initial position.
Let M be the riass of the })eudulum and ball ; m
that of tho ball ; v tho velocity of tlie ball at the instant of
impact ; h the distance of the centre of gravity of tiio pen-
dulum and ball from the axis of suspension ; a the distance
of tho point of impact from the axis of susfwusion ; w tho
angular vi'octity due to tho blow of the ball, and k the
radius of gyration of the pendulum and ball. Then since
t h(> initial velocity of tiio bullet is v, its impulse is nieaaurod
l.v iiiv, au(i therefore from (3) of Art. ^30 wu have for the
TUB UAhLlSTIC PEXDVhUM.
469
be principle of
iid. Substitut-
for T the time
interesting ajv
id pendulum is
I bullet or can-
jortance in the
y of a bulk ' as
s to deterniino
1743 invented
I a large thick
rizontal axis in
of a compound
ball projected
t rest at a cer-
locity about its
too great to be
communicated
as we please by
on being meas-
bund by calcii-
rate, is so short
3 pendulum has
L and ball; m
it the instant of
niy of the pon-
; a the distance
pension ; w the
l)all, and k the
,11, Then since
iilse is meuaurcd
ve have for the
initial angular velocity generated in the pendulum by this
impulse,
mva
6) =
M-a'
(1)
and from (I) of Art. 237 we have for the subsequent
motion
0^ qh . ^
(8)
Integrating, and observing that, if « bo the angle through
which the iiendulum moves, we have ^ = w when © = 0,
and ^ = when = «, (2) becomes
cos a).
Eliminating w between (1) and (3) we have
%Mh r-r . a
V = Vffh sm - ,
ma " 2
(3)
(4)
from which v becomes known, since all the quantities in
the second member may be observed, or are known.
Wo may determine a as follows : At a point in the jwn-
dulum at a distance h from the axis of suspension, attach
the end of a tape, and let the rest of the tape be wound
tightly round a reel ; as the pendulnm ascends, let a length
c be unwound from the reel ; then c is the chord of tlio
angle « to the radius h, so that c ~ 2h sin |, which in (4)
givce
Mkc fa
The values of k and // may bo dot«rmmed as in Art. 241.
If the mouth of the gun is placed near to the pendulum.
470
ROTATION OF A HEAVY BODY.
m
the value of v, given by (5), must be nearly the velocity of
projection.
The velocity may also be determined in the following
manner : Let the gun be attached to a heavy pendulum ;
when the gun is discharged the recoil causes the pendulum
to turn round its axis and to oscillate through an arc
which can be measured ; and ihe velocity of the bullet can
be deduced from the magnitude of this arc. (See Price's
Anal. Mech's, Vol. II, p. 231.)
Before the invention of the balliatic pendulnm by Mr. Robins in
174.'}, but little progress bad been made in the true theory of mLitary
projectiles. Robins' New Principles of Gunnery was soon translated
into several languages, and Euler added to his translation of it into
German an extensive commentary ; the work of Baler's being again
tranalatSd into English in 1784. The experiments of Robins were
all conducted with musket balls of about an ounce weight, but they
wore afterwards continued during several years by Dr. Button, who
used cannon-balls of from one to nearly three pounds in weight.
Ilutton used to suspend his cannon as a pendulum, and measure the
angle through which it was raised by the discliarge. His experi-
monts ar<) still regarded as some of the most trustworthy on smooth-
bore guns. See Routh's Rigid Dynamics, p. 04, also Encyclop»dia
Britannica, Art. Gunnery.
243. Motion of a Heavy Body about a Horizon-
tal Axle through its Centre. — Let the body be a spliere
whose radius is R, and weight W, and \z'c a weight P be
attaclied to a cord wound round the circumference of a
wheel on ^ho same axle, the radius of the wheel being r ;
rofiuircd the dif;taiice ptissed over by P in t seconds.
From (4) of Art. 230 wo have
cPd _ Prg
d^ ~ Wkf+ Pr^'
I Multiplying by dt and integrating twice, we have
Prgt^
$ =
i{Wk* + Pr^y
(1)
m
>r.
y the velocity of
n the following
!avy pendulum ;
cs the pendulum
through an arc
ii the bullet can
re. (See Price's
1 by Mr. Robi-s in
e theory of mLitary
eras soon translated
ranslation of it into
Euler's being again
its of Robins were
ce weight, but they
by Dr. Hutton, who
I pounds in weight,
m, and measure the
charge. His experi-
stworthy on smooth-
;, also Encyclopsedia
)OQt a Horizon-
e body be a sphere
1; a -weight P be
ircumfercnoe of a
le wheel being r ;
t seconds.
ii|gB;i^ <lW»Vlwy> j MI|> ! «j ! MMW
!e, we have
(1)
FXAMPLBS.
471
the constants being zero in both integrations, since the body
starts from rest when t = 0. The space will be rd.
EXAMPLES.
1. Let the body be a sphere whose radius is 3 ft. and
weight 500 lbs.; let P be 50 Ibe., and the radius of the
wheel 6 ins.; required the time in which the weight P will
descend through 50 ft. (Take g = 32.)
Ans. 21 seconds.
2. Let the body be a sphere Whose radius is 14 ins. and
weight 800 lbs.; let it be moved by a weight of 200 lbs.
attached to a cord wound round a wheel the radius of
which is one foot ; find the number of revolutions of the
sphere in eight seconds. (Take g = 32.) Ans. 51.3.
244. Motion of a Wheel and
Azle when a Given Weight P
RaiseH a Given Weight W.— Let
the weights P and W be atfached to
cords wound round the wheel and axle,
respectively, (Fig. 97) ; let R and r be
be the radii of the wheel and axle, and
w and w' their weights; required the
angulai' distance passed over in t
seconds.
From (4) of Art. 236, we have
I
w
Fig.S7
PR-Wr
PR' + Wt^ + itoR^ + \w'r^
</;
{PR - Wr) t'
Pm-^Wr'* + \wR? + i«)V
y-
(1)
(2)
EXAMPLE.
Let the weight P = 30 lbs., If = 80 lbs., w = 8 lbs.,
and w' = 4 lbs.; and let R and r be 10 ins. and 4 ins.;
472
MOTION AUOXTT A VERTICAL AXIS.
required (1) the space passed over by P in 12 seconds if it
starts from rest, and (2) the tonsiona Tand T' of the cords,
supporting P and W. (Take g = 32.)
Ann. (1) 97.79 ft.; (2) T := 31.28 lbs.; T' = 78.64 Iba.
T
\
-e-
Fig.SS
245. Motion of a Rigid Body
about a Vertical Aada— Let AB
be a vertical axis about which the
body 0, on the horizontal arm ED,
revolves, under the action of a con-
stant horizontal force F, applied at
the extremity E, perpendicular to
ED. T^t M be the mass of the body whose centre is C,
and r and it i^he distances ED and CD, respectively. Then
from (4) of Art. ii'i". we have
dp - jn (k,» + w\' :
ft
Multiplying by di and integrating twice, obsorviaj that
the constants of both integrations are zero, we have " '
d =
FrP
which is the angular space passed over in t seconds.
(1)
EXAMPLE.
Let the body be a sphere whose radius is 2 ft, whose
weight is 600 lbs., and the distance of whose centre from
the axis is 8 ft., and let i' be a foree of 40 lbs. acting at the
end of an arm 10 ft long; find (1) the number of revolu-
t\on8 which the body will make about the axis in 10
minutes, and (2) the time of one revolution. (Take
9 - 32.) Am. (1) 9316.3 ; (2) 6.2 sees.
xrs.
L3 seconds if it
T' of the cords,
" = 78.64 lbs.
e
Fis.M
ose centre is C,
ectively. Then
5, observiK J that
wo have
<1)
seconds.
3 is 2 ft., whose
iiose centre from
lbs. acting at the
imber of revolu-
the axis in 10
olation. (Take
; (3) 6.2 sees.
Fifl.99
BOOr ROLLTNO DOWN AN INCLTNBD PLANE. 473
246. Body Rolling down an Inclined Plane.— ^
homogeneous sphere rolls directly down a rough inclined
plane under the action of gravity. Find the motion.
Let Fig. 99 represent a section
Oi Ae sphere and plane made by a
vertical plane passing through 0,
the centre of the sphere. Let « be
the inclination of the plane to the
horizon, a the radius of the sphere,
the point of the plane which
was initially touched by the sphere
at the point A, P the point of contact at the time /,
ACP = 0, which is the angle turned through by tho
sphere, m = tho mass of the sphere, F = the friction
acting upwards, R = the pressure of the sphere on the
plane. Then it is convenient to choose for origin and
OB for the axis of x ; hence OP = x.
The forces which act on the sphere are (1) the reaction,
B, perpeu'licular to OB at P, (2) the friction, F, acting at
P along PO, and (3) its weight, mg, acting vertically at
■^bo centre. Now evidently moves along a straight line
paralk' tq the plane ; so that for its motion of translation
we have, by ic=plving along the plane,
a*X
in^Ts — '^P sin « — F.
d^ ' '
(1)
The sphere evidently rotates about itK point of contact
with the plane; but it may be considered as Tot^ting at
any instant about its centre G as fixed ; and the miliar
velocity of at that instant in reference to P is the same
as that of P in reference to C. From (4) of Art. 236, we
have for the motion of rotation
»"V^ = i^
(8)
.^"smm^^mfiijitsmim-x^iimsif;:
:im.a-x-!<iiiefw^sr'msmim' t
i.C
i:
474 BODr ROLLTNO DOWN AN INCLINED FLANS.
and since the plane is perfectly rough, so that the sphere
does not slide, we have
X = ofl. (3)
Multiplying (1) by a and adding the result to (2), we get
(4)
ma^ + mki* ^ = mag sin a.
Differentiating (3) twice we get ^^
united to (4) gives
rj2
cPx
Since the sphere is homogeneous, A?,»
becomes
dh; .
^=^^Bin«
d»d , , ,
a^, which
(6)
f«a, and (5)
(6)
which gives the acceleration down the plane.
If the sphere had been sliding down a smooth plane, the
acceleration would have been g sin « (Art. 144) ; so that
two-sevenths of gravity is used in turning the sphere, and
five-sevenths in urging the sphere down the plane.
Integrating (6) twice, and supposing the sphere to start
from rest, we have
X ~ ^g • sin «5 • <'
which gives the space passed over in the time t.
Resolving perpendicular to the phne, we have
R =. mg sin a.
Cor. — If the rolling body were a circular cylinder with
< its axis horizontal, then k^^ = \a*, and (.5) becomes
(7)
)
W PLANE.
that the sphere
(3)
ult to (3), we get
(4)
Q a.
a j^^, which
(5)
» = |<?a, and (5)
(6)
smooth plane, the
Irt. 144) ; so that
ig the sphere, and
he plane,
he sphere to start
me t.
Bre have
ular cylinder with
(5) becomes
(7)
IMPULSIVE FORCE.
476
BO that one-third of gravity is used in turning the cylinder,
and two-thirds in urging it down the piano.
From (7) we have
a; = ^ sin a • /'
which gives the space passed over in the time t from rest.
(8)
247. Motion of a Falling Body under the Action
of an Impulsive Force not Directly through its
Centre. — A string is wound round the circumference of a
reel, and the free end is attached to a fixed point. The reel
is then lifted w/> and let fall so that at the moment when
the string becomes tight it is vertical, and tangent to tJie reel.
The whole motion being supposed to take place in one plane,
determine the effect of th^ impulse.
The reel at first will fall vertically without rotation.
Let V be the velocity of the centre at the momer^t when the
string becomes tight ; f ', « the velocity of the centre and
the angular velocity just after the impulse ; T the impul-
sive tension ; m the mass of the reel, and a its radius.
Just after the impact the part of the reel in contact with
the string has no velocity, and at this instant the reel
rotates about this part; but it may be considered as
rotating about its axis as fixed, and the angular velocity of
its axis, at this instant, in reference to the part in contact
is the same as that of the latter in reference to the former.
The impulsive tension is
T — m {v — v').
(1)
Hence from (4) of Art 236, we have for the motion of
rotation .
myfcj'w =z m{v — v').
(8)
»Mmi
lUtfian
■ i/i-'>-:l.<,if^^-^-/r\^-,',.:'
478
IMPULSIVE FORCE.
u
lit'
Sinco the part of the rod in contact with the string has
no velocity at the instant of impact, we have
V = aw.
Solving (3) and (3) we have
6) =:
av
a> + *,»
(3)
(4)
«.a
If the reel bo a homogeneous cylinder, ft^' = -, and we
«
have from (3) and (4)
w = f-, v' = Iv, (5)
I*
and from (1) we have for the impulsive tension,
T = ^mv.
Cob. — To find the subsequent motion. The centre of the
reel begins to descend vertically ; and as there is no hori-
zontal force on it, it will continue to descend in a vertical
straight line, and throughout all its subsequent motion ?;ho
string will be vertical. The motion may therefore be
easily investigated, as in Art 246, since it is similar to the
case of a body rolling down an inclined plane which is
vertical, the tension of the string taking the place of the
friction along the plane. Hence putting « = S' '^^^
letting the friction F = the finite tension of the string, we
have, from (1) and (7) of Art. 346,
■ ' F= \mg, and ^ = fer ;
that is, the finite tension of the string is one-third of the
jjggg^^jIMi
li the Btring has
(8)
(4)
,» = -, and we
(6)
lion.
he centre of the
bere is no hori-
ind in a vertical
aent motion ^ho
ay therefore be
is similar to the
[ plane which is
the place of the
ig « = I' ^^^
of the sti-ing, we
one-third of the
EXAMPLES.
477
weight, and the reel descends with a nniform acceleration
Since the initial velocity of the reel from (5) is \v, we
have, for the space descended in the time t after the impact,
from (8) of Art ''46,
\v + \gfl. (See Kouth's Rigid Dynamics, p. 131.)
X
EXA.MPL.BS.
1. A thin rod of steel 10 ft long, oscillates about an axis
passing through one end of it ; find (1) the time of an
oscillation, and (3) the number of oscillations it makes in a
day. Ans. (1) 1.434 sec. ; (3) 60254.
2. A pc-ndulum oscillates about an axis passing through
its end ; it consists of a steel rod 60 ins. long, with a rect-
angular section i by J of an inch ; on this rod is a steel
cylinder 2 in. in diameter and 4 in. long; when the ends of
the rod and cylinder are set square, find the time of an
oscillation. Am. 1.174 sees.
3. Determine the radius of gyration with reference to
the axis of suspension of a body that makes 73 oscillations
in 2 minutes, the distance of the centre of gravity from the
axis being 3 ft 2 in. Ans. 5.267 ft.
4. Determine the distance between the centres of suspen-
sion and oscillation of a body that oscillates in 2J^ sec.
Ans. 20.264 ft
5. A thin circular plate oscillates about an axis jfassing
through the circumference ; find the length of the simple
equivalent pendulum, (1) when the axis touches the circl?
and is in its plane, and (2) when it is at right angles to
the plane of the circle. Ans. (1) Jw ; (2) |«.
6. A cube oscillates about one of its edges; find the
length of the simple equivalent pendulum, the edge being
= ^«- Ans. |« ^2.
m
:t3i!i"'
liil
■■m
m
■■W'[
478
EXAMPLES.
7. A prism, whose cross section is a square, each sido
being = 2n, aud whose length is I, oscillates about one of
its upper edges; find the length of the simple equivalent
pendulum. Ans. f V'to* + P.
8. An elliptic lamina is such that when it swings about
one latus rectum as a horizontal axis, the other Intus
rectum passes through the centre of oscillation ; prove,
that the eccentricity is ^.
9. The density of a rod varies as the distaiice from one
end j find the axis perpendicular to it about which the
time of oscillation is a minimum, I being the length of the
rod.
Ans. The distauce of the axis from the centre of gravity
is \ Va.
10. Find the axia about which an elliptic lamina must
oscillate thai the time of oscillation may bo a minimum.
Alls. The axis must be parallel to the major axis, and
bisect the semi-minor axis.
11. Find the centre of percussion of ■\ cube w'lich rotate?
about an axis parallel to the four parallel edges of the cube,
and equidistant from the two nearer, as well as from the
two farther edges. Let 'ia bo a side of the cube, and let ■:.
be the distauce of the rotation-axis fn m its centre of
gravity.
I = c ■\- -5-, where / is the distance from the rota-
/1ns.
3c'
tion-iixis to the centre of percussion.
12. Find .the centre of ix^rcussion of a sphere which
rotates about an axis tangent to its surface.
A,h9. I ss \a.
' 13. Tift the body in Art. 243, be u sphere whoso radius is
17 ins. and weight 1200 lbs. ; let it be moved by a weight
of 260 Iba atf'"^hcd to a cord, wound round a wheel whose
KXAMPLES.
479
quare, each sido
,C3 about one of
imple equivalent
. f Vi^T" P.
it swings about
the other Ifitus
BciUation; prove.
istapce from one
ibout which the
the length of the
centre of gravity
ptic lamina must
:)e a minimum.
3 major axis, and
lube w'lich rotates
edges of the cube,
well as from the
10 cube, and let :7
« m its centre of
ice from *;he rotsi-
F a sphere which
ce.
A, IS. I = \a.
ere whose radius is
noved by a weight
nd a wheel whose
revolutions of the
Ans. 5b V.
radius Is 15 ins.; And the number of
sphere in 10 seconds, (g = 33.)
• 14. Let the body in Art. 243 be a sphere of radius 8 ins.
and weight 500 lbs. ; let it be moved by a weight of 10(^ lbs.
attached to a cord wound round a wheel whose radius is
6 in.; find the number of revolutions of the sphere in
5 seconds, {ff = 32^.) 4»s. 28.09.
15. In Art 244, let the weight i^ = 40 lbs., W = 100
lbs., w = 12 lbs., and w' = lbs.; and let E and r be
12 ins. and 7 ins. ; required (1) the space T>assed over by P
in 16 sees, if it starts from rest, and (2) the tensions T and
T' of the cords supporting P and W. {(/ = 32).
Ans. (1) 926.5; (2) T = 49.04 lbs., T' = 86.81 lbs.
16. In Art. 344, let the weight P - 25 lbs., W - 60
lbs., w = 6 lbs., and -lo' = 2 lbs. ; and let R and r be
8 in. and 3 in.; required (1) tlie space passed over by P in
10 sees, if it starts from rest, and (2) the tensions T and
T' of the cords supporting P and W. {g = 32|.)
Ans. (1) 109.92 ft; (2) T = 23.29 lbs.; T'= G1.54 lbs.
17. In Art 245, let the body be a sphere whose radius
is 3 ft, whose weight is 800 lbs., and the distance of whose
centre from the axis is 9 ft. ; and let F he a force of 60 lbs.
acting at the end of an arm 12 ft long; find (1) the num-
ber of revolutions which the body will make about the
axis in 12 min., and (2) the time of one revolution.
{g = 32.) Ans. (1) 14043.6 ; (2) 6.07 sees.
18. In Ex. 17, let the radius = one foot, the weight =
100 lbs., the distance of centre from axis = 5 ft., and
F ■= 25 lbs. acting at end of arm 8 ft long; find (1) the
number of revolutions which the body will make about the
axis in 5 min., and (2) the time of one revolution.
(<, = 32|.) Ans. (1) 18139.09 ; (2) 2.23 sees.
19. If the body in Art 247 be a homogeneous sphei-e,
the s*.riiig being round the circumference of a great circle,
tmme&am^sss
3S£££S
480
EXAMPLB8.
find (1) the angular velocity just after the impulse, and (2)
the impulsive tension. ^ Sy
2" A bar, I feet long, falls vertically, retaining its hori-
zon^l position till it strikes a fixed obstacle at one-quarter
the length of the bar from the centre ; find (1) the angu-
lar velocity of the bar, (2) the linear velocity of its centre
just after the impulse, and (3) the impulsive force, the
velocity at the iustan . of the impulse being v.
12«
21. A bar, 40 ft. long, falls through a vortical height of
60 ft., retaining its horizontal position till one end strikes
a fixed obstacle 60 ft. above the ground ; find (1) its anjju-
lar velocity, (2) the linear velocity of its centre just after
the impulse ; (3) the number of revolutions it will make
before reaching the g;ound, (4) the whole time of falling
to the ground, and (5) i\& linear velocity on reaching the
ground.
^ Ann. (1) 2.12; (2) 42.43; (3) 0.345; (4) 2.79; (5)
I '
1
npulso, and (2)
aining its hori-
I at one-quartor
(1) the angu-
ty of its centre
Isive force, the
|v ; (3) Htnv.
rtical height of
one end strikes
id (1) its angu-
entre just after
18 it vill make
I time of falling
)U reaching the
(4) 2.79; (5)
CHAPTER VHI.
MOTION OF A SYSTEM OF RIGID BODIES IN SPACE.
248. The EqnationB of Motion of a System of
Rigid Bodies obtained by D'Alemberf s Principle.—
Let {x, y, z) be the position of the particle m at the time /
referred to any set cf rectangular axes fixed in space, and
X, r, Z, the axial components of the impressed accelera-
ting forces acting on the same particle. Then -ttj , j^f > ;^.
are the axial components of the accelerations of the parti-
cle ; and by D'Alembert's Principle (Art. 235) the forces,
«(x-'^). ,„(K-g). ,«(z-f;).
acting on m together with pimilur forces acting on every
f article of the system, are in equilibrium. Hence by the
principles of Statics (Art G5) we have the following six
equations of motion :
(1)
(2)
21
482
TRAXSLATION AND ROTATIOIT.
w\
\M
m
By means of these six equations the motion of a rigid
body acted on by any finite forces, may be determined.
They lead immediately to two important propositions, one
of which enables us to calculate the motion of translation
of the body in space ; and the other the motion of rotation.
L49. Independe&oe of the Motion cf Translation
of ths Centre of Gravity, and of Rotation about an
Axis Passing through it.- -Let (i, y, i) be the position
of the centre of granty of the body at the time t, referred
to fixed axes, {x, y, z) the position of the particle m referred
to the same axes, («', y, z') the position of m referred t^ a
Bystcui of axes passing through the centre of gravity and
parallel to the fixed axes, and M the whole mass. Then
1. x=zi + x', y = y + y', z — l + z'. (1)
Since the origin of the movable system is at the centre of
gravity, we have (Art. 59)
Ima;' = 7.my' = Smz' = ; (2)
^ <ftB' ^, «Py' ^ (Pz'
Ki;
Also "Lmas = Mx, ^my = My, Swa = MZ;
^"*5^ = ^5^' ^""dt^
M
df»'
d>z „<?>«
Substituting these values in (1) of Art. 248, wo have
dr
(4)
ON.
notion of a rigid
be determined,
propositions, one
n of translation
otion of rotation.
cf Translation
;aiion about an
) be the position
e time /, referred
(article m, referred
I m referred t.:. »
re of gravity and
3 maas. Then
5 + «'. (1)
is at the centre of
0; (2)
=:a (3)
. 248, we have
(4)
MOTION OF A BO or.
488
These three eqnations do not contain the co-ordinatos of
the point of apphcation of the forces, and are the same as
those which would be obtained for the motion of the
contre of gravity supposing the forces all applied at that
point. Hence
The motion of the centre of gravity of a system acted on
by any forces is the same as if all the mass were collcded at
the centre of gravity and all the forces were applied at that
point parallel to t/ieir former directions.
2. Differentiating {] ) twice we have
W ~ dt^ ■*" rf^«' di^ ~ dt» "^ dt^'
dt* ~ dt^^ dP'
Substituting these valras in the first of equations (2) of
Art. 248, we have
2m[(y+>')Z-(5 + On
Performing the operations indicated we get
Uii>-
f
II
11
484
TRAySLAriON AND ROTATION.
I
Omitting the Ist, 2d, 4th, 5th, Gth, and 8th terms which
vanish by reason of (2), (3), and (4), .we have
similarly from the other two equations o' (2) \ e have/
(•^)
j = I.mix'r-y'X).
These three equations do not contain the co-ordinates of
the centre of gravity, and are exactly the equations we
would have obtained if we had regarded the centre of
gravity as a fixed point, and taken it as the origin of
moments. Hence
ne motion of a body, acted on by any forces, about its
centre of gravity is the samp, as if the centre of gravity were
fixed and the same forces acted on the body. That is, from
(4) the motion of translation of the centre of gravity of the
body is independent of its rotation ,• and from (5) the rota-
tion of the body is independent of the translation of its
centre.
These two important propositions are called respectively,
the principles of the conservation of the motions of transla-
tion and rotation.
Sen.— -By the first principle the problem of finding the
motion of the centre of gravity of a system, however com-
plex the system may be, is reduced to the problem of
finding the motion of a single particle. By the second
principle the problem of finding the angular motion of a
fiw body in space is reduced to that of determining the
motion of that body about a fixed jmint
ON.
3th terms which
ive
•2) y e have^
x'Z),
CONSBltVATION OF CENTRE OF ORAVITT.
485
(&)
y'X).
Rem.— In using the first principle it should be noticed
that the impressed forces are to be applied at the centre of
gravity parallel to their former directions. Thus, if a rigid
body be moving under the influence of a central force, the
motion of the centre of gravity is not generally the same
us if the whole mass were collected at the centre of gravity
and it were then acted on by the same central force. What
the principle asserts is, that, if the attraction of the central
force on each element of the body be found, the motion of
the centre of gravity is the same as if these forces were
applied at the centre of gravity parallel to their original
directions.
le co-ordinates of
le equations we
d the centre of
as the origin of
forces, about its
e of gravity were
t. That is, from
I of gravity of the
from (5) the rota-
ranslation of its
lied respectively,
Hions of transltt-
n of finding the
m, however com-
the problem of
By the second
ular motion of a
determining the
250. The Principle of the Conservation of the
Centre of Gravity.— Suppose that a material system is
acted on by no other forces than the mutual attractions of
its parts ; then the impressed accelerating forces are zero,
which give
2X=SF=SZ=0;
thereiore from (4) of Art. 249, we get
0. ^ = 0;
^^ = r„cos«,
^ = »o COS ft ^ = «o COS y.
(1)
where i;„ is the velocity of the centre of gravity when
f = 0, and «, ft y, are the angles which its direction makes
with the axes. Therefore, calling v the valocity of the
centre of gravity at the time /, we liave
V=:^J'■
iV? + dy^ + d?
dC^
= v,
0*
(3)
lohich is evidently coiistunt.
486
CONSERVATION 'OF AREAS.
If Vj = 0, the centre of gravity remainx at rest.
Integrating (1) we get
i =z v^t COB a ■+■ a, y = v^t 008/3 + 4,
i = v„f COS y + c ;
a — a _ y — b _ z — c
cos a "~ cos /3 ~ cos y
(3)
(ff, b, c) being the platje of the centre of gravity of the
system when / = 0. As (3) ar: the equations of a straight
line it follows that the motion of the centre of gravity is
rectilinear.
Hence token a material system is in motion under the
action of forces, none of which are external to the system,
then the centre of gravity moves uniformly «i a straight line
or remains at rest.
REM.^Thns the motion of the centre of gravity of a
system of particles is not altered by their mutual collision,
Avhatever degree of elasticity they may have, because a
reaction always exists equal and opposite to the action. If
an explosion occurs in a moving body, whereby it is broken
into pieces, the line of motion and the velocity of tlie
centre of gravity of the body are not changed by the
explosion ; thus the motion of the centre of gravity of the
earth is unaltered by earthquakes ; volcanic explosions on
the moon will not change its motion in space. The motion
of the centre of gravity of the solar system is not affected
by the mutual and reciprocal action of its several members ;
it is changed only by the action of forces external to the
system.
( 251. The Principle of the Conservation of Areas.—
If X, y be the rectangular, and r, the polar co-ordiuatcs
of u particle, we have
1
I.
at rest.
113 + *,
m
[)f gravity of the
tions of a straight
ntre of gravity is
motion under the
'tial to the system,
in a straight line
e of gravity of a
mutual collision,
have, because a
to the action. If
loreby it is broken
le velocity of the
; changed by the
e of gravity of the
;anic explosions on
pace. The motion
em is not affected
6 several members ;
cea external to the
ation of Areas. —
3 polar co-ordinates
coNsesvATioy of axsas.
r» co8» e ^ (tan e) = /^^.
dt^ ' dt
487
(1)
Now \i/^dO is the elementary ai-ea described round the
origin in the time dt by the projection of the radius vector
of the pai-ticle on the plane of xy, (Art. 182.) If twice
this polar area be multiplied by the mass of the particle,
it is called the area conserved by the particle in the time dt
round the axis of z. Hence
is called the area conserved by the system.
Let dAx, dAy, dAg be twice the areas described by the
projections of the radius vector of the particle m on the
planes olyz, zx, xy, respectively ; then from (1) we have
^ I dy dx\ ^ dAg
and differentiating we get
„ / d^/ €Px\ „ d^A,
(2)
If the impressed accelerating forces are zero the first
member of (2) is zero, from (5) of Art. 24i/; therefore the
second member is zero. Hence
v^.^-^' A.
dt^
aM
488
CONSEBVATTON OF VIS VIVA.
Similarly . 1?» -~ — 0, Sw -^ = ;
and therefore by integration
at dt at
«",
h, h', h" being constants.
. • . ^mAx = ht, ^mAy = h't, ImAg — h"t ;
the limits nf integration being snch that the areas and the
time begin simultaneously. Thus, the sum of the products
of the mass of every particle, and the Drojection of the area
described by its radius vector on each co-ordinate plane,
varies as the time. This theorem is called the principle of
the conservation of areas. That is,
When a material system is in motion under the action
offerees, none of which are external to the systetn, tJten the
sum of the products of the mass of each partich by the pro-
jection, on any plane, of the area described by the radius
vector of this particle measured from any fixed point, varies
as the time of motion.
252. Conservation of Vis Viva or Energy.'*'— Let
{x, y, z) be the place of the particle m at the time t, and
let X, Y, Z be the axial components of the impressed
accelerating forces acting on the particle, as in Art. 348.
Tlie axial components of the efEective forces acting on the
same particle at any time / are
dhi d^ ^
"^d?' "'Sp' "^di^-
' If the efEective forces on all the particles be reversed,
* See Art. 189.
"hU
1
h"t',
e areas and the
of the products
stion of the area
i-ordinate plane,
the principle of
under the action
system, tJien the
iiclo by the pro-
id by the radius
Ixed point, varies
Energy.*— Let
b the time t, and
f the impressed
I, as in Art. 248.
3s acting on the
cles be reversed,
?«r?Ri
I WW Iiy i « ^ l| JM il MPM;. l i W^i '.t pM^J t t l HMIIf m J-
CONSSRVATION OF VIS VIVA.
489
they will be in equilibrinm with the whole gronp of im-
pressed forces (Art. 236). Hence, by the principle of
virtual velocities (Art. 104), we have
.».[(x-*)^^(r-g)., + (z-g).«]=„.,:,
where dx, 6y, dt are any smaU arbitrary displacements of
the particle m parallel to the axes, consistent with the con-
nection of the parts of the system with one another at the
time t.
Now the spaces actually described by the particle m dur-
ing the instant after the time t ijarallel to the axes are
consistent with the connection of the parts of the system
witli each other, and hence we may take the arbitrary dis-
placements, Sx, 6y, 6z, to be respectively equal to the
actual displacements, ^ 6t, ^ 6t, j^ 6t, of the particle.*
Making this substitution, (1) becomes
^"^ w dt ^ d(^ dt'^ m^ It)
-(^f+4f+4)-
Integrating, we get
£»««;» - 27«v„» = 2Sm / {Xdx + Ydy + Zdz), (2)
vhere v and v^ are the velocities of the particle m at the
times t and t^.
The first member of (2, is twice the vis viva or kinetic
euL '^ of the system acquired in its motion from the time
^0 to the time t, under the action of the given forces.
, * '^i^' /"• t'"»>"'8'» «« ^ •«»» eqoal to <te, yel the nwto of &e to dit is equal lo tho
rutio 01 it Unit.
ii
iif
490
OONBEHVATION OF VIS VIVA.
Tho second member expresses twice the work done by
these forces in tho same time (Art 189).
If the second member of (2) be au exact differential of a
function of x, y, z, so that it equals df{x, y, z) ; then tak-
ing the definite integral between tho limits z, y, z and x^,
y^i Xq) corresponding to t and t^, (2) becomes
2mv» - £mvo» = 2/(«, y, z) - Zfix^, y^, z^). (3)
Now the second member of (2) is an exact differential so
far as any particle m is acted on by a central force whose
centre is fixed at (a, b, c), and which is a function of the
distance r between the centre and {(jr., y, r) the place of m.
Thus, let P be the central force =/{r), say; then
X — a
fir), r =
:r_y-i
fir),
Z=^'-^fir)i
r» = (« - a)« + (y - t)» + (z - c)»;
, • , rdr = {x — a)dx + (y — b)dy + {z — c)dzi
.'. m (Xdx + Ydy + Zdt) = m/{r) dr-,
which is an exact differential ;
second member of (2), it
substituting this in the
2m
rfir)dr,
where the limits r and r, correspond to t and /j.
Also, the second member of (2) is an exact differential,
so far as any two particles of the system are attracted
towards or repelled from er oh other by a force which varies
as the mass of each, and is a function of the distance
between them. Let m and m' be any two particles ; let
(a;, y, z), («', y', z') be their places at the time t ; r their
distance apart; P = f{r), the mutual action of the unit
mass of each particle. Then the whole attractive force of
'r'i£«ne#a:<,««3i»i$^&wM«%i#^.^«««£.
ijii i riill,,.
work done by
differential of a
y, z) ; then tak-
i x,y,z and x^,
nes
«o» yo' «o)- (3)
MJt differential so
ntral force whose
\ function of the
r) the place of m.
say; then
z = ^VW;
- (z — c)dz',
f(r)dr',
ating this ui the
t and /g*
exact differential,
rstem are attracted
I force which varies
ion of the distance
two particles ; let
he time t ; r their
action of the unit
e attractive force of
" Vg".y ' ™ V!t y ' gT !* ;gwT >~y«!»^g g i'i;« ^^^ ^
CONSERVATION OP VIS VIVA,
491
m on m' is Pm, and the whole attractive force of »? ' on to
is Pm' ; and we have
X
— *M ^ iL
Z-lif
X=.m~LF, Y=m^-=^P, Z=m^-P;
r=-m^^P, Z'=z.
m J
r
r ■ r
Also r* = (« - «')» + (y - y')* + (« - «')'•
Therefore for these two particles, we have •
m (Xdx + Ydy + Zdz) + to' (X'da;' + F'rfy' + Z'dz')
= ~^ [(« - *') ('^a; - rfo;') + (y _ y') (rfy _ rfy')
+ {z- z') {dm — </«')]
= tnm'f{r)dr;
which is an exact differential. The same reasoning applied
to every two particles in the system must lead to a similar
result ; so that finally the second member of (2)
= Stoto' //(r) dr,
where the limits r and r, correspond to t and <„, so that
the integral will be a function solely of the initial and final
co-ordinates of the particles of the system.
Hence, wJisn a material system is in motion under the
action of forces, none of which are external to the system,
then the chaise of the vis viva of the system, in passing
from one position to another, depends only on the two posi-
tions of the system, and is independent of the path described
by each particle of the system.
This theorem is called the principle of the conservation of
vis viva or energy.
PntNClPLE OF VIS VIVA.
Gob. 1. — If a system be under the action of no external
forces, we ha?e X = V = Z = 0, and hence the vis viva
of the system is constant.
OoB. 9.. — lict gravity bo the only force acting on the
system. Let the axis of c be vertical and positive down-
wards, then we have X = 0, Y = 0, Z = g. Hence (3)
becomes
But if e and s, are the distences !W)m the plane of xy to the
centre of gravity of the system at the times t and t^, and if
M is the mass of the system, we have
Mi — I.mz, ilfio = Xm«o;
(4)
That is, the increase of vis viva of the system depends only
on the vertical distance over which the centre of gravity
passes : and therefore the vis viva is the same whenever the
centre of gravity passes through a given horizontal plane.
Rem. — The principle of vis viva was first used bj Hayghens in
M» doternilnation of tuo centre of oscillation of a body (Art. 237,
llcm.).
Tlio ndvantag« of tliie principle ia that It gives at onro a rnlution
between the velocities of the bodias considered and the coordinates
which determine their (tositions In space, so that when, from the
nature of the problem, the position of all the bodies may bo made to
depend on one variable, the equation of vis viva is sufficient to deter-
mine the motion.
Suppose a weight m^ to be ])lBced nJ. any height A above tho nur-
face of tlie earth. As it falls through a height c, the force of gravity
does work which ia measured by mgt. Tho weight has acquired %
Velocity V, and therefore its vis viva is imc' wldch is wjual to t/ig$
(Art. 21 ). If the weiglit falls through the remainder of tho height
/(, gniviiy diKw nuirn wtirli wliich iu monsuKHl by mff (A — «). Wliun
tlie weight huu nuclii'd tli>< ground, it huu fallen aa far as the ciruuui-
)f no external
36 the vis viva
acting on the
positive dowii-
g. Hence (2)
)•
me of xy to the
! and t^, and if
,).
(4)
m depends only
ntre of gravity
m whenever the
izontal plane.
d bj Huyghens in
a body (Art. 337,
at onco a relation
id tbe coordinates
»t when, from the
38 may be made to
I sufficient to deter-
it h above tl:n mr-
the force of gravity
gbt has acquirwl a
ich is wjual to »/»</»
inder of the height
mt/(h-s). When
a far as the circuui-
"li
COMPOSITION OF ROTATIONS.
403
stance.! of the case permit, and gravity has done work which is meas-
ured by ii.^h, and can do no more work until the weight has been
lifted up again. Hence, thionghout the motion when the weight has
descended through any jpaoe (s, its vis viva, \mx?{=mgt), together
with the work that can be done during the rest of the descent,
mg (h — e), is constant and equal to mgk, the work done by gravity
daring the wliole descent h.
It we complicate the motion by making the weight work some
machine during its dewcent, the wamo tlieorem ia still true. The vis
viva of the weight, when it has deflcen(*ed any space i:. Is equal to the
work mgz which has bren done by gravity during this descent, dimin-
ished by the work done on the machine. Hence, as before, the vis
viva together with the diflf&rence between the work done by gravity
• and that done on the machine during the remainder of the descent is
constant and equal to the excess, of the work done by gravity over
that done on the machine during tho whole descent. (See Routli's
Rigid Dynamics, p. 270.)
253. Composition of Rotations.— It is often neces-
saiy to compound rotations about axes which meet at u
point. When a body is said to have angular velocities
about three diiferent axes at the same time, it is only meant
that the motion may be determined as follows : Divide tho
whole time into a number of infinitesimal intervals each
equal to dt. During each of those, turn the body round
the three axes successively, through angles ^idi, Wgdt, u^dt.
The result will be the same in whatever order the rotationn
take place. Tho flnal displacement of the body is tho
diagonal of the parallelepiped described on these throe linos
as sides, and is therefore independent of the order of the
rotations. Since then the three successive rotations are
quite independent, thoy may be s»»'..I to take place simul-
taneously.
Hence we infer that angular velocities and angular acnel-
erationi* may be compounded and resolved by the saiuo
rules and in the same way as if they were linear. Thus, an
angular velocity w about any given axis may be resolved
into two, w cos a and w sin «, about axes at right angles to
494
MOTION OF A RIGID BODY.
oach other and making angles a and ^ — « with the given
axis.
Also, if a body have angnlar yelocities 6)^, &>,, 6), about
three axea at right angles, they are to /ether equivalent to
a single angalar velocity a», where w = Vwi'-j-Wi'+Wj^
about an axis inclined to the given axes at ungles whose
cosined are respectively --,-*,
-1.
254. Motion of a Rigid Body referred to Fixed
Ax- '■- — Let U8 suppose that one point in the body is flxed.
I.,: ..... point be taken as the origin of co-ordinates, and
let tlie axes OX, OY, OZ be any directions fixed in space
and at right angles to one another. The body at the time
t is turning about some axis of instantaneous rotation
(Art. 240). Let its angular velocity about this axis be <•>,
and let this be resolved into the angular velocities <>>,, 6>„
fa>, about the co-oixlinat^ axes. It is required to find the
roaolved linear velocities, jft ^■>-jf> parallel t) the axes of
co-ordinates, of a particle wi at the point P, {x, y, z), in
terms of the angular velocities about the axes.
These angular velocities are sup-
posed positive when they tend the
same way round the axes that
positive CO' .; - *'jnd in Statics
(Art. 06). 'i' i the positive
direction^ Oi ■• - ,, w, are re-
spectively froi. ;^ to z about x,
from « to a; about y, and from x
to y about t ; and those negative
which act in the opposite direc-
tions.
Tjot us determine the velocity
of P parallel to the axis of t. Let PN bo the ordinate z,
na.ioo
with the given
, w,, w, about
• equivalent to
A angles whose
red to Fixed
e body is fixed.
>-ordinateB, and
i fixed in space
ody at tht time
ineons rotation
this axis be <•>,
jloeities <•>,, w»,
red to find the
el i > the axes of
P, («, y, «)» »»
Les.
ria.iQO
the ordinate z,
AXIS OF INaTAJfTANSOUS BOTATION.
495
and draw PM perpendicular to the axis of x. The velocity
of P doe to rotation about OX is u^PM, Resolving this
parallel to the axes of y and z, and reckoning those linear
velocities positive which tend from the origin, and vice
verm, we have the velocity
along MN = — t^^PM cos NPM =—«,«;
and along NP = UiPM sin JVPJf = w,y.
Similarly the velocity dne to the rotation about OF par-
allel to OX is iifZ, and parallel to OZ is — u^x. And that
due t3 the rotation about OZ parallel to CX is — «jy, and
parallel to OF is <o^x.
Adding together those '^looities which are parallel to
the same axes, we have for the velocities of P p^trallel'to
the axes of x, y, and z, respectively.
dx
It
■ = 6).2
dy
"sy.
"i«»
(1)
dz
^^u>,y-u>,x.
255. Azi« of Xnatantaneoiu Rotatioa-^Every par-
ticle in the axis of instantaneous rotation is at rest relative
to the origin ; hence, for these particles each of the first
members of (1) in Art. 264, will reduce to zero, and wc
have
«i« — w«y = 0, '
w,« " «,« = 0,
(1)
496
ANOULAK VELOCITY.
which are the equations of the axis of instantaneous rota-
tion, the third equntion being a tiecessary consequence of
the first tvio; hence.
(li. u.
« = --«, y = -a « ;
u.
w.
(2)
that is, tho instantaneous axis is a straight line passing
through the origin which is at rest at the instant con-
sidered ; and the whole body must, for the instant, rotate
about this line.
Cob. — Denote by a, /3, y the angles which this axis
makes with tho co-ordinate axes », y, «, respectively,
then (Anal- Geom., Art. 175) we have
cos « =
«4
Vw,» + Wg* + w,»'
COS |3 s=
(t).
V<V + Wg'-I- w,«
COS y =
(0.
V'wi» + V + '^s*'
which gives tht position of the instantaneous axis in terms
of the atigular velocities about the co-ordinate axes.
256. The Angular Velocity of the Body abont the
Axis of Inatantaneous Rotation. — The angular veloc-
ity of tho body about this axis will be the same as that of
any single particle chosen at pleasui-e. Lei the particle be
taken on the axis of a; ; if from it wo draw a perpendicular,
p, to the instantaneous axis, then tho distance of the par-
ticle from' tho origin being x, we have
tntaneoua rota-
QQimquence of
(8)
ht line passing
le instant ^on-
I instant, rotate
vhich this axis
«, respectively,
m axis in terma
te 0X68.
Body aboat the
le angular veloc-
e same as that of
}t the particle be
r a perpendicular,
tance of the par-
SULER'a XQUdTlOIfS,
m
p = XBm a
= X Vl — cos» a = a; A / — '^i "+• "^ —
Since, for this particle, y = 0, « = 0, we have from (1)
of Art. 254, for the absolute velocity,
and hence, for the angular velocity ?>, we have
V = — = V«i* + w,* + w,»,
wAtcA »« /Ae angular telodty required.
257. Euler'B Equations. — To determine the general
equations of molio^n of a body about a fixed point.
Let the fixed point be taken as origin ; let {x, y, f) be
^he place of any jiarticle m, at the time /, referred to any
rectangular axes fixed in spaoe, a.. ^ let Ox^, Oyi, Ozi be
the rrincipal axes of the body (Art. 231). Piffereutiating
(1) of Art. 254 with respect to t, we have
d?~'^~d~^~dt'^^* <"»y ~ ^^'^^ "■ "' (<^8«~'^i«)j
<Ptf rfw. dot. , , . . V
d^ ~'^'di ~'^~d ^^* ^^** ~ "»^^ ~ *^* ^"'^ ~ "»^^'
<P« dill. dii), . , , , V
dfl-^~di W'^'^^ ^"»* ~ "»*^ ~ *"» ^"•* ~ ''•^^•
Denoting by L, M, N, the first terms respectively of (2),
(Art. 248), and substituting the above values of ^ and t^
in the last of these equations, wo get
4t)8
BULSR'S SqUATIONS.
= jv: (1)
The other two eqnatioDB may he treated in the same way.
The coefficients in thiq equation are the moments and
products of inertia of the body with regard to axes fixed in
space (Art 224), and are therefore variable as the body
moves about. Let Uj,, tjy, u, be the angular velocities about
the principal axes. Since the axes fixed in space are per-
fectly arbitrary, let them be so chosen that the principal
axes are coinciding with them at the moment under con-
sideration. Then at this moment we have (Art 232),
litnxy = 0, ^myz = 0, "Lmzx = ;
also 6), = Wj,, 6)g = 6)y, 0), = u, ; and likewise -^ = ~^^,
etc.* Hence, denoting by ^, B, G, the moments of inertia
about the principal axes (Art 231), (1) becomes
''^-(^
B) 6)a,w^ = N,
in which all the coefficients are constants ; and similarly
for the other two equations.
Hence, uniting them in order, and retaining the letters
b)j, (i>„ fa),, since they are equal to Uo;, (^, (•>,, the three
, for the chanRM In the two angolar Tclocltloe, u, and <■<«, dnrinir a
W ~' of
tfiven BDitll time after the axle of a), colneide* with the axis of x, will dUhr only by
a unantlly which depends npon the angle passed throngh by the axis of x, dnring
timt given khisII time ; the dUferenoo between u, and oi* will therefore be an
Inflnitesimal of the second order and thc.cfore their derlratWes will bo eqnal. (Hee
Prntt's Mech., p. 438. For flirthor demoi'stratlon of this equality, tbesindcnt la
roftorred to Ruuth's BiKid Dynamics, pp. 181' and 188.)
r
n
= N. (1)
le same way.
moments and
axes fixed in
e as the body
velocities about
1 space are per-
it the principal
ent under con-
^rt 232),
= 0;
meuts of inertia
>me8
; and similarly
oing the letters
iy, tOt, the three
OS, w, and <<>«, darins a
of z, wiU dUftr only by
)y the axil of x, during
I* will therefore be an
veg will bo eqnal. (See
jquallty, the (indent ta
SULES'S SqUATTONS.
499
equations of motion of the body referred to the principal
axes at the fixed point are
dt
/?!!■;;»_ (C - 4) UjO), =2 M,
(2)
These are called Eulor's Equations.
SoH. — ^If the body is moving so there is no point in it
which is fixed in space, the motion of the body about its
centre of gravity is the same as if that point were fixed.
It is clear that, instead of referring the motion of the
body to the principal axes at the fixed point, as Enler has
done, we may use any axes fixed in the body. But these
are in general so complicated as to be nearly useless.
258. Motion of a Body about a Principal Asia
through its Centre of Gra^ty. — If a body rotate about
one of its principal axes pamng through the centre of
gravity, this axis mil suffer no pressure from the centrifu-
gal force.
Let the body rotate about the axis of « ; then if u be its
angular velocity, the centrifugal force of any particle m
will be (Art. 198, Cor. 1)
nw^p — WW* V'*' + y',
which gives for the x- and y-components muh; and tmchf ;
and the moments of these forces with respect to the axes of
y and x are for the whole body
S,tnuh;x, and £m<>>>y«.
600
AXIS OF PERMANENT ROTATION.
But these are each equal to zero when the axis of rotation
is a principal axis (Art. a32) ; hence, the centrifugal force
will have no tendency to incline the axis of z towai-ds the
plane of xy. In this case the only effect of the forces mtJ^x
and mul^y on the axis U to move it parallel to itself, or to
translate the body in the directions of x and y. But the
sum of all these forces is
^Lmurhi and I.miJ'y,
each of which is equal to zero when the axis of rotation
passes through the centre of gravity ; hence we conclude
that, wfien a body rotates about one of its principal axes
passing through its centre of gravity, the rotation causes no
pressure upon the axis.
If the body rotates about this axis it will continue to
rotate about it if the axis be removed. On, this account a
principal axis through the centre of gravity is called an
axis of permanent rotation.*
SoH. — If the body be free, and it begins to rotate about
an axis very near to a principal axis, the centrifugal force
will cause the axis of rotation to change continuidly, inas-
much as the foregoing conditions cannot obtain, and this
axis of rotation will either continually oscillate about the
principal axis, always remaining very near to it, or else it
will remove itself indefinitely from the principal axis.
Hence, whenever we observe a fVee body rotating about an
axis during any time, however short, we may infer that it
has continued to rotate about that axis from the beginning
of the motion, and that it will continue to rotate about it
for ever, unless chocked by some extrajieous obstacle. (See
Young's Mechs., p. 230, also Venturoli, pp. 135 and 160.)
♦ Prstt's Mochs., p. 489. Called also a natural axU qf rotatUm, seo Tonng's
Hectw., p. sao ; aim (?n intartailt aicU, we Price'« Mechs., Vol. n, p. 997.
m.
ixis of rotation
ntrifugal force
if z towards the
the forces m(>)^x
I to itself, or to
id y. But the
ixis of rotation
ce we conclude
( principal axes
lation causes no
will continue to
this account a
ity is called an
to rotate about
entrifugal force
ontinuaUy, inas-
obtain, and this
cillate about the
to it, or else it
) principal axis.
>tating about an
lay infer that it
m the beginning
» rotate about it
18 obstacle. (See
.. 136 and 160.)
r rotation, «eo Tonng'B
Vol. n, p. m-
VBLOCmr ABOUT A PRINCIPAL AXIS.
501
259. Velocity about a Principal Axis when there
are no Accelerating Forces. — In this case L — M =
JV = in (2) of Art. 257 ; also A, B, axQ constant for
the same body ; and if we put
B-o^^^ £^ = 0, ^
G
= H,
(1)
A -" B
(2) of Art. 257 becomes
dui^ = Htii^u^dt.
Put u^fti^u^dt = dip, and we have
a^dt'ix = Fdtp., <>),fifu>g = Od^, (^idu^ = ffdip;
and integrating,^ we get
w,» = 2i?V> + o', w,» == 2(70 + **, w,« = 2Zr0 + c«. (2)
where a, &, c are the initiil yalues of w„ w,, w,; hence
from (1) and (2)
dt =
d^
V'(2/l^ + a») (2(70 + IP) {2H<p + c»)
(3)
Suppose now the Irody begins to tuni about only one of
the principal axes, say the axis of x, with the angular
velocity a, then * = 0, c = 0, and (3) becomes
<?< =
d<t>
2 'S/^OS V^Fp + a»
Replacing 2i^ + a' by its value W|«, and rf0 by its value
F
i-, we have
<« =
dt^^^
Voir Wi'
i8'
602
THB INTEORAL OF EULER'a XQUATIONS,
and integrating, we get
ti. —a
0+ tVGff=~ log^^~.
w.
e*i(7 . giat ^oa — rii
w, +o
(4)
The constant (7 mnst be determined so that when ^ = 0,
Wj is the initial velocity a ; hence ^^ = or (7 = — oo ,
which makes the first member of (4) zero for every value
of L Hence, at any time t, we mnst have u^ = a; and
therefore from (2) ^ = 0, and w, = u, = 0. Conse-
qumily tits impressed velocity about one of the principal
axes of rotation continues perpetttal and uniform, as before
shown (Art 258).
260. The Integral of Euler's Equations.—^ body
revolves about its centre of gravity acted on by no forces but
such as pass through that point; to integrate the equations
of motion.
As the only forces acting on the body ai'e those which
pass through its centre of gravity, they create no moment
of rotation about an axis passing through that centre; and
therefore (2) of Art. 257 become
A^-(B-C)u>,o>,=0/
B^-iC-A)o>,o>,=0,]
(1)
the principal axes being drawn through the centre of
gravity.
Tiosa.
(4)
at when ^ = 0,
or C = — oo ,
for every value
i u^ = a', and
= 0. Conse-
f the principal
iform, as before
dona— ^ body
>y no forces but
le the equations
!u'e those which
ate no moment
bat centre ; and
(1)
the centre of
SraW INTEQBAL OF EULSR'S EQUATIONS.
503
Multiply these equations severally (1) by Wj, w,, Wj ;
and (2) by Au)^, -Bw,, Cwg, and add ; then we have
di»>t
(?6),
du>t
dt
= 0;
^ + (^.^ = 0,1
(2)
integrating, we have
where A« and ** are the constants of integration.
Eliminating Wj' from (3), we have
A{A-C)i^* + B{B- C) 0),' = *» - a»;
(8)
I
"•' ~ £ (5 - C)
[ifc9_oa»_^(^_C)V]; (*)
anda,,» = ^^^^[i»-5A»-^M-P)«,«]. (5)
Substituting these values of u, and Wj in the first of equa-
tions (1), we have
dt ^L
dt
(A- O)iA-B)
EG
r-"- A{A-C) )
G^ISj-')]*- <«)
which is generally an elliptic transcendent, and so does not
admit of integration in finite terms. In certain particular
cases it may be integrated, which will give the value of <«>,
in terms of /, and if this value bo substituted in (4) and (5),
604 APPLICATION OF TBS OSNERAL EqUATIONS,
the values of w, and w, in tonus of I will be known, and
thus, in these caaes, the problem admits of compiute solu-
tion.
Cob. — I^t Wa, jwy, 0)8 bo the axial components of the
initial angular velocity about the principal axes when
< = 0; then integrating the first of (2), and taking tiie
limits corresponding to / and 0, we have
^w*» -I- 5a»,a + Cw3» = Jw,2 + 5a),» + Cio?. (7)
Ijet a, /J, y be the direction-angles of the instantaneous
axis at the time t relative to the principal axes ; so that, if
6) is the instantaneous angular velocity, and I.mr^ is the
moment of inertia relative to that axis, we have (Art. 253),
b), = u cos a, 6), = (i) cos /3, 0), = (i> cos y, which sub-
-tuted in (7), gives
J + B(^y* + Cw," = 6)« ( J cos^ « + 5 cos" /3 + C COS* y)
' ' ? = <JiZmr» (Art. 332, Cor.)
w ;(>l'^ ■'.:■
= the vis viva of the body ;
from which it appears that the vis viva of the body is con-
stant throughout the whole motion. ,
Rem.— An application of the general equations of rotatory
motion (Art. 267), which is of great interest and impor-
tance, is that of the rotatory phenomena of the earth under
the action of the attracting forces of the sun and the moon,
tlie rotation being considered relative to the centre of
gravity and an axis passing through it, just as if the centre
ef gravity was a fixed point (Art. 249, Sch.) ; and the
problem treated as purely a mathematical one. Also, in
addition to the sun and the moon, the problem may be
qVATIONS.
be known, and
)f complete solu-
nponenta of the
jipal axes when
, and taking tl)c
•>? + Ci^.\ (7)
the instantaneous
axes ; so that, if
and 'Lmr^ is the
have (Art. 253),
los y, which eub-
coB'/S + Cco^y)
, Cor.)
e bodj ;
)/ the body is con-r
ations of rotatory
crest and impor-
»f the earth nnder
in and the moon,
to the centre of
9t as if the centre
Sch.) ; and the
al one. Also, in
problem may be
'•■^rssiz iiiimi t ri.iKiv.i.mi/fnj^'f « ir",) ^ «^v •
I ' ^^/i- i ii , . " i i ^yfJi
SXAMPLES.
605
extended so aa to include the action of all the other bodies
whose influence affects the motion of the eaith's rotation.
In fact the investigation of the motion of a system of bodies
in space might be continued at great length ; but such
investigations would be clearly beyond the limits proposed
in this treatise. The student who desires to continue this
interesting subject, is referred to more extended works.*
EXAMPLES.
1. A hollow spherical shell is filled with fluid, and rolls
down a rough inclined plane ; determine its motion.
Let M and M' l/C the masses of the shell and fluid
respectively, h and h' their radii of gyration respectively
alHJut a diameter, and a and a' the radii of the exterior and
interior surfaces of the shell ; then using the same nota-
tion aa in Art. 246, we have
(Jf + if ') S = (-^ + M') gAna- F.
df>
(1)
As the spherical shell rotates in its descent down the plane,
the fluid has only motion of translation ; so that the equa-
tion of rotation is
JfP^ = F«. (3)
Multiplying (1) by a' and (2) by a, and adding, we have
[{M + M') d> + JfF] H = (J/ + M') a\j sin «. (3)
If the interior were solid, and rigidly joined to the shell,
the equation of motion would be
• See Price's Mech'g, VoL II, Pmtt'a Xeob's, Bonth's Rigid Dynamics, La Place's
Hgcanique C61este, ete.
asB
506
KXAKPLMS.
[{M+M') fl«+^*»+if' /fc'»] g ^ {M+M') ct-g sin «. (4)
Integrating (3) and (4) twice, and denoting by s and «' the
BIM1C08 thr tugli which the centre moves during the time /
in these two cases respectively, we have
(6)
80 that a greater space is described by the sphere which has
the fluid than by that which has the solid in its interior.
If the densities of the solid and the fluid are the same,
we have fiom (5), by Art. 233, Ex. 14,
~, = y . _ q ,, • (Price's Anal Mechs., Vol II, p. 368).
2. A homoguneons spliere rolls down within a rongh
spherical bowl ; it is ro<{uired to determine the motion.
liOt a be the radius of the sphere, and b the radius of the
bowl ; and let us suppose the sphere to be placed in the
bowl at rest. Lot OCQ = ^,
QPA = e, BCO = a, u> =
the angular velocity of the
ball about an axis throagh its
centre P, k = the correspond-
ing radius of gyration ; 0M=:
X, MP = y ; «i = the mass of
the ball. Then Fig. mm
TO jTj =r — 7? sin + /* cos ;
♦» T^ = /< 008 ^ -f Fnn ^ — mg\
(1)
')(^ga.na. (4)
by s and s' the
ing the time t
k^
(6) ,
ihcre which has
its interior.
) are the same,
3l II, p. 368).
itliin a rough
he motion.
le radins of the
I placed in the
i»;
mg
(1)
BXAMPhES.
at
507
(8)
Also X — (J — a) sin ^ ; if — b — (b — a) cos ^
... g=(J-a)co,*g-(S-.)«.*(f); (4)
g = (»- a) eiu ** + (»-.) 00. *(*)•. (6)
— CO80 + ^8in^ = (i-fl)^.
m
(d — o) ^ = JT— fw^- sin ^
(7)
Now to determine the angnlar velocity of the ball, we mnst
estimate the angle described by a fixed line in It, as PA,
from a lino fixed in direction, as PM, and the ratio of the
infinitesimal increase of this angle to that of the time will
be the angular velocity of the ball.
ti
dMPA d4> . de
dt
~ dt "^ dt
Since the sphere does not slide, aO = ft (a — ^) ;
a — b d^
a dt
dit> a — b d^
'** dt ~~ir dp'
from (3), (7), and (8) we get
(*-o)^~ -^8in^;
(•)
(»)
S08
BXAMPTjBB.
(10)
(11)
. • . {b-a) (^p = — ^ (cos - COB «).
Substituting (9) in (7) we have
F = ^mg Bin 0.
Substituting (4), (9), (10), (11) in (1) we have
5 = ^ (17 COB ^ — 10 COB «) J
therefore the pressure at the lowest point
= ^(17-10coH«);
and the pressure of the ball on the bowl vanishes when
cos ^ = ^ cos a.
Cob. — If the ball rolls over u small arc at the lowest part
of the be ivl, 80 that « and <f> are always small, cos «, and
cos </) may be roplaced by 1 — — and I — ?■ respectively ;
and from (10) we have
(«» _ ^)i L^ (* - «)J
••• * = «''««[7-(/i:^)J''
thus the ball comes to rest at points whoso angular distance
is rt on l>oth sides of 0, the lowest point of the bowl ; and
the periodic time is
iS a).
(10)
(11)
ave
•)i
'<
mishcs wben
A tho lowest part
small, COB «, and
- ^ respectively ;
dt;
t\
) angular distance
of the bowl ; and
SXAMPLSS.
509
therefore the oscillations are performed in the same time aa
those of a dimple pendulum whose length is \ {b — a),
(Art. 194). (Price's Anal. Mech's, Vol. II, p. 369.)
3. A homogeneous sphere has an angular velocity w
about its diameter, and gradually contrs^cts, remaining
constantly homogeneous, till it has half the original
diameter ; required the final angular velocity. Ans. 4w.
4. If the earth were a homogeneous sphere, at what point
must it be struck, that it may receive its prtsent viilocity
of translation and of rotation, the former being 68000 miles
per hour nearly ? Ana. 24 miles nearly from the centre.
&. A homogeneous sphere rolls down a rough inclined
plane; the inclined plane rests on a smooth horizontal
plane, along which it slides by reason of the pressure of the
sphere ; required the motions of the inclined plane and of
the centre of the sphere.
Let m = the mass of the sphere,
M = tho mass of the inclined
plane, a = the radius of the sphere,
rt =. the angle of tho inclined
plane, Q its apex ; the place of Q
when t z=z 0; 0' the point on the
plaiio which was in contact with
the point A of the sphere when
/ = 0, at which time we may sup-
pose all to be at rest ; A CF = 6, the angle through which
tho sphere has revolved in the time i.
Let be the origin, and let the horizontal and vertical
lines through it be tiio axes of x and y ; OQ = x' ; and -let
{x, y) {h, k) be the places of the centre of tho sphere at the
times ( — t and / = respectively. Then the equations
of motion of the sphere are
w j^ = F cos « — li sin a,
FI8.I02
610
BXAMPLES.
«» ^ = ^ein a + /J COS o — mgt
and the equation of motion of the plane is
. M -^ = — F co^ a ■{■ R sax a.
From the geometry we have
a; = A + «' — aO cos o,
y = ifc — oO sin a.
From these equations we obtain
, wi cos « -
__ 6ct sin a cos a gfi _
~ 7 (m + My— bm cos*^ ' T '
« = A —
y = *-
Slfsinacosa gfi
6{m + M) sin* a gfi
7 (m + if ) — 5»« co8» a ' aT
which give the values of a; and y in' terms of /.
Also wo obtain
(m + M) {x — h) sin o — if (y — /fc) cos a = ;
which is the equation of the path describeti by the centre
of tlie sphere ; and therefore tliis path is a straiglit line.
6. A heavy solid wheel in tlie form of a right circular
cylinder, it) composed uf two substunces, whuau volumes are
-mg.
u <c
COB a
9?.,
bm cob' a "i
of/.
fc) COS « = ;
bed by the centre
a straight line.
)f a right circular
whusu vuluiuea are
BXAJtPLES.
611
equal, and whose densities are p and p' ; these substances
are arranged in two diflferent forms; in one case, that whose
density is p occupies the central part of the wheel, and the
other is placed as a ring round it ; in the second case, the
places of the substances ai-e interchanged ; t and t' are the
times in which the wheels roll down a given rou^h inclined
plane from rest; show that
<»:<'»:: 6p + 7p' : 5p' + 7p.
7. A homogeneous sphere moves down a rough inclined
plane, whose angle of inclination « to the horizon is greater
than that of the angle of frictiou ; it is required to show (1)
that the sphere will roll without sliding when ^ is equal to
or greater than f tan «, and (2) that it will slide and roll
when n is less than f tan a, where |u is the coefficient of
friction.
b. In the last example show that the angular velocity of
. ,■ .■ A M^ i. 5/10 cos a .
the sphere at the time t from rest = -• ■ - — t.
9. If the body moving down the plane is a circular
cylinder of radius = a, with its axis horizontal, show that
the body will slide and roll, or roll only, according as a is
greater or not greater than tan~^ 3/i.