. * .4 ELEMENTARY MEGHMICS INCLUDING HYDROSTATICS AND PNEUMATICS BY SIR OLIVER J. LODGE, D.Sc., LL.D., F.R.S. i PRINCIPAL OF THE UNIVERSITY OF BIRMINGHAM, LATE PROFESSOR OF PHYSICS IN UNIVERSITY COLLEGE, LIVERPOOL, AND EX-PRESIDENT OF THE PHYSICAL SOCIETY OF LONDON (Ebition COMPLETELY REVISED BY THE AUTHOR AND BY ALFRED LODGE, M.A. PROFESSOR OF PURE MATHEMATICS AT THE ROYAL INDIAN ENGINEERING COLLEGE, COOPERS HILL ANSWERS REVISED BY CHAELES S. LODGE, B.A. OF THF UNIVERSITY D. V NEW YORK J PKEFACE TO THE 1896 EDITION. THE present book aims at giving a clear knowledge of the 'principles of the subject, in as elementary and even popular a manner as is consistent with careful accuracy, and without assuming any mathematical knowledge beyond the most rudi- mentary algebra. At the same time it is hoped that students who use this manual will be able to master the elements of the science in such a way that they may rise from it to more advanced treatises, not only without having anything to un- learn, but with a very sound knowledge of principles. Copious illustrations and explanations have been inserted, and the needs of students who are without the aid of a teacher have been kept steadily in view. The subject is treated as an introduction to Physics, and its laws are deduced from the first principles of familiar experi- ence rather than from special experiment. Experiments in Mechanics have a subordinate though most useful part in illustrating and emphasising the facts, but the author has no faith in making the establishment of principles depend on special experiments. So also in Geometry : drawing, measuring, and weighing may well be used for purposes of instruction and illustration, but propositions should be otherwise proved. The early examples at the ends of the chapters are typical ones, and are intended not only to be worked without looking PREFACE. at the answers, but also to be read almost as part of the book, because they frequently direct attention to important details. A large number of examples for practice have now been added to these, and the text has been thoroughly revised. In this work, as stated on the title-page, the author has had the collaboration of his brother. The statements made in a book should be carefully criticised, and not taken for granted ; and all kinds of special cases should be thought of or tried, to see if an exception cannot be found. It is by thinking one's self on a subject that it becomes really knoivn to one's self; it will never be really known if we only try to under- stand and remember what the book says. The author thinks that students will derive benefit from referring to Part I. of Deschanel's Natural Philosophy, trans- lated by Dr Everett, as a supplementary well-illustrated work introductory to general Physics, and reference is accordingly made to it or to the corresponding portion of Ganot's Physics for details which would unduly swell the size of the present book. From the more engineering side, Professor Perry's Practical Mechanics is also to be recommended. The book, as now revised, is intended to be not only an easy introduction to the subject, but, as far as it goes, a philo- sophical work. If at any place it is unable to stand the test of hostile criticism, the failure is a defect which the author will gladly utilise the aid of the critic to remove. From friendly critics he has already received several welcome minor corrections. OLIVER J. LODGE. SUGGESTIONS FOR READING. BEGINNERS are recommended to omit the following sections on a first reading : 17, 18, 31, 40-43, 53, 55, 56, 76, 80-82, 104, 105, 123 ; and then to return and read the omitted portions together, and finally to read the whole book carefully through without omitting anything. Students preparing only for London U-ni- versity matriculation, or for the elementary stage of the Science and Art Department, may with safety omit any of the above sections over which they experience much difficulty until the examination is over. The introduction being harder than many other parts of the book, its complete reading may be deferred. It is inadvisable to begin the study of either Mechanics or Physics without a knowledge of the Greek alphabet. ANALYTICAL TABLE OF CONTENTS. INTRODUCTION. SECT. PAGE 1. Definition of mechanics 1 2. Definition of force 2 3. Dual nature of force Stress 2 4. Relative rest 5 5. Effects of force on matter 5 6. Measure of force 6 CHAPTER I. ON MOTION (Kinematics). (1) Translation, or Motion of a Point. (a) Rectilinear Motion. 7. Uniform and variable motion 7 8. Definition of velocity 8 9. Definition of average velocity, and of actual velocity 9 10. Definition of acceleration 11 11. The unit of acceleration 13 12. Negative acceleration, or algebraically decreasing velocity.. 13 (6) Curvilinear Motion. 13. Change in direction of motion, and curvature 15 14. Circular measure of angles 16 15. Acceleration normal to the direction of motion 17 (2) Rotation, or Motion of an Extended Body. 16. The translation of an extended body T. 18 17. Rotation of an extended body Angular velocity 18 18. Definition of angular acceleration 20 ELEMENTARY MECHANICS IX CHAPTER II. RECTILINEAR MOTION continued. SECT. PAGE 19. Discussion of the equations separately 22 20. Discussion of the equations together ; 23 21. Discussion of the equations when the initial velocity is zero. 24 22. Return to the more general case 25 23. Geometrical illustration of the algebraical equation 25 Composition of Motions. 24. Polygon of motions 32 25. Triangle and parallelogram of motions 34 26. General statement of the parallelogram law 34 27. Composition of velocities 34 28. Composition of accelerations 35 29. Combination of a uniform velocity with a velocity uni- formly accelerated in a constant direction 36 30. Resolution of motions 37 31. Relative motions ,...39 CHAPTER III. ON QUANTITY OF MATTER AND QUANTITY OF MOTION. (1) Motion of a Particle (Inertia and Momentum). 32. Motion of matter, and quantity of motion 43 33. Meaning of 'inertia,' and Newton's first law of motion 43 34. Quantity of matter measured by inertia 45 35. Definitions of density and mass 46 36. Unit of mass Popular confusion of masses and weights.... 47 37. Definition of momentum 49 38. Unit of momentum 49 39. Definition of particle and rigid body; composition of momenta..., ...50 (2) Spinning Motion of an Extended Body (Moment of Inertia and Moment of Momentum). 40. Definition of the terms moment and second moment 51 41. Notion of the quantity of motion of a rotating body 51 X CONTENTS. SECT. PAGE 42. Moment of momentum or angular momentum 52 43. Definition and list of moments of inertia 53 CHAPTER IV. ON FORCE AND MOTION (Dynamics). (1) On the speed of Motion as affected by Force; or, Force and Rectilinear Motion. 44. The measure of a force 56 45. Force is rate of change of momentum, or mass acceleration. .56 46. Unit of force The poundal The dyne 57 47. Illustration of the use of the fundamental equation F=ma... 59 48. Newton's second law of motion in Maxwell's language 60 49. Aspects of this law 62 50. Composition of forces 62 51. Specification of a force, and representation of forces hy lines..63 52. Newton's third law of motion 64 53. Impact, direct and oblique 66 (2) On Angular Velocity as affected by Force ; or Force, and Rotation. 54. Effect of force on an extended but rigid piece of matter. ...73 65. The moment of momentum equals the moment of the impulse 73 56. Or the angular acceleration is the ratio of the moment of the force to the moment of inertia of the body 74 57. Moment of a force represented geometrically 74 58. Action of a force on a particle perpendicular to its motion... 75 (3) On the Direction of Motion as affected by Force ; or, Force and Curvilinear Motion (Centrifugal Force). 59. Centripetal force 76 60. Determination of the magnitude of this force 78 61. Illustrations of this force ...80 CHAPTER V. ON FORCE AND MOTION continued. The Force of Gravitation. 62. Gravity is a practically uniform force in mechanics 84 63. The weight of a body is independent of its material 84 ELEMENTARY MECHANICS. XI SECT. PAGE 64. Weight is proportional to mass ; the ratio being g 85 65. Laws of falling bodies starting from rest 87 66. Modes of diluting gravity A t wood's machine 88 67. Measure of g by Atwood's machine 90 68. Experimental verification of the laws of falling bodies 91 69. Geometrical illustration of the laws of falling bodies 91 70. Experimental verification of the velocity of falling bodies... 94 71. Generality of the equation ~F=ma 95 72. Tension in ropes which are raising or lowering weights 92 73. The composition of motions illustrated by falling bodies. ...98 74. The composition of motions illustrated by projectiles 98 75. And by Morin's machine 99 76. Projectiles 100 77. The curvilinear motion of a particle under gravity 102 78. Simple pendulum ..103 79. Conical pendulum 104 80. Compound pendulum 106 81. Centre of oscillation or percussion 107 82. Ballistic pendulum 109 CHAPTER VI. WORK AND ENERGY. 83. Object of the chapter...., Ill 84. A force produces an effect only when allowed to move Ill 85. Definition of energy 112 86. Definition of work, and measure of it Unit of work 112 87. Enumeration of the forms of energy 113 88. Impossibility of creating fresh energy. 114 89. Conservation of energy 115 90. Transference of energy from one body to another 115 91. Energy considered as a power of doing work 116 92. Dissipation of energy 117 93. All energy is actual energy and possible work 118 94. When work is done, energy is always transformed 119 95. Measure of kinetic energy Unit of kinetic energy 119 96. Gain of kinetic energy by a particle not originally at rest.. .121 97. Illustrations Railway truck ; rifle bullet 121 98. Force is time rate of change of momentum, but distance rate of change of energy 123 99. Measure of potential energy 123 Xll CONTENTS. SECT. PAGE 100. Meaning of expression a foot-pound 124 101. Transformation of energy illustrated by a falling weight. 124 102. Energy method of treating machines 125 103. Potential energy of strain illustrated by a bow 126 104. Kinetic energy of a rotating body ....126 105. Energy of a rolling body 128 CHAPTER VII. COMPOSITION AND RESOLUTION OF FORCES. 106. Advantage of reducing forces in number or compound- ing them 133 107. Advantage of splitting up one force into more, or resolving it .- 134 Composition of Forces acting on a Particle. 108. Parallelogram and polygon methods of compounding forces 134 109. Examples of the parallelogram of forces 1 35 110. Difference between parallelogram and triangle of forces.. 137 111. An example of the polygon of forces 138 Resolution of Forces. 1 12. The components of a given force are indefinite 139 113. Constrained motion 140 114. Illustrations of the resolution of forces Inclined plane... 140 115. .. .. Friction 141 116. Illustrations of the resolution of motions 143 117. Resolution of motions and of forces combined 144 Composition of Forces on a Rigid Body. 118. Reason why position is important 147 119. General statements 148 120. Case of two forces in a plane 150 121. Case of two forces in different planes A wrench 150 122. Case of any number of forces in a plane 151 123. Construction for finding the resultant, Funicular polygon. 151 Composition of Parallel Forces. 124. Magnitude and direction of resultant of parallel forces ...154 125. Position of the resultant of parallel forces 155 ELEMENTARY MECHANICS. Xlll SECT. PAGE 126. Constructions for finding the resultant of two parallel forces 156 127. Statements applicable to two parallel forces 159 128. Definition of a 'couple' 159 The Composition of Parallel Forces as illustrated by Gravity Centre of Gravity. 129. Experimental mode of finding resultant of parallel forces. 160 130. Experimental determination of the centre of gravity 161 131. Arithmetical n ,, .. ...162 CHAPTER VIII. ON EQUILIBRIUM (Statics). 132. Object of the chapter 170 133. Definition of equilibrium 170 Conditions of Equilibrium. 134. For two forces 171 135. For three forces 172 136. For a particle : 173 137. Fora rigid body 173 138. Equilibrium of a rigid body acted on by forces in a plane. 174 139. Illustrations. A ladder, by construction 176 140. A ladder, by calculation 177 141. A stick, rail, and wall 179 142. A body sliding on an inclined plane 180 Stability of Equilibrium. 143. Stable equilibrium and measures of stability 182 144. Unstable and neutral equilibrium , . 184 145. Equilibrium of things balanced on a point 185 146. Equilibrium of things rolling on a plane Metacentre 186 147. Method of virtual velocities. ... ... 188 CHAPTER IX. ON MACHINES, OR THE RUDIMENTS OF APPLIED MECHANICS. 148. Use of a machine Mechanical advantage 192 149. Pulley; inclined plane; lever; wheel and axle 194 XIV CONTENTS. SECT. PAGE 150. Combinations of these machines 196 151. Weston pulley block 199 152. Definition of the term power 199 CHAPTER X. PROPERTIES AND STATES OF MATTEB. 153. The effects of a force depend on the sort of matter it acts on 203 154. Elasticity of volume and of figure 203 155. Solids may be elastic or plastic 204 156. Limits of elasticity Tenacity 205 157. Strains of solids 205 158. Definition of elasticity 206 159. The three elements of any strain 206 160. Measure of strain and of stress 206 161. Viscosity All plastic solids are more or less viscous 207 162. Limpidity A non-viscous plastic body is called a fluid... 207 163. Compressible and incompressible fluids 208 164. Special mechanics required for special states of matter ...209 165. Mode in which fluids transmit pressure 209 CHAPTER XI. (Hydrostatics). ON THE PRESSURE OF GRAVITATING LIQUIDS AT REST. 166. Definition of a perfect liquid 211 167. Fluid friction depends on velocity, and vanishes with it. 211 168. Laws of hydrostatic pressure The hydraulic press 212 169. Pressure of gravitating liquids 214 170. Pressure on the base of a cylindrical vessel full of liquid.214 171. Weight of a cutiic unit, or specific weight 215 172. Pressure on the base of any vessel full of liquid 216 173. Pressure on any area immersed under a liquid 217 174. Hydrostatic bellows 219 175. Meaning of ' intensity of pressure ; ' variation with depth. 220 176. Level of liquids in communicating vessels 221 177. Centre of pressure 221 ELEMENTARY MECHANICS. XV CHAPTER XII. FLOATING BODIES (Hydrostatics continued}. SECT. PAGE 178. A solid immersed in a liquid displaces its own bulk 224 179. And is forced upward by the weight of this liquid dis- placed 224 180. Loss of weight of a solid in a liquid 226 181. Floating of a light solid 227 182. Definition of specific gravity as relative density 228 183. Methods of obtaining the relative specific gravities of liquids 229 184. Determination of absolute specific weight of sub- stances 231 185. Four methods of obtaining the specific gravity of solids 231 186. Hydrometers of both classes 233 187. Equilibrium of floating bodies as regards rotation 235 188. General conditions for stability Metacentre 238 CHAPTER XIII. ON THE PRESSURE OF THE ATMOSPHERE, AND ON THE PROPERTIES OF GASES (Pneumatics). 189. Many properties of liquids are possessed also by gases... .242 190. Intensity of pressure of the atmosphere 243 191. Modes of removing air from vessels Pumps 244 192. Modes of measuring the pressure of the air Barometers 246 193. Pressure gauges in general 249 194. Modes of raising water House-pumps 250 195. Pressure on the piston of a pump 252 196. Modes of lowering water Pipette ; Siphon 252 197. Buoyant power of air Specific weight of air Bal- loons 254 Properties peculiar to Gases. 198. Law that product of pressure and volume is constant 257 199. Decrease of atmospheric pressure with elevation 260 200. Imperfection of actually existing gases 260 Xvi CONTENTS. SECT. PAGE 201. Air-pump 260 202. Mercury gauge for pump 262 203. Compressed-air manometers Diving-bells 263 MISCELLANEOUS EXERCISES 267 SPECIMEN EXAMINATION PAPERS 275 ANSWERS TO EXAMPLES 286 ANSWERS TO MISCELLANEOUS EXERCISES 303 APPENDIX 307 C.G.S. SYSTEM OF UNITS... ,...308 ^m^^v* r OFTFiF 8 UNIVERSITY OF ELEMENTARY MECHANICS. INTRODUCTION. ON FORCE. 1. Physics is the comprehensive science which deals with the general relation and properties of the three funda- mental facts or phenomena Space, Time, and Matter. Mechanics is the foundation of Physics, and deals with the simplest and most direct relations among these same phenomena. (There .are other branches of Natural Philosophy, such as Chemistry, which discusses particularly the properties whereby forms of matter differ from each other, and Astronomy, which deals with the motions and constitutions of large and distant masses of matter.) Metaphysics, on the other hand, attempts to solve problems as to the ultimate nature of the above phenomena, seeking to express them in terms of mind and consciousness, or vice versd ; and it also considers how far the things called space, time, and matter really exist. Physics silently accepts their existence, and seeks to express all their properties and relations in the simplest terms. The science of Mechanics or Dynamics embraces that part of Physics in which this attempt has been so far successful. This is the scientific use of the term Mechanics ; it used to mean chiefly the science of Machines, and this is still a part of the F abject ; usually now this part is specially distinguished as Applied Mechanics, and the more general aspect of the science itself is A 2 ELEMENTARY MECHANICS. called Dynamics, which- {signifies a treatment of the action of the fact or conception which links together the three fundamental phenomena already specified namely, the very important fact or conception of * Force.' 2. By the term Force we are to understand muscular exertion, and whatever else is capable of producing the same effects. Muscular action impeded gives us our primitive idea of force ; our sense of muscular exertion itself is a primary one for which we have special nerves, and it is not resolved into anything simpler. When any inanimate agent pro- duces an effect on bodies exactly similar to that which would be produced by muscular exertion on the part of an animal, it also is said to exert force. Thus, a steam-engine exerts force when propelling a carriage, or pumping water, or turning a mill ; gunpowder exerts force on a cannon-ball during the time the ball is passing from the breech to the muzzle of the gun. But in order that an agent may exert force it must meet with some resistance ; in other words, force is always the mutual action of two bodies against one another, and the amount of the force is precisely equal to the amount of resistance. Thus, a flying meteor or a cannon-ball is not exerting force (except, indeed, on the earth, by reason of the fact of gravitation) unless it meets with some resistance : but if the air rub against it and resist its motion, it will exert a force against the air ; and when it strikes a target, it meets a very great resistance, and therefore exerts a very great force, possibly smashing the target. A running stream exerts very little force unless it meets with an obstacle ; but if you resist its motion with your hand, it will press against your hand ; or if you dip in the vanes of a water-wheel, it may force the wheel round. 3. Stress. Forces, then, always occur in pairs, con- stituting a mutual action, a pulling or pushing between two bodies, and the action (pull or push) of the one on the other INTRODUCTION. 3 is always precisely equal to the reaction (pull or push back) of the other on the one. In other words, action and reaction are equal and opposite. This pair of forces which always go together it is convenient to have a name for, and it is called a stress, It is a tension if the forces are acting away from each other, and a pressure if they are acting to- wards each other. If the stress be directly between two dif- ferent bodies, it is always a pressure ; if between different parts of the same body, as at any section of a rod, it may be either a pressure or a tension, and is in either case called an internal stress. (The effect which a stress produces in an ordinary solid before rupture is called a strain ; see sect. 5.) There are indirect actions between bodies, such as Gravi- tation, Cohesion, Magnetic and Electric Forces, &c., which are not yet thoroughly understood, but which prob- ably arise from internal stresses in some energetic connecting medium which thus exerts equal pressures on both bodies, forcing them towards or away from each other, as the case may be. So far as the bodies are concerned, therefore, these forces may be classed under the head of Pressures, not between the two bodies, but between the medium and each body. When the earth is one body and a stone the other, the gravitational pressure which is driving them together is commonly called the ' weight ' of the stone. The peculiarity of gravitational pressure" is that it acts on every atom of both bodies throughout their entire mass. Where two contiguous particles of the same body are being con- sidered, the pressure which holds them together is called cohesion: and it is the existence of this remarkable molecular stress which permits the possibility of Tension of any kind in material bodies. It is often convenient to isolate one of the components of a stress between a pair of bodies, and to consider only the force acting on one body ; but we only do so by attending to this one and neglecting the other component, which always necessarily 4 ELEMENTARY MECHANICS. exists and is acting on the other body of the pair. Moreover, which of the forces we choose to call the direct action, and which the reaction, is merely a matter of convenience ; but it will be obviously convenient to speak of that component which acts on the piece of matter we are dealing with as the force, or the action of the other piece of matter, while that component which affects this other piece of matter will of course be the reaction of the first piece on it. An absurd puzzle is sometimes made out of the fact that a cart always pulls back the horse with precisely the same force as the horse pulls forward the cart. It is asked, ' How, if that be so, can they ever start ? ' The puzzle can only be felt by those who forget that each force acts on a different body. In the simplest case, there is no equilibrium or balance of forces acting on the cart ; there is only one force acting on the cart namely, the pull of the horse, and this force is quite unbalanced ; the reaction or pull back of the cart does not act on the cart, but on another body altogether namely, the horse. It needs there- fore no knowledge of mechanics to see clearly the way out of this puzzle ; it needs only a little thought and some common-sense. To explain why a horse or a steam-engine is able to exert force at all that is, to explain how the system of horse- and -cart is able to progress is more complicated. We have spoken of force as exerted by matter. Of inert matter this is hardly correct. Matter does not of itself exert force ; it must be set in motion, or have some other form of Energy conferred upon it, before it can exert force. Remembering this, however, we shall do no harm by habitually using the convenient phrase, ' the force exerted by such and such a body.' Even in the case of the indirect actions between two bodies, such as gravitation, &c., above mentioned, it is customary to talk of the force as if it were exerted by one body on the other, although, strictly speaking, the forces are between each body and the con- necting medium. Thus we say that the weight of a stone is due to the pull of the earth upon it ; but, since the action is really a stress in the surrounding medium, it follows that a stone exerts precisely the same force on the earth as the earth exerts on the stone. INTRODUCTION. 5 4. Equilibrium. A book lying on a table is at rest. Why ? Not because no force is acting on it, for the earth is pulling it ; but because another and equal force is also acting on it in the opposite direction namely, the resistance of the table. This is the condition of all bodies at rest near the surface of the earth ; they are subject to two or more forces which neutralise each other as far as motion is concerned, though they do not neutralise each other as regards strain. Indeed, when we wish to produce strain and not motion, we must subject the body to the action of two equal opposite forces ; for example, if you want to tear a piece of paper, or break a string, or stretch a piece of elastic, or crack a nut, it is no use pulling or pushing at one side only ; you must apply a force to both ends or sides that is, you must apply a tension or a pressure ; you must subject the body to an internal stress. Strictly speaking, motion appears to be the normal condition of matter at present ; all known bodies are moving through space with considerable speed, and no such thing as absolute rest is known. We can, therefore, only consider the motion of liodies relative to some body regarded for the time being as fixed. It is generally convenient, in mechanics proper, to consider the earth as a body at rest, and to leave the study of the motion of it, and of the group of bodies to which it belongs, to Astronomy, which is really a branch of mechanics in a wide sense. It may sometimes be convenient to consider the earth to be moving through space in any direction desired. It is also often advantageous to consider the motions of two bodies, or parts of a body, as if one of the two were fixed, although both or either may be moving relatively to the earth. For example, on board ship, one generally considers the motions of the various people and movables with regard to the ship, and not with regard to the water or dry land ; and in applied mechanics the study of the relative motions of the various parts of a machine is very important. 5. The effects of force on matter are :* * Whatever other effects of force there may appear to be, are studied under Physics, and physicists are hoping to reduce all of them ultimately to the above two forms. Hence Physics is constantly tending to become more and more mechanical. 6 ELEMENTARY MECHANICS. A. Change of motion, which is called Acceleration. B. Change of size or shape, which is called Strain or deformation. If only one force acts on a body, it must produce the effect A, and it may produce B also. If two or more forces act in different directions on different parts of a body which is not absolutely stiff and rigid, they must produce B, and they may produce A also. 6. The two kinds of effect, A and B, are distinct; and each would furnish a measure of force. A force may be measured by the amount of motion it can produce in a given piece of matter in a given time; and this is the measure w r e shall mostly use. Or a force may be measured by the amount of strain it can produce in a certain piece of matter : the amount it can bend a certain spring, for instance, as in a dynamometer; or the amount it can twist a certain wire, as in a torsion or spiral spring balance. If we are not concerned with measuring forces absolutely, but merely wish to compare two forces, we may of course simply balance them one against the other, as is done in a balance or steelyard. Of the two classes of effect, A and B, A is much the simpler, and constitutes the branch of mechanics of which a portion is studied in an elementary course ; it is the only branch suited to elementary exposition ^ch as the present. But before proceeding to our -actual subject, the motive effect of force (called Dynamics, from dfoa/Ms, force), it is convenient to study motion itself a little in the abstract, and without reference to either force or matter. (The subject of abstract motion is called Kinematics, from KIV-/HM, motion.) We may conceive a geometrical point or surface moving about in all sorts of ways without troubling ourselves with the cause of the motion, and the propositions which we so discover will bs useful when we come to the motion of an actual piece of matter under the influence of a force. CHAP. I.'] RECTILINEAR MOTION. CHAPTEE I. ON MOTION (Kinematics). I. MOTION OF A POINT (TRANSLATION). (a) Rectilinear Motion. 7. A body is said to move when it is in different positions ' at different times. This is to be regarded as the essential characteristic of motion it involves a reference to both space and time. Geometry deals with space alone. Kine- matics deals with both time and space. Now motion has two primary properties to be studied Speed and Direction ; both of which are sometimes held to be included under the one name, Telocity. Let us take them in order. When a body moves over equal spaces in equal times, its motion is said to be uniform, or its speed is said to be con- stant* For instance, the tip of the hand of a clock has such a motion as regards speed, in spite of the fact that its direction of motion is constantly changing. The apparent motion of a fixed star across the field of a telescope is another instance of uniform motion. When a body moves over zmequal spaces in equal times, its velocity is said to be variable. As an example of variable velocity, we may take the case of a falling stone, which moves quicker and quicker as it descends ; or of a stone thrown upward, which has a decreasing velocity till it * It is probable that our idea of motion (that is, of free muscular action) pre- cedes and suggests our idea of time ; and that our notion of equal intervals of time depends on our recognition of uniform motion. Every measurer of time is simply a uniformly moving body. The most uniformly moving body we know is the earth, which rotates on its axis in a period of always the same duration ; this period is taken as our fundamental unit of time, and the ^ 5 ^th part of it is called a second of ' mean solar time,' and is used as the practical unit. 8 ELEMENTARY MECHANICS. [SECT. 7. reaches its highest point; or of the bob of a pendulum, which has a velocity alternately increasing and decreasing, as well as changing in direction. To begin with, we shall consider motion in a straight line only that is, with con- stant direction. 8. Velocity is denned as the rate of motion of a body. When uniform, it is measured by the distance travelled, divided by the time taken in the journey ; when variable, its average value is measured in the same way. Thus, if a point move over a distance s in a time t, its velocity is s/t ; orV = * 6 For example, if a train goes 80 miles in 4 hours, its average speed is 20 miles an hour. This expression may be put in the form of a fraction, and may be written 80 miles 20 miles _nr\ 4 hours 1 hour 1 hour ' the last fraction denoting a velocity of 1 mile per hour. Similarly, - - , or, shortly, -, denotes a speed of 1 foot 1 second sec. per second, which we may consider as a sort of standard British unit of speed, suited to the majority of problems with which we shall have to deal. The speed of the above train may if we please be reduced to feet per second, as follows : 80 miles 80 x 1760 x 3 feet 88 , r^i - =- - = feet per second. 4 hours 4 x 60 x 60 seconds 3 The mode of dealing with units indicated at full length in this extremely simple example will be found of con- siderable service in more complex cases. Note that a velocity is length per time, and that it is not correct to speak of a velocity of so many feet. We speak of a length of so many feet or a time of so many seconds but a velocity of so many feet per second. CHAP. I.] RECTILINEAR MOTION. 9 r .. r- T , r- unit of length ,-, . . The unit of velocity is of course ; that is, unit of time 1 foot 1 centimetre / i n r n ,- or .. (read 1 foot, or 1 centimetre, per 1 second 1 second second). Strictly speaking, it is incorrect to speak of a velocity 6 simply, but it is sometimes done when the par- ticular unit of velocity is specified by the context. 9. The above measure of velocity as the ratio of s to t is independent of the size of s and /, so that it remains per- fectly true when s and t are very small. Thus in the case of a body moving uniformly 6 feet every second, its velocity may be written either - or - or 1222 and any of these 1 TT FOW fractions represents its velocity equally well so long as it be uniform. But if the velocity were variable, the body might still go 6 feet in a second, so that its average velocity would still be 6 ; but its actual velocity at each instant might take all kinds of values, some greater and some less. Thus a train which had gone from London to York, 200 miles, in 5 hours, would have had an average speed of 40 miles an hour; but its actual speed would have varied greatly; some- times rising to 60 perhaps, sometimes falling to 0, as at a station. The whole distance travelled, divided by the whole time taken, will always give us the average velocity for that distance; and in the case of uniform motion, the average velocity coincides with the actual velocity at each instant. But to get information on the actual velocity, at any one place, of a thing whose speed varies continually, it is neces- sary to suppose a small distance taken at that place, and divided by the time taken to traverse it. The smaller the distance taken, the less possibility is there of variation, and the more exact will the specification be ; hence the actual velocity of any moving body at a given instant is the infinitely small distance then being described divided by the infinitely small time required for the purpose. 10 ELEMENTARY MECHANICS. [SECT. 9. (The facts are often expressed in a form which appears more simple, but which involves less important ideas namely : Uniform velocity is measured by the space de- scribed in unit time. Variable velocity, by the space which would be described in a unit of time if at the given moment the velocity were to cease to vary.) So then, using little v to stand for actual velocity at any instant, v '- is true when s and t are small ; but, using big t V to stand for average velocity throughout any time, V = - T is always true unconditionally. EXAMPLES I. (1) A train is travelling at the rate of 25 miles an hour. How long will it take to travel between two telegraph poles 100 yards apart ? There are usually about 24 telegraph poles to the mile, hence the speed of a train may be roughly estimated by a traveller. It may be shown in fact that in that case the speed of the train, in miles per hour, is 150 divided by the number of seconds taken to travel between two poles. Because if x in. _1 tel. pole 1 hour"" s seconds ' it follows that x _ Ihour x 1 tel. pole_3600_150 ~s seconds 1 mile ~24s~ s Another way of putting the result is to say that the speed, in miles an hour, is 2 times the number of telegraph poles passed per minute. (2) A man walking from A to B at 3^ miles an hour arrives at B in 1 hour 35 minutes. A cyclist starting from A an hour later arrives at B at the same time ; at what rate per hour was he travelling ? (3) If 100 inches =254 centimetres, how many centimetres per second is equivalent to 3 miles an hour ? (4) A man walks at the rate of 2 yards a second. What is his speed in miles per hour ? The result is a convenient fact to remember : 2 yards a second is 4 miles an hour, roughly. (5) How many feet does the tip of the minute hand of a clock travel in 24 hours if it is 4 feet long ? How much does it CHAP. I.] VELOCITY. 11 move for each complete swing of its pendulum if it ticks 40 to the minute ? (6) If a snail crawl at the rate of | inch a second, how far will it go in an hour ? (7) How long would a train take to go 100 yards at the rate of 20 miles an hour ? (8) With what velocity must I walk in order to go half a mile in five minutes ? ANSWERS TO I. (Worked in full to show the mode of dealing with units.) 3x30 J seconds^ second. 11 3i miles __ {2} $=-=-, x 9o minutes 1 hour 1 hour 19 55 mileg hour = 9i miles per hour. (3) . 3 miles _3 x 1760 x 36 inches 5280 i hour ~ 60 x 60 seconds 3 x 1760 x ^ 254 centimetres 134 2 1120 =^= 134 centimetres per second. 90 * " = >TV)?M TT if* ( ' 22 ' 10. Acceleration. The rate of change of velocity is called acceleration. Velocity may change in magnitude and 12 ELEMENTARY MECHANICS. [SECT. 10. in direction, and the rate of either change is called accelera- tion. When uniform, the acceleration in any direction is measured by the velocity gained in that direction in a certain time, divided by the time taken to gain it. When variable, its average value is measured in the same way. Thus, if a falling body acquire a velocity of 96 feet per second in three seconds, its average acceleration is said to be ^ 6 - or 32 units of velocity per second. Hence acceleration bears the same relation to velocity as velocity did to distance ; and denoting it by a, we have "f as the algebraic statement of the measure of acceleration ; remembering that v stands for the velocity gained by the body in the time t, and need not stand for any velocity actu- ally possessed by the body. Thus the above falling body, instead of simply falling from rest, might have been thrown Nlown from a balloon with an initial velocity of 100 feet a second ; but if at the end of three seconds its velocity were 196, then its gain of velocity would be precisely the same as before, and its acceleration therefore still -^- or 32. Hence, generally, v may be said to stand for the difference between the final and the initial velocities, which are conveniently denoted by v l and v respectively, so that v = v l v^ and _v l -v 11. When a body moving in a straight line acquires equal increments of velocity in equal intervals of time, its acceleration is said to be constant : for instance, a falling stone has constant acceleration ; its velocity uniformly in- creases. It gains in fact a velocity 32 feet per second during every second of its motion. In all that follows, the acceleration is supposed to be constant, unless it is otherwise stated. CHAP. I.] ACCELERATION. 13 Note that acceleration is velocity per time, and that it is absurd to speak of an. acceleration of so many feet, or even of an acceleration of so many feet per second, for this last is a velocity. An acceleration may properly be specified as so many feet-per-second per second. The expression for acceler- ation can be put in a fractional form ; thus the acceleration 96 96 feet per sec. ^_sec. = 32 ft/sec.. which ^ ^ com . 3 sec. 3 sec. sec. monly treated like an ordinary fraction, and briefly written -. Vo j though it must be admitted that the notion of a (sec.) 2 squared second is absurd. There is, however, no principle involved in this mode of writing ; it is merely an abbrevi- ation, and it must always be interpreted as above. The unit of acceleration appropriate to the C.G.S. system of 1 centim./sec. cm. units (see page 308) is - 1 gec> - or y^-y 2 ; and the acceleration already expressed above in British units trans- lates itself easily into C.G.S. units, by the knowledge that a foot equals 30*48 centimetres, thus : ft. 30 '48 centim. centim. 32 7 r 9 = 32-x 7 T2 = 975-7 ro-- (sec.) 2 (sec.) 2 (sec.) 2 Similarly, 981 centim. -second units equal 32-2 foot-second units very nearly. 12. If then the velocity of a body increases, its accelera- tion is the gain of velocity in each second of time ; but if its velocity decreases, then the acceleration is really a retardation, and it must be reckoned negative, but as numerically equal to the loss of velocity in each second. Thus, suppose that in 3 seconds the velocity of a body changes from 196 to 100, its acceleration is -32. If the velocity of a body is constant, then of course its acceleration (or rate of change of velocity) is zero. The Use of the Negative Sign. It is a well-known method to 14 ELEMENTARY MECHANICS. [SECT. 12. distinguish between opposite directions by opposite signs. Thus, if all distances measured to the right of any point be reckoned positive, any distance to the left will be negative, so that - 30 feet will mean 30 feet to the left. It is usual to reckon distances up as positive, and hence distances down as negative. The same may be extended to velocities, and a velocity upward may be called a positive velocity, a velocity downward a negative one. Thus, the velocity of a falling stone may be called negative, and it is continually getting numerically greater (though algebraically less) : so the accelera- tion produced by gravity ought on the same convention to be called negative, because it is negative velocity which is added by it every second. In fact, an increasing negative quantity corre- sponds in algebra to a decreasing positive one, and vice versa. EXAMPLES II. * (1) A body starts from rest and acquires a velocity of 600 feet per second in half a minute. What is its acceleration ? (2) A body starts with a velocity 50 feet per second, and in 6^ seconds has acquired the velocity 102 feet per second. What is its acceleration ? (3) A body moves with acceleration 32 ft./(sec.) 2 , starting with a velocity of 20 feet per second. What is its velocity in 1 , 2, 3, 6 seconds respectively ? (4) A train acquires a velocity of 20 miles per hour 5 minutes after leaving the station. What was its average accelera- tion during this time ? (5) How many miles an hour per hour is 32 feet a second per second ? (6) A train going 40 miles an hour is brought up in 40 seconds by the brakes. What is the rate of retardation in feet-per- second per second ? * All these are merely profit and loss questions. Velocity corresponds to capital, and acceleration to rate of gain. Thus Question 3 may be paraphrased thus : ' A man starts in business with 20, and gains 32 every year. How much has he got in 1, 2, 3, 6 years respectively?' And No. 8 thus : ' A man starts with 128, and loses 32 annually. How soon will he have lost all? and what will he have in 1, 3, 5, 7 years?' Obviously he will have lost all in four years, and in seven years he will be 96 in debt/ A less simple kind of question is one that involves distance; for some examples, see Ex. IV. p. 29. CHAP. I.] CURVILINEAR MOTION. 15 (7) A body starting with velocity 100 feet per second has only a velocity 52 in 4 seconds. What is its acceleration ? (8) A body with acceleration - 32 foot-second units starts with velocity 128. How soon is its velocity zero ? and what is its velocity after 1, 3, 5, 7 seconds respectively? (9) A body dropped from a stationary balloon falls with ac- celeration 32, and hits the ground with a velocity 512, the units being feet and seconds. How long was it in falling? (10) The acceleration of a moving point expressed in terms of centimetres and seconds is 200. Explain exactly what this means. Find what number expresses the same accel- eration in terms of metres and half-seconds. (11) The acceleration of a falling body is 981 when referred to centimetres and seconds. If 1 foot were used as unit of length, and 1 minute as unit of time, how would this acceleration be represented ? 1 foot=30'48 centimetres, (12) A train starts from rest, and 24 seconds later is moving at 240 yards per minute. Find its average acceleration. (13) What acceleration is needed in order to get up a speed of 60 miles an hour in 2 minutes ? (14) What rate of retardation will destroy this motion in 5 seconds ? (b) Curvilinear Motion of a Point. 13. Besides change in the magnitude of velocity or rate of motion, there is Another thing to be considered namely, change in its direction. Hitherto we have only considered motion in a constant direction 'that is, in a straight line ; but when the direction of a point's motion is constantly changing, the path described is a curved line, or the motion is curvilinear. The rate of change of direction per unit length of a curve is called its curvature ; and this again may be constant or variable. Most curves (the parabola, sect. 29, for instance) have variable curvature. A circle or helix has constant curvature. A straight line possesses zero curvatoe. The curvature of a circle is inversely propor- tional to its linear dimensions ; because the angle which the 16 ELEMENTARY MECHANICS. [SECT. 13. direction of motion turns through in going once round any circle is four right angles, which in circular measure is 2?r (see sect. 14), and the curvature will be this angle divided by the distance travelled that is, by the circumference, 2irr ; hence the curvature of a circle is numerically equal to the reciprocal of the radius, for = = curvature of a circle. 27T7* T And the curvature at any point of any other curve is denned on the strength of this, as the reciprocal of the radius of that circle which coincides most closely with the curve at the point. 14. The circular measure of an angle at the centre of a circle is obtained by measuring or estimating the arc subtended by the angle and dividing this arc by the radius. The value of this ratio, arc -f- radius, is independent of the size of the circle, and depends only on the angle to be measured. The angle subtended by an arc which equals the radius would, on this system of measurement, be denoted by 1, and is called a radian. It is about 57 17' 45", being equal to 360 -j- 2ir. The circular measure of any other angle is equal to the number of radians it contains. Four right angles, expressed in circular measure, circumference 2irr = = = JTT, radius r where TT denotes the ratio of the circumference of the circle to its diameter, and is approximately equal to 3y, or, more nearly, 3-1416. Hence, 4 right angles = 2ir radians. We can now more fully state what is meant by curvature. For, as we have seen, the curvature of a circle equals angle turned through in going round the circle 4 right angles distance travelled circumference 2-n- radians 1 radian CHAP. I.] CURVATURE. 17 That is, the angle turned through by the direction of motion is 1 radian for each portion of the circumference travelled whose length equals the radius of the circle. For example, if the radius is 4 feet, a radian is turned through for each 4 feet of arc travelled, or J radian for each foot, so that the curvature of this particular circle is J radian per foot of arc travelled. The numerical value of the curvature is - when r' the angles are measured in radians, but this expression must be considered as an abbreviation for 1 radian per arc-equal- to-radius, which finally in any given case becomes such and such a fraction of a radian per foot, or per inch, or per whatever unit of length is used. 15. A point moving in a curve, besides any acceleration it may have along the curve increasing its velocity, pos- sesses an acceleration at right angles to the curve, or normal to the direction of its motion ; this acceleration being pro- portional to the curvature of the curve, and affecting only the direction and not the magnitude of the velocity. Its magnitude is the rate at which velocity normal to the curve is gained by the point. This normal acceleration is called centripetal acceleration, and is further discussed in sects. 58- 61, where it will be founcl to be proportional to the square of the velocity of the point as well as to the curvature of the curve ; to be equal, in fact, to v 2 x Although the point is always gaining velocity normal to the curve or along its radius at this rate, it does not follow that it ever possesses any such velocity. It is in fact im- possible for a point to possess any velocity except that along the curve, or at right angles to the radius of curvature ; for as fast as velocity along the radius is generated, so fast does the direction of the radius change ; in the same sort of way that a promise for to-morrow need never be fulfilled, because 'to-morrow never comes.' B 18 ELEMENTARY MECHANICS. [SECT. 16. II. MOTION OF AN EXTENDED BODY (ROTATION). 16. A point can only move along, it cannot spin; OT rather, spinning makes no difference whatever to it or to its motion : but an extended body, whether it be a line, surface, or solid, may not only move bodily along or be translated ; it may also turn round or rotate. The most general motion of an extended body is a combination of translatio'n and rotation, but it is simpler to consider them separately. All that we have said about the motion of a point is equally- true of the motion of an extended body so far as its trans- lation is concerned ; because, in simple translation, if we know the motion of any single point, we know that of the whole. Its rotation involves different ideas, which must now be considered briefly. 17. When a body rotates, every point of it describes a circle round s"ome point or line which is the centre or axis of rotation. The velocity of a point far from the axis is greater than that of a point nearer the axis ; and in general every point has its own velocity, which is proportional to its distance from the axis, only points at the same distance having the same velocity ; hence the ' yelocity of a rotating body ' is a meaningless expression. The number of times the body turns round in a second, however, is perfectly characteristic, and we must define some kind of rotational or angular velocity proportional to this. To express the speed with which a body rotates, it is sufficient to specify the velocity of any one point together with its distance from the axis; for the velocity of the point, divided by its distance from the axis, is a constant quan- tity that is, is the same for all points of the body at each instant, and is called the angular velocity of the rotating body. The velocity of every particle of the body is known in terms of this, for, being proportional to its distance from OFTHF { UNIVERSITY CHAP. I.] '. ANGULAR VELOClW. 19 the axis, it is equal to the 'angular velocity' multiplied by this distance ; or, denoting the angular velocity by the letter w, as is customary, the velocity of any particle at a distance r is v = rco. Angular velocity in rotations takes the place of ordinary velocity in translations. The name 'angular velocity' is given because it really represents the angle (expressed in circular measure, or radians) turned through per second by the whole body. An example may render this more clear. The circle described by a particle at a distance r from the axis of a rotating body (say a nail on the circumference of a fly-wheel of r feet radius) is 2r feet in diameter, and hence 2irr feet in circumference. . If the wheel turn round in T seconds, the velocity of the nail is -- feet per second ; hence the angular velocity of the wheel is - radians per second, which is w. The most generally useful specification of angular velocity is in radians per second, as above, but it is usually first measured in revolutions per minute; and so a rapid mode of conversion from one measure to the other is often needed. Now, 1 revolution per minute equals radians per second, and a good approximation to this number is yV+slJo-; which may be further improved by deducting -J- per cent. from the result. Thus, to convert 1264 revolutions a minute into radians per second, the work is as follows : 126 -4 + 6 -32 =132 72, or, deducting J per cent, of this, 132 - 39 radians per second. (A more exact value is 132 -37.) To perform the inverse operation, multiply by 10, and subtract 4^ per cent, of the result. 18. Of course angular velocity may be uniform or vari- 20 ELEMENTARY MECHANICS. [SECT. 18. able ; and if the latter, its rate of change, or increase per second, is called the angular acceleration of the body. Denoting this by a, we have a= - (just as we had a = in t t sect. 10). But to = -. Hence a = = -L. = _ that is, the r rt r r angular acceleration of a body is the acceleration of any particle divided by its distance from the axis. In other words, angular acceleration : acceleration : : angular velo- city : velocity : : angle turned : distance travelled ::!:?*. EXAMPLES III. (1) What is the curvature of a circle 14f yards in circumference ? It is numerically equal to the reciprocal of the radius in feet that is, 2 - = - of a radian per foot nearly ; or your compass-bearings change about 8 for every foot you travel round such a circle. (2) A point moves in the above circle with a constant velocity of 6 feet a second. What is its acceleration in magnitude and direction ? Its acceleration is always along that radius of the circle which passes through the moving point, and its magnitude is -\ 6 . (3) A point moving in a circle 8 feet in diameter has a velocity increasing by 18 every 3 seconds. What is the acceleration in magnitude and direction at different times ? There is a constant tangential acceleration equal to 6. The normal acceleration is zero at starting ; at the end of the first second of motion it is 3 ? 6 = 9 ; in two seconds it is 36 ; in three seconds, 81 ; and in t seconds it is i(6) 2 = 9 2 - The actual acceleration at any instant is the square root of the sum of the squares of the tangential and normal ac- celerations at that instant ; hence its direction, which at first is tan- gential, gradually swings round, so that in a few seconds it nearly coin- cides with the radius. This explains what happens when we whirl a stone at the end of a string : it is necessary to start it with some purely tangential ac- celeration, obtained either by the help of gravity, or by a tangential push or pull. When once started, however, the speed may be increased to any extent by simply pulling the string a little to one side of the centre of the circle of motion, so that the tension Th the string has both a tangential and radial component; and since the faster the stone is going, the smaller need the former be in comparison with the latter, it follows, that at a high speed the hand remains very nearly CHAP. I.] ANGULAR VELOCITY. 21 steady in the centre of the circle ; but it is really travelling round a small circle about a quadrant in advance of the stone thus supplying the tangential force necessary to overcome the resistance of the air, even if the motion is not being accelerated. Verify all this experimentally whirling a weight in a horizontal circle on a flat table, in order to simplify matters by eliminating gravity. (4) If the latitude that is, the elevation of the N celestial pole changes by 3 degrees while travelling a distance of 208 miles due north from Greenwich, what is the circum- ference of the earth ? (5) What is the angular acceleration of the moving point in No. 3? (6) If a point describe a circle 5 feet in radius with an angular acceleration of 2 radians-per- second per second, what is its [linear] velocity at the end of 5 minutes from rest, and how many revolutions will it continue to make per minute if the acceleration then ceases ? (7) A wheel makes 20 revolutions per minute. What is its angular velocity in radians per second ? (8) A wheel possesses an angular velocity of 2 radians per second. How many revolutions per minute does it make ? (9) A point in the rim of a revolving wheel whose radius is 5 feet moves with a velocity of 6 feet per second. Find the angular velocity of the wheel, (1) in radians per second, (2) in revolutions per minute. (10) A dogcart is travelling 16 miles an hour, and its wheels are 5 feet high. Find their angular velocity. Find also the speed Avith which the tire of the wheel is passing the elbow of the driver. (11) A humming top, 6 inches in diameter, is started by a string of which 1 yard is wrapped round a spindle |-inch thick, and pulled off by a steady pull in 3 seconds. Find the initial angular velocity of the top, and the speed of its humming aperture. (12) The wheel above the shaft of a coal-pit is 8 feet in diameter, and makes 90 revolutions per minute while letting down the cage. What is the speed of descent of the cage? Find also the angular speed of the drum in the engine-house, off which the rope is being unwound, if its diameter is 18 inches. 22 ELEMENTARY MECHANICS. [SECT. 19. CHAPTER II. CONTINUATION OF THE SUBJECT OF RECTILINEAR MOTION. DISCUSSION OP THE STATEMENTS MADE IN CHAPTER I. 19. We have now obtained two definite statements, each of the nature of a definition namely : Average velocity = r distance travelled or y s ^ time taken in the journey t velocity gained v average acceleration = -7--, *rir rr- ora = r . time taken in the acquisition t And we can proceed to reason on them, and trace their logical consequences, which will all be certainly consistent. The treatment is very simple in the case of uniform accelera- tion. We desire to find the distance travelled during any interval of time when the initial velocity and the acceleration are given that is, when the velocity at each moment is known. To calculate this distance s, we have to find the average velocity V, and then the first formula above gives us s=Vt. In the case of uniform acceleration, the average velocity is the arithmetic mean of the initial and final velocities. This can, perhaps, best be shown by an example. Thus, if the initial velocity had been 7 feet per second, and the final velocity, in t seconds, had been 19 feet per second, the average velocity according to the rule just stated would be (7 + 19) = 13 feet per second, and the distance travelled would be I3t feet. Now suppose the t seconds to be divided up into three equal intervals, the velocities > at the beginning and end of each interval would be 7 and 11, 11 and 15, 15 and 19 respectively, since the velocity increases uniformly, and therefore, applying the rule, the average velocities during the intervals are |(7 + 11 ), i(ll + 15), $(15 + 19) that is, 9, 13, and 17 feet per second respectively, and the duration of each interval is, by supposition, \t seconds ; . . the distance travelled is $(9 + 13 + 17) =\t x 39 = 13* feet as before. Hence the rule gives CHAP. II.] RECTILINEAR MOTION. 23 us the same value of s at the end of the t seconds however \ve choose to divide it up. This would not have been true if the acceleration had not been uniform ; as may be seen by repeating the process with different intermediate values, say for the set 7, 9, 13, 19. However, in all cases we shall have to deal with for the present, the velocity will increase regularly (i.e. the acceleration will be constant), consequently the average velocity is obtained at once by halving the sum of the initial and final velocities, V= 1 9 ; using v to stand for initial, and v l for final velocity (sect. 10). Hence our first equation, which may be put into the form s = Vt, may be written more fully thus : which signifies that in cases of uniform acceleration the, distance travelled is equal to the half sum of initial and final velocities multiplied by the duration of the motion. Similarly the second equation may be written in the form v = at, or more fully, v 1 -v Q = at, which signifies that the excess of the final velocity over the initial velocity is equal to the gain per second multiplied by the duration of the constant rate of gain. In case the final velocity is less than the initial, the gain becomes a loss, or v^ - v is negative, and therefore also a is negative that is, it is really a retardation, but it may still be called an acceleration, only a negative one. 20. Now let us study the two equations together, and see what we can get from them by any algebraical operation ; remembering that algebra, like all other reasoning, never gives us anything absolutely and essentially fresh ; it only brings out explicitly what is already contained implicitly in the physical statements which we subject to reasoning. The physical statements must be the results of the observation of nature, which is the only way of arriving at fundament- 24 ELEMENTARY MECHANICS. [SECT. 20. / ally new truths. Mathematical reasoning will, however, 1 serve to bring out and make manifest what is really in- volved in the statements themselves when put together, if only we had sufficient insight to perceive it. Our two statements or equations, written out fully, are v i + V QJ. v i - v a s = 1 % and a = -~T--- 2i t First multiply the two left-hand members together and double the product, then do the same with the two right- hand members, then write the two products equal to each other (as of course they must be), and you get the new equation, . 205 = (V 1 + V Q ) (Wj - V Q ) = V* - V Q \ This is a relation between a, s, and v, without explicit refer- ence to time, and it will often be useful. Now try again, and this time get a statement not involv- ing v v which we can do by substituting in the first equation the value of v 1 obtained from the second equation namely, and we get s = v Q t + Similarly we can get a relation excluding v Qt and it is 21. But before proceeding to study the two equations together, we might have first made a simplification. An obvious simplification would occur if the initial velocity were made zero (v = 0) ; in other words, if we agreed to consider only bodies starting from rest. In this case the gain of velocity v is equal to the final, v v and the average velocity V is equal to %v v which is now the same as %v ; and so the two fundamental equations reduce to v s = vt, and a = -7 ; and the three derived from them simplify in like manner. CHAP. If.] CONSTANT ACCELERATION. 25 We thus obtain the following four equations between the distance travelled by a body from rest, the time taken in the journey, the acceleration, and the final velocity gained ; v = at. s = \vt. s = Of these, any two are independent statements, and the other two are logical consequences of them. The first of the four reads thus : The velocity gained in t seconds equals t times the velocity gained in each second. The second one thus : The distance travelled over in t seconds equals t times the average distance travelled over in one second (for this last is the meaning of average velocity). Both these statements are perfectly obvious. The other two statements cannot be put in quite so obvious a form. Observe that there are only four quantities involved, s, v, a, t, and that one of them is absent from each of the four equations. 22. The meaning of the second derived equation in sect. 20 is now clear. The space described by a body with the constant velocity v is v t, and by one with the uniform acceleration a is ^at 2 ; so the whole space described by the body possessing the initial velocity v^ and also subject to the acceleration , is This may be regarded as a case of the composition of motions in the same direction. See sects. 24 and 73. 23. The results expressed by these equations may be -made to appeal to the eye more directly, and thus be rendered easier to grasp, if illustrated by their analogy with geometrical diagrams. If a horizontal line be considered as representing by its length a definite lapse of time that is, if it be divided into a number of equal parts, each part representing say one * If a and VQ are of opposite sign, the subtraction is to be performed when the letters are arithmetically interpreted. The sign + means algebraical addition, which includes subtraction. 26 ELEMENTARY MECHANICS. [SECT. 23. second ; and if a vertical line represent by its length a certain velocity, by being divided into a number of equal parts, each part representing 1 foot-per-second ; then if a body move with that velocity for that time, the distance travelled will be represented by the area of the rectangle contained by these two lines : that is to say, the number of feet travelled will be equal to the number of unit rectangles in the area, the height of each of which represents a foot- per-second, and the breadth of each a second. In a similar way a diagram can be made giving the velocity at each instant, and the distance travelled by a point moving in an accelerated or retarded manner. Fig.l. Thus in fig. 1, OT is the line of time, with the seconds marked off upon it. OA is a vertical line, and represents a velocity, say of 12 feet a second. If a body moved with this constant speed for 8 seconds, the distance travelled could be represented by the area of a rectangle constructed with base OT and height OA, because this area would be 12 x 8 = 96 appropriate units of area, and the distance travelled would be 96 feet. Let a body start with this velocity 12, and lose 2 of it every second, then in 1 second its velocity will be repre- sented by the line cl, in 2 seconds by the length, of the line d2, and so on. Consequently, in 6 seconds the body will be at rest. The diagram thus represents, in a conventional and utterly non-pictorial fashion, a body starting with initial CHAP. II.] GEOMETRICAL REPRESENTATION. 27 velocity 12, and going with a uniform negative acceleration - 2, till it stops. The average velocity would be 6, and would be represented by the length of the vertical line drawn in the middle of the time -namely, e3. The distance traversed would be ;this average velocity multiplied by the time. That is, geometrically, e3 multi- plied by OP, which is the area of the triangle GAP ; for the area of a triangle is equal to the product of base and average height in other words, to the product of half its height into its base. Areas then in this figure represent distances. Or, more correctly, the number of units of area in one of these figures equals the number of linear units in the distance travelled. On this scale the area OAcl represents the distance travelled in the first second; Icd2, that travelled in the second second; 5#P, that travelled in the last second. The distance travelled in the three seconds between the first and fourth is represented by the area lc/4, and so on that is, each of these distances is equal to the number of units of area con- tained in the respective spaces. (Observe that the vertical scale and the horizontal scale in these, as in so many other diagrams, are quite independent of each other. The appro- priate unit of area is therefore not necessarily a unit square, but a rectangle with unit sides.) Fig. 2. The representation of a body starting from rest with a positive acceleration is given in fig. 2. 28 ELEMENTARY MECHANICS. [SECT. 23. The line of time is divided to represent seven seconds. The velocity gained in one second is represented on some convenient scale by the line marked a, which therefore represents the numerical value of the acceleration. The velocity gained in the whole time is marked v\ it is obvi- ously equal to 7a or at. The dotted line in the middle of the time is the average velocity, and it is evi- dently \v. The area of the whole triangle represents the whole dis- tance travelled, and it is half the height multiplied by the base, or ^ v . t, or, what is the same thing, at . t, that is, * at\ The little left-hand triangle is numerically equal to \ a in area (its base being unity), and it represents the distance travelled in the first second. The velocity possessed by the body at any second or fraction of a second is found at once, simply by measuring, and interpreting on the proper scale, the vertical height of the triangle at the place defined by the time. The whole problem is in fact geometrically represented. If the body started with an initial velocity, and then went on with increasing velocity, its motion would be represented by fig. 3, which is supposed to represent what ft happens in three seconds. The initial velocity is marked v , and at the final v 1 ; the latter being made up of two parts, the gain of velocity at, and the original velocity v . v The rest is marked as before, the p base represents t, and the whole area represents the whole distance, s, travelled in the three seconds namely, v t, the rectangle, plus i at 2 , the triangle on the top of it. The dotted line in the middle is of height v + % at, or, what is the same thing, v 1 \ at; and therefore it is CHAP. II.] RECTILINEAR MOTION. 29 ^(VQ + V^, or the average velocity. Any one of these three expressions for average velocity when multiplied by the time will give the distance travelled (cf. equations of sect. 20). - If the initial velocity be negative, the line representing it must be drawn down from the line of time, instead of up. These diagrams will be found exceedingly useful and conducive to clear ideas, as soon as a little practice has made you familiar with them. For some more illustrations of their use, see sect. 69, which can be read now ; for the commonest example of uniform acceleration is that caused by the earth's attraction, which causes all freely falling bodies to acquire a speed of 32 feet a second in every second of their fall in other words, which causes a uniform acceleration of 32 feet a second per second. EXAMPLES IV. (1) In Examples II., 1-9, find the distance travelled by the different tilings in the times given. (2) A body starting from rest, and travelling 63 feet in a straight line, gains a velocity of 81 feet per second. What is its acceleration ? (The most direct way to get the Answer is to use the formula v z =2as.) (3) What is the acceleration of a body whose velocity changes from 7 to 21 while it travels 100 feet ? ( Use the formula N.B. The arithmetic is often simplified by taking the difference of two squares in. the form of the product of sum and difference. Questions (2) and (3) may also be solved by dividing the velocity gained by the time taken in gaining it; the time being found by dividing the distance travelled by the average velocity. (4) Find the accelerations of the following bodies : A, whose velocity changes from 15 to 5 in going 50 ft. B, ,. it 15 to -5 M 50ft. C, M .. -5 to 15 50ft. B, ii n 120 to n 640 ft. (Remember that the square of a negative number is positive. ) The extreme distance from the starting-point attained by B is 56J feet, but 6i feet of this is retraced. It therefore takes longer in the journey than A did, but its acceleration happens to be the same. , 30 ELEMENTARY MECHANICS. [EXS. IV. Similarly with C, the first thing it does is to go 6 feet backwards and come to rest for an instant ; then it retraces its path and goes 50 feet forwards, where the question leaves it ; but it is still going on with a speed increasing by 2 in every second. (5) Find the time of the motion in all these cases, and draw a diagram for the several motions. Begin by drawing the line of time ; then draw verticals for the initial and final velocities, paying attention to sign, and join the extremities of these lines ; then study every part of the diagram, and note its connection with the equations. (6) A train with the brakes on, moving with acceleration - 3, has a velocity 78 when passing a particular station. How much farther will it go? (7) A point moves 16 feet in 1 second and 20 feet in the next. How long has it been moving with uniform acceleration since it started from rest, and what is the rate of the acceleration ? Also, how far would it go in the next 12 seconds of its motion, and when will its velocity be 128 ? (8) A body slackens speed from 50 to 30 feet a second in going 20 yards. Find how soon it will stop if the same rate of retardation continues. (9) The velocity of a train, moving with uniform acceleration, is, at three points A, B, and C, 40, 50, and 60 miles an hour respectively. The distance AB is 27 miles. Find the distance BC. (10) A train going 40 miles an hour is brought up in 200 yards by the brakes to avoid a collision. What is the acceleration in miles-per-hour per second, and also in feet-per-second per second ? (11) What acceleration is needed to get up a speed of 60 miles an hour in a half-mile run ? What is the brake retardation that could destroy this motion in a train-length, say 100 yards? What retardation would a stoppage in 5 yards represent ? The following examples are cases of uniform acceleration under gravity. The acceleration may be taken as 32 feet-per-second per second. The resistance of the air is neglected. (12) A body falls from rest. Find the distance travelled in 5 seconds, and how far it will go in the next second. (13) A falling body describes 100 feet in the last second of its motion. Find how far it must have fallen, and also the time taken. CHAP. II.] CONSTANT ACCELERATION. 31 (14) A bullet is dropped at a place where the intensity of gravity is only 20. How many feet will it have fallen in 4 seconds; and how far will it go in the next second ? (15) Express by means of a diagram, connecting velocity and time, the motion of a body which is thrown up into the air, rises 256 feet, and falls to the earth again. (16) A stone is thrown vertically upward with a velocity of 40 feet a second. How high will it rise ? and how long will it be before it returns to your hand? If you let another stone drop down a well, at the instant the first is within 20 feet of your hand on its return journey, at what distance below your hand will the two bodies meet ? (17) A bullet is dropped from the top of a tower 100 feet high, and at the same instant another bullet is thrown vertically from the bottom of the tower with velocity just sufficient to carry it to the top. Show where and when the bullets will pass each other. (18) If you throw up a cricket ball and catch it after 5 seconds, how high will it go, and with what velocity will it return to you ? Composition of Motions in General. 24. When a body has several motions given to it at the same time, its actual motion is a compromise between them, and the motions are said to be com- pounded, the actual path taken being called the resultant. Thus, suppose a fly to crawl along a tea-tray from A to B (fig. 4), while at the same time some one pushes the tray along a table a distance PQ ; the fly will then have two motions, and its actual motion with reference to the table is the resultant of the two motions, AB and PQ. To find where the fly is at the end of the two motions, we must observe where Fig. 4. 32 ELEMENTARY MECHANICS. [SECT. 24. the point B of the tray has gone to, for the fly has crawled to B; but B has been moved to a point C, such that BC = PQ. Hence the fly is at C, and its actual motion must have been along some path AC Fig. 5. (not necessarily a straight line); and AC is therefore called the resultant of the two motions, AB and BC. If, besides these two, the table itself had been pushed in the direction ST, or what is the same thing, CD, then we should have had three motions to compound; and, as the fly would have got ultimately to D, AD would have been the resultant of the three motions. The order in which the steps are added evidently does not matter, for the same point D is arrived at by taking the table motion Y before that of the tray, as in 1, fig. 5 ; or the fly's proper motion after both the others, as in 2 ; or the fly's motion between the other two; or in any other of the six possible orders in which three motions can be compounded. And so we readily see the rule for compounding any number of motions. Draw lines, or cut pieces of stick,* representing each motion in magnitude, direction, and sense,! and lay these * See footnote to page 135. t That is, make some difference between the two ends of the line, indicating by an arrow-head or otherwise which way the motion takes place in the given direction. CHAP. II.] COMPOSITION OF MOTIONS. 33 lines or sticks in any order, with the end of one coinciding with the beginning of the next (the lines may be moved into any positions, provided each is kept parallel to itself) ; then some line joining the first point of the first with the last point of the last, must be the resultant of the whole set of motions. Thus, some line AG is the resultant of the six motions, AB, EC, CD, DE, EF, FG. This proposition is called the polygon of motions, because the resultant is represented by the line required to complete a polygon. As a matter of fact, however, the sides of the polygon need not necessarily be straight lines. The end points of the line are the only essential matter when one is dealing with simple change of position without regard to time or speed. 25. The composition of two motions, AB, BC, into a third, AC, requires only a three-sided polygon, so it is often called the triangle of motions. Or if we choose to represent the V\^ two component motions, AB, BC, by lines, AB, AB', drawn from the \ same point, we get the parallelo- gram of motions, which is merely a less simple, but sometimes convenient, way of regarding the triangle of motions. The resultant motion is the diagonal of a parallelogram whose two adjacent sides represent the component motions. 26. This law, by which two motions are compounded, is of very frequent occurrence in all parts of mechanics, and is referred to as the parallelogram law. It may be stated thus : If two causes act on a body at once, or if a body experience two simultaneous effects in different directions, then if these effects are represented in magnitude and direction by two adjacent sides of a parallelogram, the effect experienced by the body is called the resultant effect, and is represented, on the same scale, by the concurrent diagonal c 34 ELEMENTARY MECHANICS. [SECT. 26. of the parallelogram that is, the diagonal which passes through the point of intersection of the two sides ; or, it is the same effect as would be produced by a resultant cause represented in magnitude and ' direction by the similarly situated diagonal of a parallelogram whose two sides repre- sent the component causes ; provided always that the causes or the effects can be shown to be of such a nature that this law is applicable to them. Composition of Uniform Velocities. 27. So far we have only studied the composition of changes of position ; now let us study the composition of velocities first, when uniform. Remember that a velocity is numerically equal to the space described in 1 second. Let a body start from (fig. 8) with two velocities, one horizontal and of magnitude Oa, the other vertical and of magnitude Ob. Then Oa and Ob represent the distances travelled in 1 second in the respective directions, and con- sequently at the end of 1 second the body is at the point c, the opposite corner of a parallelogram with sides Oa and Ob ; hence the body must really have travelled the distance Oc in 1 second, therefore Oc is its resultant velocity in magnitude and direction. In 2 seconds it will have travelled horizontally to a', and vertically to &', and therefore it will really have reached c'. And it is easy to see, by drawing or otherwise, Fig 8 that the straight line Oc' passes through the point c, and that Oc' equals twice Oc (because Oa' = twice Oa, and Ob' = twice Ob) ; or the distance travelled in 2 seconds is twice the distance travelled in 1 j and so, generally, the resultant of two uniform velocities is another uniform velocity along the diagonal of the parallelogram whose adjacent sides represent the components. CHAP. II.] COMPOSITION OF VELOCITIES. 35 Hence the resultant of two velocities is obtained by pre- cisely the same parallelogram law as the resultant of two simple motions. Similarly the * polygon ' law is applicable for compounding any number of velocities greater than two. Composition of Uniform Accelerations. 28. Accelerations may evidently be compounded by the same law as velocities, because acceleration is the velocity gained per second. Thus let a body be subject to any two accelerations, say a horizontal one Oa, and a vertical one Ob (fig. 8) ; then Oa and Ob represent the velocities gained per second in these' two directions respectively, and there- fore the actual velocity gained in the time is Oc; in other words, Oc, the diagonal of the parallelo- gram, measures the result- ant acceleration. Hence accelerations are com- pounded by the same law as velocities. 29. Combination of a uniform Velocity witli a Velocity uniformly ac- celerated in a constant ^^ \ direction. Let a body start from O with a uniform velocity u in some direction 0V (fig. 9), and a uniform Fig. 9. acceleration a in some other direction, such as OL vertically downwards ; then in succes- sive seconds the distances traversed in the first direction will be numerically equal to u, 2u, 3u, 4u, &c.; so that, if this constant velocity u were the only one 36 ELEMENTARY MECHANICS. [SECT. 29. .-by the body, the body would be at T after 1 second, at U after 2, at V after 3, and so on (fig. 9.) But the uniform acceleration is acting at the same time, and causing the body to descend a height proportional to the square of the time (| at 2 ) ; hence in successive seconds the vertical distances traversed will be bringing the body to the level N in 1 second, M in 2, L in 3, and so on, if it had acted alone. The actual position of the body, therefore, at the end of successive seconds will be found by completing the parallelograms, OTPN, OUQM, OVRL, &c.; the result being that the body reaches the point P in 1 second, Q in 2, K in 3, and so on. Now the simplest continuous curve which can be drawn through these points, OPQR, &c., is a parabola, and this is the actual path of the body. It will be shown later (sect. 74) that this is the' path of a projectile thrown in vacuo in the given direction with the velocity u, and subject to gravity, which causes a uniform downward acceleration. It is well seen in the curve of a steady jet of water, for each drop of water takes this path. It is also illustrated by Morin's machine (sect. 75). -. This example illustrates the fact, noted at the end of sect. 23, that the resultant motion (the diagonal of the parallelogram) need not be represented by a straight line. It will be straight if the two things compounded are both uniform ; otherwise it will in general be curved. EXAMPLES V. (1) A point has two motions, one east with a uniform velocity 30, the other north with a uniform velocity 40. What is its actual motion ? CHAP. II.] RESOLUTION OF MOTIONS. 37 (2) A boat is rowed at right angles to the banks of 'a- straight river, at a pace half as fast again as the stream flows : it reaches the opposite bank 2 miles below the starting-point. Find the breadth of the river and the distance rowed. (3) A point describes a circle with a constant velocity v, and at the same time the centre of the circle moves forward in a straight line with the same velocity. What is the motion of the point ? N.B. This is the case of a nail on the circumference of a coach- wheel. The point describes a curve with cusps, called a cycloid ; its velocity when at the top of the wheel is 2v, and when on the ground is zero ; its velocity at the extreme right and left points of the wheel is v \/ 2 ; its velocity is v at two points whose distance from the ground is half the radius. Resolution of Motions. 30. The inverse process to that of 'compounding' is called resolving, and is an operation which, in practice, is found extremely useful. "vVe have seen that a pair of motions (or velocities or accelerations), one in a .vertical and the other in a horizontal direction, compound into a single motion in a slant direction, wherefore it follows that, when we choose, we may analyse or resolve a slant motion (or velocity or acceleration) into a pair of com- ponents, one of which may be horizontal while the other is vertical, or both of which may have any definite directions we please provided we affix to each its appropriate magni- tude. All that is essential is that the two components shall be represented by the sides of a parallelogram, of which* the diagonal represents the thing whose resolution is sought. We shall study this process in greater generality hereafter (see sect. 112 and fig. 25) ; for the present, it will be best to illustrate the use of resolution by an example. Suppose a body thrown in a direction slanting upwards at an angle of 45 with a velocity of, say, 141*42 feet a second. This velocity may be considered as the diagonal of a square whose sides each represent 100 feet per second, and the whole motion may be considered as analysed into 38 ELBMENTARV MECHANICS. [SECT. 30. two parts a horizontal part, which continues uniform but for friction ; and a vertical part, which is constantly subject to the downward accelerating force of gravity, which destroys its upward velocity, and generates downward velocity at the rate of 32 feet a second every second. Hence such a projectile will ascend in a curved path for y^ or about 3 seconds, attaining a height of about 150 feet and tra- velling horizontally about 300 feet in the same time ; then its curve will begin to slant down, and ultimately it will arrive at its original level, about 600 feet from its starting-point, after the lapse of about 6 seconds. If the ball were shot in any other direction, its motion could be treated and its velocity resolved in like manner, but its initial vertical and horizontal components would no longer be equal. We may return to this subject in sect. 76. Relative Motions. 31. So far the pair of motions compounded or combined have really belonged to the body under consideration, but there are cases where a practical problem is simplified by the device of attributing to a body motion which it does not really possess. The most frequent example of this device is the assumption that the earth is at rest; for this is attributing to it a motion which it does not really possess (for rest is of course a particular case of motion), and prob- lems are certainly simplified thereby. For instance, when we say, as above, that the path of a projectile is a parabola, we are ignoring the motion of the earth ; we are not think- ing of what is called the absolute motion of the projectile that is, its velocity through space or through the ether; we are thinking of its velocity relative to the earth. Our ignorance of absolute velocities makes this notion of relative velocity not only convenient but essential; and CHAP. II.] ABSOLUTE AND RELATIVE MOTIONS. 39 being thus compelled to employ the notion, we may pro- ceed to make further use of it. Now, in treating of the simultaneous motion of two bodies, say of two ships at sea or of two impinging balls, it may be sometimes convenient to ignore their motions with respect to water or ground, and attend only to their motion relative to each other. This relative motion can be brought out and displayed most clearly by the artificial device of imagining both bodies to be affected with a fictitious velo- city which is to be compounded with their real velocities; for a fictitious velocity may be chosen so as to be equal and oppo- site to the real velocity of one of the bodies, and in that case the resultant velocity of that one becomes zero, while the re- sultant velocity of the other body becomes its relative velocity with respect to the one thus imagined to be stationary. There is no important principle underlying all this; it is a mere practical device for simplifying problems in relative motion which otherwise would be more com- plicated for clear mental grasp. As an illustration, consider two ships sailing, one (B) due north with velocity v, and the other (A) due east with velocity u; let us observe them in some definite position, say at distances b and a respectively from the point at which their paths will hereafter cross (fig. 10), and let us ask when and where they are in most danger of collision that is, when and where and what is their shortest distance. If the velocities are in simple proportion to the distances from the crossing-point that is, if u:v a:b, they will certainly collide, unless one of them alters her course ; but if the speeds are in some other proportion, they will pass within a certain minimum hailing distance. To find this distance, the neatest way is to suppose ja, fictitious velocity, - v, impressed upon the whole system ; by this means B is brought to imaginary rest, and A is sent moving (imaginably) with velocity J(u z + v*) in another 40 ELEMENTARY MECHANICS. [SECT. 31. track indicated in the figure by the line AP; and a perpendicular on to this line from B namely, BP represents the nearest ap- proach of A to B. This is the shortest distance re- quired, but it is not in the true locality. The real condition of the problem "I/ limits B to the north course pX and A to the east course, ^ hence their real positions when hailing are B' and A' respectively, where A'B' is a line drawn equal and parallel to PB. Thus the problem is solved in all essential features, even for those who do not care to take >the trouble to work out the arithmetical details. By drawing the above figure to scale, the solution can be simply constructed; whereas, if both bodies were contemplated as moving in their true paths, a direct geometrical construction for the result would be hardly possible, and arithmetic would be necessary. Applying kinematical considerations to the above figure, and denoting AA', BB' by x, y respectively, we see that if a time t elapses between the initial positions A, B, and the desired positions A', B', x = ut, (1) t V = vt; (2) and the distances of A' and B' from 0, the crossing-point of the tracks, are, in the figure, a - x and y - b respectively. (If A' had been beyond 0, and B' ftot up to 0, the distances would have been x - a, and b - y respectively.) We may further see that the triangle A'OB' is similar to the triangle PA'A, because the corresponding sides are perpendicular ; therefore (a - x) : (y - 1) = y : x. (3) CHAP. II.] RELATIVE MOTION. 41 These three equations express all the essential kinematic and geometrical facts ; and simple algebra will now extract the information required namely, the place, time, and dis- tance corresponding to the positions A' and B'. The solutions are : shortest distance A'B' = 5^-5 -- ^ = s, say : V ^ / 7 \ a-x = s = (av - uu) ; ^ av-bu); EXAMPLES-VI. (1) The courses of two steamers are at right angles to each other, and their speeds are 12 and 16 miles an hour respectively. If they are at first, both of them, 1 mile distant from the point where their tracks will cross, find how near they will approach each other and how soon. (2) A lady wishes to cross a muddy road, 30 yards wide, by the shortest path, at 3 miles an hour, in front of a vehicle which is coming along the middle of the road downhill at 9 miles an hour, but which has still a distance of 50 yards to go before reaching her level. By how much will she escape the vehicle ? . Also, what is the least speed at which she need walk if she ignores the mud and takes a slanting direction ? (3) Two trains are moving, with velocities of 25 and 40 miles an hour, along two lines inclined at an angle of 60, and are respectively 300 yards and 200 yards from the crossing- point. Represent completely the motion of either train as it appears to the passengers in the other. (4) Find by a graphical construction drawn to scale, and also by calculation, the resultant of the following velocities which are communicated to a point namely, 10 feet per second in an easterly, 20 feet per second in a north-easterly, and 30 feet per second in a northerly direction respectively. 42 ELEMENTARY MECHANICS. [fiXS. VI. (5) If a point has a velocity of 1 foot per second to the east, and also a velocity of V3 feet per second to the north, deter- mine the velocity which must be compounded with these to bring it to rest. (6) A steamer, travelling at 20 miles an hour, has to travel north-east through a current flowing south at 5 miles an hour : show, in a diagram, the direction in which it should be steered to allow for the current, and measure, or calculate, how far it will have travelled from its starting- point in 3 hours. (7) If the steamer, through ignorance of the current, is steered north-east, find how far it will be carried out of its course in 3 hours, and how far it will be from its starting-point. CHAP. III.1 INERTIA. 43 CHAPTER III. ON QUANTITY OF MATTER AND QUANTITY OF MOTION. A CHAPTER OF DEFINITIONS. (I.) MOTION OF A PARTICLE, OR TRANSLATION. (Inertia and Momentum.} 32. We have so far studied motion in the abstract, witlr reference to its direction and its rate, but without reference to the body moving, or to the amount of motion possessed by it. Let us now consider what is meant by this last phrase 'amount' or 'quantity of motion.' First, it is plain that in any actual case of motion there must be some matter moving ; and it will be sensible and consistent with the ordinary use of language to consider the quantity of motion in a body as proportional, first, to its speed, and, secondly, to its quantity of matter ; and this is the scientific custom. 33. Now we understand what is meant by speed, but what do we mean by quantity of matter 1 First of all, of course, a large solid ball contains more matter than a small one of the same material; but quantity of matter does not depend on size alone, it depends also on the closeness or density of the substance. A small iron ball may contain more matter than a large cork one. Now matter possesses a certain characteristic property called 'inertia,' or power of reacting against a force applied to change its state of motion. It is on account of this pro- perty that force is required to move matter or to check its motion the passive resistance or reaction of the matter 44 ELEMENTARY MECHANICS. [SECT. 33. itself against whatever is forcing it to move being called its inertia-reaction or inertia-force. Thus a railway truck has great inertia, because considerable exertion is required to stop it or to set it going, even on a level line, with friction reduced to an insignificant item. This fact of ' inertia' was expressed by Newton in the following ' law ' or axiomatic statement : ' Every body perseveres in its state of rest or of moving uniformly in a straight line, except in so far as it is made to change that state by external forces.' This is often referred to as Newton's FIRST LAW OF MOTION, or as the Law of Inertia ; and it is equivalent to defining force as that which causes change of motion in matter. Its essence can be briefly expressed by saying that without force there can be no change of motion ; pr the motion of a body is constant both in magnitude and direction except when a force is acting on it; or, again, that wlien resultant force is zero, acceleration is also zero. That a body should thus persist in its state of motion (of which rest is only a particular case) unless some cause acts to change its state, is in accordance with the fundamental or common-sense axiom that no effect happens without a cause. It is not likely that a piece of matter should itself be able to change its own state, whether of rest or motion : some external cause or influence from other bodies is necessary. The above law asserts that the sole cause of change of motion in matter is force. It may be taken as a definition of force in terms of matter, or a definition of matter in terms of force. Since we have a direct sense of force (the muscular sense), as stated in the Introduction, the latter is the most useful definition : hence matter is defined as that which requires muscular action or its equivalent to change its motion ; in other words, matter is that which possesses inertia. This is the sole Newtonian test of matter ; any reference to gravitation, attraction, or weight is beside the mark. If it be asked whether elec- CHAr. III.] INERTIA. 45 tricity or ether is matter, we must inquire whether either of these entities requires force to change its state of motion. The answer may or may not be an easy one to obtain by experiment, but there is no haze about the question. If the ether possesses inertia it is a form of matter, if it does not possess inertia it is something else. Heat possesses no inertia ; it is propelled by difference of temperature, not by mechanical force, hence heat is not a form of matter ; and so on. 34. Since inertia then is a characteristic property of all matter, it will serve to measure the quantity of matter in any given mass, and it is always used for this purpose in Dynamics. Suppose you have a number of smooth cubes or blocks, each made of a different material but of the same size, resting on a perfectly smooth horizontal table, and you give them each a little push of exactly the same strength ; the push will have the least effect on those which contain the greatest quantity of matter. Thus imagine four of the cubes to be of cork, wood, iron, and gold respectively, and that you give each a sudden knock. The cork block would be considerably affected, and would slide off the table ; the block of wood would be affected next in extent ; while the iron and gold blocks would perhaps hardly be stirred, but whatever movement there were would be greater in the iron than in the gold. We should hence conclude that the gold block contained most matter, the iron next, and the cork least. This is a perfectly direct and scientific method of comparing the masses of bodies, and more than comparing, for it is capable of affording a definite measure of the quantity of matter in a body. Thus either apply the same force for * the same time to each body, and measure the velocity imparted (if the same velocity is imparted to a number of bodies by the same shock or impulse, they have all the same inertia, and therefore the same quantity of matter) ; or graduate the forces applied to the different bodies, so that 46 ELEMENTARY MECHANICS. [SECT. 34. each may move with the same acceleration, the forces required will measure the inertia of the several bodies. The forces themselves must he measured hy the strain method (see Introduction, sect. 6), as the other method would lead to reasoning in a circle. Fig. 11 shows the experiment carried out as far as it is possible to carry it out without a perfectly smooth table. The blocks are mounted on rollers to diminish friction, and are attached each to a stiff spring balance, such as some of those made in dial form by Salter, which will yield in a very small but yet a measurable degree, since the yield is magnified by the index. The simple form shown in the figure yields too much to be suitable. These balances are then all tied to a rod, and are pulled quickly along, so that all the blocks have practically the same acceleration imparted to them. The springs indicate by their stretch the inertia-reaction of each body. Fig. 11. 35. One often actually applies this method of comparing masses in common life. Suppose you see a cask lying on level ground, and Avisli to know whetlier it is full or empty ; you give it a kick or a push with your foot, and if it yields and moves easily, you conclude that it contains very little matter that is, that it is empty ; whereas if it almost refuses to move, it must contain much matter ; and if it contains dense, matter, such as iron or lead, it will be harder to move than if it contained, say, eartherf- ware, and this again harder than if it were full of straw. Hence we find that the quantity of matter in a given body, as measured by its inertia, depends first on the density of its material ; and secondly, on its size or volume. And we might define quantity of matter as the product of volume and density, giving this product the name of mass. The ' mass ' of a body hereafter, then, shall stand for the quantity of matter in it, and shall equal its volume CHAP. III.] MASS. 47 multiplied by its density. This last serves strictly for a definition of density rather than of mass, as thus : quantity of matter in body Density -^ * : volume of body or, more shortly, density mass per unit volume. The simplest unit of volume is the cube of the unit of length say a cubic foot or a cubic centimetre ; and density will be expressed as so many pounds per cubic foot, ^ 3 , or so many grammes per cubic centimetre. The numerical value of the density of water on the first system of units is about 62 ; on the second system it is about 1. We see, then, that mass is measured, and must be held to be denned, by the property of inertness possessed by- matter that is, by its requiring force to move it if at rest, and to stop it if in motion. This idea of the muscular effort needed to set a body moving or to stop it, must be held to be the primitive idea of inertia. The greater the effort required to produce a given motion, the greater the inertia ; and as every particle of matter possesses this property, the more particles there are the greater is the inertia, and inertia is the only direct measure of mass in mechanics. To recapitulate, then, mass means quantity of matter, and is measured by inertia. 36. Just as the standard of length is an arbitrary distance, called in this country a standard yard, and denned as the distance between two marks on a certain bar of metal at 62 F., so the standard of mass must be an arbitrary quantity of matter. In this country the standard mass is one of several equal masses of platinum kept in the Houses of Parliament, the Mint, and other places, from which copies are taken for general use, and it is called a pound avoirdupois. The metric standard of mass is a similar mass of platinum kept at Paris, and called a kilogramme. A kilogramme is 48 ELEMENTARY MECHANICS. [SECT. 36. about 2 '2 pounds. The thousandth part of this (a gramme) is commonly used as the practical unit of mass for scientific work all over the world, and the system of units based on the centimetre, the gramme, and the second is called the C.G.S. system. Similarly, the common British system of units, based on the foot, the pound, and the second, may be called the F.P.S. system, and units in that system may be called F.P.S. units. The student must avoid confusing the mass of a body with its heaviness or weight. A pound is the British unit of mass, but because the pound happens to pull downwards with a certain force (avoir, in fact, du pois), people often think of this pull, force, or weight as the essential thing, whereas it is quite a secondary thing. When we speak of this force, we shall call it the pound.- weight, or the weight of a pound : it is not the pound itself (see sect. 64). Suppose you wish to leave some flowers to be pressed all night in a book, and you put on the book for the purpose a few pound or other weights ; what you are then con- cerned wit!) is the weight of the pounds, or their pull downwards. But suppose you buy six pounds of sugar or of soap ; what you are then concerned Avith is the quantity of matter or mass which you obtain, and the force with which the matter tends downward is a secondary, and sometimes a burdensome consideration. This confusion has arisen from the fact that the shopman measures the mass out to you, not by a direct method like that shown in fig. 11, but by an indirect, though practically simpler method, founded on the attraction of gravitation, which is proportional to the masses of the attracted bodies within the limits of experimental error. So that the shopman compares not the mass but the weight of your purchase with his standards. We must try, however, to avoid confusing mass with weight, even at the risk of a little pedantry, which may be necessary until we are quite clear on the subject. 37. Now we have already seen (sect. 32) that it is reasonable to define quantity of motion as directly propor-. tional to the quantity of matter (or mass) moving, and to its rate of motion (or velocity). Hence let us at once define quantity of motion as equal to the product of the mass in motion and its velocity. The name given to quan- CHAP. TII.l MOMENTUM. 49 y tity of motion is momentum; so we have now the de- finition : Momentum mass x velocity, or M = mv, where m stands for mass, and M for momentum. , Momentum means quantity of motion, and is measured by the quantity of moving matter multiplied by its velocity. 38. The unit of mass being a pound, the unit of momen- tum must be that quantity of motion possessed by a pound of matter when moving with a velocity of one foot per second. The momentum of a J-lb. cricket-ball moving at the rate of 56 feet a second, is J x 56 = 14 that is, four- teen British units of momentum, as just denned. The. C.G.S. unit of momentum is that of 1 gramme moving at the rate of 1 centimetre per second. Thus, the jet of a fire-engine which is delivering a cubic metre of water (1000 kilogrammes) every minute, at a speed of 6 metres per second, possesses momentum to the amount of ten million C.G.S. units in each six-metres length of it, and transfers this amount of momentum per second to any fixed obstacle against which it impinges. The result, we shall hereafter learn (sect. 48), is a steady pressure of ten megadynes, a little more than ten kilogrammes weight. The momentum of a 50-lb. cannon-ball moving with a velocity of 1612 feet per second, is 80,600 foot-pound-second (F. P. S.) units. That of a three- ton truck (1 ton =2240 Ib. ), moving at the rate of 12 feet per second (roughly about eight miles an hour, see Ex. I. 4), would be 80,640 F.P.S. units, or nearly the same as that of the above cannon-ball. Now we shall find in the next chapter that an impulse or propulsion is proportional to the quantity of motion it causes ; hence we see that in some sense or other the same motive power was required to set the above cannon-ball going as was required to set the truck, for both possess the same quantity of motion. Yet the force exerted by the powder in the cannon was undoubtedly greater while it lasted than the force exerted by the horse or engine, or whatever started the track ; but then the former acted for the fraction of a second only, while the latter took perhaps a minute. What is called the impulse or propulsion of the force D 50 ELEMENTARY MECHANICS. [SECT. 38. was the same in the two cases. If you put an obstacle in the path of each body so as to stop both in the same time, they would each deal the same blow. Suppose, for instance, that the cannon-ball and the truck were to meet each other end-on, and the ball were to remain imbedded in the material of the truck, both would be stopped dead by the impact. 39. Before passing on to the action of force on matter, it will be well to explain that now we have come to deal with the motion of actual pieces of matter, we shall, if we wish to consider a piece so small that its parts may be neglected, use the term particle instead of 'point;' meaning by particle a point possessing inertia, or ' a material point. A ' particle ' may have any finite mass ; its size, indeed, is to be small (or at any rate negligible), but its density may be anything infinite if we like. A body whose parts are taken into account may be called an ' extended body,' but if stress is wished to be laid on the fact that these parts are immovable relatively to each other, it will be called a rigid body. An extended body whose parts are capable of relative motion is either an elastic or else a plastic body. (Chapter X.) Also it will be well to point out that the parallelogram and polygon laws apply to the composition of momenta just as they do to the composition of velocities (sect. 27). For the momentum of a given mass is simply proportional to its velocity, and the resultant velocity of a particle when multiplied by its mass must be its resultant momentum. (II.) SPINNING MOTION OF AN EXTENDED BODY, OB ROTATION. (Moment of Inertia and Moment of Momentum.) 40. We have already partly seen (sect. 17) that when we come to consider the motion of a rotating body, the distance of each particle from the axis of rotation is always coming in as a factor, multiplying the term which previously had CHAP. III.] MOMENT OF MOMENTUM. 51 been sufficiently expressive. As this product so often occurs, it is convenient to have a name for it, and the name employed is moment. The moment of any physical quantity is the numerical measure of its importance. [This must not be confounded with momentum, with which it has nothing to do.] So the actual velocity of a particle of a rotating body might be called the moment of the angular velocity, for it equals wr, r being measured from the particle to the axis of rota- tion. The actual acceleration, again, is the moment of the angular acceleration that is, it equals ar (see sect. 18). It often happens that the distance from the axis comes in as a factor twice, so that we have a moment of a moment, which is called a second moment. Thus for some purposes it is convenient to speak of the moment of the velocity of a particle of a rotating body that is, vr ; and this is the second moment of its angular velocity, being equal to tor 2 . The moment of momentum of such a particle is, of course, mv-r, or, as it may also be written, ??ir 2 o>. 41. These terms being understood, we will proceed to consider how we must define the quantity of motion of a rotating body, or a system of circularly moving particles. Simple momentum, or product of velocity and quantity of matter, will not do, for the effect produced by a given shock depends not only upon this, but also upon how far distant from the axis the bulk of that matter is. For con- sider a flywheel ; which you know is a large heavy wheel fixed to the crank-shaft of stationary engines and driven at a high speed, not for the purpose of communicating its motion to a lathe-band or anything, but simply for the purpose of storing up a certain quantity of motion sufficient to carry the engine over its ' dead points,' and also over any accidental shocks or sudden impediments which the machinery may experience : it is made massive so as to have 52 ELEMENTARY MECHANICS. [SECT. 41. great inertia, it is also made to go fast so that its parts may possess great momentum ; but besides this it is made large, and nearly all the mass is placed in its rim, which not only increases the momentum but causes that momentum to have a great leverage; so that altogether the motion stored up in the wheel has a great moment of momentum. For just as the power or moment of a force depends not only on its magnitude but also on the place at which it is applied not only on its strength but on its leverage being equal to the product of the force into its distance from the fulcrum (for example, the longer a crowbar is, the more power it gives you ; the more unequal the length of the arms of a steelyard, the bigger the weight which can be balanced by a little one ; and so on, see sect. 1 48) ; so with the flywheel, the effect or power of its stored-up motion depends not only on the actual quantity of motion or momentum of the rim, but also on the distance this rim is from the axle that is, on the radius of the wheel. It depends, in fact, on the moment of its momentum, Mr. 42. Now if the wheel were a simple infinitely thin rim, the meaning of this would be simple enough ; r would stand for the radius of the rim, and M for the product of its mass and velocity, mv (sect. 37); but any actual wheel must have a rim of some thickness, as well as some spokes and a nave, so the meaning of neither M nor r is quite clear without further definition. The moment of momentum of a rotating body is the sum of the moments of momenta of its several particles. Let a wheel turn with the uniform angular velocity w. A particle of mass ??i 1 , at a distance r x from the axis, and moving with velocity T^CO or v lt has a momentum m-jV^ and therefore a moment of momentum m-p^r^ or what is the same thing, m 1 r 1 2 w. Similarly with a particle of mass ??? 2 at a distance r 2 ; and with one of mass m 3 at distance r 3 , and so on ; hence the moment of momentum of the whole CHAP. III.] MOMENT OF INERTIA. 53 wheel is the sum of these terms for all the particles in the body, or, as it is often written, 2 (mvr). Since v = rw, and since o> is the same for every particle as for the whole body, we may write the above expression for the moment of momentum in this equivalent form, w(??z 1 ?- 1 2 + m 2 r 2 2 + wy 3 2 4- ......... ) = o>2(rar 2 ) ; or in words, the moment of momentum of the wheel is the angular velocity multiplied by the sum of the second moments of inertia of every particle in the wheel. In this last form, w?* 2 u>, the moment of momentum is often called the angular momentum ; because, instead of being simply the product of inertia and velocity (as momentum is), it is the product of a moment of inertia and angular velocity. 43. In the last paragraph we have the occurrence of the second moment of mass or inertia, w?' 2 , and indeed this occurs in Dynamics so much more frequently than the first moment (mr\ that it is usually called the moment of inertia. The moment of inertia of any rotating body about its axis of rotation is the sum of the second moments of the masses of all the particles in it about that axis ; and we will denote it by I, so that I = 2(m?* 2 ). The angular momentum, or moment of momentum, of the above flywheel is thus simply Iw. The value of the moments of inertia of bodies of regular shape is obtained by actual calculation of the above sum, W 1 r 1 2 +w 2 r 2 2 + ...... , for any required axis. The process is easy to those who have learned how to integrate : to others the following list of results may be useful : LIST OF MOMENTS OF INERTIA. A. About an Axis of Symmetry. (1) For a thin ring or hollow drum (of mass m and radius r) .................................................... mr*. 54 ELEMENTARY MECHANICS. [SECT. 43. (2) For a thick ring or drum of internal radius i\ and external r 2 fyn( r \ + r i/ 2 + m(H) 2 (10) The moment of inertia of an isosceles triangular area about its median line, (base = b) (11) The moment of inertia of a triangular area of height h, about its base as axis (12) The moment of inertia of any plane lamina or plate about two axes through its centre of gravity, at right angles to each other and to the axis of symmetry, add up to that about the symmetrical axis. Thus : for a circular plate spinning about a diameter, the moment of inertia is (see No. 3) (13) The moment of inertia of a rhombus about a normal axis through its middle (its longest diagonal being I and its shortest b) EXAMPLES VII. (1) Find, in foot and pound units, the moment of inertia of a rod of iron 5 feet long and 1 square inch in cross-section, about an end, given that the mass of a cubic foot of iron is 480 Ib. (2) Find the moment of inertia of the rim of an iron flywheel of 11 feet mean diameter, the cross-section of the rim being 12 square inches. (3) Find the moment of momentum of the above rim when the wheel is making 200 revolutions a minute. CHAP. III.] MOMENTS OF INERTIA. 55 (4) Find the moment of inertia of a thin rectangular plate measuring 4 feet by 3 feet and weighing 5 Ib. : (a) About an axis through its centre normal to its jplane. (6) About a parallel axis through one corner. (c) About a long edge. (d) About a short edge. (e) About a median line drawn lengthways. (/) About a median line drawn breadthways. (5) Find the moment of inertia of a hollow sphere 50 centimetres in diameter, weighing half a kilogramme, and spinning like a teetotum. (6) Compare the moments of inertia of a hollow sphere and a hollow cylinder of the same diameter and weight, also of a solid cylinder and a solid sphere of the same diameter and weight. (7) Find the moment of inertia, about its point of suspension; of a solid sphere 1 foot in diameter and weighing 50 Ib., swinging at the end of a string 6 inches long. (8) Find the moment of inertia of a triangular iron plate, -inch thick, about its base ; the base being 5 inches long, and the height 6 inches. (9) Find the moment of inertia of the same plate about an axis passing through its centre of gravity and parallel to the base. In the above list of Moments of Inertia, (10) Show that No. 13 can be deduced from Nos. 11 and 12. (11) Deduce No. 10 from No. 11. (12) Deduce (2) from (3), and (1) from (2). (13) Deduce (5) from (4) ; the volume of a sphere being ^Trr 3 , and its surface being 47rr 2 . 56 ELEMENTARY MECHANICS. [SECT. 44. CHAPTER IV. ON FORCE AND MOTION (Dynamics). It was stated in the Introduction that force produced two kinds of effects on matter ' acceleration ' and ' strain.' In the present chapter we will consider only the first or motive effects of force that is, the effects of force on rigid bodies or particles (see sect. 39) ; and first on particles moving in the direction of the force (I.) ON THE SPEED OF MOTION AS AFFECTED BY FORCE ; OB, FORCE AND RECTILINEAR MOTION. (Dynamics of a Particle. ) 44. When a single force F is applied to a certain quantity of matter or mass, m, for a given time, a certain quantity of motion or momentum is generated in the mass. If the same force (for example, a piece of elastic stretched to the same extent as before) is applied to a greater quantity of matter for the same time, it will move with less velocity, but the product of the quantity of matter and the velocity that is, the quantity of motion or the momentum will be found to be the same ; so the force may be measured by the momentum generated by it per second, since this is constant, and depends on nothing but the force. If the same force be applied for a greater time, a proportionally greater quantity of momentum will be generated ; hence the measure of the force is the momentum generated per second, and is obtained by dividing the whole momentum gener- ated by the time taken to do it ; or in symbols, F = 5L; CHAP. IV.] SECOND LAW OF MOTION. 57 and the unit force will be that which can generate unit momentum in unit time. 45. Force, then, by this definition comes to be rate of change of momentum, just as acceleration was defined to be rate of change of velocity : a = - (sect. 10). ~c Hence force bears the same relation to acceleration as momentum does to velocity : each, in fact, equals the other multiplied by m, or F = ma. This last is a very convenient form of the definition, and may be expressed thus : A force is numerically equal to the acceleration it can produce in unit mass; and in general a resultant or un- balanced force is measured ~by the product of the mass moved into the acceleration produced, being proportional to the two conjointly ; or concisely, FORCE = MASS ACCELERATION. This is, indeed, the fundamental relation of Dynamics, for it makes all that we have learned about motion in the abstract (Kinematics) available for dynamical problems that is, for all problems involving force. It is called Newton's SECOND LAW OF MOTION. 46. The unit of force may be expressed in these three different but equivalent ways : The unit of force is that which causes unit acceleration in unit mass (one pound or one gramme of matter) j Also, unit force is that which generates unit momentum in unit time, as said above ; Also, it is that which, acting on unit mass for unit time, causes it to move with unit velocity. So, if the British unit of force act on a pound for a second, the pound at the end of that second will be mojdng at the rate of one foot OF THF 58 ELEMENTARY MECHANICS. [SECT. 46. per second. If the C.G.S. unit of force acts on a gramme, the gramme will be moving at the rate of n centimetres per second after the lapse of n seconds. It is often convenient to have a name for the unit of force as defined in any of these equivalent ways. The name poundal has been suggested for the British unit of force in order to indicate a connection between it and the British unit of mass (not by any means to signify that the force unit equals the weight of a pound : it is nearer the weight of half an ounce). A poundal is also called the British absolute unit of force, to distinguish it from the unit founded on the metric system, which involves grammes and centimetres instead of pounds and feet. This C.G.S. unit of force is now very frequently called a dyne. It is, of course, that force which, acting on a gramme for a second, generates in it the velocity of one centimetre per second. It is a very small force indeed, only about the thousandth part of the weight of a gramme, which is itself only about 15 grains. One poundal equals 13825 '38 dynes. A poundal equals a pound foot per second per second, or briefly - = . A pound weight is about 32 poundals sec. 2 (see next chapter) ; a gramme weight is about 981 dynes. These standard weights are frequently used as practical units of force, and in statical problems are very convenient. The load of a ton, a hundredweight, or a kilogramme, is an easily imagined quantity, and forces so expressed are said to be stated in gravitational measure, since weight depends on the earth's gravitative attraction. There is no difficulty in translating these gravitational units into absolute* units whenever the problem ceases to be statical and we need to enter on dynamical considerations. * The word 'absolute* is not very appropriate in this connection, but it is constantly so employed. The meaning to be expressed is that the unit is a com- pletely specified one, not depending on the properties of any concrete piece of matter. For instance, the pull of a spring, stretched by a certain amount, might CHAP. IV.] THE GRAVITATION CONSTANT. 59 The dyne being a very small unit, a megadyne (or million dynes) is often employed. A kilogramme weight is nearly equal to it, being equal to 981,000 dynes j or 2 per cent, short of the full megadyne. A pound weight is 11 per cent, less than half a megadyne (being about 445,000 dyneij). The Gravitation Constant. But though the force of gravitation thus appears to be so great, this is only because the earth is so massive. Newton found that the same force exists between all material bodies, that it is proportional to the masses, and that, if the masses are spherical, it is inversely proportional to the square of the distance between their centres. These facts may be expressed by the equation F = y , where ct y is a constant, called the gravitation constant, whose value must be determined by direct observation or experiment. The result of such experiments is that, very approximately, _ 666 (cm.) 3 JL_ (ft.) 3 7 10 1 " gm.(sec.) 2 ~ 10 9 ' lb. (sec.) 2 ' which may serve as a memorandum till it is intelligible. The gravitation force between ordinary pieces of matter is so small that if a couple of lead globes, each weighing a pound, were placed with their centres one foot apart, their mutual attraction would be only a thousand-millionth part of a poundal ; so that, if they were perfectly free to move in an undisturbed way, they would each have moved three-quarters of an inch in 3 hours that is, they would be an inch and a half nearer to each other, after the lapse of 3 hours, under the influence of their mutual attraction. be used as a unit or standard of force, but it would be by no ineans an absolute one. So also, the weight of a pound is not an absolute unit, for it depends on the neighbourhood of the earth. 60 ELEMENTARY MECHANICS. [SECT. 47. 47. The fundamental connection between force and ac- celeration, F = ma (sect. 45), may be written, of course, in two other forms ; and this one, =* m is an abbreviated statement of the fact that when a force F acts on a mass m, the acceleration produced in it is the ratio of the force to the mass. Let us take an example to illustrate the application of this. Find the distance travelled in 8 seconds by a mass of 2 Ib. which starts from rest, and has a force of 6 poundals acting on it all the time. The acceleration, or velocity acquired per second, is g= J = 6ppundal8 =3 F p g units> in 2 pounds The whole velocity acquired in the 8 seconds is therefore 24, and hence the average velocity is 12 feet per second. The distance travelled is the average velocity multiplied by the time, that is, 96 feet ; which is the answer. Or we might, without troubling about the velocity, have applied the formula s = ^at 2 as soon as we knew the value of the ac- celeration, a = 3, and of course we should have arrived at the same result. But all this latter part is simple Kinematics : the only dynamical part was the finding of the acceleration from the given force and mass namely, a=F-f m. Whether the body is in motion or not when the force begins to act, matters nothing the acceleration produced is precisely the same. Of course the distance travelled in a given time will be different, because of the initial velocity, (v t will have to be added to the ^at 2 ) ; but all that was considered in Kinematics, Chapter II. 48. The following is Newton's statement of the above connection between force and motion : ' Change of motion is proportional to impressed force, and takes place in the direction in which the force acts ;' CHAP. IV.] SECOND LAW OF MOTION. 61 Or as it has been restated by Professor Clerk Maxwell, in equivalent modern language : ' The change of momentum of a body is equal to the impulse which produces it, and is in the same direction.' By impulse is meant the product of the force acting, and the time it lasts ; for it is on both these that the motive power of a force depends. Thus the blow of a hammer is a very great force while it lasts ; but as it is only mo- mentary, its impulse (or motive effect) may not be so great as a much smaller force applied continuously for some time * (cf. sect. 38). The motive effect or impulse is proportional both to the strength of the force, F, and to its duration, t ; and hence it is defined as the product F^. So the first" portion of the above statement is, in symbols, mv = Ft; where v represents the velocity gained by the mass in owing to the action of the force F for a time t ; it is, in fact, simply the fundamental relation of sect. 44 in another form. It is convenient at this stage to refer to the numerical examples given in sect. 38, and to realise that to start the cricket-ball with momentum 14 F.P.S. units would require a force of 14 poundals lasting for a second, as when thrown; or a force of 14,000 poundals (say 2 tons) lasting the thousandth part of a second, as when struck with a bat. Also, that to start the cannon-ball in the thousandth part of a second would require a force of 80 million poundals, which is about equal to the weight of a thousand tons ; a force which, applied to the sectional area of such a cannon-ball (say 33 sq. inches), would need an average pressure in the gun of 30 tons to the square inch. All this on the assumption that the thousandth part of a second represents fairly the time between the ignition of the powder and the ejection of the ball. The railway truck if pushed by three men each exerting a force of one hundredweight, would get up its given speed in * This is best observed by first striking sharply, and then pushing steadily, a thing on wheels where the friction is small. The advantage of a blow is felt, not when you want to move a massive body, but when you have a great force of friction to overcome, as in hammering a nail. 62 ELEMENTARY MECHANICS. [SECT. 48. 80640 -=-(336x32) that is, in about 1\ seconds; supposing there was no friction and that the line was perfectly level. As for the jet of water, it represents a momentum of ten million C.G.S. units per second, and hence corresponds to a steady force of ten million dynes. This force must be operating, to propel the jet, over the sectional area of the nozzle ; and if this area is a circle half a centimetre in diameter, the pressure to which the water is pumped must be about 50 megadynes per square centimetre that is, about 50 atmospheres, or 700 Ib. to the square inch. 49. The above statement, in italic or black type (sect. 48 and sect. 45) is often called, as has been already stated, the second law of motion : it might with propriety be called the law of motion, or the law of force and motion. It is very general, and involves a great deal. It is the fundamental law of mechanics. First, it shows that where there is no force there is no change of momentum that is, that a body not acted upon by any external force, if in motion, will continue with that motion unaltered, and, if at rest, will remain at rest ; a fact often stated separately as the law of inertia, or the first law of motion (sect. 33). It further declares implicitly that if a force act on a body in motion, it produces just the same effect as if it had acted on the same body at rest that is to say, the state of the body on which the force acts is immaterial, as nothing is said about it in the statement. Moreover, the second law implies that if two or more separate and independent forces act on a body, each pro- duces its own change of motion in its own direction without regard to the others. 50. This last is an important aspect of the law, and tells us that the operation of compounding together a lot of independent forces is just the same as that of compounding together the motions which each force separately tends to produce in the same time. CHAP. IV.] SECOND LAW OF MOTION. 63 Thus if AB represents the quantity of motion (that is, the momentum) which would be produced by one force by itself in a second, and BC the motion which would be produced by another force by itself ; then AB and BC may also be taken to represent the two forces themselves. But we learn from Chapter II., and from sect. 39, that the resultant of the two motions AB and BC is the single motion AC, hence AC may be taken as representing the resultant force that is, a force which, if acting by itself, would produce precisely the same effect as the other two forces acting together ; provided they are independent of each other. Hence all that we have said about the composition of motions applies equally well to the composition of independ- ent forces. In other words, forces are compounded by the parallelogram and polygon laws just as motions are com- pounded (see sect. 24 and Chapter VII.). 51. Moreover, we learn that in order to specify the translating power of a force, it is only necessary to specify the velocity it is able to produce in unit mass in a second ; which is readily done by drawing a straight line anywhere of definite length in a definite direction. But we shall soon learn (sect. 55) that, as force has rotating as well as trans- lating power, it is necessary, for the complete specification of a force, to assign also its position or line of action ; it is not necessary to assign it any definite place in that line. Hence three things determine a force Direction (with sign), Position, and Magnitude. As these things are pos- sessed by an arrow-headed line of given length, such a line is often used to symbolise a force. This |, for instance, would be one force, and this -H a force of the same magni- tude as the first, but in a different direction ; while this other one, equal and parallel to the first, | would be equiva- lent to the first in translating power, for it has the same magnitude and direction, but different in rotating power, having a different position, that is, line of action. The only 64 ELEMENTARY MECHANICS. [SECT. 51. defect of this mode of representation is that it is a little too expressive that is, it expresses a little more than is wanted. For -i- and -H-, though two distinct lines, represent the same force in every respect, having the same direction, magnitude, and line of action the rotating and translating powers are the same (see end of sect. 57). For further development of this, see Chapter VIII. 52. There is one more thing about force which is very important, but in the present stage its full meaning can scarcely be appreciated, and that is the fact, mentioned in the Introduction, that force is always due to the mutual action of two bodies or systems of bodies ; that every force, in fact, is one of a pair of equal opposite ones one com- ponent, that is, of a stress either like the stress exerted by a piece of stretched elastic, which pulls the two things to which it is attached with equal force in opposite directions, and which is called a tension ; or like the stress of a pair of compressed railway buffers, or of a piece .of squeezed india- rubber, which exerts an equal push each way, and is called a pressure (see sect. 3). Newton's law concerning this is what is called his THIRD LAW OF MOTION : ' Reaction is always equal and opposite to action that is to say, the actions of two bodies upon each other are always equal and in opposite directions. 1 This may be called the law of stress, and it has been shown by Professor Tait to be susceptible of considerable development (see Thomson and Tait's Natural Philosophy, art. 269, and see also Chapter VI. of the present text-book). It is deducible from the first law of motion (see Maxwell, Matter and Motion, art. Iviii.), for if the forces exerted by two parts of the same body on each other were not equal and opposite, they would not be in equilibrium ; and con- sequently two parts of the same body might, by their mutual action, cause it to move with increasing velocity for ever, the possibility of which the first law denies. The CHAP. IV.] THIRD LAW OF MOTION. 65 same proof holds without modification for the mutual forces between any two or more bodies; for those bodies may be regarded as a single system or complex body, within which all internal forces must balance, else would there be the impossible result of an internal force capable of accelerating the system. We have already shown (sect. 49) that the first law is a special case of the second, and now we have deduced the third from the first ; hence all are really included in the second, which is therefore excessively important. EXAMPLES VIII. (1) What is the acceleration when a force of 36 units acts on a mass 4 ; and how far will the mass move in 10 seconds ? (2) What is the least force necessary to cause 15 Ib. to move 30 feet from rest in 5 seconds ? (3) If a mass of 7 Ib. is acted on by two opposite forces of magnitudes 56 and 42 respectively, what is the accelera- tion ; and what will be the momentum generated in 5 seconds ? (4) How long must a force of 8 poundals act on a 20 Ib. mass to change its velocity from 2 to 26 feet per second ? (5) In what distance will a force of 2 poundals be able to stop a mass of 30 Ib. , which at the time the force begins to act is moving 50 feet every second ? (6) A mass of 20 Ib., which has been going 40 feet per second, is now retarded by a constant force equal to the weight of 4 pounds. How soon will its velocity be 24 feet per second in the reverse direction ? (7) Find the force which, acting on a mass of 2 kilogrammes, gives it a velocity of 98*1 cm. per second in 5 seconds. Compare this force with the weight of the body. (8) If a mass of 6 Ib. is propelled so as to gain a velocity of 10 feet a second every second, what is the magnitude of the force urging it ? (9) If a mass of 6 Ib. be pushed by a force of 2 poundals without friction for 5 minutes, how much will the momentum of the mass be altered ? (10) If a certain force acting on a mass of 6 Ib. for 4 seconds B 66 ELEMENTARY MECHANICS. [EXS. VIII. gives it' a velocity of 40 feet per second, through what distance would the same force move a mass of 10 Ib. in 5 seconds ? (11) What weight is equal to a force of 1200 absolute foot-pound- second units ? (12) What weight is equal to a force of a million dynes? How many dynes will support an ounce against gravity ? How many dynes can support a gramme ? How many corre- spond to the weight of a ton ? (13) A force equal to the weight of a cwt. acts on a ton for 2 minutes. What velocity will it produce? If a force equivalent to the weight of a ton operated on a quies- cent hundredweight for 2 minutes, how far would it push it ? (14) If a force equal to the weight of a gramme pull a mass of 1 kilogramme along a smooth level surface, find the velocity when the mass has moved 1 metre. (15) A mass of 100 grammes acquires a velocity of 30 cm. per second in 10 seconds. Find the force acting on it. (16) A load of 50 Ib. is being lowered by a cord from a height. Find the tension in the cord : (a) When the speed is increasing at the rate of 8 feet-per- second per second ; (b) When it is decreasing at the same rate ; (c) When the speed is uniform. (17) What steady force must act on a mass of 10 Ib. initially at rest in order to move it 144 feet in 3 seconds ? How does this force compare with that of the earth's attraction for the mass ? Impact. 53. When two or more bodies in free motion impinge on one another, their action and reaction are equal and opposite, or, in other words, there is no outstanding or resultant force tending to move the system of bodies as a whole in any direction. Whatever was their average motion before im- pact, that same will continue to be their average motion after impact. The momentum of the whole can only be changed by a force exerted by something outside the system ; internal forces can rearrange the distribution of momenta, CHAP. IV.] IMPACT. 67 but are incompetent to affect total momentum. Hence, w x m 2 being the masses, if u v u 2 are their initial velocities, and flj, v 2 their final velocities, the above deduction from Newton's third law may be written : The resultant of m^ and w 2 w 2 = the resultant of m l v 1 and m 2 v 2 ; that is, the initial momentum and the final momentum are equal in both magnitude and direction. The impact of particles or homogeneous spheres may be either direct or oblique. The impact is direct when they approach each other along the same line and when their surfaces at the point of contact are perpendicular to this line, so that the bodies also recede along the same line after impact. In this case the equation of momenta is m^ + m 2 u 2 = m^ + m 2 v 2 . . The line through the point of contact perpendicular to their surfaces at that point may be^calFedlEeTine of the blow, or the line of impact. If the velocities of the approaching bodies are inclined to this line, they may be usefully resolved into components one along this line, and the other at right angles to it the first being the component in the direction of impact, and the other the transverse component of the velocity of each body. The direct components of momentum obey the above law. The transverse momenta not only obey the same equation of total equality, but are individually absolutely unchanged by the impact. Another mode of stating the law of constancy of total momentum is to say that the motion of the centre of gravity of the two bodies continues unchanged by the impacts, or by any other exertion of internal forces. For, as will be shown in sect. 131, the centre of gravity of the two bodies is a point between them such that its distance, x, from any line in the same plane with them is given by (m 1 + m 2 )x = mfa + m 2 x 2 , whence it follows that its velocity, u, perpendicular to that same line, is given by (m 1 + m 2 )u = m^ + m 2 w 2 before im- 68 ELEMENTARY MECHANICS. [SECT. 53. pact, and that its velocity, v, after impact, is given by (w*! + m<>)v = m^ + m 2 v 2 . Hence, taking the line of reference perpendicular to the line of impact, that is, attending to the component velocities in this line, and remembering that m^ + m 2 u 2 = m l v l + W 2 v 2 , we see that v = 21, so that this component of the velocity of the centre of gravity is unchanged by the impact ; and, a fortiori, the transverse component is unchanged, therefore the motion of the centre of gravity is entirely unaffected by the blow. And this is general : for instance, when a shell explodes in mid-air, the centre of gravity of the whole of its materials continues its parabolic orbit unaltered until some of the pieces strike external objects. Or, again, when a shot is fired from a gun, the total momentum after explosion is the same as it was before namely, ; that is, the gun recoils with a momentum which is equal and opposite to that of the shot and the powder gases forward. This fact is utilised in a rocket in an emphatic manner. All this represents the first fundamental law of impact or of internal forces in general, and it is called the law of ' conservation of momentum.' It is to be entirely dis- tinguished from the conservation of energy ; indeed, energy is by no means obviously conserved in cases of impact. Heat and sound, as well as rotation, have to be taken into consideration before the conservation of energy can be asserted (see Chapter VI.), but the conservation of mo- mentum is a simple variant or extension of Newton's first law of motion. The next fact that has to be stated is of a more empirical character namely, that, in the case of direct impact, the relative velocity of recoil always bears a fixed ratio to the relative velocity of approach for a given pair of bodies, provided that, in general, the shock is not so great as to permanently deform or break them. Consider two spheres moving in the same line and one CHAP. IV.] IMPACT. 69 overtaking the other. Before impact their relative velocity is u^ u% ; after impact their relative velocity is v 2 - V L ; and these two velocities are proportional, so that the ratio -2 - - = a constant. This constant, which is usually denoted u^-u^. by e, is called the ' coefficient of restitution,' which we may shorten into the recoil-ratio of the two bodies. In a few simple cases, such as perfectly elastic spheres or equal rods impinging ' end-on,' this coefficient is practically unity; but in general it is less than 1 by reason of the setting up of rotations, and also by reason of some energy taking the form of vibrations, whether of sound or heat ; and in the extreme case of perfectly inelastic bodies, like putty, or dough, or wet clay, the value of the coefficient is zero. In these latter cases the bodies move on together after impact without any recoil. The following four state- ments are now easy to verify for cases of direct impact ; though, like the rest of this section, they will not be found easy till more has been read : (1) Two equal bodies with e= 1 interchange their original velocities after impact. (For example, if one had been stationary at first, the other will be left stationary at last.) (2) When e = 1, the initial and final energies are equal. (3) When e = 0, the loss of visible energy at impact is 1 1 1 where ^ = + ; M TT&J m 2 the loss being equal to the energy of an imaginary particle moving with the relative velocity of approach of the two impinging bodies, and of mass equal to half their harmonic mean. (4) In all cases of direct impact the loss of energy is 1 - e 2 times the above amount. 70 ELEMENTARY MECHANICS. [SECT. 53. Oblique Impact. If the impact is oblique, the component velocities in the direction of impact are changed in accord- ance with the above equations namely : and v 2 -v 1 = e(u l - u 2 ) ; (2) while their components perpendicular to this direction are, as previously stated, wholly unaffected. If one of the masses (ra 2 ) is infinite, equation (1), though still true, is useless, but equation (2) is then sufficient. Moreover, since the velocity u 2 of the infinite mass is prac- tically unaltered by the collision, and will most naturally be taken as zero, the equation reduces to that is, the component velocity of m 1 in the direction of im- pact is reversed and diminished in the ratio of e:l. The transverse component is, of course, unaltered by the collision. It is interesting to note that, since the path of the centre of gravity of two colliding masses is absolutely unaltered by the collision, the circumstances of impact are the same as if the two masses each collided with an infinite mass moving with the velocity of their centre of gravity. If we consider their motion relative to their centre of gravity (obtained by compounding with the actual velocity of each mass a velocity equal and opposite to the velocity of their centre of gravity), the component velocity of each in the direction of impact is reversed and diminished in the ratio of e : 1, in accordance with what always happens when a finite mass collides with an infinite mass; and their transverse relative velocities continue unchanged. These ideas will assist us to represent in a diagram the change of velocities caused by a collision under given conditions. The following general construction can be given, and its proof left as an exercise. If AO represents in magnitude and direction CHAP. IV.] IMPACT. 71 the velocity of some particle A, and BO the velocity of any other particle B, and if G is the centre of gravity of the two particles A and B, then GO is the velocity of their centre of gravity ; and AG, BG are the velocities of the particles relative to their centre of gravity. After impact has taken place, in some line to which MN is drawn normal, the velocity of the centre of gravity continues unaltered, but the veltf- cities of the particles become A'O and B'O respectively, such that, if e = l, the line MN is equally inclined to AB and to A'B', while A'G = AG and B'G=BG. (If e is not unity, then the tan- gent of the angle between A'B' and MN is e times the tangent of the angle between AB and MN.) It should be noticed that AM, BN are the direct components, and MG, NG the transverse components, of the velocities AG, BG ; consequently in all cases MG, NG remain unchanged, while the direct components are reversed and diminished in the ratio e : 1, so that A'M = - e AM, and B'N = - e BN. (In the figure, 6=1.) A' Fig. 12. EXAMPLES IX. (1) Two balls of masses 4 Ib. and 8 Ib. are moving towards each other with velocities 25 and 3 feet per second respec- tively. After impact they move on together. Find the common velocity. (2) Two perfectly inelastic bodies of masses 120 and 150 Ib. respectively, moving with equal opposite velocities of 18 feet a second, impinge directly on each other. Find the subsequent velocity. (3) A lump of clay when thrown horizontally against a mass 20 times as great, resting on perfectly smooth ice, makes it move 3 feet in 2 seconds. What speed had the clay at the moment of impact ? (4) Two masses of 6 and 10 Ib., moving in opposite directions with velocities of 10 and 8 feet per second respectively, collide. If the velocity of the smaller mass be exactly 72 ELEMENTARY MECHANICS. [EXS. IX. reversed by the impact, find the coefficient of restitution for the two bodies. (5) Find the velocities of two bodies after collision, their masses being 18 Ib. and 32 lb., their velocities (in the same straight line and in the same direction) 12 and 7 feet per second, and their coefficient of restitution \. (6) A 1-oz. bullet fired from a 20-lb. rifle pressed against a mass of 180 lb., kicks the latter back with an initial velocity of 6 inches per second. Find the bullet's initial velocity. (7) A bullet of mass ^ lb. is fired with velocity 300 feet per second into the middle of a wooden block weighing 30 lb. hung by a very long cord. What is the common velocity of pendulum and bullet just after the collision ? (8) Two masses of 10 and 20 lb. , moving in the same direction with velocities of 25 and 15 feet per second respectively, collide. Find their velocities after impact, assuming them to be perfectly elastic. (9) Masses of 4 and 6 lb. collide when moving in opposite direc- tions with velocities of 8 and 12 feet per second respectively. If their coefficient of restitution be , find their velocities after impact. (10) Two bodies, for which the coefficient of restitution is $, approach each other with equal velocities of 20 feet per second. After collision one body, the mass of which is 5 lb., returns with velocity 15 feet per second. Find the mass of the other body, and its velocity after collision. (11) Two masses of 5 and 3 lb. collide when moving in opposite directions with velocities of 4 and 10 feet per second respectively. If the smaller is just brought to rest by the impact, find the coefficient of restitution for each body. (12) A bullet, weighing 50 grammes, is fired into a target with a velocity of 500 metres a second. The target is supposed to weigh a kilogramme, and to be free to move. Find, in kilogrammetres, the loss of energy in the impact. N.B. Remember that $mv 2 gives a result in absolute, not in gravita- tional, units. (13) A ball let fall on a stone slab from a height of 16 feet bounces the first time to a height of 9 feet. What is the coefficient of restitution, and how high will the ball bounce next time? Find also the total distance it will travel before coming to rest. (14) Two smooth spheres whose masses are proportional to 10 CHAP. IV.] RIGID DYNAMICS. 73 and 8, moving in directions perpendicular to each other, with velocities of 20 arid 30 feet per second respectively, collide so that the line of centres makes angles of 30 and 60 respectively with the directions of the motions. Cal- culate their velocities after impact ; taking the recoil ratio as unity. Also lind them by means of a diagram. (15) A cannon on an armoured train fires a 10-lb. projectile rearwards with a velocity of 2000 feet a second. If the truck with the cannon weighs 2 tons, what initial velocity is imparted to it ? (16) A Maxim gun delivers 300 one-ounce bullets per minute with a speed of 1600 feet a second. What force is necessary to hold the gun still ? (17) A jet of water from a fire-engine delivering 30 gallons of water per minute through a quarter square inch opening impinges directly on a brick wall. What force does it exert ? (18) Why does a foul gun kick worse than a clean one, notwith- standing that it ejects the shot with less velocity ? (II.) ON ANGULAR VELOCITY AS AFFECTED BY FORCE; OR, FORCE AND ROTATION. (Dynamics of a Rigid Body.) 54. When force acts on an extended piece of matter, it produces in general both motion and strain (sect. 5). The latter we do not want to consider at present ; so to exclude it, we suppose the body to be rigid all its parts rigidly bound together and incapable of distortion or relative dis- placement. The effects of force on such a body are translation and rotation. If the effect is translation only, the body acts like a particle (sect. 39), as if all its mass were concentrated at a point (called its centre of inertia^ or sometimes its centre of gravity), and "the second law of motion as stated for particles applies to the rigid body ; so that if E is the resultant of all the external forces acting on the body, and if m is its mass, the acceleration of its centre of inertia is T> . This is true anyhow, whether there is rotation or not : m J 74 ELEMENTARY MECHANICS. [SECT. ^4. but when rotation is allowed the subject becomes much more complicated, especially if translation is possible as well. We can, however, consider rotation by itself, by supposing one line or point in the body to be fixed in position, so as to constitute an axis or centre of rotation. 55. All we can say about the subject here is, that in estimating the rotating effect of a force, one must not only consider its impulse that is, its magnitude multiplied by its duration (sect. 48) but we must also consider its position ; how far its line of action is from the fixed line or axis of rotation : the farther it is off, the more effect it has ; its moment (sect. 40) is greater. Suppose a force acts on a body only capable of rotation, at a distance R from its fixed axis : the moment of momentum, or angular momentum, generated [2(wwr), or 2(?w 2 a being the centripetal acceleration. We have then a case of composition of motions very like that discussed in sect. 29, where a uniform rectilinear motion was compounded with a uniform acceleration in a constant CHAP. IV.] CENTRIFUGAL FORCE. 79 direction that is, always parallel to itself ; and where the path of the resultant motion was found to be a parabola. But, in the present case, we have to compound a uniform motion with a uniform accelera- tion at right angles to the path of motion at each instant, in fact along the radius of the circle, and by no means parallel to itself. Drawing a figure similar in principle to that of sect. 29 (fig. 9, which see), let OP be the very minute portion of the circular path described in an infinitesimal portion time t with the constant velocity v, so that Fig. 15. of and complete the figure as shown in fig. 15, letting fall PN perpendicularly to the diameter of the circle OD. Then OP is the diagonal of an infinitely small parallelo- gram* with sides OT and ON; wherefore the motion along OP may be regarded as compounded of two motions one with the constant velocity v along OT, which the particle would have if left to itself; the other, due to a constant pull of the centre C, and therefore uniformly accelerated, along ON, which is the distance travelled in that direction in the above small time t ; wherefore It only remains to determine, from the geometry of the figure, the relation between ON and OP, in order to find * The quadrilateral ONPT is not really a parallelogram, but it is more nearly one the smaller it is that is, the nearer P is taken to O ; and it is accurately one in the limit when it is infinitely small that is, when P is the next consecutive point to O, which is supposed to be the case ; for of course OPQ, &c. are really consecutive points of the circle, only they have to be spread out in the diagram. la the limit also OP and OT are equal, and hence OT is also equal to vt. 80 ELEMENTARY MECHANICS. [SECT. 60, the value of the centripetal acceleration for a point moving with given velocity in a circle of given size. The angle OPD, being an angle in a semicircle, is a right angle (Euc. III. 31), and so is the angle at N; moreover, the angle at is common to the two triangles ONP and OPD ; wherefore these triangles are similar (that is, one is like the other magnified), and their corresponding sides are therefore proportional ; so ON : OP : : OP : OD ; or in symbols, if r is the radius, CP, of the circle, -I at 2 : vt : : vt : 2r ; or, a : v = v : r, whence v 2 = ra, or v is a 'mean proportional' or 'geometric mean ' between a and r. The value of the centripetal acceleration is then v 2 a = ; or (writing v = ior) a o>V = vat. The centripetal force is of course simply in times this, m being the mass of the revolving particle of matter, or F = - = mo> 2 r = mv So , the centripetal acceleration, is 3374 x 2?r = -Q0898. v 2^360,000 This is the value of the acceleration above denoted by g' t and the centripetal force is 00898m. Now if this force be really due to gravity, and if gravity really diminishes with the square of the distance, then, the distance of the centre of the earth from the moon being sixty times as great as its distance from the surface of the earth (that is, the earth's radius), it would follow that g at the distance of the moon should be the 3600th part of the, value of g at the surface of the earth. But the value of g is 32*2 (see next chapter), and the 3600th part of this is -00894 ; so the weight of the moon should be '00894m ; and this is as nearly equal to the necessary centripetal farce, '00898???, as our rough data can be expected to give it. This is the sort of calculation which Newton went through when he proved that the force required to keep the moon in her orbit was just the same as would be exerted by the gravitative pull of the earth; supposing that the force which pulls down stones and apples extended so far, and decreased regularly all the way with the square of the distance from the centre of the earth. Whence he concluded that this force does so extend, and is the actual force in operation. EXAMPLES-X. (1) Half a pound is whirled at the end of a string 18 inches long 3 times round per second. What is the tension in the string ? CHAP. IV.] CENTRIFUGAL FORCE. 83 (2) If a string can stand a force of 1000 units without breaking, what is the greatest length of it which can be used to whirl a 5-lb. mass once round a second ? (3) What is the smallest length of the same string which can be used to whirl a 5-lb. mass with a velocity of 10 feet a second ? (4) A string just able to carry 40 Ib. is used to whirl a |-lb. weight in a horizontal circle 5 feet in diameter. At what speed will the string break ? (5) Find the tension in each spoke of a six-spoked flywheel, 8 feet in diameter and weighing 12 cwt., when making 200 revolutions per minute, assuming all its mass collected at its rim, and that by reason of cracks in the rim, the spokes have to bear the whole of the strain. As a matter of fact the rim of a flywheel ought to stand its own strain, and the spokes need add but little to the strength. (6) A stone weighing 6 oz. is attached to the end of a cord 3 feet long, the other end being fixed. If the cord breaks when the stone is whirled round at the rate of 10 turns per second, find the greatest weight it could have been able to support. 84 ELEMENTARY MECHANICS. [SECT. 62. CHAPTER Y. ON FORCE AND MQTHON-Contimted. THE FORCE OF GRAVITATION. 62. Before proceeding further, it will help our ideas to apply some of the general laws to a few special cases. The most universal force known is the force of gravitation, and it will be convenient to take illustrations from the action of this force ; but we will, in the present stage, only consider it as a uniform action exerted by the earth, tending to pull every piece of matter down to the earth's surface with a force varying with the mass of the piece of matter, but with nothing else. This is practically true in all common cases, for though the force really varies inversely with the square of the distance from the centre of the earth, yet the variation for ordinary heights is very small. For there is scarcely any difference in the distance of the centre of the earth from the sea-level and from the top of a mountain one is say 4000, and the other perhaps 4001 miles. 63. This force is what is known as weight ; it is measured like every other force by the acceleration it can produce in unit mass, or, in other words, by the momentum it can generate in a second. To measure the force, and sefc how it depends on the nature of the attracted body, we will first take the same mass of different bodies, and compare the accelerations which gravity is able to produce in them. Thus take a pound (see sects. 34 and 36) of lead, of iron, of stone, of wood, and of cork, and drop them all at the same instant from a high tower ; then if every disturbing cause were absent that is, if they were subject to no other force CHAP. V.] GRAVITATION. 85 but that of gravitation they would all be found to reach the ground at precisely the same instant, having all acquired the same velocities. If, however, the experiment took place in air, they would be subject to disturbing causes, and the falling together would be only approximate. The wood and cork would be retarded by the air more than the others, partly from the same cause that enables us to winnow chaff from grain, and partly for a reason which may be rendered more obvious by dropping the different things under water. The falling of the wood and cork would be then not only retarded but reversed into a rise. The air has a floating power, only it is less than that of water. The air must therefore be removed, and the bodies dropped in vacuo, if the observation is to be precise and extended to all sorts of bodies such as cotton-wool or feathers ; the experiment is then often called the- guinea and feather experiment, and for a description of it you may refer to Ganot or Deschanel. The above experiment, if carried out accurately, would prove that the pull of gravity has nothing to do ivith the material or nature of the substances, for all equal masses are equally accelerated whatever the material; and, since the masses are equal, this means that they are all pulled with equal force (sect. 45). 64. Next take unequal masses (either of the same material or not), say a swan-shot and a cannon-ball, and drop them from a height at the same instant. They will both reach the ground at the same time that is, they each receive the same acceleration. This experiment was carried out by Galileo from the Tower of Pisa. It shows that the earth's pall on a body is directly proportional to its mass. For since force is equal to the product of mass and acceleration (sect. 45), and since the acceleration is found experimentally to be the same for all masses, it follows that the force is necessarily proportional to the mass. If we denote by g the acceleration produced by gravity that is, the velocity gained per second by a freely 86 ELEMENTARY MECHANICS. [SECT. 64. falling body the force pulling it down is measured by its mass multiplied by y, and this force is termed its weight; so "W" = ing ; or, the weight of a body is g times its mass ; in other words, g = weight -f- mass, that is, weight per unit mass. Hence g is often called the intensity of gravity. This is simply a special case of the general relation F = ma, weight being a particular case of force, and g being a par- ticular case of acceleration. The acceleration due to gravity is the same for all material bodies, and if one thing is ever observed to fall more slowly than another, it is because of the disturbing effect of the air. In a vacuum all things fall at the same rate if they start fair; and this experi- mental fact, combined with the second law of motion, proves that weight is proportional to mass. The ratio of weight to mass is often called the intensity of gravity, and denoted by g ; and a knowledge of its value at any place enables us to translate gravitational units of force there into invariable absolute units. In this country, g is found by experiment (see next section) to be about 32-2 F.P.S. units, or about 981 C.G.S. units; so that the weight of a pound is about 32*2 poundals, and the weight of a gramme about 981 dynes. If the earth were a stationary sphere, g would be constant all over its surface, and it would only vary on ascending above or descending below the sur- face ; as the inverse square of the distance from the centre as you ascend, as the direct simple distance from the centre as you descend (supposing the density uniform). But inas- much as the earth is a rotating spheroid, the intensity of gravity is different at different parts of the surface, as shown in the following table, whose figures are calculated from the CHAP. V.] FALLING BODIES. 87 known shape of. the earth, which, again, was calculated by Newton from its period of rotation. INTENSITY OF GRAVITY AT DIFFERENT LATITUDES. Latitude. g in C.G.S. units. # in F.P.S. units. Equator 978-10 32-090 Paris 45 48 '50' 980-61 980-94 ' 32-173 32-184 Greenwich. 51-29 981-17 32-191 JBerlin 52-30 981-25 32-194 Liverpool 53-29 981-34 32-197 Edinburgh North Pole 55-57 90 981-64 983'H 32-207 32-255 Thus, whereas a stone falls 16 '045 feet in the first second at the equator, it falls 16 '09 9 feet at Liverpool, and 16-128 feet at either pole. The weight of one pound is therefore about 32 '2 absolute F.P.S. units of force or poundals. (Eead sect. 36 again.) Falling Bodies. 65. To express all the laws of falling bodies, we have simply to apply all the kinematics we know about bodies moving with constant acceleration. Thus a stone let drop is found to fall about 1 6 feet in one second (more accurately 16-09), so that 16 is its average velocity during that second ; but the average velocity is half the final velocity; hence the velocity acquired in one second, or the acceleration, is 32 (more accurately 32'18), and this is the value of g. (Since 16*09 feet equal 490*5 centimetres nearly, the value of g is 981 in centimetres-per- second per second; and this therefore is the weight of a gramme in dynes.) A more accurate method of determining 88 ELEMENTARY MECHANICS. [SECT. 65. , g is to let a body fall again and again, and take the time of a long succession of falls. This can be practically carried out with a pendulum (see sect. 78) ; but for many purposes an approximate value of g in round numbers is sufficient, and the number 32 happens to be very con- venient, because both its half and its double have an easy square root. Hence the following handy rules : The velocity 32 is gained in every second of the fall, so the velocity gained in t seconds is 32t (feet per second). The distance travelled, being proportional to the square of the time, is IQt 2 feet (h = gt*). The velocity gamed while falling from a height of h feet from rest is (by equation v^ = 2gs), in feet per second, S,Jh; 8 being the square root of 64 or 2g. The time taken to fall from rest at a height of h feet is, in which follows at once from the equation s = These expressions, 32, 16 2 , 8\/h, and V^> only apply when everything is expressed in feet and seconds, and when the falling body starts from rest ; but given these conditions, they are very useful for rapid mental estimates, especially the last two. For instance, the time taken to drop 144 feet is \A44, or 3 seconds, and the velocity acquired is 8V144, or 96 feet a second. To drop 400 feet, 5 seconds are needed, and the speed attained is 160. 66. Modes of diluting the Intensity of Gravity. The acceleration is equal to g for all bodies only on con- dition that they fall freely that is, that the weight of each has only its own mass to move and nothing else ; for then a = F/w, but F = mg, so a = g. If, however, by any arrangement, we make a weight move CHAP. V.] ATWOOD8 MACHINE. 89 Fig. 16. another mass as well as its own, the acceleration must be less. Thus suppose we Jae a falling weight P(say|jlb.) to a mass Q of 18 Ib. resting on a smooth iiat table, 'as in fig. 16 ; then the force causing the motion is the weight of the 6 Ib. that is, Qg but the total mass moved is 18 + 6 = 24 Ib. ; hence the ac- celeration is lH^ =8 feet-per-second per second. Hence in the first second the com- bination would move 4 feet, and in t seconds 4t z feet, while the velocity acquired in t seconds would be 8t feet per second. In a similar way, we can find the ac- celeration if two weights are connected Ss^ by a string passed over a frictionless pulley, without inertia, as in fig. 17. Let the masses Q and P be 7 and 9 Ib., their weights will be Ig and 9 or tae acceleration observed is -^- F.P.S. units ; but it has just been calculated as also ^, hence g=31$. An actual experiment with At wood's machine would be hardly likely to give g so nearly correct as this. There are other methods of finding the value of g which are much better practically, though not so theoretically simple. The CHAP. V.J FALLING BODIES. 91 most' accurate method consists in observing the time of oscillation of a pendulum of measured length, which may be considered as a heavy body constrained to fall along a circular path, and having its fall repeated automatically again and again, so that the time of fall may be accurately measured namely, one quarter of the period of a complete swing to and fro (see sect. 78). 68. It is easy to understand how experiments may be made with Atwood's machine on the laws of uniform acceleration. Thus, to take the case when the weights are in the ratio of 21 to 13 23, and the acceleration therefore -^, we should find that the distances travelled in 1, 2, 3, 4, 5, 6 seconds respectively were, in fact, 13 4x13 9x13 16x13 25x13 36x13 IS' 18 ' 18 ' 18 ' 18 ' 18 that is, always half the acceleration multiplied by the square of the time (^cit 2 ). The distances travelled during each second would follow another law. They are easily obtained from the preceding numbers, for if we subtract the distance travelled in three seconds from the distance travelled in four, we obtain the distance travelled during the fourth second namely, 16x13 9xl3_7xl3. 18 18 " : 18 ' and similarly, we get for the distance travelled in the first, second, third, fourth, and fifth seconds respectively, 13 3x13 5_xl3 7x13 9x13, 18' ~~18~' 18 ' 18 ' 18 ' a series ascending by the odd numbers ; the distance travelled in the nth second being that travelled in the first second multiplied by the nth odd number, a(2/t- 1). 69. All this may be readily remembered by observing its analogy with a simple geometrical diagram, as in sect. 23. Draw any right-angled triangle, OPC (upside down does best for falling bodies); divide its base, OP, into any 92 ELEMENTARY MECHANICS. [SECT. 69. number of equal parts, and draw a vertical line at each division. You will thus cut up your triangle into trapeziums, of which the left-hand one degenerates into a triangle ; and it is plain to simple inspection that, whatever be the area of this small triangle, the trapezium next to it has three times that area, the next five times, the next Fi 18 seven times, and so on, as may be seen from the dotted lines drawn in fig. 18. Hence, if the first area represent the distance travelled by a uniformly accelerated body in the first second, the second area will represent that described in the second second ; and the sum of the two figures will be the space described in the two seconds together, and so on. Moreover, the whole area of the triangle will represent the space travelled in the whole time, as measured by a number of seconds equal to the number of segments of the base. Thus, in the above figure, the whole area is the space described in four seconds. (Read sect. 23 again.) The vertical height of the figure being nothing at its left- hand point, corresponds with the fact that the falling body starts from rest that is, is dropped. But if the body is tliwwn either down or up with an initial velocity, this velo- city must be represented by a line drawn at the left-hand point, either down or up, and the figure becomes as in fig. 19 or as in fig. 20, where OA represents a velocity downwards, and OA' a velocity upwards. In the first case the velocity continually increases, until in four seconds it becomes equal to PC. , In the second case it at first decreases, becoming zero at the point E two seconds after starting, and then increases downwards until it becomes P'C'. This second case exactly corresponds with that of a ball thrown up in a vacuum against gravity. In both cases the CHAP. V.] FALLING BODIES. 93 whole area of the figure represents the whole space travelled. In the second case we see that the area OA'E is the space or height the ball rose through, and EP'C' the height it afterwards fell through. The ball was at its highest Fig. 19. Fig. 20. point two seconds after being thrown up, having then no velocity. In both cases the ball must have been thrown from the top of a tower or some other height, or it could not fall for so much as four seconds without striking the ground. The area OACP may represent the height of this tower, and OP the time taken to fall, in the first case that is, when the ball was thrown downwards ; but in the second case, when the ball was started upwards, the height of the tower is the difference of the areas EP'C' and OA'E. OP' is the whole time taken by the ball, first to rise a height above the tower equal to the area OA'E, and then to fall from this height to the ground. The lines AC and A'C' are necessarily parallel, since the slope of each represents the rate of numerical gain of velocity, 32 feet-per-second per second. (It might, how- ever, be anything less than this if Atwood's or some other 1 diluting ' machine were used.) 94 ELEMENTARY MECHANICS. [SECT. 69. Supposing it is 32, and that OE is two seconds, and EP' six ; then, of course, the initial velocity OA' must be 64, and P'C' must be 192, feet per second. The area OA'E will be 64 units (its height being 64, and its base 2), and therefore the height the ball rises is 64 feet. The area EP'C' is 576, and so the height of the tower is 512 feet. In the diagrams, the time represented by OP' in the second diagram is greater than the time, OP, in the first, by twice OE ; and the initial velocity OA is numerically equal to OA' ; hence also the final velocit} r PC is equal to P'C', and the area OACP represents 512 linear feet. In each of these figures (neglecting dashes), OP is the line of time; OABP represents the space described due to the initial velocity alone ; and ABC the space described due to gravity. Also BC represents the gain of velocity at ; and PC the actual final velocity. Refer to sect. 23 for some more statements concerning these diagrams and practise drawing diagrams for all kinds of cases of rectilinear motion. Thus, draw diagrams for the motion of a railway train, which gets up speed, goes uniformly, slackens, stops, and goes on again, several times, and then comes back; for the motion of an india-rubber ball thrown down to the ground and then bouncing; for the motion of the bob of a very long pendulum ; for the motion of a tilt hammer, &c. ; and remember in drawing these diagrams that time never retrogrades, and hence that no part of a diagram can be vertically under or over another part, but the drawing must progress continually forwards. Journeys back are represented by areas below the line of time instead of above. The curve for a bouncing ball is practically a discontinuous set of parallel straight lines, because at the moment of bounce the velocity is suddenly reversed. 70. To actually experiment on the velocity acquired by the falling weights in Atwood's machine, we must remember the definition of variable velocity at any instant (given in Chapter I., end of sect. 9) namely, the distance the body would go in the next second if at that instant the accelera- CHAP. V.] FORCE AND MOTION. 95 tion ceased. Now, the cause of the acceleration in this machine is the force (P Q)#. If this force were suddenly removed that is, if P and Q were suddenly made equal, there would be no further acceleration, and the masses would continue to move uniformly forward with the velo- cities they had already acquired, until they were checked either by striking something or by friction. This sudden removal of the inequality in the two weights is practically accomplished by making the extra weight by which P exceeds Q (2 ounces in the experiment of sect. 67), a loose metal bar too big to pass through a certain fixed ring placed in the path of P. When P passes through this ring the bar is removed ; P and Q become equal, and move a distance in the next second which is numerically equal to the velocity they had acquired at the instant the bar was taken off. For a fuller description of Atwood's machine, and for many details of its actual construc- tion, you may refer to Deschanel or Ganot. 71. Further Illustration of the Fundamental Equa- tion. This example of the two weights, one pulling up the other, illustrates the statement in sect. 54 that the second law of motion applies to other cases than those where the motion is perfectly free and unresisted ; in fact, that it is quite general, if we always consider the force F as the re- sultant of all the forces acting on a body, and not simply that -force which happens to be most obviously apparent to us. Thus, go back to the mass of 18 Ib. resting on a table, and pulled along by a weight of 6 Ib. hanging over the edge by a string (fig. 16, sect. 66). The acceleration we saw ought to be 8, but suppose it was observed to be only 3, we should at once know that all the forces had not been taken into account. The table, perhaps, is rough, and retards the motion of the 18 Ib. with a force sufficient to reduce its acceleration to- 3, and the force of friction may from these data be calculated. So again when a 56-lb. bucket is dragged up a well with a force of 1920 poundals (the weight of 60 Ib.); if this were the only force acting, the acceleration of the bucket upward would be 56 96 ELEMENTARY MECHANICS. [SECT, 71. F.P.S. units that is, it would gain this velocity per second ; but if the experiment be tried, the velocity actually gained per second will be found to be nothing like so much as this it will be only about 2|- units. The reason obviously is that there is another force left out of account, opposing the pull of the rope namely, the pull of the earth, which is 56 x 32 = 1792 poundals ; and the resultant force is the difference of these two, or 128. Hence the actual acceleration is 72. To take another very similar example, a cage of mass m, or say 1000 lb., is lowered by a rope down a coal-pit; what is the tension in the rope, at a time when the cage is gaining downward velocity at the rate a, or say 24, feet a second every second ? Well, tlfe resultant force must equal the mass acceleration, but this resultant force is the difference between the weight of the cage, mg, and the pull of the rope, T, hence mg-T=ma, or T=m(g-a), which is the tension required. Numerically, in the present sooo illustration, the tension is 1000x8 poundals = -=- lb. weight = % of the normal weight of the cage. The tension in the rope is A. always y - - times the weight of the cage if a represents its downward acceleration. If the cage is being accelerated upwards, the tension in its rope is times its weight. If the tension in the rope were equal to the weight of the cage, the cage would necessarily have a constant velocity ; it might be moving either up or down, but there could be no acceleration (cf. sect. 133). Such a body is obeying the first law of motion; it is subject to no resultant or unbalanced force. A locomotive dragging a train at constant speed on a straight line, even uphill, is in the same case ; the forces on it are balanced ; and until the rails curve, or the steam is shut off, or the brakes are put on, the motion is perfectly uniform, the effective force is zero. EXAMPLES XL (1) What is the weight of 20 lb. at a place where a falling body travels 4 feet in the first second ? CHAP. V.] FORCE AND MOTION. 97 (2) At what height above the earth's surface could such a place be found ? (3) A curling weight is thrown on ice with a velocity 50 ft. per second. Supposing the force of friction to be T Vth of the weight, how soon will it stop ? (4) In an Atwood's machine a 40-gramme weight on one side is drawn up by a 50-gramme weight on the other, 2'18 metres in two seconds. What is the value of g in centimetres-per-second per second ? (5) In the preceding question find the tension in the cord in grammes weight, and in dynes. (6) When a 3-lb. weight hanging over the edge of a smooth table drags a 45-lb. mass along it, find the acceleration and the tension in the string. (7) Find also the acceleration if the coefficient of friction between the table and the weight is *05. (8) From a balloon which is ascending with a velocity of 80 feet a second a stone is dropped, and seen to strike the ground in 7 seconds. Find the height of the balloon at the moment of letting go the stone. (9) A cage is hauled up from the bottom of a mine with an acceleration of 12 feet-per-second per second. After rising 96 feet a stone is dropped from it. How soon will the stone reach the bottom of the mine ? , (10) If a body fell at a certain place 50 feet in the 3d second of its fall from rest, what would be the intensity of gravity at that place ? (11) A body weighing 30 Ib. placed on a smooth horizontal table has a string attached which runs parallel to the table, passes over a smooth peg at the edge of the table, and has a weight of 2 Ib. hanging at its end. Find the tension in the string, and the acceleration of the system. (12) Two bodies of 17 and 15 Ib. respectively are connected by a string which passes over a smooth pulley. Find the acceleration with which they will move, and the tension of the string. (13) Determine the acceleration of an Atwood's machine if the masses at the ends of the thread are 40 and 50 grammes, and the pulley is equivalent to an additional mass of 10 grammes. (14) In an Atwood's machine the weights on either side are 5 and 4 Ib. respectively, and the wheel is equivalent to an extra G 98 ELEMENTARY MECHANICS. [fiXS. XI. inertia of 3 Ib. Find the distance descended by the heavier weight in 3 seconds, and the velocity acquired. Find also the tension in each portion of the light cord. (15) If the weights are respectively 3^ and 4^ Ib., and if the motion is 12 feet in 3 seconds from rest without friction, what inertia must the wheel be equivalent to ? (16) A mass of 20 Ib. is being dragged off a table by a 4-lb. weight attached by a string which passes over a smooth pulley at the edge of the table. The friction between the weight and the table is T V of that weight. Determine how far the whole moves in 3 seconds from rest, and find the tension in the string. (17) If a stone is dropped out of an ascending or descending balloon, how long a time will elapse before it is 16 feet below the balloon ? If a stone and a light fleck of cotton- wool are released simultaneously ami are in one second 30 feet apart, how rapid is the vertical motion of the bal- loon ? If the stone and the wool are observed to meet and pass each other 3 seconds after their simultaneous release, what is the balloon doing ? Assume in each case that the wool remains practically stationary, suspended in air, and that the balloon's motion, whatever it is, is steady. (18) An engine winds a three-ton cage up a coal-pit shaft at a uniform pace of 11 yards a second. What is the tension in the rope ? (19) Instead of a uniform velocity, the above cage is wound up with a uniform acceleration 6 feet-per-second per second. What is the tension in the rope ? (20) A monkey clings to a light flexible rope passed over a large fixed pulley without inertia or friction, and is balanced by a precisely equal weight at the other end of the rope. What happens if it now begins to climb the rope ? 73. That aspect of the second law of motion which says that it makes no difference to the effect of a force on a body whether that body was in motion or not (sect. 49) is well illustrated by falling bodies (see sect. 22). 74. But the law is more strikingly illustrated when the direction of the initial velocity of a falling body (now called a projectile) is inclined at some angle to the force of gravity. CHAP. V.] PROJECTILES. 99 The path of a projectile is shown in fig. 9, sect. 29 ; the simplest case being where the initial velocity is at right angles to the force of gravity, or horizontal. Thus a rifle-bullet, starting with an initial horizontal velocity u, retains this velocity unaltered, if we neglect friction against the air, and therefore in t seconds it travels a horizontal distance ut ; but its vertical velocity, which at first was zero, continually increases, and in t seconds is gt ; the vertical space fallen through being \gt\ or just the same as if the gravity had acted upon the body at rest. The whole circumstances of the motion of such a projectile have therefore been already worked out in sect. 29 ; which see, and read again. If the rifle is fired horizontally from the top of a cliff of given height, say 144 feet, it is easy to find how far the bullet will go before striking level ground, its initial velocity being known. Let the initial horizontal velocity be 1200 feet per second. We must first find t, the time the bullet takes to fall from the top of the cliff to the ground, from the equation 144 = 16t 2 (for it takes just the same time as if it had no horizontal velocity. Law II., sect. 49); this gives 2 =9, or = 3. It goes therefore three seconds before striking the ground, so evidently the horizontal distance it travels is 3 x 1200 = 3600 feet. And generally if h be the height of the cliff, and u the initial horizontal velocity of the bullet, its range, or horizontal distance, is \u JJi. 75. The composition of these two motions, a uniform horizontal velocity with a uniform, vertical acceleration, is well illustrated by Morin's machine, for a description of which see Deschanel or Ganot. It consists of a long drum or cylinder, capable of rotating by clockwork about a vertical axis. Down one side a weight can fall between guides, and can, by means of a pencil, mark a line on the drum as it falls. If the drum is stationary, the line drawn is, of course, straight and vertical ; but if the drum rotates, it is spread out into a curve. This curve, when un- 100 ELEMENTARY MECHANICS. "SECT. 75. ^\\ \ \ \ \ \ \ \ \ v \ \ \ \ \ \ 1 \ wrapped from the drum, is precisely the same as that which is described by a projectile shot out horizontally in vacua with a velocity equal to that imparted to the sur- face of the drum by its clockwork. The drum is usually covered with paper, ruled into squares or oblongs, which can be detached and unrolled. The Hue traced on it may then present the appearance shown in fig. 21. In successive seconds the horizontal distances are as 1, 2, 3, 4, 5, the vertical as 1, 4, 9, 16, 25, and so on. A curve with this property is called a parabola. It is the path of a projectile in a vacuum (compare sect. 29). 76. The simplest method of dealing with projectiles is to resolve their initial velocity of projection into a horizontal and a vertical component, and to treat them separately. The horizontal motion is not subjected to any force except the resistance of the air ; the vertical motion Fig< 21 * is subject to gravity as well. It is a complicated matter to take the resistance of the air into account ; especially since, if the projectile is spinning, the air resistance directly alters its path as well as its speed. Suffice it to say that the simple parabola could only be really attained in a vacuum, and that the path of a simply thrown cricket-ball is an unsym- metrical curve with its descending portion shorter and steeper than its ascending portion. In the case of a golf-ball, the spinning motion so complicates matters that it may travel straight for some distance, and then actually rise upon the air resistance and drop fairly dead. Or, if cut sideways, like a racquet-ball, it may describe a curved path not by any means in one plane. We shall limit ourselves to the case of a body shot from level ground in a vacuum, with a certain initial velocity inclined at a certain angle to the horizon. Given these data, it is easy, by constructing a parallelogram, or otherwise, to determine its horizontal com- ponent, which we will call u, and its vertical component, v. Then the following statements should be proved, and examples on them should be worked : The time, T, taken for a projectile to reach its highest point is given by v = T. The centripetal force which must be acting on P in the direction PN to keep it moving in the circle (sect. 60) is wiwV, where r is the radius PN of the circle in which P moves ; and if the rotation were to cease, this is the force which must be applied in the opposite direction PF, in order to keep the ball in its position without letting it fall back to the axis AB. Hence in the diagram (fig. 22), we may regard P as stationary and in equilibrium under the action of three forces the force F=wzw 2 r, its weight W=mg, and the tension in its supporting arm. The triangle APN has its sides parallel to these forces, and hence represents them; so, calling the vertical distance AN, h, we have : nig : : r : h ; that is, the vertical distance of the governor ball below the pivot A is inversely proportional to the square of the angular velocity of rotation. The time of one revolution is T = -=27r y ./-; and such an arrangement is sometimes used as a measurer of time, when it is called a ' conical pendulum,' because the arm AP traces out a cone. If the radius of the circle in which P moves is very small, the height h is practically equal to the length of the pendulum, AP, which we will call I. Moreover, if you try swinging a weight at the end of a string, you will find that the time of a complete small motion is the same whether the pendulum simply oscillates in a nearly straight line or whether it revolves in a horizontal circle or in any other elliptical curve ; that is, the time of an CHAP. V.] PENDULUMS. 105 oscillation (to and fro) of a simple pendulum equals the time of rotation of a conical one, prqvided the motion of both is small ; and each period is very approximately By a simple pendulum is meant one about whose length there can be no ambiguity. It is a heavy particle, swinging at the end of a perfectly light cord attached to a fixed point. EXAMPLES XIII. (1) Find the time of beat (half an oscillation is called a beat) of a .simple pendulum 39 inches long at the equator, Avhere # = 32-09. (2) Where the length of a pendulum which beats seconds (called the seconds pendulum) is 39 inches, find the value of g. (3) If gravity is ^ T greater at the north pole than at the equator, how many seconds a day will a seconds pendulum at the north pole lose when taken to the equator ? (4) How many seconds a day will a clock lose whose pendulum (intended to beat seconds) is a metre long, at a place where the intensity of gravity is 981 dynes per gramme ? Find the length of the seconds pendulum at the same place. (5) If a simple pendulum 39 inches long beats seconds, what should be the length of one which shall beat 40 times in one minute ? (6) Calculate the number of beats per day made by a simple pendulum 40 inches long at a place where the length of the seconds pendulum is 39 inches. (7) If a pendulum 39 inches long is gaining a minute a day, how much should it be lengthened to keep correct time ? (8) What is the length of the seconds pendulum at the latitude of Greenwich ? (See sect. 64.) (9) What is the value of g at a place where a simple pendulum 2^ inches long makes two complete oscillations a second ? (10) Show that the time of beat (in seconds) of a simple pendu- lum is, approximately, 0'16V(l en gth in inches). Find the value of g for which this formula is accurate. (11) Show that, if # = 981 dynes per gramme, the time of a beat (in seconds) = l'00303\/(length in metres). Find the number of beats per minute of pendulums whose lengths are respectively 9 cm., 25 cm., and 16 metres. 106 ELEMENTARY MECHANICS. [EXS. xin. (12) Find the number of beats per minute of a pendulum 3 feet long, using the formula of Ex. (10). .(13) If a weight be attached to a string 4 feet long, and is then caused to describe a horizontal circle, so that the string is inclined at 60 to the vertical, find its angular velocity, its actual velocity, and the time of one revolution. 80. Compound Pendulum. The time of oscillation of a com- pound pendulum, that is, of a rigid body of any size pivoted on an axis through O, and swinging slightly under gravity, may now be calculated. Let G be the centre of gravity of the mass, and call the distance OG, a ; the small angle of displacement from the vertical, NOG, call 6 ; and the distance NG, call x ; the latter is prac- tically equal to ad, the arc of a circle with centre 0. Then, if m be the mass of the whole body, the force restoring the body to its position of equilibrium is mg acting at G, so that its moment about is mgx ; and the angular acceleration produced by this is (see sect. 56) _mgx_mga6, where I is the moment of inertia of the body about the point O. For the particular case of a simple pendulum, when the whole mass is concentrated into a particle at G, and when a=l and I=mP, this equation becomes vnulQ aQ *=- = - Fig. 23. Now we can choose a simple pendulum of such length that its angular acceleration at every instant, and therefore its whole motion, is the same as for the compound pendulum. Let L be the length of such an equivalent simple pendulum, then the equation mc/a6 qd -T- = -L is satisfied ; and the length of the equivalent simple pendulum (sometimes called the * length ' of the compound pendulum itself) CHAP. V.] PENDULUMS. 107 But the time of a small oscillation of this simple pendulum is therefore the time of a small oscillation of the compound pen- duluni is mga where I stands, as already stated, for its moment of inertia about the centre of suspension O, and a is the distance between this point and its centre of gravity. The above equation l = inaL gives a simple means of experi- mentally determining the moment of inertia of any body about any axis. Hang it up by this axis and measure a, the distance from it to the centre of gravity ; then set it swinging slightly, and observe the length of a simple pendulum which keeps time with it : multiply the product of these two lengths by the mass of the body (in Ibs. or grammes), and you have its moment of inertia under those circumstances. 81. Centre of Oscillation or Percussion. A point O' in a swinging body, situated in OG produced, at a distance L (the length of the equivalent simple pendulum under the circum- stances) from the centre of suspension O, is called the centre of oscillation, because the body oscillates as if a portion of its mass were concentrated there, the rest of it being at O, in such a way that G remains the centre of gravity ; this may be verified as regards mere time of swing by the experiment of swinging the body and a simple pendulum or plumb-bob together, and observ- ing that, when of the proper length, the motions of the simple and of the compound pendulums are identical. It may be shown that the body will swing in just the same period if suspended at this point O' as if it were suspended at O. This depends upon the fact that if I be its moment of inertia about a point O, at a distance a from G, and I' its moment of inertia about a point O', at a distance a', such that a + a'=L, then I:a = I':a'; or the length L is the same for both points. A line through O', perpendicular to OG, and to the axis of suspension, is sometimes called the ' line of percussion ' or the centre of percussion, because this is the place where the body strikes things best without any jar on its support. A cricket-bat drives the ball best if the ball strikes it at a point on this line, and it does not then jar the hand. 108 ELEMENTARY MECHANICS. Fig. 23ct. [SECT. 81. To find all the points about which a rigid body (say for in- stance a flat plate or board pierced by a pin which supports it, fig. 23) will swing in the same time as about any point O, determine experimentally the length of the equivalent simple pendulum, say by adjusting a plumb-bob to swing in the same time as the plate, and measuring its length ; then mark this length upon the plate, as OO', and draw a couple of circles with centre G, through O and through O' respectively ; the plate will swing in the same time if pivoted at any point on either of these circles. If pivoted at an axis between the circles, it will swing more quickly; if pivoted inside the smaller circle or outside the bigger circle, it will swing more slowly. The experiment should be tried; and if "some care is bestowed upon it, and a series of pairs of such circles recorded, the result will be instructive to a student. The product of the radii of every such pair of circles will turn out the same ; and, when multiplied by the mass of the body, it will represent the principal moment of inertia of the body namely, that about the centre G. The circumstances of a swinging body pivoted on a line or axis at any point A may be stated in terms of figure 24, where G is the centre of gravity and AKB is a right angle (or semicircle). The moment of inertia of the body, about an axis parallel to the pivot but drawn through G, is the mass ;-K multiplied by GK 2 . The moment of inertia about the pivot A is the mass multiplied by AK 2 , and that about B is the mass multiplied by BK 2 . A is the centre of suspension, B is the corresponding centre of oscillation, and a line through B perpendicular to AB is the line in which a blow must act to Fig. 24 spin the body automatically about A without any force from the pivot. CHAP. V.] PENDULUMS. 109 The length of the equivalent simple pendulum is AB, and the axes A and B are interchangeable. Expressing these facts algebraically : k z =ab, R 2 II is called the swing-radius of the body about the axis A, and is such that I. =mR? while L =mk\ The whole dynamic and static behaviour of the body is like that of two heavy particles rigidly connected, the one at A and the other at B ; the mass of the one at A being m t and the mass of the one at B being m, so that a+b a + b their centre of gravity is the same as that of the body. 82. Ballistic Pendulum. A heavy block of wood hung up as a pendulum by two strings, so that it can swing without any rotation, is sometimes used to measure the impulse (mv) of a blow, such as that of a rifle-bullet fired into the wood. The block will be displaced and will rise a vertical height, h, which must be observed (either directly or by calculation from the angle of swing) ; and the velocity v imparted to it is calculated as J(2gh). The velocity v with which the rifle-bullet struck the wood can then be found, if the mass m' of the block is known, from the equations, mv = and v' = EXAMPLES XIV. (1) A uniform rod 3 feet long is swung as a pendulum about one end. Find the length of the equivalent simple pendulum. (2) Find the point about which the above rod should swing so that the time of oscillation may be a minimum, and find that minimum time. In this case the centres of suspension and oscillation must be equidistant from the centre of the rod see end of sect. 81. 110 ELEMENTARY MECHANICS. [EXS. XTV. (3) A uniformly thick rigid door on smooth hinges is shot at with a hullet. Find where and how the bullet must strike the door so as to cause no jar on the hinge. (4) A one-ounce bullet fired horizontally into a 20-lb. block of wood suspended by two strings displaces it so as to rise 3 inches. Find the speed of the bullet. (5) A 1-foot sphere hanging by a 6-inch string oscillates like a pendulum. Find its time of swing and the length of the equivalent simple pendulum. (6) Find the period of a rectangular thin plate 4 feet x 3 feet swinging in its own plane about one corner. (7) Find where else it may be suspended to swing equally fast, and find how it must be hung to swing fastest. (8) A triangular plate of height h and mass ra swings about its base. Find the length and mass of the equivalent simple pendulum. (9) The same plate swings about an axis parallel to its base, bisecting the sides. Find the length and mass of the equivalent simple pendulum. (10) ShoAv that a triangular plate has the same moment of inertia about any axis as a system of 3 equal particles (each = J of the mass of the plate), situate at the middle points of its sides, and find the time of swing of a triangular plate, with sides each 4 feet long, swinging in its own plane about a vertex. (11) Find the mass of the equivalent simple pendulum in Ex. (1), if the mass of the rod is 4 Ib. (12) A chair weighing 20 Ib. is hung by a point 2J feet from its centre of gravity, and is found to oscillate in precisely the same way as a simple pendulum 3 feet long. Find the moment of inertia of the chair about the point of suspen- sion. (13) Find the time the chair would take to complete a small oscillation. (14) A one-ounce rifle bullet is fired into a suspended block of wood weighing 30 Ih. If the blow causes the wood to rise a vertical height of 1^ inches without any rotation, find the velocity of the bullet just before it struck the wood. CHAP. VI. 1 WORK AND ENERGY, 111 CHAPTER VI. WORK AND ENERGY. 83. The present chapter is to indicate a method of treating the effects of force on matter in a perfectly general manner; all consideration of how the force acts, or what it acts on, being regarded as accidental and secondary. Whether the body acted on is a particle, or a rigid solid, or an elastic solid, or a liquid, or a gas, matters nothing; and whether the effect produced is motion, or strain, or both, or neither, also matters nothing. It is to treat of the effects of force in general on any body whatever. 84. Now, in order that an agent exerting a force may produce any effect on the body to which it is applied, it is necessary that the body shall yield somewhat that is, that the point of application of the force shall move in the direction of the force ; and whenever this happens when- ever the point of application of the force does move along its line of action some effect is necessarily produced. Thus either the body is set rolling, or swinging, or moving in some way, or its motion is checked, or it is squeezed into smaller compass, or bent out of shape, or it is lifted up against gravity, or it is merely shifted along against friction, or it is warmed or electrified ; no matter what the effect is, some effect is always produced, and the force, or more pro- perly the agent exerting the force, is said to have done ivork. Moreover, a body upon which work has been done is found to have an increased power of doing work itself that is, of producing physical changes in other bodies; and it is therefore said to possess more energy than before. This 112 ELEMENTARY MECHANICS. [SECT. 84. increase of energy is indeed the most essential part of the effect produced in a body by an act of work. 85. Energy therefore is that part of the effect produced when work is done upon matter which confers upon the body possessing it an increased power of doing work. 86. The work done in any case is proportional both to the magnitude of the force and to the distance through which its point of application moves in the direction of the force. Unless the point of application moves, no work is done and no energy is produced, however great the force may be ; for instance, a pillar supporting a portico is doing no work, though it is manifestly exerting great force. Work, tlien, is the act of producing an effect in bodies by means of a force whose point of application moves through a distance in its own line of action, and it is measured ~by the product of the force into the distance* or Such a force is conveniently called an ' effort ' if the motion is in its own direction, and a ' resistance ' if the motion is against it. (If the force and motion are inclined to each other, only one component of the force is the effective effort.) The work is reckoned positive, and is called simply 'work,' when the body acted on is moved in the same sense as the force ; if, however, by any means, it be caused to move in opposition to the force exerted by an agent, the work done l>y that agent must be reckoned negative that is, work is done upon it. Thus if a force of five poundals acts through a distance of six feet in its own direction, it does thirty times the work which would be done by one poundal acting through one foot. This latter work may be called a foot-poundal, and represents the F.P.S. absolute unit of work. British engineers use as their practical unit the work * The inoment of a force was also defined as a force multiplied by a distance, but by a distance measured at right angles to the force. It is therefore an entirely different thing from work. CHAP. VT.] WORK AND ENERGY. 113 done by an effort equal to the weight of a pound acting vertically through a foot, and they call this a foot-pound. It is of course equal to 32 foot-poundals approximately. French engineers use the kilogrammetre as' their practical unit of work, being equivalent to the weight of a kilogramme raised a metre. It is equal to 98,100,000 absolute C.G.S. units. The C.G.S. unit of work is called an erg (from fpyov, work), and is the work done by a dyne effort acting through a centimetre. There are 421,390 ergs in one foot-poundal. Since the erg is so small a unit, it has recently become customary to employ ten million ergs as a more convenient unit of energy for some purposes, and to call it a Joule. It is approximately f of a foot-lb. The multiplicity of units is admittedly troublesome to present-day students, but it is one of the penalties they pay for living in an age of transition and activity, The algebraic expression or ' dimensions ' of an absolute unit of work is P" ntl (f " t)2 or g 1 " 116 ( m ' )8 ; being essentially (second) 2 (sec. ) 2 a momentum multiplied by a velocity, or, what is the same thing, a mass multiplied by an acceleration and a length. The gravitational unit, or foot-pound, is a weight multiplied by a vertical height, and is frequently convenient; though it is liable to be hastily misinterpreted as an incomplete specifica- tion, the 'dimensions' of acceleration necessarily involved in it ( 2 ) being ignored, instead of being only taken as suppressed \S6C. / or ' understood.' 87. The effects produced in material bodies when work is done upon them are various, and constitute the different forms of energy. The full discussion of the subject of energy belongs to the science of physics, so we can here only just roughly enumerate its principal forms. (1) Motion (whether translation or rotation). 114 ELEMENTARY MECHANICS. [SECT. 87. (2) Strain (whether extension, compression, or dis- tortion). (3) Vibration (including the particular kinds called Sound). (4) Heat (sensible and latent). (5) Radiation (including the particular kinds which are able to affect the eye, and which are therefore called Light). (6) Electrification. (7) Electricity in motion. (8) Magnetisation. (9) Chemical separation. (10) Gravitative separation. To these we ought perhaps to add vital energy, only that it may be held to be included under head 9. It is quite possible that many of these may reduce to simpler forms; in fact, all but Nos. 9 and 10 are already pretty well known to be special cases of Nos. 1 and 2 (cf. sect. 5). It is usual to consider those forms of energy which are more directly connected with large and visible masses of matter as more particularly the province of mechanics'; and we shall here discuss only these more mechanical forms of energy, Nos. 1, 2, and 10. The essential nature of No. 10 is at present unknown (see Introduction, sect. 3), but for most practical pur- poses it comes under the class indicated by No. 2. 88. Now the question arises When work is done and energy produced, is it created out of nothing, or is it only manufactured from previously existing materials'? The latter is the truth, for it has been found, as the result of innumerable experiments on the subject of 'perpetual motion' and others, that it is as impossible to create energy as it is to create matter, and that whenever energy appears as the result of work, it is always at the expense of some other form of energy which was previously existing. This fact is popularly expressed by saying that ' perpetual motion is impossible ' a statement which requires inter- pretation, because if there is one thing more universal than CHAP. VI.] WORK AND ENERGY. 115 another it is perpetual motion (see sect. 4). The statement, however, is understood to be an abbreviation for the follow- ing : It is impossible for us to construct any machine which shall move and do work (and therefore generate energy) of itself without consuming at least an equal quantity of pre-existing energy. 89. All this indeed, in a much more complete and accurate form more complete, because it involves the non-destruc- tion of energy as well as its non-creation follows from Newton's third law of motion, sect. 52, provided we assume that energy is to be measured by work done that is to say, that when a body does work, it loses a precisely corre- sponding quantity of energy, and that when a body has work done upon it, it gains an amount of energy equal to that work. For the third law tells us, that whenever force is exerted, and therefore (a fortiori) whenever work is done, two bodies are always concerned there is the body which acts, and the body which is acted upon or re-acts and these two bodies exert .equal and opposite forces ; hence whatever quantity of work one body does, the other has done upon it ; or the positive and negative works are equal (see sect. 86, small print). The ' agent,' or body which does the positive work, loses a certain quantity of energy. The body which has the work done upon it gains the same amount. Hence, on the whole that is, taking both bodies into account no energy is lost, and, algebraically speaking, no. work is done. The energy is merely transferred, and the act of transfer involves two equal opposite works. The law that, on the whole, no energy is ever created or destroyed by any forces which we know of and have experimented upon, is called the law of the ' Conservation of Energy' 90. Just in the same way then that a force is the partial aspect of a stress, so work is the partial aspect of a some- 116 ELEMENTARY MECHANICS. [SECT. 90. thing which consists of action and re-action, in the sense of work and anti-work, but which neither has, nor as yet perhaps needs, any name ; and whenever we speak of * work done,' it will be by attending to the action of one body on another, and neglecting the reaction of that other on the one. To summarise then : "Work creates energy, anti-work destroys it, so both together simply transfer it. If it were possible to have a force without its anti-force, it would also be possible to get work done without its anti-work, but as a fact of experience it is not possible. 91. The fact that work is done whenever energy is trans- ferred, taken in connection with the experience that energy often manifests a tendency to transfer itself from one body to another, and thereby to do work, has caused energy to be denned as the power of doing work. Now certainly a body possessing energy thereby possesses the power of doing an equivalent amount of work, provided the energy is of such a sort that it can be transferred to some other body ; and in this sense energy and power of doing ivork are equivalent, though it is more precise to say that the possession of energy confers upon a body the power of doing work, than to say that energy is the power of doing work. It is quite pos- sible, however, for a body to possess energy and yet have practically no power of doing work, for energy is not always available. Thus, a stone lying on the ground may be said to possess an amount of energy corresponding to its fall to the centre of the earth, but this energy confers on it no power of doing work, for it would be impossible to let it fall without first expending a great deal more energy in digging a hole. Again, energy is indestructihle, and a given quantity may be transferred from one body to another, from A to B, from B to C, from C to D, and so on and back again, each time conferring upon its possessor a power of doing work, which work is done at each transfer by the body losing it. Hence, if it were correct to speak of work as being done by the energy, instead of by the body CHAP. VI.] WORK AND ENERGY. 117 possessing the energy, the working power of a given quantity of energy might be unlimited, and at any rate would be wholly in- commensurate with the quantity of energy. The power of doing work, in fact, does not depend solely on the absolute quantity of energy in a body or system, but on its capability of being trans- ferred to other bodies or systems. We seldom have to deal with the total or absolute energy in a body, but only with its variations. 92. There are, however, practical difficulties in effecting such a series of transfers of energy without loss of working power, for though the quantity is unalterable, yet the quality has a tendency to deteriorate. These practical difficulties are very similar to those which you would experience if you attempted to transfer a given quantity of water down a series of vessels. For yoii might spill some, some would evaporate, some of the vessels might leak, and all would remain wet. The quantity of water would be unchanged it would be all there but some of it would be unavailable. It would be not destroyed only useless. Just so with energy, whenever it is transferred from one body to another that is, whenever work is done, some of it is pretty sure to pass into a less available and more useless form. Its quantity is not altered, but its availability is less. This tendency of energy to become less available is called the law of the Dissipation or Degradation of Energy. It may be expressed thus : When energy is transferred from one body to another, it is also always transformed from one of its forms to another, and some portion of the new form is pretty sure to be lower in the scale of energy than the original form ; because of friction, imperfect elasticity, and so on. It is, in fact, impossible by any known process to raise energy in the scale of availability on the whole. Any given quantity, indeed, may be raised, but some other greater quantity will in the operation be degraded. The average is usually lower, and cannot be higher. The energy of the earth in its orbit is not available to us. The energy of a flying molecule is almost unavailable, because we have as yet no means of dealing with molecules singly ; if we could see and handle them, their motion would be as high a form of energy to us as the motion of other visible masses. Hence the distinction between high and low forms of energy is a purely relative one. Energy falls in availability usually by becoming molecular 118 ELEMENTARY MECHANICS. [SECT. 92. that 1% by being transferred from visible masses to their ultimate molecules. This transfer is effected by friction and viscosity ; hence friction and viscosity are the chief practical causes of the dissipation of energy which is perpetually going on. No known means exist whereby energy is automatically raised in the scale of availability, but it has been surmised that perhaps this power appertains to certain forms of life. 93. Energy and work are not to be confounded together ; and all such phrases as ' accumulated work,' ' conservation of work,' ' work consumed,' &c., should be eschewed, or else regarded as permissible colloquial inaccuracies. Energy is not work, but work can be got out of it if the proper condition be supplied. Energy might therefore be called possible work. For consider the two fundamental forms of energy : (1) The free motion of masses of matter relatively to one another; and (2) The separation of masses of matter from one another against stress. In the first case, the body possessing the energy is moving through a distance, but is not exerting any force. Supply a resistance, and work is immediately done. In the second case, the body possessing the energy is exerting force or pressure, but it is stationary. Allow it to move, -and work is immediately done. The two fundamental forms of energy, therefore, corre- spond to the two factors in the product called work namely, F and s. The first form corresponds to s; there is motion through space, but no force. The second corresponds to F ; there is force, but no motion. The first is called Kinetic Energy, or the energy of motion ; the second might be called Dynamic Energy, or the energy of force (properly stress) ; or it might be called Static Energy, to distinguish it from Kinetic. As a matter of fact, however, it is generally called Potential Energy, which is not a bad name so long as it is not misunderstood j 1 C UNIVERSITY CHAP. VI.] WORK AND ENEKGY. 119 to moan, possible energy a phrase without sense. Neither is Kinetic ever to be called Actual Energy. All energy is actual and real potential just as much as kinetic ; and both represent possible toork that is, work that will become actual as soon as the other factor is supplied. * Possible work ' merely means possible transfer of energy ; just as money and goods, which might be called forms of mercantile energy, represent a possible transfer or ex- change. 94. Whenever work is done, both factors, and therefore both kinetic and potential energy, must be present ; and the energy is always passing from one of these forms into the other while the work is being done. For if the motion is ivitli the force, the speed must increase, and if it is against the force, it must decrease ; while in the first case the distance through which the force can act, or the range of the force, is decreasing, in the second increasing. The energy of a vibrating body is continually alternating from one form to the other. Enough has now been said to show' that the energy method of treating forces and their effects is a very general one, and extends to the whole of Physics. But the branch of the subject concerning which we can here enter into any detail will be a very small one, and will only extend to giving some examples of the transformation of energy from form 1, that of motion, to some other form, especially that of gravitative separation, and back again. Measure of Kinetic Energy. 95. First consider how to measure the energy of motion in the case of simple translation of a particle ; remembering that its energy (more strictly its gain of energy, over and above any other forms of energy, such as heat, &c., which it may retain constant all the time) is defined as^ equal to the work done by the force which caused the motion. 120 ELEMENTARY MECHANICS. [SECT. 95. Now when a force F is applied to a mass m, the accelera- tion is a = - (Chapter IV.), m x r and the velocity generated when a body moves, from rest, a distance s, with the acceleration a, is given by p v* = 2as (Chapter II.), that is, v 2 = 2-s ; 111 an equation readily written in the form J?s = fynv 2 . But Fs equals the work done by the agency of the force while it acts through the distance s; and as energy is measured by the work done in its production, it follows that the energy of a body of mass m moving with velocity v, is ^mv 2 , because v is the velocity generated in the body during the performance of the amount of work, Fs. This expression, mv 2 , is a most important one, and it is called the kinetic energy of a particle due to its motion relatively to the body which is supposed to be at rest usually, of course, the earth. It equals the work that has been done upon the body in setting it in motion, and also the amount of work Avhich it must do in order to stop itself that is, to transfer its energy to some other body, either to the earth or to anything else which happens to' come in its way. When one suspended elastic ball impinges directly on an equal one at rest, the first stops dead, and the other receives the whole motion ; the energy has been here obviously transferred. The transference takes place just as really, though not so obviously, in every case where a body comes to rest or starts moving. The unit of energy is the equivalent of the unit of work, and usually goes by the same name (sect. 86). For instance, the British unit of energy would be a foot-poundal, being the energy produced, or, rather, transferred, by the action of CHAP. VI.] WORK AND ENERGY. 121 unit force through unit distance; the C.G.S. unit of energy would be a dyne-centimetre, or one erg. 96. If a body, instead of being at rest when the force acted on it, had been moving with velocity V Q , it would have already possessed the energy ^mv Q 2 ; and so the gain of kinetic energy, equivalent to the work done, would have been where v 1 represents the final velocity possessed after the force has acted for a distance s. This immediately follows from the old equation v-f v Q 2 = 2os, if we write ~F/m for a, and leave the term F by itself on one side of the equation. ILLUSTRATIONS. 97. A truck of mass 2000 Ib. running along a level line at the rate of 20 feet a second, has an amount of energy equal to (90 ft \ 2 - ) =400,000 foot-poundals, SGC. / or 12,500 foot-pounds. If it were required to stop it in a distance of 500 feet, we should have to apply a brake exerting 800 ponndals, equivalent to a retarding load of 25 pounds-weight ; for the work done by the truck against this force in the given distance would be 25 pounds-weight x 500 feet, or 12,500 foot-pounds, which is precisely the energy of the truck required to be destroyed, or rather to be transferred to something else. One can always find the force necessary in any such case by dividing the work required by the distance given ; for, of course, Again, to propel a one-ounce rifle-bullet dVth Ib.) with a velocity of 1200 feet per second, will require work to be done upon it equal to the energy generated namely, = 45,000 foot-poundals, or about 1400 foot-lb., or f- of a foot-ton. (This energy, and a good deal more, was contained in the charge of powder in the form 122. ELEMENTARY MECHANICS. [SECT. 97. of chemical separation, No. 8 (sect. 87) ; a quantity is always wasted in the useless noise and flash attending the explosion, Nos. 3, 4, arid 5.) This work must have been done by the powder while the bullet was travelling from the breech to the muzzle of the gun, a length of say four feet j hence the average force exerted by the powder must have been 45,000 foot-poundals divided by 4 feet, or 11,250 poundals, equivalent to about 3 cwt. Suppose now in passing through the air it loses 400 of its velocity by friction, so that it reaches the target with the velocity of only 800 feet per second, then the energy of the blow will be ^ x T V x (800) 2 , or 20,000 units ; while that which has been ' lost ' by friction (that is, transferred, some to the air and some to the molecules of the bullet, but in any case debased into the form of heat) is 45,000-20,000, or 25,000 F.P.S. units of energy ; and this must be the number of units of work which have been done by the flying bullet against the resistance of the air. Hence if its range, or distance travelled, were 1500 feet, the average resistance exerted by the air must have been 25,000 divided by 1500, or 16f units of force, equivalent to the weight of about half a pound. Finally, let a target stop the bullet dead in the space of inch (^th of a foot), then, since the whole (negative) work it has to do is numerically equal to the energy of the blow namely, 20,000 units it follows that the average force of the blow on the target is 20,000 divided by ^, that is, 960,000 units of force, or about 13| tons weight, a much greater force than even the powder exerted ; and this is apparent in the results, for the bullet is flattened out by the target, while the force of the powder had but a slight effect upon its shape. Very likely an iron target would not yield so much as \ inch ; if it only yielded half as much, the force of the blow would be doubled. Whether the bullet bounces off or not, matters nothing ; it must have been stopped before its motion can be reversed. The reverse motion would not alter the force required to stop the bullet, but it would increase its impulse (sect. 48) by lengthening the time during which the force was exerted against the target. Thus, if the bullet bounced off with its original speed, the time and therefore the impulse would be double what they would have been if it had stopped dead like dough. CHAP. VI.] WORK AND ENERGY. 123 98. Notice the distinction between the energy of a blow, the impulse of a blow, and the force of a blow. The energy equals Fs, or ^mv 2 . The impulse equals F, or mv. rpi t i i7 ^ mv<2 mv , N ihe average torce equals .b, or , or (ci. sect. 45). 8 t/ It will be a good exercise to find from this last equation, in all the above cases, the time taken to do the work that is, to transfer the energy. For instance, find the time of flight of the bullet, and also how long it took to travel the length of the gun, and so on. It is worth noticing that work = force x distance moved = impulse x average velocity = momentum generated x average velocity v ) x $( Vj_ + v ) These two expressions for an average force, and t mv 2 2 , are worth comparing. The first we know is expressed in words by saying that force is rate of change of momen- tum ; rate here having a reference to time, and meaning the increase per second of time elapsed. Similarly the second may be expressed by saying that force is rate of change of energy, only rate here has a reference to distance, and means the increase per linear foot of distance travelled. The whole subject of the rates of variation of things with respect to different variables, considered as a branch of pure mathematics, is called the differential calculus, a science the foundation of which was laid by Newton and developed by Leibnitz for the purpose of treating questions concerning velocity, acceleration, and the like. Measure of Potential or Dynamic Energy. 99. Now let us consider how to measure potential energy, or the energy of stress, especially in the form of gravitative 124 ELEMENTARY MECHANICS. [SECT. 99. stress exerted between the earth and a raised weight. This is a very simple matter, for, supposing a stone is at a height h, we have a constant force mg exerted on the body, and a distance h through which it can act, so the work it can do while the stone falls is simply m 25. alent' of the heat generated. If /= 0, that is, if the plane be smooth, the velocity v acquired in descending the vertical height h down the plane is the same as that found for a freely falling body in the last section, and it has no connec- tion with the slope of the plane ; showing that the path of a falling body has no influence on the velocity acquired by it, provided everything be smooth. (The time of descent is greatly influenced by the path.) 103. The simplicity of gravitation examples is due to the fact that the force acting (the weight of the raised body) is constant and does not alter as the weight descends. But in every case, if s be the range that is, the distance through which the force can act and if F be the average value of this force, the potential energy is Fs. (See Appendix, p. 307.) Energy of Rotation. 104. So far we have only considered energy of motion in the form of translation, or the motion of a particle ; but the energy of a rotating body can now be easily expressed, since CHAP. VI.] WORK AND ENERGY. 127 it is made up of particles, and the energy of the whole is the sum of their separate energies.* Any particle of mass m, at a distance r from the axis of a body rotating with angular velocity o>, is revolving round and round a circle with velocity v = ro>, and its energy is \mv 9 ', or, as it may be also written, Jw? i2 a> 2 . Now the energy of the whole body is the sum of the energies of all the particles in it ; it is therefore for, since the to is constant, it may be taken outside the sign of summation ; but 2(r/M' 2 ), the sum of the second moments of inertia of all the particles in the body, is the quantity we have called the moment of inertia of the rotating body Fig. 26. (sect. 43), and denoted by I ; hence the simplest expression for the energy of a rotating body, like a flywheel, is Jlto 2 . A flywheel mounted on a vertical axle can be started spinning by a descending weight, as shown in fig. 26. * Notice that the parallelogram law (sect. 26) does not apply to the composition of energies. Energy is not a directed quantity, and simple arithmetical addition applies to it. 128 ELEMENTARY MECHANICS. [SECT. 104. By the time a weight of 14 Ib. over and above what is needed for friction has thus descended 6 feet, it will have done 84 foot-pounds of work and generated this amount of kinetic energy, of which a portion will belong to the wheel, and a portion to the weight. If it has taken 10 seconds to descend, its acceleration has been '12, and its kinetic energy is 14 x -12 x 6 = 10-08 foot-second-units, or, what is the same thing, Jxl4x(l*2) 2 . This is less than one-third of a foot-lb., and all the rest of the energy belongs to the wheel. If the winding pulley were *6 foot in circumference, it would now be making two revolutions a second, therefore its angular velocity would be 47r, and so its moment of inertia would be determined by the energy equation JI(47r) 2 = 84x32 -10-08. 105. A rolling sphere or cylinder has a motion compounded of a translation forward and a rotation round the centre of its rim, and consequently its energy is similarly compounded. Its translational energy is ^mv 2 , where v is the speed of its centre forward ; and its rotational energy is -Jlto 2 , where I is the moment of inertia of the body about its centre ; but a simple relation holds between v and o>, since the speed of rolling advance is the same as the speed the rim would have if the centre were stationary ; wherefore the above two terms may be added together, making -|(I + mr 2 )w 2 . This suggests that the body is really rotating at each instant about the point where its rim touches the ground, and that its mo- ment of inertia about any point on its rim is greater than that about its centre by m?* 2 ; at any rate these statements are true ones, as may be seen by referring to the list of Moments of Inertia, sect. 43. EXAMPLES XV. (1) A body slides down a rough plane, travelling 20 feet along the plane, but only descending 12 feet vertically. If the force of friction is equal to th of the weight CHAP. VI.] WORK AND ENERGY. 129 of the body, find the velocity gained during the descent. (2) What is the work that must be done in order to propel a 3-lb. stone at the rate of 40 feet a second ? (3) A simple pendulum is pulled aside till its heavy bob is raised 3 inches, and is then let go. Find its velocity when it passes its lowest point. (4) What initial velocity is necessary to make a rifle bullet strike a target placed 300 feet high vertically above the gun with a velocity of 600 feet per second, neglecting the resistance of the air ? (5) What would be the answer to the last question if the bullet weighed an. ounce, and if the resistance of the air were taken to be equivalent to a drag-back equal to the weight of 5 ounces ? (6) Tf a projectile were started in any direction with the velocity 80, and arrived at another point on the same level with the velocity 30, after having travelled 150 feet, what must the average resistance of the air have been equal to ? (7) What force would be necessary in order to stop the projectile of Question No. 2 in the space of 6 feet, and how long would it take ? (8) Find the mechanical equivalents of the heat generated by friction in the motions considered in Questions 1, 5, 6, and 7 ; assuming the mass to be 3 Ib. in each case. (9) What is the energy of a hollow globe 2 feet in diameter which is swinging round in a horizontal circle at the rate of 90 revolutions per minute, the mass of the ball being 5 Ib., and the radius of the circle described by its centre 3 feet? Consider the string so nearly horizontal that practically the ball rotates once during each revolution. (10) What is the energy of a uniform steel disc a yard in diameter and an eighth of an inch thick rotating about a vertical axis 3000 times a minute, if a cubic inch of iron weighs 4 Ib.? (11) What is the kinetic energy of a 5-cwt. projectile moving with a velocity of 2000 feet per second ? (12) A body whose mass is 12 Ib. moves from rest with a uniform acceleration of 100 inches-per-second per second. Calculate its kinetic energy after it has moved a distance of 20 feet. (13) Find what work is being done per minute that is, find the I 130 ELEMENTARY MECHANICS. [BXS. XV. activity or the power of an engine which is raising 2000 gallons of water an hour from a mine 300 feet deep. A gallon of water weighs 10 Ib. (14) A train of 150 tons is running at 60 miles an hour. What force is required to stop it in a quarter of a mile ? (15) A small heavy hody weighing 20 Ib. slides down a rough circular arc 10 feet in radius whose plane is vertical. It begins to move from one end of a horizontal diameter, and is found to reach the lowest point with a velocity of 12 feet a second. How many foot-pounds of work have been done against friction during the motion? And if the same proportionate loss of energy occurred in the next portion of the same circle, how high would it ascend ? (16) A massive slow-moving flywheel, mounted on a horizontal axis 1 foot in diameter, possesses 1500 foot-pounds of kinetic energy, which is used to raise a weight of 25 Ib. by means of a rope coiled round the axis. Assuming that a weight of 5 Ib. is able to overcome the friction, how many times will the wheel revolve before it comes to rest ? How many revolutions in the opposite direction must be made before the original energy is restored to the wheel ? (17) A railway carriage of 4 tons moving at the rate of 5 miles an hour strikes a pair of buffers which yield to the extent of 6 inches. Find the average force exerted upon them. (18) A three-ton truck sliding down a plane rising 1 in 20 acquires a speed of 30 feet a second after travelling 500 feet down the plane. Find the average force of friction acting on it. What velocity would it have acquired if there had been no friction ? (19) A train of 50 tons moves up a rough incline of 1 in 10, the resistance caused by friction being 16 Ib. weight per ton. What horse-power must the engine exert in order to main- tain a uniform speed of 3 miles an hour ? A horse-power was defined by James Watt to mean 33,000 foot-pounds of work per minute. (20) If a horse walking once round a circle 10 yards across raises a ton weight 18 inches, what force does he exert over and above that necessary to overcome friction ? (21) Calculate the work done in turning a wheel ten times round against a load of 15 Ib. applied by means of a string CHAP. VI. J WORK AND ENERGY. ] 31 wrapped round its axle, which is 8 inches in diameter. If the wheel starting from rest makes 20 revolutions in the first quarter minute after being let go, what is its moment of inertia, supposing no friction ? (22) A mass of 224 Ib. falls from a height of 10 feet upon a pile. Express its energy and its momentum when it reaches the pile. (23) A man cycles up a hill, whose slope is 1 in 20, at the rate of 4 miles an hour. The weight of man and machine is 187^ Ib. What work per minute is he doing ? (24) At the top of the hill the cyclist is met by a strong head- wind, and he finds he has to work twice as hard to keep the same rate of 4 miles an hour on the level. What force is the wind exerting against him ? (25) If a train moving at 80 feet per second, up an incline of 1 in 64, slips a carriage, how far will the carriage move before beginning to run back, ignoring axle-friction ? (26) Find the horse-power of a locomotive which draws a train at 10 miles per hour up an incline of 1 in 40, the weight of train and engine being 400 tons. (27) Find the horse-power of an engine which draws a train of 100 tons up an incline of 1 in 60 at a speed of 30 miles an hour, the friction being equal to a drag of 20 Ib. weight per ton. (28) A sledge left to itself slackens speed from 30 to 20 feet a second while going 15 yards. Assuming the coefficient of friction constant, find its value ; also find how soon and in what distance the sledge will stop. (29) A train weighing 60 tons, and running at the rate of 40 miles an hour, is stopped by an obstacle in 10 yards. What is the average force applied by the obstacle ? (30) How much work has a man, weighing 16 stone, done in walking twenty miles up a slope rising 1 in 40 ? What force could drag a dead load of the same weight up the same hill (a) if the friction be negligible, (b) if the friction be of the weight ? (31) How much energy is expended in winding up the hour- striking part of a turret clock each day, if its weight of 2 cwt. descends 1 inch for each stroke of the bell, and if the friction is equivalent to of the weight? On the same hypothesis as to friction, how much of the energy is available for the production of sound ? 132 ELEMENTARY MECHANICS. [EXS. XV. (32) A sliding and a rolling body start down similar slopes to- gether ; which travels fastest, if one plane is smooth, and the other only just rough enough to ensure rolling? (33) What must be the coefficient of friction of the plane pre- viously smooth in order that the roller and the slider may travel at the same pace and arrive together? Take the case of a plane whose height is th the base, and the rolling body a solid cylinder. (34) Show that all solid and homogeneous spheres in rolling down an incline will travel together if they start together, whatever their size or material. (35) Show the same for cylinders, and find by how much the spheres will beat the cylinders on a given slope. (36) What is the energy of a pendulum bob weighing half a ton and swinging past its equilibrium position at the rate of 1 foot a second ? (37) What energy is stored in a cross-bow whose cord has been pulled 15 inches with a maximum force of 2 cwt. CHAP. VII. I COMPOSITION OF FORCES. 133 CHAPTER VII. COMPOSITION AND RESOLUTION OF FORCES. (Introduction to Statics. ) 106. Hitherto we have only considered the effect of a single force when it acts on a particle or on a rigid body, and we find that it may either pull the body along (trans- lation), or turn it round (rotation), or do both at once. But in very few cases in practice do we have only one force acting in this way ; often there are a great number of different forces, so that it becomes necessary to consider how the motive effect of a number of forces may be deduced. The simplest way is to reduce the forces in number; and the following statements may be recorded here as memoranda : When any number of forces act on a particle they may always be reduced to one that is, they may be replaced by a single force which produces precisely the same effect as them all. This single force is called the resultant ; and the operation of reducing the number of forces is called the composition of forces. If a number of forces act on different points of a rigid body that is, an assemblage of particles connected rigidly together they cannot in general be reduced to one force, but they may always be reduced to two forces in different planes (sect. 121). They can, however, always be ..reduced to a single force (of which what is called * a couple ' is a special case) if they either all pass through the same point (that is, virtually act on a particle), or else all lie in the same plane (cf. sect. 137). 134 ELEMENTARY MECHANICS. [SECT. 106. A picture hanging by a cord over a nail furnishes us with an example of a rigid body acted on by several forces, and the tension in the two parts of the cord is equivalent to the weight of the picture. A weight resting on a tripod-stand is another example, and the three stresses in the legs are equivalent to the one weight. Again, a table or chair is supported by as many forces as it has legs, unless some are too short (which they often are). A teetotum is spun by forces which may be reduced to two equal and parallel ones in opposite directions, and we have here a case of pure rotation without translation. A kite in the air is acted upon by the wind pressing it, by a tension in the string, and by the pull of gravity ; and the kite moves about according to the direction of the resultant of all these forces. 107. Again, for some purposes, it is convenient to analyse or split up a single force acting on a body into two or three components so as to study their effects separately. This operation is called the resolution of forces ; and it is carried out in the same way, and for the same sort of object, as the resolution of motions and velocities (see sect. 30). Thus, suppose a body resting on an inclined plane, we may resolve its weight into two forces, one perpendicular to the plane, and therefore balanced by its resistance; the other acting along the plane and producing motion, except in so far as it is balanced by friction. Again, in a windmill, it is convenient to resolve the wind's pressure on the sails into two components one the effective one in the direction of motion ; the other a useless one in the direction in which, by the construction of the machine, no motion is allowed.* This last component, therefore, only produces strain. Composition of Forces acting on a Particle. 108. The method of compounding forces into a resultant, or resolving them into components, is a very simple one, being the same as that by which motions were compounded and resolved. For if several forces act on a particle, each tends to accelerate its motion in its own direction, and the resultant acceleration is the resultant of these several com- * N.B.A. windmill always faces the wind. CHAP. VII. J COMPOSITION OP FORCES. 135 ponent accelerations. The single force which would cause this resultant acceleration is equivalent to the several forces combined, and is called the resultant force; while the separate forces are called its components. Hence forces are compounded in the same way as accelerations or motions (sect. 28), being represented by lines in their respective directions proportional to the accelera- tions they could produce in unit mass. And they can be resolved also in the same way. JSTo further proof of the triangle or the polygon of forces is necessary. The above deduction from Newton's second law, F ma, does not, however, establish the position of the resultant, but in the case of a particle this is obvious. The further condition neces- sary for an extended body is given in sect. 119. The rule, then, is Draw a set of lines one after the other, without taking the pen* off, parallel to, and in the same sense as the successive forces acting on the body, and proportional to them in magnitude; then the line required to complete the polygon, taken in the reverse sense (that is, drawn from the starting-point, not to it), will be the re- sultant in magnitude and direction. The forces may be taken in any order just as the motions might (sect. 24). Since we are only dealing with a particle, this is the full and complete solution ; for the resultant, of course, acts on the particle, and therefore its position is known ; and the three things, magnitude, direction, and position, completely specify a force (see sect. 51). The resultant of two forceb is often more conveniently ex- pressed as the diagonal of the parallelogram whose sides represent the forces, than as equal to the third side of a triangle. 109. Examples of the Composition of Two Forces. A particle of mass m is pulled along by two strings one * If the forces do not all lie in one plane, the polygon cannot be drawn on paper, but it may be constructed in wood. 136 ELEMENTARY MECHANICS. [SECT. 109. always pulling east, with a force P ; the other always north, with a force Q. What is the acceleration and direction of motion 1 Drawing the two forces P and Q (fig 27), one finds the resultant R at once as equal to V(P 2 + Q 2 ) by Euclid, I. 47 ; and since this in p Fig. 27. is the resultant force, the acceleration is along the dia- m gonal of the parallelogram. If the two forces P and Q were equal, then R 2 would be simply 2P 2 ; that is, 11= P \/2 a result worth remembering. Suppose now that the two forces act at some acute angle, say 60, then to find R we may use Euclid, II. 12, which says that AD' 2 exceeds AB 2 + BD 2 by twice the rectangle AB-BN (fig. 28). The angles CAB and DBN are always equal (I. 29), and if each equals 60, BN is easily seen to be half BD, because the triangle BND is then half an equilateral triangle ; so putting in this value for BN, and noting that BD = AC = Q, and therefore BN = Q, we can write the general relation AD 2 =AB 2 + BD 2 + 2AB-BN in the form R 2 =P 2 + Q 2 + PQ for the case when the angle between P and Q is 60. If the angle CAB between the forces had been an obtuse angle, such as 120, we should have proceeded similarly, only using Euc. II. 13, and we should have arrived at R 2 =P 2 + Q 2 -PQ. Similarly we might proceed for angles between P and Q of 45 or 135, of 30 or 150; but for angles in general, though the relation AD 2 = AB 2 + BD 2 + 2 AB BN, will always apply regard being paid to sign in the last term (see sect. 12) yet it is not so easy to express the side CHAP. VII.] COMPOSITION OF FORCES. 137 BN in terms of the side BD (or Q) the subject of the mensuration of triangles, or Trigonometry, not being sup- posed known at this stage. Our resource is then to find the resultant by construction ; and this indeed is often a very good way, even when one knows some trigonometry. You lay off on paper the two given forces to any scale, and inclined at the proper angle ; then you complete the parallelogram, and measure the diagonal on the same scale this gives you its magnitude ; and its direction referred to the given forces you get also from the figure. 110. Notice that in the parallelogram of forces, you really have two diagrams drawn together as one : a representation of the forces, and a geometrical construction; but they should be understood to be essentially distinct. The propo- sition of the triangle of forces is really the geometrical part of the parallelogram by itself. An example will render the meaning of this clearer. Let two forces, 6 and 8, act on a particle with an angle of 60 between them. Find then- resultant. Fig. 29. On the left of fig. 29 is a picture of the given forces. On the right is the geometrical figure namely, a triangle in which AB represents the force 8, BD the force 6, and AD their resultant, in magnitude and direction. (AD equals 12'17 nearly, as maybe found either by drawing and measuring, or by calculating it from sect. 99 as V(8 2 + 6 2 + 8 x 6.) Its position is known, for of course it acts on the given particle; so we return to the left-hand diagram, draw through the point of intersection of the two given forces a line equal and parallel to AD, and this will be the result- ant. Obviously it is the diagonal of the parallelogram of forces the triangle ABD is simply half the parallelogram ; compare fig. 28. 138 ELEMENTARY MECHANICS. [SECT. 110. Observe that the geometrical construction is based upon only magnitude and direction : it does not give you position ; this must always be determined from the positions of the given forces in the force diagram. It is not usual to separate the two figures in simple cases, but as a matter of principle it is best always to keep them distinct. 111. The diagrams for one instance of the polygon of forces may be also given, just to make sure it is fully understood. Forces in a plane, of magnitudes 4, 5, 3, 8, act on a particle, their directions making angles with each other of 75, 45, and 120 respectively. Find the resultant. In actual cases the Fig. 30. angles between the forces are not specified numerically, but are indicated directly. In artificial questions, however, like the above, when the angles are specified. in degrees, a protractor maybe used to lay off the directions.* It turns out to be '504, so that the given forces are very nearly in equilibrium, or their resultant is very small. Observe the reciprocity of these diagrams. In one the lines meet in a point, in the other they enclose an area. Try drawing the sides of a polygon in some other order, and * In all these figures the lines are drawn parallel to the forces ; this is the easiest, though not the essential plan. What is essential is, that the lines shall represent the direction of the forces in some understood manner. It is usually said that they will do either parallel or perpendicular ; but they would do equally well if all were inclined at 45, or at any other angle, to the forces which they respec- tively represent, provided this angle were the same for all. The interior angles of the polygon are supplementary to the angles between the corresponding forces. CHAP. VII.] COMPOSITION OP FORCES. 139 see that you always get the same result. Especially try the order 4, 3, 8, 5 ; for the polygon then happens to be a crossed one. The bits of forces represented by the lines surrounding the enclosed space in a crossed polygon are in equilibrium (Chapter VIII. ), and may be removed from the particle without disturbance. In the above case the enclosed bit will be found to be an equi- lateral triangle, and the forces which may be removed are three threes namely, three parts from force 8, three from 4, and all of 3 ; the forces left being 1, 5, 0, 5. Construct the polygon for this mutilated set, and see that you still get the same resultant. Notice the reason why the three removed forces were in equilibrium namely, that they were equal and lay symmetrically, making angles of 120 with each other. You are strongly recommended at once to get out your instruments and a sheet of drawing-paper, and verify all this by careful drawing, as well as some of the examples in Ex. XVI. The instruments needed are a graduated scale of equal parts, a couple of set squares to act as a parallel ruler, perhaps also a T square, and a pair of compasses. A protractor for measuring angles is also convenient. Resolution of Forces. 112. Every force may be split up into two definite com- ponents acting at given angles with it ; but, if the angles are not given, a force may be resolved into two components in an infinite number of ways; in other words, the same line may be the diagonal of an in- finite number of parallel- ograms (fig. 31). One chooses in each problem the* particular pair of components which are Fig. 31. most convenient, the most convenient being usually at right angles to each other. Often one is in the direction of possible motion, 140 ELEMENTARY MECHANICS. [SECT. 112. and the other perpendicular to it; again, in cases where gravity is concerned, one is often horizontal and the other vertical. Verify, by drawing, the following : A force of 8 units is equiva- o lent to two components of 7- each, acting one on each V(* + Vw side the given force at angles of 15 with it; also to two of 8 each, o if the angles be 60 ; also to two of -- each, if the angles be 75 ; also to a component 4, acting at an angle of 60, and another of 4 \/3 at an angle of 30 ; and so on. 113. Constrained Motion. When a body is constrained, as by a line of rails, to move in some fixed direction, and when the propelling force acts partly athwart the con- straint, its treatment is simplified by resolving it into two components, such that one acts along the line of possible motion, and the other across it. The first is the effective or working force, and it either accelerates or retards the motion, while the other is the lateral force exerted against the rails and balanced by their constraining pressure. 114. To illustrate the use of this, take a mass m, or say J Ib. for those who like numbers best, on a smooth inclined plane inclined to the horizon at an angle say of 30. Such a body is constrained by the plane not to fall vertically, and the forces acting on it are the force w ( = mg) due to gravity acting downwards, and the pressure of the plane, say K, acting normal to the plane. Now, resolve the downward force into two one along the plane, that is, in the direction of motion, and call this effective component p the other normal to the plane, as q (fig. 32). CHAP. VII.] FRICTION. 141 This you do by drawing a parallelogram with sides in these directions, and of such size that w (in the present example, \g or 8) is the diagonal. The angle between p and w is 60, that between w and q is 30, and we have just found (end of sect. 112) that a force of eight units is equivalent to two forces, 4 and 4\/3, acting at 60 and 30 respectively; so then p = 4 and q = 4\/3. The motion is in the direction of p, the acceleration being E. or - = 16, or half its unconstrained value ; whereas there m J is no motion in the direction of q, because it is balanced by the constraining force R, hence E, = q - 4 v /3. 115. Friction. The reaction of rough surfaces will afford us other examples. When anything exerts pressure on a plane surface, the reaction of the surface is in general inclined in some direction or other to the surface usually in that direction most likely to oppose relative motion ; but it is convenient to resolve this reaction into two one normal to the surface (normal merely means perpendicular), which is called the normal pressure ; the other along the surface, which is called the friction. If either surface be perfectly smooth, this last component is absent, and all the reaction is normal. And even for rough surfaces this may be so too, as in the case of a ball resting on a level floor ; but if any forces are tending to cause motion over rough surfaces, then there is some component along the surfaces, or friction, which always opposes the motion. The force of friction is precisely equal and opposite to the resultant force tending to cause the motion, so long as the body does not move ; but if the applied force gradually increases, it will, at a certain instant, become too much for the friction, which reaches a maximum, and can increase no further ; so then motion ensues, the effective or accelerative force being the applied effort, minus the friction. For instance, in the above example of the inclined plane, suppose the force of friction to be called /, it would act up 142 ELEMENTARY MECHANICS. [SECT. 115. the plane in exact opposition to p. If the body were at rest, it would be so because/--^; if it were in motion, the acceleration would be 1L-~L m It is found experimentally that, the maximum or critical value of /is proportional to the normal pressure E between the surfaces, the ratio between / and R depending on the nature of the surfaces in contact, and being called the Coefficient of Friction.* Say that this coefficient in the above case of the inclined plane is , so that t /=|R, then the acceleration would be The actual pressure or reaction between the surfaces in con tact is, of course, the resultant of the two forces R and /, the normal pressure and the friction that is, it is the square root of the sum of their squares ; and the angle which its direction makes with the normal when the two surfaces are on the point of sliding over one another is called the limiting angle of friction. When a body is resting on an inclined plane, supported only by friction, the resultant reaction of the plane must be vertical, in order to balance the weight ; the greatest tilt that can be given to the plane without causing slip is therefore equal to the limiting angle of friction that is, the maximum angle between vertical and normal; and so this angle is called the angle of repose. On any plane with less tilt than this there will be a margin, since the reaction is not as * The coefficient of friction when the surfaces are in actual relative motion is usually less than when they are just going to move ; hence there are two coeffi- cients of different value, one when the bodies are on the point of slipping, called the static friction, or ' stiction ; ' the other when the surfaces are actually sliding over each other, called the kinetic friction. The latter is the smaller of the two, and frequently depends somewhat upon the speed of the relative motion. Lubrication not only lessens friction but tends to abolish the difference between static and kinetic friction. Resin, on the other hand, exaggerates this difference. Vibrations, with accompanying noise, are liable to be set up whenever there is a marked difference between static and kinetic friction, because the slipping is apt to become intermittent, being alternated with moments of adhesion. CHAP. VII.] RESOLUTION OF MOTIONS. 143 much inclined to the normal as it can possibly be (see, for further information, sect. 142). 116. It is often convenient to resolve motions and velocities. Thus, as we saw (sect. 76), a projectile shot up at any angle has a certain initial velocity imparted to it, which may be conveniently resolved into two one a hori- zontal one unaffected by gravity, which therefore remains constant except for the resistance of the air ; the other a vertical one, which is gradually diminished by gravity at a definite rate, until it is converted into a negative, that is, a downward, velocity which increases at the same rate till the body strikes the ground. Again, take a north-east wind. This may be considered as made up of a north and an east wind, each -^th of the /2 actual strength, and on any thin, flat, smooth surface facing the north only the northerly component can exert any pressure, the easterly component simply gliding over it. Or suppose the surface faced NNW., and we wanted to find the pressure on it ; the wind might be resolved into an NNW. component, \/2 - \/2 times its strength, and an ENE. one 4V2 + V 2 times its strength, and the surface would experience the pressure of the NNW. component only, the other being useless. This is how one deals with kites and windmill- and boat- sails. They are all surfaces exposed in a skew fashion to the wind, so that the pressure on the surface is a component only of the whole available force of the wind. The sails of - a windmill are set so as to be inclined both to the direction of the wind and to the direction of possible motion ; so also usually are the sails of a boat. It is convenient to remark and remember that, disregarding viscosity or fluid friction, the pressure of a fluid is always normal to surfaces immersed in it. 144 ELEMENTARY MECHANICS. [SECT. 117. 117. In the case of a kite the normal pressure of the wind is balanced by two other forces, the pull of gravity and the pull of the string, otherwise the kite would be blown about, scarcely experiencing any pressure at all. The sails of a windmill are not blown in the direction of the normal pressure on them, but in some other direction deter- mined by the way they are set on the axle and by the sole direction in which this can turn ; the axle is always pur- posely set so as to face the wind, and so the sails can only move in a plane perpendicular to the wind. So also with a boat ; the reason why it is not blown in the direction of the normal pressure on its sails is that it is more easily moved through the water lengthways than breadthways because of its shape. Hence the normal pressure of the wind requires again resolving into two components, one along the direction of easy motion, the other at right angles to it. The first component is the active one in the case of both windmill and boat ; the other component is entirely counteracted in the case of the windmill, but in the case of the boat it does cause a slow broadside motion, which is called leeway. Thus if BR (fig. 33) represents the plan of a boat, MS its sail, and W the relative direction and strength of the wind (represented also by the arrow i), P is the normal pressure and Q the useless -X W\ /M "*" H V \ Fig. 33. component or tail-wind. Producing P for convenience, and re- solving it along and across the boat, H is the effective component producing headway, and L is the leeway component. The arrow CHAP. VII.] RESOLUTION OP FORCES. 145 ii shows the direction in which the boat might tend to sail.* The rudder R is represented as turned in the direction required to counteract the leeway and make it sail along the line RB pro- duced. The rudder also affords an illustration of the present subject. When turned, there is a normal pressure on its front surface due to its motion through the water, and this pressure is resolvable into two forces one in a direction opposite to the boat's motion, which simply acts as a drag (hence in racing, the coxswain uses the rudder as little as possible), the other at right angles to the length of the boat, which pushes the stern round. It is obvious that no force can directly exert pressure at right angles to itself, and yet it is easy for a ship to sail at right angles to the wind. The reason is, that the sails act as a mediary, being inclined to both wind and boat. The force directly urging the boat is a component of the pressure on the sails, this pressure again being due to a component of the wind's motion. Remember that the effective wind the wind felt by the ship is the relative wind, which is compounded of the true wind and the speed of the ship (cf. sect. 31). This explains why a ship can sail very close to the wind. These examples will serve to illustrate the application of the principle ; but other examples occur daily, and may be worked out in the same way as the preceding cases, drawing and measuring being often sufficient. EXAMPLES XVI. (1) Find the resultant of two equal forces each equal to 10 units for each of the following cases namely, when the angle between them is 120, 90", 60, 45, 30, respectively. * This assumes that the sail is set amidships. In practice there is always a preponderance of sail towards the stern, consequently an unsteered ship gets blown round, and 'sails up into the wind's eye.' The rudder would therefore more likely have to be turned the other way, so as to counteract the action of the wind in causing rotation. The effect of the wind on the body of the boat has also to be taken into account in practice, and this may be different according as the bow or the stern is most out of the water a thing which depends on the distri- bution of the load. 146 ELEMENTARY MECHANICS. [EXS. XVI. (2) Resolve the force 12 into two forces, making angles of 45 with the given force on either side of it. (3) A picture weighing 12 Ib. is hung by a cord over a nail so that each half of the cord makes 45 with the vertical. What is the tension in the cord ? (4) Find the tension in a picture cord when the two halves of the cord make an angle of 30, 60, 90, 120, 150 with each other. What is the way to hang a heavy picture so as to get the least tension in the cord ? (5) A weight of 10 Ib. is placed on a smooth plane inclined 30 to the horizon. What force acting horizontally is required to support it? What force acting along the plane will suffice ? and what is the normal pressure on the plane in each case ? (6) A carriage weighing 2 tons is to be drawn up a smooth road by a rope parallel to the road. The road rises 4 feet in a slope of 32 feet. What must the pull of the rope exceed in order that it may move the carriage ? (7) What weight can be drawn up a smooth plane rising 1 in 5 by a force equal to the weight of 200 Ib. (a) when the force acts up the plane ? (b) when it is horizontal ? (8) A heavy ball hangs from a point by a string. A second string is attached to it, and by this the ball is drawn aside so that the first string is no longer vertical. Draw a figure showing a triangle with its sides proportional to the three forces acting on the ball, and observe how they change with the inclination when the pulling string is kept horizontal. (9) Draw a diagram to scale showing the resultant of two forces equal to the weights of 7 and 11 Ib. acting on a particle, with an angle of 60 between them ; and by measuring the resultant find its magnitude. Indicate two equal forces, at right angles to each other, which would be equivalent to the above two forces. (10) Calculate the magnitude of the resultant of two forces, of 35 and 40 units respectively, acting at the same point and making with each other an angle of 120. (11) Six forces, 3, 4, 7, 10, 9, 5, act from the centre of a regular hexagon towards the angular points. Find the magnitude and position of their resultant. (12) Find, to two decimal places, the resultant of two forces, 20 and 12, both acting from the corner of a square, the former along the diagonal, the latter along a side. CHAP. VII.] COMPOSITION OF FORCES. 147 (13) A cross wind strong enough to exert a pressure of 10 Ib. per square foot on an object placed normally to its direction acts against a sail of 500 square feet area inclined at 4,5 to it and to the boat. Find the effective component of the wind-pressure on the boat, neglecting leeway ; (a) when the boat is stationary ; (b) when the boat is travelling at 10 miles an hour. The latter case is to be solved by measurement from a diagram. (14) Draw a square ABCD, and take E the middle point of BC. Forces of 10, 15, 20, and 25 units act at A, from A to B, D to A, C to A, and A to E respectively. Find their resultant by measurement from a diagram. (15) A piece of wire 26 inches long, and strong enough to support directly a weight of 100 Ib. , is attached to two points 24 inches apart in the same horizontal line. Find the maxi- mum load that can be slung on the middle of the piece of wire without breaking it. (16) The sides AB, AD of a rectangle ABCD are 5 and 12 inches long respectively. Forces of 8 and 20 Ib. weight act at A in the direction AB and AC respectively. Find their resultant, either by construction or calculation. (17) In what time will a body slide down 4 feet of a rough incline of 30 for which the coefficient of friction is T V ? (18) What coefficient of friction will enable a weight just to rest on an inclined plane of 30 without extra support ? (19) Find the least force that will pull a hundredweight up such a plane as that in No. 18, or that in No. 17, and show that the best angle of traction is in general equal to the angle of repose. COMPOSITION OF FORCES ACTING ON A RIGID BODY. 118. For the case of a rigid body, in addition to the magnitude and direction of the resultant as determined by the polygon construction, sect. 108, it is necessary also to determine its position that is, its line of action. For observe that, though as regards translation a force in one place is as good as an equal parallel force in another, yet as 148 ELEMENTARY MECHANICS. [SECT. 118. regards rotating power its position is important. Thus, imagine a long trough of water lying on the ground with a string tied to it by which you wish to raise it. Any vertical force greater than the weight of the trough must needs raise it, wherever the string is tied ; but if the string is tied anywhere except above one definite point, the trough will also turn round as it rises, and the contents will be upset. Again, if you raise it by two parallel strings, one near each end, then when the pull of the two strings together is a little greater than the weight of the trough, it is raised ; but if you want to raise it without rotation, the pull of each string must be carefully proportioned, so that the resultant of the two forces may pass through the point above spoken of, which is called the centre of gravity. Again, in the case of a pivoted body, it is obvious that a force applied close to the pivot has much less effect than an equal one far off; and if applied at the pivot, it can have no motive effect whatever. 119. Now, the fundamental dynamical idea in rotation is the moment of a force (read sect. 57 again) ; and the following general statements are true, with their converses. (1) The moment of the resultant must equal the sum of the moments of the components about any point in every possible case, otherwise the resultant would not be truly the resultant, because unable to replace the components in rotating power. That this condition is fulfilled by the diagonal of a paral- lelogram whose sides represent the component forces may be proved among other ways as follows : To show that the resultant given by the parallelogram offerees is equivalent to its components in rotating as well as in translating power that is, that its moment about any point in the plane is CHAP. VII.] COMPOSITION OF FORCES. 149 equal to the sum of the moments of the two components. The moment of the force AB about a point O (fig. 34) is (see sect. 53) geometrically representable by twice the area of the triangle OAB ; the moment of AD is similarly proportional to twice the triangle OAD, and that of AC is twice OAC : hence what we have to prove is the following equality between the areas, OAB+OAC=OAD; the point being in the plane of the parallelogram. Now O AC = OBD + ADB, because the bases Fig. 34. are equal, and the height of the single tri- angle is equal to the sum of the heights of the others (this is an easy extension of Euc. I. 38 analytically obvious, thus and by inspection, OAD = OAB + OBD + ADB ; therefore OAD = OAB + OAC ; which was to be proved. (2) The algebraic sum of the moments of any number of forces about a point on their resultant is zero ; in other words, the sum of the positive moments equals the sum of the negative. (The moments of tivo forces about a point on their resultant are therefore numerically equal, but of opposite sign.) For their resultant can have no rotating power about such a point, neither therefore can the components. (3) If the body on which the forces act has one point fixed, and if their resultant passes through the fixed point or pivot, it will not be rotated by them. For instance, to keep the beam (fig. 37) steady, C is the point to fix. The pressure of the pivot or fulcrum is then equal and opposite to the resultant of all the forces. (4) The resultant of two forces acting on a rigid body passes necessarily through their point of intersection. If they do not intersect even when produced (which can only be by reason of their lying in different planes), then they have no resultant; and they cannot be further reduced. The effect of such an irreducible pair of forces is to carry a 150 ELEMENTARY MECHANICS. [SECT. 119. body along and twist it round at the same time. They are not to be confused with a simple 'couple,' which results from the intersection of given forces at infinity : quite a different thing geometrically from not intersecting at all. Composition of Two Forces in General. 120. If the two forces are in one plane, the parallelogram is a complete solution, whether they act on a particle or a rigid body, for the forces must intersect somewhere, and the point of intersection fixes the position of the resultant. Thus fig. 35 is the physical part of fig. 29 repeated for a rigid body, say a stone pulled by two strings. The geometrical part applies just as well as before. The direction of the resultant must pass through E, the point where the given forces produced backwards intersect, and it may be applied to the body at any point in a line EP parallel to AD (fig. 29). It may happen, however, that the point of intersection of the two forces is in- conveniently distant, as off the paper for instance, or even at infinity when the forces are parallel. In such cases the general construction of sect. 123 is resorted to. Fig. 35. 121. If the two forces are not in one plane they cannot intersect, and our construction for finding the resultant fails. The fact is they have no resultant, and cannot be further reduced ; they can only be put into the more convenient form of a force and a ' couple ' (sect. 128) in a plane perpendicular to the force ; so they tend to carry the body along and turn it round at the same time. This pair of forces is called a wrench, because it tends to twist the body about a certain screw ; but the subject now becomes too complicated for us in this stage. (This is what was meant in sect. 106 by the two forces to which any forces CHAP. VII.] COMPOSITION OF FORCES. 151 whatever acting on a rigid body can always be reduced, even when no more can be done.) If, however, all the forces lie in one plane, no ' wrench ' is possible, and they may then always be reduced to one simple resultant ; though it may be a resultant zero at infinity as one special case, in which case it is most easily treated as ' a couple. ' Composition of any number of Forces in a Plane. 122. The parallelogram construction may be applied several times in succession, reducing the number of forces by one each time. This is a complete but cumbrous solution. The polygon construction is a solution as regards magni- tude and direction, but requires supplementing in order to determine position. The supplementary construction em- ployed is such an important one, that it seems well to intro- duce it here, although its full discussion would lead us beyond our present mark. It will be best understood by an example, and the case of only three forces will afford a sufficient illustration of the method. It depends on the fact that a single force may be resolved into a pair of com- ponents in an infinite variety of ways (fig. 31) ; so that, if the given forces are not convenient to find the resultant from, we can choose a more convenient pair out of the set which have the same resultant, and then draw the resultant of these. Expressed in another way, it may be said to depend on the fact that forces in equilibrium produce no disturbance, and hence may be introduced or removed at pleasure. Construction for finding the line of action of the Resultant of any number of Forces whose directions all lie in one Plane. (Illustrated by the case of three forces.) 123. Let P, Q, S (fig. 36) be the forces. Draw the sides of the polygon ABCD parallel to, equal to, and in the 152 ELEMENTARY MECHANICS. [SECT. 123. same sense as the three forces ; then the completion of the polygon, AD, is the resultant E, in magnitude and direction. Where is it to be placed 1 Choose any point 0, join OB, and draw in the other diagram a line PQ parallel to it across the forces P and Q (the line is to be drawn across P and Q, because B is the Fig. 36. fLines parallel to each other in the two diagrams are labelled similarly ; but the simplest way in practice of indicating the correspondence of lines is to mark the areas in one figure and the points in the other. Thus, for instance, the letters P, Q, R, S, in the right-hand figure, may be supposed to denote the areas containing these letters, and the letter E may be affixed to all the external space. In the left-hand figure these same letters belong to points, especially if R be understood to apply to all the arrow-heads ; and all lines in either figure are denoted by two letters, in the one by points, in the other by areas which they separate. Thus the line QS is the line joining (or separ- ating) the points Q and S in one figure, and is the line separating the space Q from the space S in the other.] meeting-point of the sides of the polygon which represent P and Q). Then join OC, and draw a line QS parallel to it (C being the meeting-point of the sides representing Q and S). Then join OD, and draw a line through S parallel to it, CHAP. VII.] COMPOSITION OF FORCES. 153 say SE ; also join OA, and draw a line through P parallel to it. The point E, where these last two lines intersect, will be a point on the resultant, and its position is therefore determined. Q. E. F. Proof (This may be omitted till after Chapter VIII. has been read). A force represented by AB is the resultant of two forces represented by AO and OB, because it forms a triangle \vit\i them ; but P is a force represented by AB, hence P is equivalent to two forces acting along the lines EP and PQ, equal to AO and OB respectively, and may be replaced by them. Let it be so replaced. Similarly the force S may be replaced by two forces acting along SQ and ES, equal to CO and OD respectively. But Q is in equilibrium with two of these forces namely, those along SQ and PQ, since their representative lines, BC, CO, OB form the sides of a triangle taken in order (sect. 135) ; hence this set of three forces may be removed ; and there now remain, as the equivalents of the original forces, P, S, and Q, only forces along EP and ES, represented by AO and OD respectively. Hence these two forces have the same resultant as the three original forces had ; but the resultant of these two forces passes through E, their point of intersection (sect. 120) ; therefore the resultant of the three original forces, P, Q, and S, passes through the point E. Q.E.D. It will be seen, therefore, that what we really do in the construction, is to compound with P, Q, and S, two sets of equilibrating forces namely, two equal opposite forces in the line PQ of magnitudes OB and BO, and two in the line QS of magnitudes OC and CO ; and by their help to replace the given forces by two intersecting ones, ES and EP, the position of whose resultant is obvious. This construction applies equally well to parallel forces, only then, of course, the polygon ABCD shuts up, the points B and C being 011 the straight line AD ; but everything else remains without modification. The use of the above construction may not be quite 154 ELEMENTARY MECHANICS. [SECT. 123. apparent perhaps, but it is put here as an indication of quite a large art namely, graphical statics which may well occupy the student's attention at a later stage. The quadrilateral ABCD on the right of fig. 36 is called * the force-polygon,' and determines the magnitude and direction of the resultant; the quadrilateral i, ii, iii, iv on the left of fig. 36 is called ' the funicular polygon,' and determines its position. It is called the funicular polygon or sometimes the link polygon, because it represents the equilibrium directions of a weightless string or linkage of rods subject to the given forces. An arch or framework composed of jointed bits of wood, two of them placed like i and ii and two of them like iii and iv produced, would be in equilibrium under the given forces PQS if its terminal ends were fixed. Or if the forces were reversed in direction, it would repre- sent the shape taken by a string with fixed ends subject to the given forces. Another but less obvious statement which can be made is that the polygon i ii iii iv in the left- hand figure represents the distribution of bending-moment in a beam subjected to the force K, and to the forces PQR reversed ; it should have a thickness graduated according to this manner if it is to be as stiff as possible without waste of material. Composition of Parallel Forces. 124. Parallel forces can only act on an extended body : forces which act on a particle, of course, cannot be parallel. The direction of the resultant of parallel forces is the same as the common direction of its components, while its magni- tude is their algebraic sum that is, their sum paying regard to sign adding all that act in one direction, subtracting any that pull the other way. This is all that is required to be known for translation (sect. 118); but to discuss the rotation of a body under the influence of parallel forces, we must learn the position of the resultant, and this requires CHAP. VII. J PARALLEL FORCES. 155 either a geometrical construction or an arithmetical cal- culation. The general construction of sect. 123 applies to parallel just as well as to other forces, so we have only to give the method of calculating its position arithmetically. 125. The fact (No. 1, sect. 119) that the moment of the resultant equals the algebraic sum of the moments of all the components, though universally true, is most useful in its application to parallel forces, and it affords a ready method of finding the position of their resultant arithmetically. Thus imagine a weightless beam acted on by any parallel forces, say weights, 4, -5, 6, -2, &c., arranged anywhere on the Fig. 37. beam (as shown in fig. 37), at distances 4, 8, 16, 22 inches from some fixed point of reference O ; then the resultant R is, in magnitude, 4-5 + 6-2-3, and is at a distance x from O, such that 3x^4 x4-5x8 + 6x!6-2x 22=28 ; wherefore x=ty inches. Mark off OC equal to this ; then R acts at the point C, as shown ; and, to keep the bar in equilibrium, another pulley and string must be arranged to exert a force 3 upwards at this point. 156 ELEMENTARY MECHANICS. [SECT. 125. And, generally, if the forces be w v w 2 , iv 3 , at respective perpendicular distances x v x 2 , x 3 , from any point 0, then the distance of the resultant from the same point is w This is a constantly occurring form of fraction, and is a more general sort of average. If ti\ = tv 2 = tv B = &c., then it would be the ordinary expression for finding the average of the distances x lt x 2 , x & , &c. that is, it would give the average distance of all the weights from ; for it would add all the distances together and divide by the number of them. To take another example, let weights of 8, 6, 4, 2 pounds respectively be hung at the following inch divisions of a one-foot rule: 0, 3, 9, 12. Find the position of the resultant. Let it be at the division x, and take moments about one end. The moments of the weights are respectively 0, 18, 36, and 24; the moment of their resultant is 20^; wherefore, equating this to the sum of the separate moments, we get x = 3 '9. Once more, let there be only two weights, say 4 Ib. at one end of a rod a foot long, and 8 Ib. at the other. Then calling the distance of the resultant from the smaller weight x, its distance from the bigger weight is 12 -x, and taking moments about the resultant, we have 4# = 8(12 -x), whence x = 8 inches. That is, it divides the rod in inverse ratio to the two weights, and this is a general result. Composition of Two Parallel Forces. 126. When we have only two forces to deal with, the general statements and constructions are of course equally applicable, but they may be put into a more simple form. The resultant is equal to the sum of the forces if they act Fig. 38. CHAP. VII.] TWO PARALLEL FORCES. 157 in the same direction, and is equal to the difference of the forces if they act in opposite directions ; it is of course parallel to either force, and it only remains to find its position. The following three simple constructions may be given for find- A ing the position of the resultant geometrically. First Construction (fig. 38). Take a point M half-way between the forces P and Q, and draw two lines through it; one parallel to the forces, the other not, but cutting them in A and B respectively. Lay off from M two lengths in the former of these lines, in the same sense as the respective forces, MC equal to P, and MD equal to Q, and join AC and BD ; the resultant shall pass through E, the intersection of AC and BD. Proof. We may suppose that we have here compounded P with a force AM, and Q with an equal, opposite, and therefore equilibrating, force BM ; and AC, BD are the diagonals of parallelograms, and have the same resultant as P and Q have. Second Construction (fig. 39). Anywhere on the line of P take a length equal to Q, and on the line of Q a length equal to P. Then draw straight lines joining the extremities of these two lengths ; they will intersect in a point on the resultant, and so determine its line of action. If the forces have the same sense, they are to be joined crosswise, and E is the point. If they have contrary sense, they are to be joined without crossing, and F is the point. o. Fig. 39. 158 ELEMENTARY MECHANICS. [SECT. 126. Proof. Observe that the two triangles with common vertex E (or F) are similar (their bases being parallel), hence their heights and bases are proportional. But their bases are Q and P ; so, calling their heights/? and q, p _Q 7~P' or Pp = Q2. This fact proves the proposition; for, by (2) sect. 119, the moments Pp and Qq of the two forces about their resultant must be equal and opposite ; but they are evidently opposite, from the figure, and this equation states their equality, about the point E, and similarly about the point F. Wherefore the resultant passes through E if the forces have the same sense ; and through F if they have contrary sense. It is worth noticing that the moment of R about any point is equal to the sum of the moments of its components ; for, taking the circumstances as depicted in fig. 39, we can easily show that ~R(p+r) = Pr + Q,(p + q + r), by remembering that I1=P + Q, and that Pp = Q,q. Third Construction (fig. 40). A more general construc- tion, applicable whether the forces are parallel or not, Fig. 40. and practically useful when the forces are nearly parallel so that their point of intersection is inconveniently far off, is based upon the method of the funicular polygon (sect. 123). Draw a line AB representing the force P, and a line BC representing Q. Choose any point 0, and join OA, OB, OC. Then draw across the given forces a line parallel CHAP. VIT.] TWO PARALLEL FORCES. 159 to OB, and at its points of intersection with P and Q respectively draw other lines parallel to OA and OC. The point E where these last two lines meet is a point on the resultant, and determines its line of action ; its magnitude and direction are given by AC. Proof. The force P has been virtually replaced by forces AO, OB acting along the lines ao, ob ; the force Q has been virtually replaced by forces BO, OC acting along the lines bo, oc. Hence the resultant force is replaceable by AO, OC in the lines ao, oc, and therefore passes through their point of intersection. 127. The following propositions concerning two parallel forces are now at once seen to be true, being little more than repetitions in a compact form of what has gone before. (1) The distances between each force and the resultant are inversely as the forces that is, p : q Q : P. This formula, or some other moment formula, must be used when you want to find the resultant arithmetically. (2) If two parallel forces have the same sense, their resultant is equal to their sum, and lies between them, nearer the bigger one. In fig. 39 it passes through E. If, however, they are of contrary sense, their resultant equals their difference, and lies outside them on the side of the bigger one, agreeing with the bigger one in direction. In fig. 39, if one of the given forces is reversed in direction, the resultant passes through F. (3) If two forces are equal, the resultant must be equi- distant from both. If they are of contrary sense, this means that the result- ant is at infinity ; but its magnitude is zero, being equal to the difference of the components. 128. Hence, two equal contrary* parallel forces have a *The phrase 'non-concurrent' has been used to express parallel opposition not in the same line, but the word contrary is to be preferred, since the proper meaning of the word ' concurrent ' is meeting in a point. 160 ELEMENTARY MECHANICS. [SECT. 128. resultant zero at infinity ; or, as it is sometimes expressed, they have no resultant at all. (In any of the constructions the lines whose intersection gives the position of the resultant will for this case be found to be parallel.) Such a pair of forces cannot be further simplified, hence they are taken together and called a couple. The moment of the couple about any point will be easily seen to be independent of the position of that point, and to equal either force multiplied by the perpendicular distance between the two forces, this distance being called the arm of the couple. A couple is not properly to be regarded as two forces, but as a particular case of one namely, an infinitely small force at an infinitely great distance. It obviously possesses only rotating power. The fact that its moment about every point in its plane is the same causes its position to be un- important. Its moment and its plane have to be specified, but nothing else. (Read again sect. 106.) The Composition of Parallel Forces as illustrated by Gravity. (Centre of Gravity.) 129. The force of gravity illustrates the subject of parallel forces very well. A rigid body is made up of particles, every one of which is pulled towards the centre of the earth with a force proportional to its mass, and equal to its mass multiplied by g (sect. 64). Now, since the centre of the earth is such a long way off, these con- verging forces are for bodies of ordinary size practically parallel. Hence the whole pull of gravity on a table or a book is really the resultant of an infinite number of parallel forces the attractions on the several particles. To find the magnitude of this resultant, you hang up the body on a spring balance in ordinary language, you weigh it. To find its position, the easiest way is to hang up the CHAP. VII.] CENTRE OF GRAVITY. 161 body by a bit of string ; the line of the resultant is then a continuation of the string, since it must pass through the point of suspension. Or you may balance the body on your finger ; the line of the resultant is always the vertical through the point of support whenever the body is in equilibrium. Its direction is a fixed one namely, always pointing to the centre of the earth, no matter how you turn the body. Now when a rigid body exposed to the action of a number of parallel forces acting at definite points in the body is turned about, there is one point in the body through which their resultant always passes in every position and this point is called the centre of the parallel forces ; or, if the parallel forces are due to gravity, it is called the centre of gravity. The criterion as to whether there really is such a point in general is rather troublesome ; and if the forces are not accurately parallel, there is, strictly speaking, no such point for bodies of irregular shape. Nevertheless, the forces due to gravity acting on the parts of any body of reasonable size are so nearly parallel that practically every- thing likely to be experimented on has a centre of gravity. Determination of the Centre of Gravity by Experiment. 130. If this point be directly supported, the body is in equilibrium in every position necessarily ; and conversely, if a body is in such equilibrium, it must be because its centre of gravity is directly supported. (A coach-wheel, for instance, should be pivoted at this point.) Hence this gives one way of finding it. Another way of experiment- ally determining its position is to find out the line of the resultant in some two positions of the body by hanging it up twice in different ways (see fig. 41); then the centre of gravity must be the point common to the two lines K 162 ELEMENTARY MECHANICS. [SECT. 130. that is, it must be where they cross. However the body be hung up by a single point, the centre of gravity will always, when at rest, be vertically under or over the point Fig. 41. of suspension ; that is, the line of the resultant will always pass through the fixed point. The whole weight of a body, then, may be considered to act at its centre of gravity ; in other words, it behaves for purely statical purposes as if the whole mass of the body were concentrated at this point. Determination of the Centre of Gravity by Calculation. 131. The centre of gravity is always the most symmet- rical point in a body. In a sphere it is the centre ; so it is also in a cube or an ellipsoid, and in a square or circular plate. In a parallelogram or a parallelepiped (that is, solid parallelogram), it is the intersection of diagonals. In a rod of uniform thickness and material, it is the middle ; and so on. But it is easy to calculate its position in less uniform cases by any process which will determine the position of the resultant of a number of parallel forces, for it is simply the point through which the resultant always passes. In fact, if ??i l5 m^ are the masses, at the distances CHAP. VTT.] CENTRE OF GRAVITY. 163 x v % XK ...... from any line in the plane (restricting our- selves to masses distributed in a plane), the distance x of their centre of gravity from the same line is given by l1 2 ...... , b to-\ - (see sect. 12o). m 1 + m 2 + ...... And if the distances of the centre of gravity from two such lines (not parallel) are found, its position is completely determined. In some cases, one line on which the centre of gravity must lie is obvious, and then all that is further necessary is to determine its distance from any given point on that line, as in the following example. Thus let this rod with middle point M (fig. 42) be of weight two pounds, and let a ring A, weighing half a pound, be placed on it four inches to the left of M, and another ring B, weighing three pounds, six inches to the right ; then, if the rings fit tightly, the centre of gravity of the whole must lie somewhere in the length of the rod. To find whereabouts, we need only calculate the position of the resultant of the three weights (the two rings and the rod itself) in any position except the vertical one, say when horizontal. The magnitude of the resultant is plainly 5^. Take moments about any point, say about A ; let the resultant act at some unknown point C, such that AC=x. Then we have 5^0:= (3x10) + (2 x 4) + (4 x 0) = 38 ; wherefore x = 4=6H-; or tl ie point C is 2f inches to the right of M, and it is the centre of gravity. If the bar is to be supported in horizontal equilibrium, it must be pivoted or suspended by this point, or, better, by a point just above it (see sect. 145). Try now taking moments about M, also about B, also about O (anywhere), and see that you always get the same result (when interpreted properly), remembering to allow for negative moments. In this example it has happened, as stated above would often happen, that the line of the resultant in one position of the body (in this case when the rod is vertical) is perfectly obvious. The arithmetical determination of the position of the 164 ELEMENTARY MECHANICS. [SECT. 131. centre of gravity of a body, therefore, depends on precisely the same principle as the experimental method, and con- sists simply in finding the line of the resultant in any two positions of the body, and noting their point of intersection. It there- fore scarcely needs further exposition ; but it is probably necessary to show how this same principle is applicable to cases rather less obvious. For instance, to find the centre of gravity of a body made of two parts, each part having a known centre of gravity ; say two flat oblong plates, of known weights, w^ and iv 2 , joined end to end. G! and G 2 being the centre of each separately, their weights may be considered as acting here, and so the resultant passes through a point G, which divides the line GjG 2 in the ratio w 1 (sect. 127) that is, so that G X G : GG 2 : : w 2 w hence G is the centre of gravity of this combination. Or we might take moments about any point 0, and say 10 - OG 2 = whence the distance OG, and therefore the position of G, is determined. The same method applies if a bit is taken away instead of added on. Suppose, for instance, a square plate with a round hole in it anywhere (fig. 44). The operation of find- ing the centre of gravity in such a case may be regarded as the same as that of finding the position of the resultant of two contrary forces the weight W of the whole square acting downwards at G l and the weight w of the missing bit acting upwards at G 2 . The centre of gravity G must evidently be somewhere in the line G 1 G 2 ; so, taking moments about any point in this line, the equation W- OGj w OG 2 = (W - w)OG determines its position, C 2 Fig. 45. CHAP. VII. J CENTRE OF GRAVITY. 165 Or more simply thus : let x be the distance of G from the centre of the square G p and let a be the distance of the centre of the hole G. 2 from the same point G I} then write Wx = w(a + x) and solve for x. Or again, the centre of gravity of a trapezium (that is, a quadrilateral with two parallel sides), which may be regarded as a triangle with the top missing, can be found in precisely the same way. The last equation applies as it stands, in fact, provided we know the positions of G 1 and G 2 , the centres of gravity of the whole and of the missing triangles (see fig. 45). The centre of gravity of a triangular plate is in the line joining a vertex to the bisection of the opposite base (because this line bisects every line in the triangle parallel to the base). Three such lines can be drawn, because there are three vertices. Therefore these three lines, joining each vertex to the middle point of the opposite side, meet in a point, and that point is the centre of gravity. It is easily seen to divide each line in the ratio 1 : 2 that is, it is one-third of the way up from the base to the vertex. The centre of gravity of any quadrilateral can be found by dividing it into a pair of triangles in two different ways, and taking the crossing-point of the lines joining the triangles' centres of gravity. The following statements may for the present be assumed for the sake of examples. The centre of gravity *of any pyramid or cone is in the line joining the vertex to the centre of gravity of the base Fig. 46. 166 ELEMENTARY MECHANICS. [SECT. 131. and one quarter of the way up. The centre of gravity of a solid hemisphere is f ths of its radius from its flat boundary that is, from its geometrical centre ; of a hollow hemi- sphere is half-way between centre and circumference; of a semicircular area is , or approximately f of its radius O7T 2iT from the centre ; of a semicircular arc (like a wire) is , or 7T approximately T 7 T of its radius from the centre. EXAMPLES XVII. (1) Resolve a force 20 into two parallel forces, one of them 3 times as far from the given force as the other. (2) A weightless curtain rod has 4 equal rings on it, so that the 2 end rings are 5 feet apart, and the 2 middle rings are 1 foot apart, one of the end rings being 18 inches from the nearest middle one. Find the centre of gravity. (3) Where would the centre of gravity in the last question be, if the rod itself were 5 feet long, and weighed twice as much as a ring ? (4) A uniform circular disc has a circular hole punched out of it, extending from the circumference half-way to the centre. Find the centre of gravity of the remainder. (5) Forces of 1, 2, 3, and 4 Ib. weight act along the sides of a square whose diagonal measures 4 inches. Find the magnitude and position of their resultant. (6) Prove that the moment of a given 'couple' is the same about every point in its plane that is, that a ' couple ' has magnitude and direction but no position. What is the moment of a couple consisting of two equal contrary parallel forces, of 5 Ib. weight each, separated by a per- pendicular distance of 12 inches from each other? (7) A uniform beam 10 feet long, weighing 80 Ib., is suspended from two points in a horizontal ceiling, 16 feet apart, by strings each 5 feet long attached to its ends. Find the tension in each string. (8) An iron sphere weighing 50 Ib. is resting against a smooth vertical wall and a smooth inclrhed plane which is inclined at 60 to the horizon. Find the pressures on the wall and plane. CHAP. VII.] COMPOSITION OF FORCES. 167 (9) Find the resultant of parallel forces 1, -2, 4, -3, acting at equal distances, of one foot each, along a weightless beam, the negative sign indicating that the force acts upward. (10) Three forces represented in magnitude, direction, and posi- tion by OA, OB, and OC, are in equilibrium. Show that Fig. 47. O is the centre of area (the so-called centre of gravity) of the triangle ABC. Solution. OA is equivalent to OB and BA. and OC OB and BC. BO .. 20B and BA and BC. 3BO BAandBC. But BA and BC ,. 2BD if D bisects AC. 3BO equals 2BD in magnitude and direction. (11) A weight of 100 Ib. is fixed to the top of a weightless rod or strut 5 feet long, whose lower end rests in a corner between a floor and a vertical wall, while its upper end is attached to the wall by a horizontal wire 4 feet long. Calculate the tension in the wire, and the thrust in the rod. (12) A uniform rod 8 feet long, weighing 18 Ib., is fastened at one end to a vertical wall by a smooth hinge, and is free to move in a vertical plane perpendicular to the wall. It is kept horizontal by a string 10 feet long, attached to its free end and to a point in the wall. Find the tension in the string, and the pressure on the hinge. (13) A uniform ladder 20 feet long, weighing 60 Ib., is supported horizontally by two men at distances of 4 and 5 feet respectively from its ends. Find the weight borne by each man. 168 ELEMENTARY MECHANICS. [EXS. XVII. (14) A ladder 20 feet long, whose centre of gravity is 8 feet from one end, is carried horizontally by two men, who each carry the same weight. If one of them is at the heavier end, how far must the other be from the light end? (15) A uniform beam, 24 feet long and weighing 200 lb., is supported on two props, one 6 feet from one end, the other 9 feet from the other end of the beam. Calculate the pressure on each prop when a man weighing 180 lb. stands on the beam 8 feet from the first prop. (16) From one corner of a square plate, whose side is 10 inches, a small square, whose side is ,3 inches, is cut away. Find the centre of gravity of the remainder. (17) A circular hole 2 inches in diameter is cut in a uniform circular plate six inches in diameter, the centres being one inch apart. Find the centre of gravity of the per- forated disc. (18) Find the centre of gravity of weights of 7, 6, 9, and 2 lb., arranged at the corners of a square of 1 foot side. (19) Weights of 1, 3, 5, and 7 lb. are placed at the corners of a uniform square plate of 10 inches side, weighing 4 lb. Find the centre of gravity of the system. (20) Find the centre of gravity of weights 2, 3, 4, 5, 6, and 7 lb. placed at the corners of a regular hexagon whose diagonally opposite corners are 18 inches apart. (21) Show that the centre of gravity of a uniform triangular plate coincides with that of three equal masses placed (a) at its angular points; (b) at the middle points of its sides. (22) A triangular plate weighing 5 lb. is in shape an isosceles triangle with its two equal sides each 5 feet long, and its base 8 feet long. A weight of 10 lb. is hung at its vertex. Find the centre of gravity of the whole. (23) The mass of a plate in the form of an equilateral triangle 1 foot high is 4 lb. Masses of 1, 1, and 2 lb. respectively are placed at its angular points. Find the centre of mass of the system. (24) From a square plate, whose diagonal is 21 inches, a corner is cut off by a line joining the middle points of two adjacent sides. How far from the centre of the square is the centre of gravity of the remainder ? (25) If, in the previous question, the coiner, instead of being cut CHAP. VII.] COMPOSITION OP FORCES. 169 off, is folded over flat on to the plate, find the centre of gravity of the whole. (26) Weights each equal to 1 Ih. are placed at the ends of one side of a uniform plate 12 inches square weighing 2 Ib. Determine the centre of gravity of the system. (27) Two similar uniform bars, 4 and 6 feet long, are joined end to end at right angles so as to form an L. Find the centre of gravity of the system. (28) A uniform iron window-frame, 2 feet high, is in shape three sides of a square, surmounted by a semicircular arc. Find its centre of gravity. (29) A uniform iron plate, 2 feet high, is in shape a square surmounted by a semicircle whose base coincides with a side of the square. Find its centre of gravity. (30) A circular portion 5 inches in radius is removed from a circular lamina 8 inches in radius, the distance between the centres of the two circles being 2 inches. Find the centre of gravity of the remainder. (31) If the original plate was wood, and if the hole is filled up with lead of the same thickness, but 12 times as heavy bulk for bulk, where is the centre of gravity? (32.i Find the centre of gravity of a uniform plate 8 inches square containing a circular hole of 2 inches diameter, the centre of the hole being 2 inches from the centre of the plate. (33) The middle points of opposite sides of a rectangular plate being joined, one of the four parts of the rectangle is removed. By what fraction of the diagonal is the centre of gravity of the remainder distant from the centre of the rectangle ? (34) Find the centre of gravity of a uniform quadrilateral plate whose sides are 6, 4, 3, 4 inches long respectively, and whose two equal sides are equally inclined to the others. (35) Find the centre of gravity of a frame made of uniform bars arranged as above. <36) Where is the centre of gravity of a slate-frame, 9 inches by 12, of which one of the short bars has been removed ? (37) Where is the centre of gravity of a triangular frame of sides 5, 4, 3 ? Where is that of a triangular plate of same size ? 170 ELEMENTARY MECHANICS. [SECT. 132. CHAPTER VIII. ON EQUILIBRIUM (Statics). 132. Before leaving the subject of motion as affected by force, there is one important part to be considered namely, the conditions under which forces may act on a body without affecting its motion in any way whatever. One force cannot - satisfy these conditions, but a combination of any number of forces greater than one may ; and it is interesting, and for many practical purposes important, to be able to specify these conditions, and to decide in any given case whether they are satisfied or not. This part of the subject is called * Statics,' and it is a branch of the more general science of Dynamics. Its treatment will depend upon the ideas illus- trated at length in the last chapter, which may be regarded as an introduction to Statics; indeed, they are usually considered as a part of it, and often are made to follow, or are mixed up with, the subject of the present chapter. 133. When all the forces applied to any mass of matter are so balanced that they produce no acceleration in it of any kind, the forces are (or the body is) said to be in equilibrium, and the conditions which they then necessarily satisfy are called the conditions of equilibrium. Observe that equilibrium does not mean rest or zero velocity, it simply means zero acceleration that is, constant velocity. There is no occasion for the velocity to be nothing ; all that is meant is that it keeps the same value, whatever that may happen to be. Thus in the case of a bucket lowered down a well, suppose that it is descending with a constant velocity of 20 feet a second; then, its acceleration being zero, the resultant force acting on it CHAP VIII.] CONDITIONS OP EQUILIBRIUM. 171 (being equal to mass-acceleration, sect. 45) must also be zero. Now the actual forces acting on it are the pull of the earth downwards, and the pull of the rope upwards ; and the resultant of these two being zero, it follows that they are equal. The bucket is in fact simply obeying the first law of motion. Whether it is descending or ascending or standing still, matters nothing, the tension in the rope is always equal to the weight of the bucket so long as its velocity is not changing. The conditions of equilibrium are therefore the conditions under which acceleration is impossible ; or, as it is often correctly expressed, they are the conditions under which rest is possible. It must be observed that forces in equilibrium have no influence in causing rest. They have no effect on the motion at all, and the body exposed to such forces simply obeys the first law of motion. Rest is zero velocity. Equilibrium is zero acceleration. 134. This being clear, we will proceed to state the con- ditions of equilibrium for any number of forces, and first of all The Conditions of Equilibrium for Two Forces. The conditions which two forces have to satisfy in order to balance each other and have no effect on the motion of the body to which they are applied, are very simple and obvious namely : (1) The forces must both lie in the same straight line ; (2) They must act in opposite directions ; and (3) They must be equal. This is all usually expressed by saying simply that the two forces must be equal and opposite, the acting in the same straight line being understood. If any number of forces are in equilibrium, the resultant of any number of them must be equal and opposite to the resultant of all the rest. For obviously all the rest are equivalent to their resultant, and that resultant is balanced 172 ELEMENTARY MECHANICS. [SECT. 134. by a force equal and opposite to it. The statement just made is not to be quoted as a condition of equilibrium, it is merely a manifest fact which may help us to ascertain the conditions of equilibrium. 1 35. Let us see how it gives us the equilibrium conditions for three forces, for instance. Any one force must be equal and opposite to the resultant of the other two. Now any two of them, as A and B, in order to have a resultant, must lie in one plane, in other words, must (if produced) meet in a point, and through this point their resultant must pass, being the diagonal of the parallelogram of forces; the third force, C, in order to maintain equilibrium, must, by the above statement, be a prolongation of this diagonal, and hence it too passes through the same point as the other two, and is in the same plane namely, the plane of the parallelogram ; it must also be equal to the diagonal in magnitude ; in other words, it must be equal to the third side of a triangle, two of whose sides represent the other forces, such as OAR (fig. 48). Its magnitude, direction, and position are thus completely determined. Let us restate these : The Conditions of Equilibrium for Three Forces. (1) The three forces must all be in the same plane. (2) Their lines of action must all pass through the same point. (3) It must be possible to draw a triangle with sides parallel (or perpendicular; see foot-note, sect. Ill) to the forces, and proportional to them in magnitude. The sides of the triangle must all be drawn in the same sense as the forces (thus in the figure, OA, AR, RO are the senses), and CHAP. VIII.] CONDITIONS OF EQUILIBRIUM. 173 it must be possible to draw the triangle without taking the pen off. This is usually expressed by saying that the three forces must be representable by the sides of a triangle taken in order. The last two conditions together really include the first. Any number of forces greater than three need neither meet in a point, nor lie in the same plane, in order to be in equilibrium. Conditions of Equilibrium of a Particle. 136. Any number of forces acting on a particle will evidently be in equilibrium if they are representable by the sides of a closed polygon (plane or otherwise) drawn parallel to the respective forces and taken in order. This is the same as saying that the forces must have no resultant; for the line required to complete the polygon represents the resultant (sect. 108), but no line is required to complete a dosed polygon, hence there is no resultant. The converse is also true namely, that if forces acting on a particle are in equilibrium, they must be representable by the sides of a polygon taken in order. This proposition obviously includes the triangle of forces, for a triangle is only a three-sided polygon. Conditions of Equilibrium of a Rigid Body. 137. If the condition just stated for a particle is satisfied by the forces acting on a rigid body, they can produce no translation, only rotation ; hence a rigid body will evidently be in equilibrium if the above condition for a particle be satisfied, and if also the directions of all the forces pass through a single point ; for a set of forces which intersect in one point, and have zero resultant, cannot possibly rotate anything. But this last condition, though sufficient, is not necessary that is, the converse is not true : if the forces 174 ELEMENTARY MECHANICS. [SECT. 137. acting on a rigid body are in equilibrium, they must in- deed be representable by the sides of some closed polygon (plane or otherwise), but they need not meet in a point, unless there are only three of them. The more general con- dition for no rotation is that the moments of all the forces about every possible point or axis of rotation must add up to zero. If this and the particle condition are satisfied, equilibrium is complete ; and conversely, wherever there is equilibrium, these must be satisfied. So these are the necessary and sufficient conditions, though not in a very simple form to apply practically. It will be sufficient for us, however, to consider at length, and put into a more practical form, only the case where all the forces act in one plane ; and we will proceed to this from a fresh point of view in the next section ; but we may first notice a mode of expressing the conditions of equilibrium in terms of the construction of sect. 123. If the force polygon is closed there is no resultant force, but there may be a resultant couple, causing rotation. In that case the funicular polygon cannot be closed by pro- ducing its first and last sides, for they will be parallel. Whenever the funicular polygon is closed there is no result- ant moment, and the funicular polygon cannot be closed unless the force polygon is closed ; so in cases of complete equilibrium, the force polygon and the completed funicular polygon are both closed. General Conditions of Equilibrium of a Rigid Body acted on by Forces in a Plane. 138. The motions possible to a rigid body are translation or rotation or both, hence the conditions for equilibrium really involve the conditions for no translation and for no rotation (strictly speaking, for no rectilinear and for no angular acceleration ; but the words translation and rotation CHAP. VIIT.] CONDITIONS OF EQUILIBRIUM. 175 are used instead of these more accurate terms for shortness ; and the error is not great, for the conditions of equilibrium render entire rest possible, though they do not in any way enforce it). Now, having assumed that the body can only move in a plane (say a vertical plane), and that the forces only act in this plane, it is obvious that all translations must be up or down, or right or left, or else a motion compounded of the two, which may be analysed into up or down and right or left components. Hence, in order that there may be no translation at all, the forces must have no resultant either up or down or right or left : this being a practically con- venient form of saying that they have no resultant at all at a finite distance. Still, however, they might spin the body (sect. 128) ; hence, in addition to the above, the condition is necessary that the sum of their moments about any point in the plane must vanish ; and then the forces will be unable to cause any motion at all. Or the complete condi- tion for equilibrium might be stated by saying that the sum of the force-moments about every point in the plane must be zero ; since this necessitates the non-existence of either a resultant force or a resultant couple. This condition is, however, not so practically applicable, in general, as the two separate conditions just laid down, which we now repeat with emphasis : The general conditions of equilibrium for a body only able to move in a plane are : (1) That the sum of the components of all the forces in any two directions in the plane at right angles to each other shall vanish. (2) That the sum of the moments of all the forces about any one* point in the plane shall vanish. * One point is sufficient if condition 1 is satisfied, because the moment of a couple about every point is the same (sect. 128) ; hence, if it is zero about any cue point, it is zero altogether. 176 ELEMENTARY MECHANICS. [SECT. 138. (1) is the condition for no translation (properly speaking, for no rectilinear acceleration). (2) is the condition for no rotation (properly speaking, for no angular acceleration). If (1) is satisfied without (2), there is rotation, but no translation. If (2) is satisfied without (1), there would be translation, but no rotation about the particular point considered. If (2) is satisfied for every point, then (1) is also necessarily satisfied, because the moment of an unbalanced force must differ for different points. If neither is satisfied, there must be both translation and rotation. If both are satisfied, there must be complete equilibrium. The converse of each of these statements is also true. In case the body on which the forces act has one point fixed so as to be incapable of translation, the necessary and sufficient condition for equilibrium is simply that the resultant of all the forces must pass through the^ fixed point or pivot (see sect. 119, statement 3). And in general, instead of applying both con- ditions (1) and (2), it would be sufficient to apply condition (2) to three different points not in the same straight line, but it would be more troublesome in practice. ILLUSTRATIONS. 139. Consider a ladder standing on rough ground, and resting against a perfectly smooth wall. What forces are acting upon it? There is the weight of the ladder AV acting downwards at its centre of gravity G (fig. 49) ; there is the pressure of the ground R acting in some unknown upward direction at some angle with the vertical not greater than the 'angle of repose' (sect. 115), and the pressure of the wall P acting normal to the wall or horizontally ; and that is all. But the ladder is in equi- CHAP. VIII.] CONDITIONS OF EQUILIBRIUM. 177 J Fig. 49. librium, hence these three forces must pass through a point (sect. 135). Now W and P, whose directions are known, intersect when produced in the point C ; hence R also passes through the point C (fig. 50). This determines its direction. Moreover, when three forces are in equilibrium, they must be proportional to the sides of any triangle which are drawn respec- tively parallel to the forces. Such a triangle is ABC (fig. 50) ; CB is parallel to W, and represents it ; BA is parallel to P, and represents it; and AC is parallel to R, and represents it. If, then, j the position of the ladder were given us, and also its weight, we should simply have to draw the above diagram, and measure the sides of the triangle ABC, in order to determine the pressures P and R in terms of W; the direction of R being also given by measuring either the angle BAC or BCA. This would be solving the pro- blem by construction. 140. But suppose we wished to do it by calculation, applying the general conditions of sect. 1 38 : we should first consider the inclined force R resolved into two (see fig. 49), a normal pressure JNT, and a friction F (the friction being always in such direction as best hinders slipping, sect. 115), and then say that, since there is equilibrium L Fig. 50. 178 ELEMENTARY MECHANICS. [SECT. 140. as regards translation, there can be first no up or down resultant, or ]N" and W must be equal and opposite ; and then that there can be no horizontal components, or F and P must be equal and opposite. But to determine either F or P, in terms of W, we must make use of the second condition the condition for no rotation namely, that the forces can have no rotating power, or resultant moment, about any point. Take it numerically : Let us suppose that \ve are told the weight of the ladder is 60 lb., and that its centre of gravity is I of its length up, that the foot of the ladder stands six feet from the wall, and the top of the ladder thirty feet from the ground ; then, as the condition ipK no translation, we have already found S N=W, and F=P. But we don't know either F or P yet ; we must find them by taking moments about some point any point we like, for we know that since there is no rotation the sum of the moments about every point must be zero. Suppose we take moments about the point A, then neither N nor F has any moment ; so the moment of P, P x 30 feet, must be equal and opposite to the moment of W, "YV x \ of 6 feet. hence 15P = W = 60 lb. weight, or P=the weight of 4 lb. And we already know that F and P are equal ; so then N, and P are all known, and now too we know K, because R 2 =N 2 + F 2 ; that is, R =60-13 lb. weight. See if this agrees with a determination by measurement, and then repeat the whole process with the wall rough instead of the ground, and then with both wall and ground rough. If the inclination of the required E to the vertical in fig. 49 be greater than the limiting angle of friction, equilibrium is impossible, unless a wedge be placed under Fig. 61. CHAP. VIII.] CONDITIONS OF EQUILIBRIUM. 179 the foot of the ladder, which may be considered as equiv- alent to tilting the ground up. If the ground is like a sheet of ice, the required force F can be supplied by a rope tying the ladder to the wall. If the ground is level and smooth, and if no extra force is applied to the foot of the ladder, equilibrium cannot be attained by any roughness of wall short of actual attachment, as by a hook, because 1ST and W will not intersect except at infinity. 141. Next consider a weightless rod resting against a smooth wall over a smooth rail, and with a weight stuck somewhere on it, as shown in fig. 51. (A section only of the rail supporting the rod is shown, as a small circle.) To determine where the weight must be for equilibrium, the forces acting are: the weight, W; the normal pressure of the wall, P; and the normal pressure of the rail, R. Now, here again are three forces, so to be in equilibrium they ought to intersect in a point; but in fig. 51 they do not intersect in a point, produce them as much as you like ; their direction encloses a triangle CjCgCg instead. Hence there is no equi- librium, and cannot be until the three points c l5 c 2 , 3 coincide in one point C. Observe that no alteration of magni- tude in any of the forces can assist equilibrium ; a shift of either the direc- tion or the position of some force is essential. The easiest thing to shift is the load ; so to find where it oug'ht to be shifted to, draw a fresh figure, and from C, the inter- section of P and K, draw a vertical ; this will cut the rod Fig. 52. 180 ELEMENTARY MECHANICS. [SECT. 141. at the point where the weight ought to be placed for equilibrium (fig. 52). To measure the relation between the magnitudes of the forces when the weight is in this place, we can produce R till it cuts the wall in B ; then the triangle ABC has its sides parallel to the three forces, BA to W, ACtoP, CB to R ; hence the lengths of these sides will give the forces, if one of them, say W, is known. It is easy to see that R 2 =P2 + W 2 . In fig. 51 the rod would be slipping up the wall and falling over the rail ; this is because the line of W falls to the right of the point c v where P and R intersect. If the line of W fell to the left of this point, the rod would slip down the wall, and drop between it and the rail. There is just one position where it does not slip either way ; but it is unstably balanced, because motion either way would allow the weight to get lower. Of course if there was any friction, there would be a margin of stability. 142. Now consider a body on an inclined plane held still by some force P acting in any given direction, as depicted in fig. 64, with a diagram, fig. 53. There are three forces, P, R, and W, in equilibrium (R being the normal pressure of the plane), hence P must be in the plane of the other two. To find its magnitude : take off a length AB to represent the weight of the body, and from B draw a line parallel to P, till it cuts R pro- duced in the point C. Then we have a triangle ABC (fig. 53) ; BC represents P, and AC represents R; and it is easy to measure these lengths on the same scale as AB was drawn. Fig. 53. CHAP. VIII.] CONDITIONS OF EQUILIBRIUM. 181 There are in general two positions in which the same force P can hold up the body. For draw a circle with centre B and radius BC, it will cut R in two points, C and D ; hence the same force P would be just as effective if it acted in a direction shown dotted as P' parallel to BD ; but the pressure on the plane would then be greater than AC, namely, AD. There is one case when only one direction will do, and that is when the radius of the circle is so small that it only just touches R. This radius then represents the minimum force possible, and shows that it must act per- pendicularly to R and therefore parallel to the plane, and must have the same relation to the weight that the height of the plane has to its length. If the plane be rough, friction is such a force. If the applied force P is smaller than this that is, so small that the line representing it is unable to reach across from B to the line R, then there cannot be equilibrium ; and even if P is greater than this, but does not act in the best direction, there need not be equilibrium, and the body will slide down, as in fig. 54 : the accelerative or resultant effect- ive force being the component of W along the plane namely, AM, minus the component of P along the plane namely, AN. The pressure on the plane is the component of W at right angles to the plane, minus the component of P at right angles to the plane; that is, Am An. When the plane is rough its reaction R need not be normal, but may be inclined at any angle up to the limiting angle of friction on either side of the normal. There will thus be two critical cases one when the body is on the point of sliding down, the other when it is on the verge of being hauled up. The same construction as above (fig. 53) suffices to determine the needful applied force corresponding to one or other of these Fig. 54. 182 ELEMENTARY MECHANICS. [SECT. 142. two conditions namely, produce either limiting R backward and bridge the gap between it and B with a line in the desired R direction (fig. 55). If no force at all is N \ ^ * needed to hold the body up, the reaction R is vertical ; and if the body is on the verge of slipping down, the tilt of the plane is the angle of repose. The least force necessary in any given case is the perpendicular from B on to the line of , . N R ; hence the least force is inclined to 1 the plane just as much as R is inclined to the normal, or the best angle of trac- tion is the angle of repose. If the pro- duced line of R ever lies to the left of B, it means that some force is necessary to drag the body down ; the inclination of the plane being in that case less than the angle of repose. 13 Fig. 55. 143. We have here considered only the slipping of the body; but if it were a ball it would roll, and if it were a block it might topple over, before it began to slide. Let us just see how soon a rectangular block on a rough inclined plane will topple over. We know that the resultant of all the forces which gravity exerts on the particles of the body passes through the centre of gravity that is, the body acts statically as if its weight were all concen- trated at the centre of gravity. Hence if this point be sup- ported, the whole body is supported. The line of W is the vertical through G; and if this line falls inside the base,* the body cannot topple over ; it can only slide * By ' the base ' must be understood the area enclosed by a string stretched round that part of the body which touches the plane : consider, for example, the case of a retort stand with a forked foot. Fig. 56. CHAP. VIII.] CONDITIONS OF EQUILIBRIUM. 183 down. To upset the body, it must be tilted through the angle AGO (fig. 56) ; and if it be momentarily tilted through less than this, it will return to its old position. The angle AGO is a measure of the ' steadiness : ' the larger this angle, the more steady is the body. If the vertical through G fell - through 0, the body would * v be balanced in unstable Flg ' 57 ' equilibrium, and directly the vertical passed outside the point 0, the body would topple over; and this applies universally. A wagon going along with one wheel in the gutter does not upset so long as the vertical through its centre of gravity falls inside the wheel-base ; but the act of going over a stone may tilt it sufficiently to make this line pass beyond the base, and then it upsets. The two bodies in fig. 57 resting on a flat plane are both evidently steady, but the steadiness of the first is much greater than that of the second; and this for two reasons, firstly, because its base is wider, secondly, because its centre of gravity is lower.* The centre of gravity of an omnibus full outside, but with no inside passengers, must be very high up ; and a moderate shock might be sufficient to destroy its stability and upset it. A block resting on a level surface can be upset by a horizontal force applied high enough. The criterion is obtained by considering moments about the forward edge of its base; if the moment of the force is equal to the moment of the weight about this edge, the body is on * The most useful measures of steadiness are 1st, tJie moment of stability namely, the moment of the couple required to upset the body, or the weight of the body multiplied by the distance OA ; and 2d, the dynamic stability namely, the work that is required to upset it, or the weight of the body multiplied by the difference of the distances AG and OG (fig. 57). . 184 ELEMENTARY MECHANICS. [SECT. 143. the point of tilting over. If the force applied is sufficient to call out the maximum friction, the body is on the point of sliding ; and at one particular altitude both conditions can be satisfied simultaneously. It is easy to show that this critical altitude for a symmetrical block is half the base divided by the coefficient of friction. Stability and Instability of Equilibrium. 144. A body in equilibrium, with infinitely small stability, is said to possess unstable equilibrium; the least shock must upset it. Thus, if you narrow the above block till its base is nothing, there remains only a plane or line standing on its edge, and though, when vertical, the centre of gravity of this does not fall without its base, and therefore it is in equilibrium, yet the slightest breath will upset it. This is not the case with all bodies balanced on a point. A body with a rounded base resting on a plane may be stable enough though it cannot be called steady. Bodies supported by a point, whether slung like a pendu- lum or pivoted like a compass-needle or rolling like the half of a split billiard-ball, are said to be in equilibrium ; and these examples are in stable equilibrium, because, if you rock them, they will return to their original position after a few oscillations (see sect. 146). Any segment of a sphere less than the whole sphere will so rock. And a leaden hemisphere would rock in stable equilibrium even if loaded above with a bulky pith figure. It is quite possible for a body to possess an equilibrium which is neither stable nor unstable that is, the body, when disturbed, neither topples over nor returns to its original position. All that is necessary is that the vertical CHAP. VIII.] STABILITY OF EQUILIBRIUM. 185 through G shall always pass through the point of support, as in the case of a sphere on a flat table; or that the centre of gravity itself shall be supported, as in a flywheel. The body is then indifferent how you place it, and its equilibrium is called neutral. An egg lying on its side has neutral equilibrium for rolling, and stable equilibrium for ' pitching ; ' it is unstable all ways when balanced on its end. 145. In the case of a body pivoted at a point, if the point is above the centre of gravity, the equilibrium is perfectly stable ; if at the centre of gravity, it is neutral ; and if below, it is un- stable. EXAMPLES. The nearer the centre of gravity of the beam of a balance is to the point of support, the more sensitive is the balance ; but it is necessary to have the centre of gravity slightly lower than the point of support, or the equilibrium would not be stable. If the balance is to be equally sensitive for all loads, its three knife-edges, the one supporting the beam and the other two supporting the pans, must be in one straight line ; the restoring force is then simply the moment of the displaced centre of gravity of the tilted beam. If the weight of the beam W is displaced a minute horizontal distance x, by slightly unequal loads P and Q in the pans, each supported at a distance a from the fulcrum, the equation of equilibrium is (P-Q,)a = Wx ; and of course the pressure on the fulcrum is P + Q + W. A compass-needle is always made with a little central cap, into which the point supporting the needle passes from below, so as to be above the centre of gravity of the needle. See FifT 59 fig. 59. Again, it is easy to balance a curved beam on a knife-edge, while- a straight one will not remain balanced for more than a 186 ELEMENTARY MECHANICS. [SECT. 145. few seconds, unless loaded. Compare the diagrams in tig. 60. The weights in the third must be rigidly attached to the beam by rods, not by strings. Stable. Unstable. Stable. Fig. 60. 146. In the case of a body with a spherical base standing on a level plane, its centre of gravity cannot help being above the point of contact with the plane, and yet the equilibrium may be stable or neutral; as, for instance, in a sphere the equilibrium is neutral, and in a hemisphere it is stable ; or again, it may be unstable, as in an egg balanced on one end. The centre of the sphere, of which the base forms a part, is in these cases to be regarded as the real point of support, and then the former rules apply. Thus, if G be the centre of gravity of the combina- tion shown in fig. 61, and if C be the centre of the sphere of which the base forms a part, the whole will oscillate in stable equilibrium. When a body rolls along any surface, its centre of gravity in general describes a curve with crests and hollows ; every hollow corresponds to a position of stable equilibrium (the centre of gravity is then in one of its lowest positions) ; every crest corre- sponds to a position of unstable equilibrium, and a measure of the instability is the curvature (see sect. 13) of the path of the centre of gravity. For instance, in the case of a body balanced on a point, the higher the centre of gravity above the' point the less curved will be its path, and the less unstable will be the equilibrium : for example, it is easy to balance a stick loaded at one end on one's finger if the load be at the top of the stick, but if the stick be inverted it is not easy. The criterion for equilibrium, as well for its stability CHAP. VIII.] STABILITY OF EQUILIBRIUM. 187 or instability, obtained by considering the path of the centre of gravity, is a useful and very general one. When weights are in equilibrium, as on an inclined plane or on a system of pulleys, it is because the path of their common centre of gravity is horizontal if it move at all. If the path is a horizontal straight line, the equilibrium is neutral and a slight shift makes no difference. It is an instructive exercise to prove that in the case of weights in equilibrium on an inclined plane the path of their centre of gravity is a horizontal line. If the path of a centre of gravity is curved, the equi- librium is stable or unstable according as it curves upwards or downwards. An instructive example is afforded by a wooden disc loaded with lead near its circumference. Such a cylinder placed on a slightly inclined plane may easily roll uphill into a position of equilibrium with its centre of gravity vertically over the point of support. Metacentre. In cases where a small disturbance changes the point of support, like the case of a portion of a sphere on a level table, there is usually a point in the body through which a vertical through the point of support (what may be called the line of support) will continue to pass, at least if the disturbance be only small. This point is the intersection of the new line of support with the old line of support, if the latter be considered as marked in the body and rotating with it, and it is called the metacentre. In the case of a sphere, the centre of the sphere is its metacentre. The conditions of equilibrium for small disturbances are the same as if this point were a pivot, and its height above or below the centre of gravity measures the stability or the instability of the equilibrium. For bodies of irregular shape it does not follow that the suc- cessive lines of support intersect at all, and then there is no metacentre. Often there are two, one for a rolling or broad- 188 ELEMENTARY MECHANICS. [SECT. 146, side motion, the other for a pitching or lengthways motion. For further information on this subject, see sect. 179. 147. Criterion of equilibrium by zero work done. A powerful method for determining condition of equili- Fig. 62. Letter-balance. brium is sometimes called the method of ' virtual velocities,' and consists in observing that if a body subject to any forces is in equilibrium, it must be true that when slightly shifted in any way no work is done; that is to say, if every force is multiplied by the velocity of its point of application when a displacement is made, or by the virtual velocity of this point when displacement is only imagined, the products will all add up to zero. It is interesting and easy to show that in cases of levers, inclined planes, pulleys, &c., worked by gravity, only the centre of gravity of everything remains unchanged in vertical height whenever there is smooth equilibrium. Again, in the case of any parallel motion, such as is employed to support the platform of railway weighing- machines and in the ordinary scale letter-balances, the position of the thing to be weighed is unimportant ; and it is unimportant because wherever the load is, on any parallel- moving scale-pan, there will be the same vertical descent, and therefore the same virtual velocity, and the same work done. CHAP. VIII.] CONDITIONS OF EQUILIBRIUM. 189 EXAMPLES XVIII. (1) When a weight is supported on an inclined plane by a force acting along the plane, show that the ratio of the force to the weight is the same as the ratio of the height of the plane to its length. (2) And show that the ratio of the supporting force to the normal pressure on the plane is the same as the ratio of the height of the plane to its base. (3) Hence show that if a body is supported on a plane only by friction, it will begin to slide down when the ratio of the height of the plane to its base is equal to the coefficient of friction (see sects. 142 and 115). (4) A picture-frame weighing 10 Ib. is hung by a cord passing over a nail, the two parts of the cord making an angle of 120 with each other. Find the tension in the cord. (5) If the two parts of the cord included an angle of 90, what would then be the tension ? (6) If a rod rests inside a smooth spherical shell, its centre of gravity must be vertically under the centre of the sphere. Hence, if the rod be uniform, it can only lie horizontally, unless it is equal in length to the diameter of the sphere. Verify these statements. (7) It is wished to upset a tall column by means of a rope of given length, pulled by men on the ground. At what height above the base of the column will it be best to attach the rope ? (8) A uniform rod hanging from one end is pulled aside from the vertical by a horizontal force equal to half its weight applied at its lower end. At what angle will it be in equilibrium ? (9) A 'simple pendulum is pulled aside from the vertical by a horizontal force equal to half its weight. At what angle will it be in equilibrium ? (10) A bar of uniform thickness inclined at an angle of 30 with the horizontal, with one end against a wall, rests across a rail at a point 2 feet away from that end. Find the length of the bar if the rail and wall are both smooth. (11) If the bar is 8 feet long and weighs 10 Ib., and the rail is smooth but the wall rough, show by help of a diagram the direction and magnitude of the pressure of the bar against the wall. 190 ELEMENTARY MECHANICS. [EXS. XVIII. (12) Construct a diagram for the case of a bar in a given position, but with an adjustable load, supported by a rough wall and rail ; () when the bar is on the point of falling over the rail ; (b) when it is on the point of slipping down between rail and wall. (13) Given distance of rail from wall and length of bar, as a and b respectively, find its position of equilibrium if perfectly smooth. (14) A ladder 30 feet long, whose centre of gravity is 12 feet from the lower end, rests on rough ground against a smooth wall, which is 10 feet distant from the foot of the ladder. Draw a figure to scale showing the directions of the reaction of the wall and of the ground; and find by measurement the magnitudes of these reactions if the ladder weighs 80 Ib. (15) Check the above by calculating the result by the method of moments, and determine the coefficient of friction. (16) Show that a light ladder ascended by a heavy weight is not safe if inclined at an angle greater than the angle of repose. (17) A beam 10 feet long weighing 80 Ib., with its centre of gravity 4 feet from one end, is suspended horizontally by two strings attached to its ends, and fastened to two pegs in a horizontal line 16 feet apart ; the shorter one of the strings is 5 feet long. Find the length of the other string, and the tensions in the two strings. (18) A rectangular block of oak 8 inches long, 3 inches broad, and 3 inches thick, rests with one of its square faces on a horizontal oak surface. If the surface be gradually tilted, how will the block begin to move? [Take the coefficient of friction as 0'4.] (19) If it stand on a level surface with coefficient of friction , and if it weigh 28 Ib., what force would make it uncertain whether to topple over or slide along ? (20) A solid consists of a hemisphere with a cylinder standing centrally on the base of the hemisphere ; the radii of the hemisphere and cylinder are respectively 6 and 3 inches, and the height of the cylinder is 6 inches. Find the position of the centre of gravity of the solid. (21) If the above solid is placed on its hemispherical end on a horizontal table, will the equilibrium be stable or unstable ? (22) If the same solid stands on its cylindrical end, through what angle must it be turned to upset it ? CHAP. VTTL] CONDITIONS OF EQUILIBRIUM. 19] (23) A wooden cylinder, 1 foot in diameter, with some symmet- rical holes of two inches diameter already bored through it lengthways at a margin of one inch from the circumference, is found to weigh 3 lb., and one of the holes is then filled up with 3 lb. of lead. Find the position of the centre of gravity, the restoring moment when the cylinder is rolled through 90 on a flat surface, and also its position of equilibrium when on an inclined surface. (24) On what slant would this cylinder be on the verge of rolling down? Construct its position of equilibrium on a plane rising 1 in 6. 192 ELEMENTARY MECHANICS. [SECT. 148. CHAPTER IX. ON MACHINES AND OTHER CONTRIVANCES ILLUSTRATING THE FOREGOING PRINCIPLES. ELEMENTS OF APPLIED MECHANICS. 148. A machine is an instrument for transferring energy in such a manner that certain useful or desirable work is done. The effective force exerted by the agent which loses the energy used often to be called the * power,' the force exerted by the body "which receives it being called the * weight;' but the terms 'effort' and 'resistance,' intro- duced by Rankine, are better. The machine is simply a mediary by which the energy is indirectly transferred from one body to the other. The quantity of energy gained by the one body is equal to that lost by the other, except for what may become dis- sipated as heat or other non-mechanical form of energy ; in other words, no increase in quantity of energy is ever effected by any machine. Numerous attempts have been made to construct a machine able to effect this : such attempts are called the search after perpetual motion, and always result in failure (cf. sect. 88). All that one can do by means of any machine is to vary the ratio of the two factors, F and s, occurring in the product work, the product itself remaining unalterable. But just as the number 12 may be split up into various pairs of factors, 12 and 1, 6 and 2, 3 and 4, so the factors of the constant product work may be varied at will : and this is the use of a machine. Given a force, CHAP. IX.] MACHINES, 193 and a distance through which it can act, a machine can always be devised to overcome any other force whatever through some definite distance, such that the product of the second force and distance is nearly equal to the product of the first force and distance. The greater the force required to be overcome, the smaller is the distance through which it can be overcome by a given expenditure of energy. In other words, a feeble agent moving quickly may be able, by means of a machine, to overcome a great re- sistance, that is, may move slowly an obstacle of considerable strength ; and the slowness will be proportional to the force to be overcome. This is often expressed by saying that what is gained by any machine in power is lost in time (or in distance). Or again, by saying that the 'mechanical advantage ' of a machine the ratio of the resistance over- come to the least force required is the inverse of the ratio of the distance travelled by the ' resistance ' to the distance travelled by the ' effort ; ' supposing the machine to be perfectly efficient, having no friction. A brief way of stating the same thing is to say that the 'force-ratio,' which is another name for 'mechanical advantage,' varies inversely with the 'speed-ratio.' This condition may also be expressed by saying that if any system in equilibrium under the action of any number of forces receive a slight displacement, then the total work done by the whole of the forces, or the total loss of potential energy, is zero. In other words, the sum of the products of all the forces into the respective distances they have simultaneously moved, or, what is the same thing, into the respective velocities of their points of application measured along their lines of action, is zero. This is frequently a useful mode of finding the condition of equilibrium of a system, and it has already been so applied in sect. 147. It is often referred to as the principle of virtual work or virtual velocities ; the meaning of the word ' virtual ' being merely that the displacement or shift supposed to take place is an imaginary one, and need not really occur. Efficiency. If a machine could be arranged so that the M 194 ELEMENTARY MECHANICS, [SECT. 148. work done by the one body were equal to the energy gained by the other, the machine would be called perfectly efficient that is, its efficiency would be unity. But in practice there is always some loss of energy by friction, &c., and so the efficiency is a proper fraction; and the ratio of the energy recovered to the energy put in is a measure of the * efficiency' of the machine. It will be seen from this definition of efficiency that it is equal to the product of the force-ratio and the speed-ratio. Simple Machines. 149. A pulley is a simple machine by which a weight may apparently be supported by means of a force only half as great as itself; the obvious reason being that the other half of the force necessary to support the weight is supplied by the hook fixed in the ceiling, to which one end of the cord is attached (fig. 63). If the force P exceeds in the slightest degree half the weight, it must raise it ; but only half as fast as itself ascends. To raise it at the same rate would require both parts of the loop of cord in which W is slung to be lifted. If only one end is lifted, the wheel or pulley rotates, and W only rises at half the rate. The mechanical advantage or ' force- ratio ' of a simple pulley is thus 2, and its * speed-ratio ' is J. An inclined plane is another simple machine on which a weight may be apparently supported by a force less than its own weight ; the reason being that the rest of the necessary force is supplied in a station- ary manner by the pressure Fj g> 64. of the plane. If the sustain- ing force or 'effort,' P, is applied as shown in fig. 64, it is evident that a descent of P through a vertical height I, equal to the whole length of the Fig. 63. CHAP. IX.] MACHINES. 195 plane, would pull W all the way up the plane indeed, but would only raise it a vertical height h ; hence the mecbanical advantage of this machine is -j- ; and if P exceeds W-j- in the slightest degree, it must raise the weight ; provided, of course, that there is no friction. A lever, a wheel and axle, and a capstan are simple machines in which a weight applied at a great distance from an axis of rotation may apparently support a greater weight nearer the axis ; the reason being that the rest, or the whole, of the sus- taining force is supplied by the support of the axis, or the fulcrum. Thus, in the left-hand diagram of fig. 65, P and W are both really supported by the fulcrum F ; the pressure on it being always W + P ; in the right-hand diagram the weight is sustained by F and P jointly, and the pressure on the fulcrum is W - P. All that P does is to balance the rotation tendency of W ; and for this purpose its moment, Px AF, must equal the moment of Fig. 65. W, W x BF. Hence the mechanical advantage of a lever, the ratio of W to P, is always ^,, or the ratio of the ' arms ' of the lever. In the case of the steelyard (fig. 65), the weight of the 'yard,' Y, acting at its centre of gravity, helps the smaller weight P, so thatW. BF = P. AF + Y. GF. 196 ELEMENTARY MECHANICS. [SECT. 149. A lever cannot, however, be used to raise weights far; but an easy modification, securing contin- uous action, is to make the ful- crum F into a pivot, and to apply P and W at the circumferences of circles or wheels, with common centre F. Thus we get the wheel and axle, or capstan (fig. 66), of which the mechanical advantage is, as before, the ratio of the distance of P from the pivot to the distance of W from Fig. 66. the pivot that is, the radius of the wheel divided by the radius of the axle. Combinations of Simple Machines. 150. Any of these machines may be combined together, so that the resistance of one machine constitutes the ' effort ' of the next, and the mechanical advantage of the combination will be the product of their separate mechanical advantages. Thus three pulleys are shown combined in fig. 67, and the Fig. 67. Fig. 68. Fig. 69. mechanical advantage of the combination is 2 x 2 x 2, or 8, if the pulleys are weightless. If W is raised one foot, P must rise eight feet. The whole pull on W is here the pull of the beam CHAP. IX.] SIMPLE MACHINES. 197 above plus P; hence the pull on the beam is W - P. The arrange- ment may evidently be turned upside down, so that the beam becomes the weight, and the weight the beam (fig. 68). In this case the weight supported is less by P than it was in the former case. Fig. 67 used often to be referred to as the first system of pulleys, and fig. 68 as the third. If the weight of the pulleys is not small enough to be neglected, call them w^ w 2 , &c., and consider fig. 67. The lowest pulley is attached to the weight, and rises at the same rate as it does ; the second pulley rises at twice, and the third at four times this speed. Now, if any weight w be raised a height h, the work done is wh : so if W is raised one foot, represents the whole work done by P, in moving through a distance of 8 feet, that is, by the expenditure of 8 P units of potential energy ; hence, in general when there is equilibrium, the mechanical advantage W : P must be determined from the equation, if there are n pulleys. This equation expresses the fact that the algebraical total of the work done is nothing ; or, if we choose to put it so, that the common centre of gravity remains fixed in position, or at any rate does not rise or fall. The only one of the old-fashioned systems of pulleys frequently employed for hoisting is what was called the second system, where there are two blocks of pulleys, one attached to the weight, and the other to the beam ; and where the same rope passes round all (fig. 69). The mechanical advantage in this case is simply equal' to the number of strings supporting the weight: which in the figure happens to be four. 151. A combination of levers is sometimes used, but more often for the purpose of magnifying small motions than for exerting great force ; that is, for increasing the factor s in the product work at the expense of the factor F. In fig. 70 the motion of the screw is magnified, the pointer describing a considerable arc for one turn of the screw; such an arrangement is sometimes employed for measuring expansion of a rod by heat, as in the so-called Ferguson's pyrometer : the screw-support being there tno.ved far back, and the rod inserted horizontally between the screw and the lever. A screw alone may be regarded as a 198 ELEMENTARY MECHANICS. [SECT. 151. combination of a lever and an inclined plane, the inclined plane being coiled up into a spiral or screw-thread (fig. 71). For every complete revolution of the lever, the resistance is over- come through a distance equal to that between the spires of the screw-threads; hence the mechanical advantage of a Fig. 70. screw-press is the circumference of the circle traversed by the force applied at right angles to the lever, divided by the ' pitch ' of the screw that is, the distance between its successive spires. Wheels and axles are usually combined by means of cogs, as is well seen in the wheel-work of a clock. A pulley is often used in conjunction with a capstan, the rope passing round a pulley attached to the weight, and the mechanical advantage of the capstan being thereby doubled. Moreover, the free end of the rope, instead of being rigidly fixed, may be coiled round another smaller axle with the same centre F, so that its tension shall help the force P (fig. 72). By this means the mechanical advan- tage can be increased to any desired extent, for the weight is now wound up only because the cord wraps itself on to one, the larger, axle faster than it unwraps itself from the other smaller axle ; and the two axles may be as nearly the same size as one pleases. The mechanical advantage is the radius of the wheel (or the length of P's arm) divided by the difference of the radii of the two axles, the whole being multiplied by two because of the pulley. Fig. 72. Chinese Capstan. CHAP. IX.] SIMPLE MACHINES. 199 One of the most useful forms of pulley block, the ' Weston ' or differential pulley, depends on this principle. It is shown in diagram and in actuality in fig. 73. An endless chain is em- ployed, and the wheels have teeth or ridges grip- ping links to prevent slipping. The two wheels are of unequal size, and when the chain is pulled, one of its loops increases in size and the other decreases, so the weight can be hauled up. Its mechanical advantage is twice the radius of the big wheel divided by the difference of radii ; or, what is the same thing, twice the number of teeth jn the big \vheel divided by the number of its teeth in excess of the smaller one. But as the teeth differ in number usually by one only, the mechanical advantage is usually simply twice the number of teeth in the bigger wheel. It is a convenient feature of all machines with a great mechanical advantage that the friction is able to sustain the load without any applied force. A wheel and axle may be combined with a screw, as shown in the contrivance of fig. 74. When the handle is turned, the screw-thread on its axle sends the cog-wheel forward one tooth for every revolution. Such a screw, which itself does not advance in a nut, but which merely rotates in j^ 73 ordinary bearings, is called an 'endless' screw. If I is the length of the handle arm, n the number of teeth in the wheel, and r the radius of the axle on which the rope winds itself, the mechanical advantage of the whole machine is -^ or . 27JT r 152. To drive a machine an agent must expend energy upon it, and its rate of expenditure of energy is called its 'power.'* But when the agent is inanimate (like running water or compressed steam), its utilised power is often spoken of as the power of the machine driven by it. The * Or sometimes its activity. The word ' power ' is frequently used to express the maximum activity of which an engine is capable ; the actual power at any instant is best called its 'activity,' for an engine of 20 horse-power may be idle soTnetimes. The use of the term ' power ' to denote a force applied to a lever is simple misuse, and can only be tolerated as an old-fashioned usage. 200 ELEMENTARY MECHANICS. [SECT. 152. power of a machine, then, means its rate of doing work ; in other words, it equals the work done in any short time divided by that time so many foot-poundals per second. Pig. 74. A machine is said to have one ' horse-power ' when it can do 17,600 units of work every second; which is equivalent to raising 33,000 Ib. of matter one foot high against gravity every minute. EXAMPLES-XIX. (1) Apply the principle of ' virtual velocities' to determine the condition of equilibrium of a body resting on a rough inclined plane. The principle is that, if the body receives a slight displacement, the total work done must be zero. The limiting condition required is given in Ex. XVIII. (3), Chap. VIII. (2) Show that a body on a plane tilted to the ' angle of repose ' (see sect. 115) is on the point of sliding. (3) If a hundredweight be hung on to the hook W in fig. 67, CHAP. IX.] SIMPLE MACHINES. 201 what force P is required to support it, the pulleys being- weightless ? (4) If each pulley weighed 4 lb., what force would be necessary? (5) In fig. 68 show that if the pulleys are weightless the mechanical advantage is 7 ; but that if they each weigh ^Vtli as much as the weight, then the mechanical advantage is 8J. (6) If, in fig. 69, W=20 lb. and P = 6, find the velocity of W when it has risen one foot, neglecting friction. (7) Find also the accelerations of W and of P, and the time required for P to descend 16 feet. (8) Find the correct position of the weight W in fig. 68, so that the rod on which it hangs may be horizontal. (The figure is not quite correct : the load has to be nearer the string with the greatest tension. ) (9) If four men, each pushing with a force equal to 1 cwt., act at the ends of capstan bars 5 feet long, and wind a rope about an axle 8 inches in diameter, what weight anchor can they raise ? (10) Find the tension in a light flexible rope which is passed round a single movable pulley supporting altogether 20 lb. , the free end of the rope passing over a fixed pulley arid bearing a load of 12 lb. ; and find the acceleration upwards of the 20-lb. weight. (11) What is the greatest weight a man of 12 stone can raise by means of a Weston pulley block (fig. 73) if the wheels have 12 and 13 teeth respectively? If the friction is equivalent to ^th of the load, what weight hanging on the chain will raise a hundredweight load suspended on the movable pulley of the above system, and what weight hanging on another part of the chain will lower it ? (12) In the first system of pulleys, with 3 equal movable pulleys, a small weight of 7 lb. is found able to balance a large one of 49 lb. Find the weight of each pulley and the tension in each string. (13) In the first system of pulleys, with three equal movable pulleys, an effort of 20 lb. just supports 146 lb. Find the weight of each pulley, the tension of each string, and the pull on the ceiling. (14) In the first system of pulleys, with four equal movable pulleys, a force equal to the weight of 20 lb. suffices to 202 ELEMENTARY MECHANICS. [fiXS. XIX. support 245 Ib. Calculate the weight of each pulley and the tension in each string. (15) If a weight of 5 Ib. drags up a weight of 8 Ib. by means of a single movable pulley, calculate the tension in the cord. (16) A weight is to be lifted by means of a system of pulleys of the second order : a block of two pulleys is fixed to the weight, and a rope is carried from an upper fixed block of two pulleys round one of the lower pulleys, up round one of the fixed pulleys, then through the second lower pulley and the second upper pulley, and, finally, to the horizontal axle of a windlass fixed to the ground. The diameter of the axle of the windlass is 3 inches, and the length of the handle 18 inches. Find the ratio of the weight (including the lower block) to the effort necessary to lift the weight ; and find also the number of turns of the windlass requisite to raise the weight 22 feet. CHAP. X.] PROPERTIES OF MATTER. 203 CHAPTER X. ON PROPERTIES AND STATES OF MATTER. (Rudiments of Elasticity, and Introduction to Fluid Mechanics. ) 153. The particular kind of effect which a given force will produce in a given piece of matter when it does work 011 it, depends not on the nature of the force, for forces can only differ in amount and not in kind, but on the nature of the matter. Matter exists in various states, and has very different properties in each state ; and though the principal effects of work, or forms of energy, may be summed up, as stated in the Introduction (sect. 5), under the heads Motion and Strain, yet the kind of motion and the kind of strain produced in different sorts of matter may be very different; and we must now proceed to consider briefly some of the peculiar properties possessed by matter in its different states ; inertia and apparently gravitative attraction being properties common to all. 154. Hitherto we have only considered matter in a rigid form insusceptible of strain, but it is time now to summarise the most fundamental facts connected with the production of strains in non-rigid matter by the action of forces. Strain means either change of size or change of shape. Change-of-size strain is called Compression or Dilatation, and the active resistance of matter to it is called Elasticity of Volume, or Incompressibility. Change-of-shape strain is called Distortion, and the active resistance to it is called Elasticity of Figure, or Rigidity. The adjective 'rigid' is applied to all bodies which 204 ELEMENTARY MECHANICS. [SECT. 154. strongly resist any kind of strain ; but the term ' rigidity * is used to denote the measure of the resistance to change of shape, while the term * incompressibility ' represents the measure of the resistance -to change of size. 155. Bodies with high rigidity are called Solids. The incompressibility of solids such as india-rubber is much greater than their rigidity, and the same is true in a moderate degree of most solids j but with a substance like cork the reverse is true. By the term 'rigid body' in previous chapters, we have always meant a perfectly rigid solid. Such a solid it would be impossible to strain by any finite forces ; all its particles would maintain their relative positions unchanged, unless the body were broken for this would be possible ; perfectly rigid does not mean perfectly strong. Such a solid does not exist, though it is approximated to by rocks and metals. All actual solids are capable of being strained that is, they all yield somewhat to the action of external forces applied to them ; and they are divided into two extreme classes, according to the way in which they yield. They may yield actively ; the stress exerted by their particles in opposition to the distorting force continuing constant, no matter how long that force is applied, and restoring the body to its old shape the instant the distort- ing force is removed, without the least permanent strain or set; in which case they are called perfectly elastic. Glass and steel are practically so. Or they may yield passively ; passing into any shape without exerting any continuous stress in opposition to the distorting forces, and therefore not recovering their form at all when these forces are removed. In this case they are called perfectly plastic or inelastic ; putty, wet clay, and dough are practically so. Most solids (strictly speaking, all existing ones) lie between CHAP. X.] ELASTICITY. 205 these two extremes; they have a certain amount of elasticity combined with a certain amount of plasticity, partly yielding permanently and partly springing back ; as you see at once if you bend iron, wood, paper, &c. 156. A great number of things are elastic when the distorting forces are small, but experience a 'set' when they are too great. These are said to be elastic between certain limits, called the limits of elasticity. If strained above those limits, they are more or less plastic, and if still more strained, they are torn asunder or broken. The greatest longitudinal stress (sect. 160) which a material can bear is called its tenacity. 157. When a solid is strained, both its elasticity of volume and its elasticity of figure are generally called out, for both size and shape usually change. For instance, if you stretch a piece of india-rubber, it alters greatly in shape, but it also expands a little. The strains practically produced in solids may be con- veniently considered under the heads of (1) longitudinal elonga- tion or compression ; and (2) shear. The first is produced when a rod is either stretched or squeezed lengthways by a simple traction, and the elasticity involved is called longitudinal rigidity, or frequently Young's modulus of elasticity. Shear is produced by couples, as when you twist a rod or cut anything with a pair of scissors. It involves the sliding over one another of parallel planes in the body thus a book is sheared when its top cover is either pressed sideways or turned round, while its lower cover is held still. The sliding of the parallel planes (or leaves of the book) is then well seen, especially if you use a thick book like a London Directory. There is in a pure shear no change of size, only of shape. The elasticity involved in a shear is called torsional rigidity, or simply rigidity. When a beam is bent, say by a weight resting on its middle, its lower or convex surface is elongated, and its upper concave surface is compressed, hence longitudinal rigidity only is called out ; unless indeed its horizontal planes slide over one another to some extent, in which case simple rigidity will also be brought into play. If you bend a book, you will see that the leaves slide. 206 ELEMENTARY MECHANICS. [SECT. 158. 158. All resistances to strain are included under the general name Elasticity (the term elastic having a slightly different meaning from elasticity, just as rigid has from rigidity). A body which exerts a great stress when subject to a given strain, is said to have a high elasticity, but if a small stress, a low elasticity ; in fact, elasticity is denned as the ratio of the stress called out to the strain which calls it out ; or shortly, stress elasticity = -. : strain ' or what is the same thing, the elasticity of a body is meas- ured by the stress called out in it by unit strain. ' Stress ' is here short for pressure or tension per unit area. The fact that the above ratio is constant is called Hooke's Law. 159. The kind of elasticity depends on the nature of the strain ; if it is simple dilatation or compression, the ratio of the stress to the strain is elasticity/ of volume ; if it is a linear elongation or contraction, the ratio is Young's modu- lus; if it is a twist or shear, the ratio is simple rigidity ; and the most general kind of strain that can possibly be given to a body can be compounded of these three elements, or can be resolved into them. Moreover, a shear may be analysed into two longitudinal strains, a stretch and a squeeze, at right angles to one another ; similarly a shearing stress may be resolved into a pressure and an equal tension perpendicular to it. 160. Strain is always measured as a ratio; the ratio of a change to an original. The first sort of strain, simple change of size, is best illustrated by gases. See Chapter XIII. , Part ii. This strain is measured as the ratio change of volume . . . = ---. The second kind, or longitudinal strain, original volume is measured by the ratio of the change of length of a rod to the original length. The third kind, or shearing strain, is CHAP. X.] VISCOSITY. 207 measured by the angular contortion of the body in a direc- tion lying at 45 to the shearing stress. Stress is measured by the pressure or tension per unit area for instance, the force applied to either end of a rod divided by the area of the cross section of the rod. 161. Notice that elasticity is measured not by the ratio of distorting force to strain, but by the ratio of internal stress to strain ; for a body may be quite inelastic, and yet require a considerable force to distort it. You would find it hard work to flatten out or to punch a hole through a mass of wet clay, for instance ; but no active internal stress would be exerted capable of restoring the body to its old shape when the distorting force was removed the resistance would have been produced by friction between the different parts which have slid over one another, and friction, we know, is a passive force which can destroy motion, but can not generate it. Such bodies as these then, though plastic, are viscous that is, there is friction between their particles, so that energy is converted into heat when they are distorted. Elastic bodies also may be viscous that is, there may be some friction between their particles whenever shear or sliding of parts occurs. Even steel is very slightly viscous, and when bent becomes infinitesimally warmer, otherwise a tuning-fork in vacuo could go on vibrating for ever. 162. Matter, however, is known to exist in a perfectly plastic state, which is not viscous at all, but limpid; and in this state it is termed fluid. A perfect fluid is a body with zero rigidity and zero viscosity in other words, it has in- finite plasticity and infinite limpidity. No force whatever is required to alter its shape, but it takes the shape of whatever vessel contains it. Many actual fluids come very near to this, but they all have more or less trace of viscosity. Ether has a little less than water, while oil has more, treacle has more than oil, Canada balsam still more, and pitch or 208 ELEMENTARY MECHANICS. [SECT. 162. sealing-wax a great deal so much that it is practically a solid except for very long -continued forces. The only elasticity possessed by fluids is elasticity of volume; in other words, no permanent stress is called out in them by any strain except simple expansion or contraction. 163. All fluids are perfectly elastic as regards volume that is, they all regain their size perfectly when the com- pressing stress is removed. Nevertheless the values of theii elasticities vary very much, for some are nearly incom- pressible, while others are readily compressed; and they are divided into two great groups on this ground. The group of fluids which have a very high volume- elasticity, or are nearly incompressible, are termed liquids type, water. A perfect liquid might be defined as an incompressible perfect fluid. The other or compressible group have an elasticity not depending on themselves at all, but simply on the pressure to which at the time they are subject the elasticity being proportional to the pressure ; these are termed gases type, -air. From this it follows that the volume occupied by a gas also depends, not upon itself, but upon the pressure to which it is subject. Gases in fact take not only the shape, as all fluids do, but also the size of their containing vessel, no matter how large this may be. We may sum up shortly thus : Solids have both size and shape. Liquids have size, but not shape. Gases have neither size nor shape. Matter exists in all kinds of states, some approximating closely to one of these three types, others lying between them and passing almost insensibly from one type to another. 164. The only forms of matter which can be treated in a simple manner, besides perfectly rigid and perfectly elastic CHAP. X.] FLUIDITY. 209 solids, are perfect liquids and perfect gases; and also ordinary liquids and gases when at rest. It remains now to see what special mechanics is necessary for matter in these two fluid states. The special mechanics for liquids is called Hydro- dynamics ; the branch of it treating of liquids at rest being Hydrostatics. The branch of hydrodynamics relating to liquids in motion, or Hydrokinetics, is not an easy subject, though it is a profoundly instructive one. The practical application of liquids in machinery is called Hydraulics. The special mechanics for gases is called Pneumatics, or sometimes Aerodynamics. 165. The essential difference between the mechanics of solids and the mechanics of fluids is based upon the different ways in which they transmit pressure. Thus, take a rigid stick standing on the ground, and press downwards upon the upper end of it; the pressure is transmitted unchanged to the other end, which therefore presses the ground with an equal force ; but not the slightest pressure is exerted sideways, say against a tube surrounding and fitting the stick. But place some liquid in a closed tube, and press one end of the liquid with a piston ; then, though the pressure is still transmitted to the other end, it is also transmitted sideways to every part of the tube just as much ; and, moreover, the force experienced by the closed end of the tube is not now necessarily equal to the force applied to the piston, unless the area of the closed end equals the area of the piston ; if the area is greater, the force transmitted is greater, and if less, less. Every portion of the surface of the tube which exposes to the liquid a surface equal to the area of the piston, experiences a pres- sure equal to that exerted by the piston ; a fact which is briefly expressed thus : Fluids transmit pressure equally in all directions. N 210 ELEMENTARY MECHANICS. [SECT. 165. This is entirely because of their plasticity, or the perfect mobility of their particles. The structure of a liquid might be imitated roughly by a number of exceedingly small well- oiled shot. A bag full of such shot, if compressed in any way, would experience the pressure in every part of it. EXAMPLES XX. (1) A wire of sectional area 1 square millimetre and 1 metre long is stretched 3 millimetres by a load of 6 kilogrammes. What is the Young's modulus of its material ? (2) What stress would lengthen the above wire 1 . millimetre ? (3) What stress would lengthen a wire 1 per cent., if its Young's modulus is E ? (4) What stress would shorten a bar three parts in twelve thousand, if its Young's modulus is E ? (5) What load on a vertical iron rod 1 inch square would shorten it by one-thousandth of its length if the Young's modulus of iron is 1 million atmospheres ? (6) If a linear thrust of 60 tons per square foot diminishes the length of a bar by a tenth per cent., and its volume by a twentieth per cent., how much must the rod have tem- porarily increased in diameter during the thrust ? (7) With the above data, how much diminution of volume would you expect if the bar were subjected to a uniform hydraulic pressure of 60 tons to the square foot all over its surface ? (8) What then is the cubic compressibility of the bar? (9) What is the incompressibility or volume elasticity of the material of the above bar ; also what is its Young's modulus ? (10) The volume elasticity of sea-water is about 20,000 atmo- spheres. How much compressed is it at a depth of 150 times 34 feet, or say a mile ? (11) How much would an ocean, two miles deep, rise in level if its water became incompressible and resumed its surface density ? (12) If atmospheric air is squeezed one per cent, by the hundredth of an atmosphere applied for some time, what is its slow elasticity ? (13) If the same air is squeezed quickly by the same pressure, it only shrinks at the first instant five-sevenths of one per cent. : what is its quick elasticity ? CHAP. XI.] PRESSURE OF LIQUIDS. 211 CHAPTER XL ON THE PRESSURE OF GRAVITATING LIQUIDS AT REST. (Hydrostatics. ) 166. We conceive a perfect liquid as an incompressible fluid, that is, a body all whose particles are capable of free motion among themselves without the slightest friction, whose shape therefore is wholly indefinite, but whose volume it is impossible to change. Water is an imperfect liquid, partly because it is slightly compressible, but prin- cipally because it is slightly viscous that is, because its particles experience when they slide over one another a certain amount of resistance analogous to friction, called viscosity. Hence it is that a basin full of water which has been stirred round and round and left to itself, will after a time come to rest. The energy of motion will be wasted by ' friction ' against the wet sides of the vessel that is, it will be expended in warming the water. But because the friction is very small, a particle of water can travel against it a long way before its energy is ex- pended that is, before the work done, Fs, is equal to the energy to be got rid of, \ mv z . 167. The friction due to viscosity differs from ordinary friction in that it depends very greatly on the speed of the relative motions ; it seems, in fact, t6 be about proportional to the square of the velocity, and as the velocity vanishes, so does the viscosity-friction. The properties of water, or any other actual liquid in motion, are therefore very dif- 212 ELEMENTARY MECHANICS. [SECT. 167. ferent from those of the ideal perfect liquid ; but when water is at rest, there is no friction among its particles, * their reactions are all normal, and its behaviour is then identical with that of the perfect liquid. Hence it is that the mechanics of liquids at rest (even such liquids as treacle) is so simple ; the simple laws of the perfect liquid are applicable to them, for their viscosity does not come into action. Pressure of Fluids in General at Rest. 168. The general law of pressure common to all fluids, and following at once from the mobility of their particles, is that they act like perfectly smooth bodies (cf. sect. 115) ; or, The pressure of a fluid at rest is always perpendicular to every surface on which it acts. For if the reaction of the surface had any component along it, it would be able to move the fluid, which would therefore be not at rest. A second general law may also be stated thus : If a pressure is applied to any area of the surface of a fluid in a full closed chamber, that same pressure is transmitted to every portion of the walls of the chamber of equal area (sect. 165). Thus imagine a closed cistern quite full of water, with tubes or cylinders let into the sides anywhere, and plungers or pistons, A, B, C, D, fitting these tubes quite freely, but yet water-tight (fig. 75) ; and let A have an area of 1 square inch ; B, 2 square inches ; C, 3 ; and D, 4 square inches. Now push A in with a force say of 20 pounds-weight ; every square inch of the interior surface of the cistern will experience this pressure, and therefore B will experience a force of 40, C of 60, and D of 80 pounds-weight. Of CHAP. XI.] LIQUID PRESSURE. 213 the three larger pistons, let D be the only one free to move, and let a constant external load of 80 pounds be applied to it ; then if A is pressed in with a load the least exceeding 20, D will move out and raise the 80 pounds. But it would only move th as fast as A. This is evident ; for suppose A were pushed in 1 foot, it would throw 12 cubic inches of water into the cistern, and this water would go into the cylinder of the piston D, if that were the only part of the walls free to move ; but as this cylinder is 4 square inches in area, the 12 cubic inches of water would only cause D to move out 3 inches, the quarter of a foot. In other words, the work (Fs) done by the piston A, 20 x 12, is equal to the work done upon the piston D, 80 x 3. So that we have here simply a machine subject to the universal law of machines, that ' what is gained in force is lost in speed ; ' and there is no gain of energy in a hydraulic machine any more than in any other. The machine just described, put into a working form, is known as the hydraulic or Bramah press (fig. 76). It consists funda- mentally of two cylinders of different sizes, with pistons or plungers fitting them, and a pipe connecting them. Water fills Fig. 76. Hydraulic Press. both cylinders, and the mechanical advantage of the machine is the ratio of the areas of the two pistons A : , so that a 50-lb. pressure on the small piston balances 50 Ib. on the large one. 214 ELEMENTARY MECHANICS. [SECT. 168. The liquid acts only as an incompressible plastic medium for transmitting pressure. For a fuller account of the machine, see Ganot or Deschanel. 169. So far we have supposed the pressure to be produced only by pistons which endeavour to compress the liquid, but it is important to consider also pressures due to the weiglit of the liquid. Every particle of a liquid is attracted to the centre of the earth, and will tend to get there by percolation unless prevented by being enclosed in some vessel with impervious sides; in other words, water must be kept in non-porous vessels. The vessel, however, need not have a lid, for a liquid occupies an unchangeable volume, and therefore may have its upper surface free ; it keeps at the bottom of the vessel as the nearest accessible position to the centre of the earth. But it will press on the bottom and sides of the vessel with a certain force which will always be normal to those surfaces, and whose magnitude we have now to consider. Pressure of Liquids due to their Weight. The first simple law is that the upper or free surface of a liquid at rest is horizontal ; that is, is normal to the vertical force of gravity on each particle. Such a surface is said to be level, and it is practically flat or plane, because the forces on the several particles are practically parallel. Inasmuch, however, as these forces are not really parallel, but intersect at the centre of the earth, the level surface of a liquid at rest is not really plane, but is curved round the centre of the earth ; in other words, it forms part of a sphere with the radius of the earth as its radius. The curvature is too small to be appreciable in a bucketful of water, but it is apparent enough in the ocean. Another law, that the pressure of a liquid' varies directly with the depth, is what we must now establish. 170. Consider a cylindrical bucket with a flat bottom, CHAP. XI.] LIQUID PRESSURE. 215 filled with water ; the base of the vessel has to support the whole of the water, as if it were a rigid mass slipped into the bucket with its sides well oiled. For although certainly the sides are pressed, and therefore exert reactionary pressure on the water, yet they, being upright, press it horizontally only, and so can have nothing to do with sustaining its weight. The pressures of the sides simply maintain the shape of the water in opposition to the "force of gravity, which tends to flatten it out. The pressure on one side is equal and opposite to the pressure on the other, and therefore there is equilibrium, unless part of one side be removed by boring a hole through it. In that case the water will flow out, and the uncompensated pressure on the side opposite the hole will force the vessel bodily along in a direction opposed to the stream of water. This is the principle of Barker's mill, turbines, Catherine wheels, rockets, &c. See Deschanel, page 92 ; or Ganot, sect. 193. In an upright cylindrical vessel, then that is, any vessel with vertical sides the pressure on the base is equal to the whole weight of water contained in the vessel. But the cubic contents of a cylinder are obtained by multiplying its height by the area of its base always, whether that base be round or square, or any other shape; and the weight of water a vessel can contain is, of course, its contents in cubic feet multiplied by the weight of each cubic foot. Hence, the pressure on the base of an upright-sided vessel A square feet in area, filled to a height of h feet with a liquid of which a cubic foot weighs s lb., is in Ib. weight, P = sA/i. Thus, suppose an oblong-based plane-sided cylinder (also called a, prism) with base 10 inches by 5 inches, and height 15 inches ; the contents would be 10x5x15 = 750 cubic inches, and the pressure on its base when full of water would be the weight of 750 cubic inches of water ; which happens to be about 27 lb. weight. 171. If we are speaking about water, this s is often written w t meaning the weight of a cubic unit of water, just as it might be written m if we were speaking of mercury. Whether w stand 216 ELEMENTARY MECHANICS. [SECT. 171. for the weight of a cubic inch, a cubic foot, or a cubic centi- metre, is wholly immaterial, being only a matter of custom or convenience ; only we must keep to one unit all through. Hence, we use the word, a cubic unit, as expressing the cube of whatever arbitrary length happens to be taken as the unit of length in other parts of the book, or question, or problem under con- sideration. A cubic foot is found to contain 62 -33 Ib. avoirdupois of water, which is not far off 1000 ounces. A cubic inch contains the i T V^th part of this namely, 252^ grains. A cubic centimetre contains one gramme of water ; and this is one reason why the French system of weights founded on the gramme makes calculations simpler : the unit of mass, or unit quantity of matter, is defined as that of unit volume of water. Mercury is 13*6 times as heavy as water. Hence 1 cubic inch contains about VWY ounces of mercury ; and a cubic centimetre 13*6 grammes. 172. Suppose now that, instead of a cylindrical vessel, we consider a conical one, set up like a tumbler, with the wider end uppermost : then the pressure on the sides, being still perpendicular to them, is no longer hori- zontal, but has more or less of a vertical component as well as a horizontal one; Fig. 77. hence, we can no longer say that the pres- sure on the base is the whole weight of water in the vessel, for the sides may and do support some. How much the sides support, and how much the base, may be readily seen by imagining an infinitely thin circular drum of the same diameter as the base of the vessel to be let into the water, as shown by the dotted lines (fig. 77). Or you may suppose a thin circular drum of the liquid to freeze or become rigid, as indicated by the dotted lines. The pressure across the walls of this imaginary drum is hori- zontal ; and inside the drum we have what is equivalent to a rigid cylinder, with w r ell-oiled sides, resting on and entirely supported by the base (just as we had in the cylindrical vessel) ; while CHAP. XI.] LIQUID PRESSURE. 217 outside we have a ring-shaped mass of water which is not sup- ported hy the base at all, and therefore must be supported by the sides. It is, in fact, supported by the vertical component of the pressure of the sides, and therefore it has nothing to do with the pressure on the base, which is ivAh as before. So also, if we turned the conical vessel the other way up, with the wide end as base ; the pressure on the base would then be greater than the whole weight of water in the vessel, because of the vertical component of the pressure of the sides, which now acts downwards. And, as the pressure of the water on the sides would, if the sides were removed, be able to sustain the ring- shaped mass of water completing a drum set up on the base, it follows that the whole pressure on the base is still the weight of a volume of liquid filling a cylinder whose base is the actual base, and whose height is the height to which the vessel is filled j or again, wAh as before. Notice particularly that none of this reasoning is impaired or affected if the sides of the vessel, instead of being plane, are curved or zigzag, or indeed any shape whatever, as in figs. 78 and 79. The pressure on the base is always simply sAh, or the weight of a cylinder of the given liquid with Fig. 79. the given base as base, and the given height as height ; for the base supports this cylinder, the sides support the rest. 173. The vessel shown in fig. 79 is supposed to be flexible like an india-rubber tube, and its base can be turned into different positions as in fig. 80; but, since liquids transmit pressure equally in all directions, the pres- sure on it will not vary except in so far as the bending of ^** --- Fi go 218 ELEMENTARY MECHANICS. [SECT. 173. the tube alters the height of the liquid in it. The only difficulty is the knowing what point to measure the depth to. The pressure on the lower part of the base is greater than that on the upper portion; but since the pressure is simply propor- tional to the depth, the average or mean pressure will be simply the pressure at the average or mean depth (compare average velocity in sect. 23) that is, the pressure at the middle point of the base. Hence, the pressure on any surface of area A, immersed under a liquid to the mean depth Ji, is always sAh. The surface plainly need not form the base of a vessel, but may be immersed anyhow. Thus, let a rectangular plate 5 inches long by 4 inches broad be immersed slantingly under water, so that its upper edge is 8, and its lower edge 10 inches below the surface. Then evidently its mean depth, or depth of its middle point, is 9 inches; and the pressure on its surface, being equal to wAh, wx 5x4x9 = 180w;=lSOXj72g ounces weight If the liquid had been mercury, this pressure would have been 13 - 6 times as great. To find the mean depth of a bent or curved plate of irregular shape requires calculation, and the calculation required is just the same as that which would be used to find the centre of gravity of the plate (indeed, the centre of gravity is the most middle point in a body) ; hence the mean deptli of a surface is often spoken of as the depth of its centre of gravity. So we get the perfectly general result for liquids subject only to gravity : The total pressure on any plane surface whatever, due to the weight of a liquid under whicli it is immersed, is its area, multiplied by the vertical depth of its centre of gravity CHAP. XI.] LIQUID PRESSURE. 219 below the free surface of the liquid, multiplied by the weight of a cubic unit of the liquid ; or in symbols, P = sAh ,. There is nothing more to explain. This simple formula contains it all. 174. Since the pressure of a liquid does not depend upon the quantity of the liquid, but only upon its depth, we may make a small quantity of liquid exert any pressure we please by putting it in a long narrow vertical tube, and giving it a large area to press upon. This is the principle of the ' Hydrostatic Bellows ; ' which con- sists of a pair of circular boards joined water-tight by corrugated leather like ordinary blow-bellows, with a long tube opening into the cavity between the boards, the tube rising a good height, and finishing off with a funnel. See fig. 81. A man may stand on the upper board of the bellows, and raise his own weight slowly by simply pouring water down the tube. For if A be the area of the upper board of the bellows, and h its vertical depth below the surface of the water in the tube, all that is necessary to balance the man is that w Ah shall be equal to or greater than his weight, say 200 Ib. or 3200 ounces. Suppose A is a square foot, then to find the necessary height h to which the tube must be filled, we have 1000xlx7i = 3200; or h=3'2 feet, a very moderate height indeed. The man is, in fact, equal to a cylinder of water standing on A as base, and of height 3 '2 feet ; for this quantity of water would be balanced by the column of water in the tube (see sect. 176 and fig. 82), and the board and man take its place. The man rises so soon as this imaginary cylinder of water is equal to himself in weight ; and it will be equal to him in weight just about the same time as it is equal to him in bulk, for a man is just about able to float in water (see Chapter XII. ). Hence the average cross section of a man is equal to the area of the board of a hydrostatic bellows, on which he would just be supported by a column of liquid Fig. 81. 220 ELEMENTARY MECHANICS. [SECT. 174. equal to himself in height ; for instance, if his height were 6 feet, and his weight 15 stone (210 lb.), his average cross section would be '56 square foot, or S0'64 square inches, because 1000 x -56x6 = 210x16. 175. The total pressure on a surface under a liquid depends partly on itself namely, on its area; but the pressure per square inch, of surface depends not at all on itself, but on external conditions namely, how deep it is immersed, and what it is immersed in : hence it is con- venient to distinguish these, and to call the pressure per unit of surface the intensity of the pressure, and to denote p it by p, so that p = -T-; or of course, p = sh. J\. One often speaks simply of ' the pressure of a liquid ' at such and such a depth, without specifying the surface on which the pressure is exerted ; for instance, the pressure of the ocean at a depth of one hundred fathoms, and so on. In such cases the intensity of pressure is always meant, or the pressure which would be experienced by a surface of unit area if placed at that depth that is, simply sh. The pressure of an incompressible fluid (or liquid) there- fore varies directly with the depth (for s is constant) ; being nothing at the surface, and increasing uniformly as you descend. In so far as actual liquids are slightly compressible, this simple proportion between depth and pressure does not hold Fig. 82. down to great depths ; the liquid there becomes denser, or heavier bulk for bulk, and accordingly ocean pressure in- creases rather faster than the depth. CHAP. XI.] LIQUID PRESSURE. 221 176. When any number of communicating vessels are filled with the same liquid, the level of the liquid in all is the same. See fig. 82. For the intensity of the pressure at any point due to every column of liquid must be the same, or there could not be equili- brium ; and this pressure is proportional to the depth. Further, when communicating vessels contain different liquids which do not mix, the heights of the columns of liquid are inversely as their specific weights. For take any two communicating vessels, say the two legs of a U tube, one full of mercury say, the other of water. Call the area of the surface of contact of the two liquids (fig. 83) A, and let the vertical height of the surface of the water B above A be called h, while the vertical height of C, the surface of the mercury above A, is called h' ; then the pressure on each side of the area A must be the same, as soon as there is equilibrium and the columns have ceased to oscillate ; but the pressure on its upper side is wAh, and on its Fig. 83. lower, mAh', hence ivh = mh', or h : h' :: m: tv. This method of balancing columns is employed to find how much denser a liquid is when cold than when hot, the two columns being purposely kept at different temperatures and their heights measured ; and thus the expansibility of a liquid can be measured without having any regard to the size, shape, or ex- pansibility of the vessel containing it. Centre of Pressure. 177. The whole pressure or resultant force on any flat surface under a liquid may be considered as composed of a number of parallel forces namely, the pressures on each individual small -area of the surface and all these parallel forces will have a resultant equal to their sum, passing through a certain point of the surface which is called the centre of the parallel forces (cf. sect. 129), or the 'centre of 222 ELEMENTARY MECHANICS. [SECT. 177. pressure/ For a rectangular area like a dock-gate this point is -f of the way down from the surface of the water, because the pressure increases uniformly with the depth. The position of the centre of pressure of an immersed surface corresponds closely with that of the centre of oscillation of a swinging body (sect. 81). If A is the area of a plane surface immersed vertically in a liquid, and is its centre of oscillation when swinging about the line where the plane, or the plane produced, cuts the surface of the liquid, is the centre of pressure due to the liquid. In the case of a rectangular plate immersed, with just one edge in the surface, is two-thirds down the plate. EXAMPLES XXI. (1) The small plunger or pump-piston of a Bramah press is half an inch, and the large one is 8 inches, in diameter ; the pump is worked by a handle 5 feet long, the fulcrum being one inch from the point of attachment of the plunger. What is the greatest weight that a man of 15 stone' can lift by this machine if he sits on the end of the handle ? (2) Find the pressure on the bottom and sides of a cubical vessel 10 centimetres in the side full of mercury. (3) Find the pressure on one side of the above cubical vessel if half full of water and half full of mercury. (4) What is the pressure of water at a depth of 1020 feet ? (5) A couple of hemispheres 1 metre in radius are joined water- tight and placed in water with their join vertical and just submerged. What is the resultant force holding the two halves together ? (6) Find the total pressure on the bottom of a tank 10 feet square, 5 feet deep, and full of water. Find the pressure on a side of the same tank. (7) A block, in the form of an isosceles wedge, 1 foot high and having a base 1 foot square, is immersed in water with its base horizontal and uppermost at 4 feet below the surface. Determine the pressure, and the vertical component of the pressure on each face. Hence show what must be the weight of the wedge so that it may just float. (8) A dock-gate 30 feet square has 21 feet and 12 feet depths of CHAP. XI.] LIQUID PRESSURE. 223 water on its two sides respectively. Find the resultant of the pressures exerted upon it by the water. (9) A triangular plate is immersed vertically in water with the vertex in the surface and the base horizontal. The height and base of the triangle are each 8 inches. Find the pressure on a face of the plate. Find also the depth of the centre of pressure. (10) The above plate is immersed vertically with its base in the surface. Find the pressure and the centre of pressure on its face. (11) A cubical tank of 1 foot side is one-third full of mercury and two-thirds full of water. Find the pressure on one of its sides. (12) What is the pressure against one side of a cubical cistern, when full of water, which will hold 200 gallons ? (13) What is the pressure, in tons-weight per square inch, at the bottom of the sea where the water is two miles deep ? [A cubic foot of sea-water a mile deep weighs about 65 lb.] (14) In a hydraulic press the pump-plunger is a cylinder 1 centi- metre in diameter, and makes a stroke 7 centimetres long. The plunger of the press is 20 centimetres in diameter. Taking the collar friction as a quarter the load in each case, calculate (a) the pressure in the press when a load of 1 cwt. is applied to the pump-plunger; (b) the available force of the press-plunger ; (c) the number of strokes which the pump must make in order to raise the press-plunger 10 centimetres. (15) Find the work done per minute (a) by the operator, (b) by the machine, if the above press is worked at the rate of a stroke a second : the load on the press-plunger being as already calculated. 224 ELEMENTARY MECHANICS. [SECT. 178. CHAPTER XII. FLOATING BODIES (Hydrostatics continued). 178. We shall now proceed to consider what happens when a solid is wholly or partially immersed in a liquid. Most of what we shall state will be true of fluids in general, but receives its most obvious illustration in the case of liquids. When you dip your hand in the water, you displace some of the water ; in other words, a portion of space below the surface which was formerly occupied by water is now occupied by your hand. The volume or bulk of the water displaced is, of course, equal to the volume or bulk of your hand. All solids, then, when immersed either wholly or partially in a liquid, displace a volume of that liquid equal to the bulk of that part of them which is immersed. This is perfectly obvious. 179. Further, when your hand is immersed you can feel, if you attend, a certain pressure urging it up out of the water. This upward pressure is more apparent if you immerse your whole body; indeed the upward pressure is then so great as nearly ( to counteract the weight of your body altogether, consequently, in a bath you weigh apparently next to nothing. This upward pressure is what we must now discuss. Take an ordinary chemical test-tube of very thin glass, and plunge it in water with the closed end downward. You will feel a very distinct up war* pressure, and the tube will CHAP. XII.] BUOYANCY. 225 be forced up if you let go. Keep, however, the tube immersed, and slowly fill it with water. You will find that it is forced up gradually less and less, until, when the level of the liquid inside and out is the same, the tube will cease to press up, and will weigh pretty nearly the same as it did before it was immersed at all. The displaced water has been restored. If you perform this experiment accurately with a balance, you will find the tube does not quite recover its original weight, even when the level is the same inside and out. It will weigh 60 per cent, of its original weight. This is evidently because some little water is still displaced by the walls of the tube, which, however, are very thin, and in what follows will be assumed to be infinitely thin. Now imagine the glass tube annihilated; the water it contained will remain occupying the place the tube had occupied, and experiencing the same pressures as the tube did; because the same quantity of water is displaced as. before, only now not by the glass tube but by the liquid water which had been poured into i. Obviously, however, this water will be in equilibrium, as all water in water at rest is ; hence the two forces under whose influence it is namely, its weight downwards, and the pressure of the surrounding water upwards are equal and opposite. But the pressure upwards is the same as the tube experienced, before its annihilation ; therefore the pressure on the tube was equal to the weight of its own volume of water that is, the weight of the water it displaced and acted in the same straight line, namely, through the centre of gravity of the water displaced. This result is perfectly general, and is known as the principle of Archimedes, because it was the method that he invented when asked to ascertain the chemical composition of an irregular ornamental mass made of two unequally dense metals, without chemrcal analysis or damage. He 226 ELEMENTARY MECHANICS. [SECT. 179. perceived that if its weight and its volume were known its average density could be ascertained, and thus the proportion of its known component metals calculated. He determined its volume by immersing it in water and weighing it there. When any solid is immersed either wholly or partially in a fluid, it is pressed up with a force equal to the weight of the fluid displaced ; and this force may be considered to act at the centre of gravity of the fluid displaced. The fluid displaced is equal in volume to the solid, hence the upward force is the weight of an equal bulk of the fluid. To show this by means of our symbols, consider a special case, say a cubical block of stone, a inches in the side, immersed- in water, so that its upper surface is at a depth h below the surface of the water, and therefore, of course, its lower surface at a depth h+p (fig. 84). The area of any of its faces is 2 . The pressure on its upper face (sect. 171) is wa?h, and on its under* face is wc that is, since 12 cubic feet of ice weigh the same as 11 cubic feet of water, it follows that an iceberg must have f of its whole bulk im- 228 ELEMENTARY MECHANICS. [SECT. 181. mersed ; hence the visible berg is only ^ of the whole mass, there being eleven times as much underneath the water. So also a floating cork whose specific gravity is I has of its volume pro- jecting above the water. Determination of Densities. 182. The foregoing principles are all remarkably well illustrated by their practical application to the determina- tion of density or specific gravity. The density of a body is defined as its mass per unit volume, or the mass of any volume divided by that volume, similarly we might define specific gravity as the iveight per unit volume, or the weight of any volume divided by that volume, which would make specific gravity be to density as weight is to mass ; or, as weight is g times mass (sect. 64), the specific gravity of a substance would be g times its density. This, however, is not the definition of the term specific gravity as ordinarily used ; it is the definition of what is called absolute specific gravity, which for distinction has been here called 'specific weight,' whereas the ordinary or relative specific gravity is the weight of any volume of a substance compared with the weight of an equal volume of some standard substance. The relative specific gravity of mercury with reference to water, for instance, is 13 -6 ; of wood is, say '6, and so on. When one speaks of the relative specific gravity of any body, without stating the standard substance to which reference is made, it is understood that that standard sub- stance is water ; and so we may define the relative specific gravity, or the specific gravity of a substance, as the weight of any volume of it divided by the weight of the same volume of water. Its relative density is precisely the same thing, both being simple numbers of equal value, but one having a direct reference to weight, the other to mass. In all comparative methods of CHAP. XII.] SPECIFIC GRAVITY. 229 measurement we need make no distinction between density and specific gravity. We have, in the preceding chapter, used s as standing for the absolute specific gravity or 'specific weight' of substances in general, m for that of mercury, and w for that of water ; so the relative specific gravities of the three things s m w are, of course,^' ~ and^-; the relative specific gravity of water itself being, of course, unity. In the C.G.S. system of weights and measures, the absolute and relative specific gravity of a thing are represented by the same number, because the unit volume of water is defined to be the unit of mass (cf. sect. 171). The absolute specific gravity of water, or the weight of 1 cubic centimetre, is 1 gramme ; and if a thing is three times as heavy as water, a cubic centimetre of it weighs 3 grammes, and so on. To compare the Specific Gravities of two Liquids. 183. 1st Method. If they dp not mix, place them one in each of the two legs of a U tube, and measure the heights of their re- spective columns (sect. 176); then ~ = ]r. - s 2 % This method is used sometimes to compare with great accuracy the relative densities of a liquid at different temperatures. (Dulong and Petit's method for absolute coefficient of expansion of mercury ; see Ganot, art. 273, or Deschanel, Part II., page 287.) A convenient modification is to dip the open ends of an in- verted II tube into the two liquids, each in its own beaker, and to suck air out of the bend of the tube by means of a T piece, until the liquids rise and stand as measurable columns. 2nd Method. Weigh a bottle full of the first liquid, and then the same bottle full of the second ; deduct from each the weight of the bottle, and you will have the weight of the same volume of the two liquids to compare. In symbols, if b is the weight of the empty bottle, &! w 1 -b s 2 ~w z - b This method is often used. Flasks (' specific-gravity bottles ') are made for the purpose (fig. 85). They are very light, and are arranged so that they can be accurately filled always to the same extent. For this purpose their neck has a constriction with a ring drawn round it with a diamond, and they are always filled up to this ring. This is done by filling them at first too full and then extracting the surplus with a scrap of blotting- paper or a capillary tube. The stopper is then inserted to prevent evaporation, and the whole is weighed in a delicate I balance. The weight of the empty bottle, &, must have been previously ascertained. Fig. 85. 3rd Method. Take any non-porous solid heavier than both liquids and insoluble in either of them, such as glass, weigh it 230 ELEMENTARY MECHANICS. [SECT. 183. first in air (or vacuum), then immerse it wholly in one of the liquids (hanging it from the pan of the balance by a fine wire or hair), and weigh it in that. It now weighs less by the weight of the liquid it displaces note this loss of weight. Now weigh it in the other liquid, and note its loss of weight in that. The same volume of each liquid has been displaced, and the first loss of weight was the weight of this volume of the first liquid ; the second loss, the weight of the same volume of the second liquid ; so the specific gravity of the iirst liquid referred to the second, is the ratio of the first loss to the second loss. Or in symbols, if w is the weight of the solid in air, and w^ and w 2 its weight in the two liquids respectively, Fig. 86. Hydrostatic Balance. This method has been used by Matthiessen to determine the expansibility of water by heat, and it is called the areometric method. (See Balfour Stewart's Heat, page 51.) The operation of weighing a solid under a liquid is conducted by an ordinary balance with one of its pans replaced by a much shorter one with a hook under it, to which the solid can be hung by a fine platinum wire (fig. 86). When so arranged, it is often called a hydrostatic balance. 4th Method. Take an insoluble and non-porous solid lighter than all the liquids you have to compare, and float it in each of them ; ascertaining in each case the volume of it immersed. The weight of this volume of the liquid must in each case be equal to the weight of the solid, which is constant ; so we obtain a set of different volumes all of the same weight. Call these volumes v v v 2 , v 3 , &c., and let w be the weight of the solid ; then, since W=v 1 s l =v 2 s 2 =v 3 s 3 =&c. i the ratios of the specific gravities to one another are inversely as the immersed volumes. Instruments for carrying this out are made of glass or metal, and sold under the name of hydrometers (see sect. 186). 5th Method. Take a solid lighter than all the liquids, and CHAP. XII.] DENSITY. 231 float it in each, loading it so as to immerse the same volume in all ; that is, always make it sink to a fixed mark. The weight of this volume of the liquid is the weight of the solid plus the load, so the specific gravities of the liquids are as the numbers representing this total weight in the different cases. An instrument for carrying this out is called Fahrenheit's hydrometer, but it is seldom now used. Another method is given in Ganot, art. 121. To determine the absolute Density of a Liquid. 184. 1st Method. Weigh a known volume of the liquid in a gauged specific-gravity bottle (tig. 85), and divide the mass by the volume. 2nd Method. Weigh a solid of known volume before and after immersion in the liquid, say a sphere of measured diameter. Its loss of weight will be the weight of its own volume of the liquid, . , , r ., -. ,. ,, ,. .,. loss of weight of solid so the weight of unit volume of the liquid is , S-? p-, To determine the absolute Density of a Solid. Weigh a known volume of it, say a sphere or a cube or some- thing easily gauged, and divide the weight by the volume. To compare the Densities of a Solid and a Liquid. 185. 1st Method. If the solid be heavier than the liquid. Weigh it in air and in the liquid, and divide the weight in air by the loss of weight in the liquid ; the quotient is the relative specific gravity of the solid referred to the liquid ; s= , 2nd Method. Applicable only if the solid be lighter than the liquid. Float it in the liquid, and take the ratio of the volume immersed to the whole volume (sect. 181). If the solid is a cylinder floating upright, volumes are proportional to lengths; length immersed and the specific gravity is then whole length 3rd Method. If the solid be lighter than the liquid. Weigh it first in air ; then immerse it in the liquid by attaching a heavy body to it to sink it, and weigh the two together. Also weigh the sinker by itself in air and in the liquid. The loss of weight 232 ELEMENTARY MECHANICS. [.SECT. 185. of the two together gives the weight of liquid displaced by both ; the loss of weight of the sinker alone gives the weight of liquid it displaces ; therefore the difference of the two losses gives the weight of the liquid displaced by the body itself that is, the weight of an equal volume of the liquid. So the relative specific gravity of the solid is its weight in air divided by the difference of the two losses. If the body is porous like coke or pumice, it is necessary to distinguish between its average density, including pores, and the actual density of its material. In the former case it may be varnished over; in the latter case it is best to pound it in a mortar, and treat it by the method immediately to be described. A liquid must always be chosen in which the solid is not soluble. Thus, for a piece of rock-salt, one must not use water, but either some such liquid as turpentine or benzol, or a saturated solution of salt ; and the specific gravity of the salt referred to this liquid must be multiplied by the specific gravity of the liquid to give the specific gravity of the solid with reference to water. Another, though essentially similar, method is given under Nicholson's hydrometer, sect. 186, which see. 4th Method. Useful when the solid is porous or in the form of a powder. The difficulty with a powder is that it is impossible to gauge the volume of the solid particles directly, and also difficult to suspend the powder in water so as to determine its loss of weight. A specific-gravity bottle with a wider neck than that shown in fig. 85 is used. Ascertain the weight of the bottle when empty, and also the weight of water it will contain when full up to the mark. Put a known weight of the powder into the bottle, and fill up with water ; the powder displaces some water, so it will not now hold so much as before the powder was in ; but the weight of the whole, minus the weight of the powder and bottle, gives the weight of the water now in. The difference between this weight and the weight of water the empty bottle originally contained, gives the weight of water displaced by the solid powder ; so the specific gravity of the solid is weight of powder weight of water required to weight of water required to fill up fill empty bottle bottle after the powder is in. If the powder be soluble in water, of course some other liquid must be used : the result can be multiplied by the specific gravity of this liquid, if the specific gravity of the powder referred to water be required. 5th Method. A more elaborate method, also serviceable when CHAP. XII.] HYDROMETERS. 233 the solid is soluble or porous, is to use it to displace only air, in an arrangement something like Boyle's tube (fig. 103), which is called a volumenometer. Hydrometers. 186. A hydrometer is a light body loaded so as to float in stable equilibrium at the surface of a liquid, and of a shape which renders it easy to observe accurately how much of its volume is immersed ; and its use is to compare the specific gravities of liquids, or of solids and liquids. See methods 4 and 5, sect. 183, and methods 2 and 3, sect. 185. They are of two classes. 1st, Hydrometers of variable immersion or common hydrometers (Twaddell's, Beaume's, Sykes', &c.). 2nd, Hydrometers of constant immersion (Nicholson's and Fahrenheit's). 1st Class. Common hydrometers are glass cylinders or * stems,' loaded and arranged so as to float upright. This is done by making them terminate below in a couple of bulbs, one full of air, the other full of mercury or shot (fig. 87). They must be of such weight as to float in a liquid with part of the cylindrical stem projecting ; hence they are usu- ally sold in sets, say a set of three, one for heavy liquids, one for medium, and one for light. The heavier the liquid the more of the stem projects, but in a light liquid they sink pretty deep always sinking until they have displaced their own weight of the liquid. A thin stem makes the instrument sensitive, a wide stem diminishes its sensitiveness, but in- creases its range. Fig. 87. The specific gravity of the liquid is (see sect. 181), the whole volume of the instrument the weight oF the instrument the volume immersed the weight of an equal volume of water (the last fraction being the average specific gravity of the instru- ment) ; that is, the specific gravity of the liquid varies inversely with the volume immersed. The stem, however, is graduated so that the specific gravity is read off directly from the numbers on it. 2nd Class. Hydrometers of constant immersion will serve not only to compare the specific gravity of liquids, but also to deter- 234 ELEMENTARY MECHANICS. [SECT. 186. mine the specific gravity of any solid, whether heavier or lighter than water, and this is their principal use ; they will, moreover, make a very good substitute for a common balance. They con- sist, like the others, of a floating cylinder or rod, which, however, is usually made very thin (often only a wire), and instead of being graduated, has one fixed mark on it, to which it is always sunk. Its appendages are, a tray, A ; a large light bulb, B ; and a heavy bulb, or tray and cage, C. Fahrenheit's has only a shotted bulb below, and is made of glass. Nicholson's is made of metal, so that it cannot be used in corrosive liquids. It is, in fact, only used floating in water to determine the specific gravity of solids : it is the one which has the tray and cage C, and is shown in fig. 88. It will not sink down to the fixed mark m on the fine cylindrical stem, unless some extra weights are put on the tray A ; let 20 grammes be the weight required to sink the instrument by itself. This constitutes a constant of the instru- ment which must be known or determined before use. Now to use it as a common balance, you place on the upper tray the body you wish to weigh, and then add weights, say 6i grammes, till it has sunk to in ; one then knows that the body weighs 20-6^ = 13^ grammes. To use it as a hydrostatic balance, you place the body in the lower tray ; and now it requires say 3 more grammes to sink it to the mark, showing that the solid has lost 3 grammes of weight by being immersed in water, hence this is the weight of the water it displaces ; and its specific 131 gravity is therefore -^-=4^ (cf. method 3, sect. 185). (Its weight when under water is, of course, 10^ grammes.) Suppose the solid had been lighter than water, and that when it was in the upper tray 12 grammes had been required to sink the instrument, whereas, when placed in the lower tray (where, of course, it would tend to float upward, and have to be confined by the cage), 30 grammes were required ; then the loss of weight in water would be 18 grammes, and as its weight was 8, its specific gravity would be . (Its weight when under water is - 10 grammes ; that is, 10 grammes upwards.) Fig. 88. Nicholson's Hydrometer. CHAP. XII.] FLOATING EQUILIBRIUM. 235 Equilibrium of Floating Bodies as regards Rotation. 187. We have now learned that a body necessarily floats in a liquid whenever it displaces its own weight of that liquid that is, that under these circumstances the two contrary forces, its own weight and the resultant of all the fluid pressures on its surface, are equal, and are hence in equilibrium as far as translation is concerned. But in order that there may be also equilibrium as regards rota- tion, these two equal contrary forces must act along the same straight line ; in other words, since the weight of the body passes through its centre of gravity, the resultant of the fluid pressures must also pass through this point ; or, again, in other words, the centre of pressure (sect. 177) of the immersed surface must lie vertically under the centre of gravity of the body. "When this condition is satisfied there is complete equili- brium ; but there remains the question whether this equili- brium is stable or not. It is manifestly stable if the point of application of the upward force is above the point of application of the down- ward one. Now, just as the downward force, the weight of the solid, may be considered as acting at its centre of gravity, so the upward force, the weight of the liquid displaced, may be considered as acting at its centre of gravity ; and this point, the centre of gravity of the liquid displaced, is the real centre of buoyancy or flotation; the term 'centre of pressure' being commonly applied only to simple surfaces which displace no water. The centre of pressure is always a point on the surface namely, that point where the line of resultant pressure meets the surface. This line of resultant pressure, which is vertical, and which always passes through both the centre of pressure and the centre of buoyancy, may be called the line of buoyancy. If, then, the centre of gravity of the water displaced be above the centre of gravity of the solid, the equilibrium is certainly stable. 236 ELEMENTARY MECHANICS. [SECT. 187. This, however, cannot be the case with homogeneous solids ; it can only be satisfied by loading the floating body. And it is satisfied in all the above hydrometers ; their centre of gravity being down near the shotted bulb, while the centre of gravity of the water displaced is up near the centre of the air-bulb ; consequently their equilibrium is very stable. But unless the floating body is totally immersed, it is quite possible to g'et stable equilibrium without satisfying the above condition ; in other words, this condition is suffi- cient, but not necessary, for bodies floating at the surface of a liquid. For instance, in a canoe, the joint centre of gravity of canoe and occupant is much higher than that of the water displaced by it ; and so it is in ships and boats generally, though ballast is used to keep the centre of gravity of a vessel reasonably low. The higher the centre of gravity of a vessel is, the less is its stability ; and by making it high enough, the equilibrium is sure to become unstable, so that the least disturbance will cause the body to rotate or turn over into some more stable position. You will find an example of unstable equilibrium if you try to float an empty bottle or a common pencil upright. A penholder, however, or a bottle half full, will float upright one way, because loaded. A long cylinder like a pencil or wine-cork floats in stable equilibrium on its side ; but a short cylinder like a flat plate or a collar-box will float with its length vertical. A sphere rests in neutral equilibrium in any position ; and so does a totally immersed homogeneous body of the proper weight, whatever may be its shape. 188. To investigate fully the conditions of stability or instability of equilibrium, it is no use taking the body just in its position of equilibrium with the two equal forces acting along the same vertical line, any more than it was when a round-based body was standing on a flat table (in sect. 144) ; but one must imagine the body tilted a little, so that the equal forces act along different though parallel lines that is, form a couple and observe whether the effect of this couple is such as to restore the body to its original position, or whether it tends to increase the dis- placement more and more. In the former case the equili- CHAP. XII.] METACENTRE. 237 brium in the original position was stable j in the latter, it was unstable. Let 1 (fig. 89) be a hemisphere floating in water in equilibrium, and therefore with the two centres of gravity, G of the body, and C of the displaced water, in the same vertical line the line of buoyancy or resultant pressure. And let 2 be the same body dis- turbed from equilibrium into a new position, and therefore with a new centre of buoyancy, Ci. We have then the downward force iv acting at G, and the upward force equal to w acting at Cj ; the two constituting a couple of moment w x ab, whose tendency is to restore the body into its original position ; which was therefore one of stable equilibrium. In 3, this fig. 2 is repeated, but the old centre of buoyancy, C, of fig. 1 is indicated in the body as well as the new one, Cj ; and the old line of buoyancy, CG, is produced till it cuts the new one through C : in the point M ; which, in the case supposed, happens to be the centre of the sphere. This point M is the metacentre; as already explained in connec- tion with rolling bodies in sect. 146. Fig. 89. The metacentre is defined as the intersection of the old line of buoyancy, drawn in the body when in equilibrium, with the new line of buoyancy when the body is slightly disturbed from its position of equilibrium ; and the rule for stability is : If the metacentre M is above the centre of gravity G, the equili- brium is stable. If it is below G, the equilibrium is unstable. If M coincides with G, it is neutral. And the height of M "above G measures the stability. All this will be seen at once if one just considers the couple as in fig. 2 above. For consider the upward force acting through the point M on the line GC fixed to the body (fig. 3) ; if M is above G, the upthrust will tend to restore the body and to bring GC upright again, the moment of the couple being proportional to the length MG ; whereas, if M is below G, it tends to topple the body over more and to turn the line GC more and more from the vertical. The position of M depends on that of the new centre of buoy- 238 ELEMENTARY MECHANICS. [SECT. 188. ancy, and this depends on the shape of the floating body about the water-line. The shape of a ship or boat is devised so as to make the metacentre as high as possible, see fig. 90. Strictly speaking, the disturbance from equi- librium ought to be infinitely small in order to give the correct position of M, and the correct measure of the stability MG. If the disturbance be great, the metacentre will in general be in a different position. If a ship lurches too much, the meta- centre comes down very low, and may even pass below G ; in which case, unless all the men can Fig. 90. rush to one side, so as to alter the position of G, or unless an opportune wave comes to right the vessel, it must heel over, like the Captain. Thus, in a floating body in equilibrium, there are four points vertically over one another (see fig. 89, No. 1) : M, the metacentre ; G, the centre of gravity of the floating body ; C, the centre of gravity of the fluid displaced ; and P, the centre of pressure of the immersed surface. Of these P is always the lowest ; and M is always above C (hence if C happens to be above G, much more is M) ; and the stability or instability of the equilibrium depends on whether M is above or below G. As a matter of fact, a ship, like many floating bodies, has two metacentres ; one, the one ordinarily spoken of as the metacentre, concerned in rolling ; the other, very high up and of no practical account, concerned in pitching. It would be next to impossible to upset a ship by tilting it at the bows. In the circular Russian ironclads the two coincide. In an ordinary wine-cork floating on its side, one metacentre, the rolling one, coincides with the centre of gravity of the cork ; the other, the pitching one, is a good height up. In bodies of irregular shape the two lines of buoyancy, CG and the vertical through C 1$ need not intersect at all, for they may lie in different planes : such bodies have no metacentre at all. The whole subject of the metacentre, however, is not one that can be treated in an elementary book like the present ; and it Avill be sufficient to have indicated the sort of ideas connected with the stability of equilibrium of floating bodies. EXAMPLES XXII. (1) Find the force with which a sphere one metre in radius is urged upward, if it is totally immersed in water. N.B. Observe that the depth to which it is immersed is now immaterial. CHAP. XII.] RELATIVE DENSITY. 239 (2) Find the apparent weight of a decimetre cube of stone in water, if its specific gravity is 2 '5. (3) How much of this block of stone would project above the surface of mercury in which it was floating ? (4) A solid which weighs 35 grammes in vacuo weighs when immersed in water only 5 grammes, while in another liquid it weighs 14 grammes ; find the specific gravity of this liquid. (5) The stem of a common hydrometer is graduated into 100 equal parts. The bulb and immersed portions, when it is sunk to the division 0, are equal to 3 times the stem in bulk. If it sinks to 20 in water, what will be the specific gravity of liquids in which it sinks to 80 and to respectively ? (6) How deep would the hydrometer of the last question sink in a liquid of specific gravity -8 ? (7) If a floating body projects i of its bulk above water, what will be the specific gravity of a liquid from which $ of its bulk projects ? (8) If a centimetre cube of metal weighs 8 '5 grammes under water, what is its true weight ? (9) A Nicholson hydrometer which will sink to the fixed mark if 20 grammes be placed on the upper tray, requires 5 grammes more if the weights are placed on the lower tray beneath the surface of the water instead of on the upper one. What is the specific gravity of the metal of which the weights are made ? (10) A body A weighing 3 grammes is attached to another body B weighing 6 grammes, and the whole immersed under water, when they are found to weigh 2 grammes. The body B alone weighs 4 grammes under water. What are the specific gravities of A and B ? (11) A specific-gravity bottle, when empty, weighs 15 grammes; when full of mercury, it weighs 151 grammes ; and when full of another liquid, it weighs 33 grammes ; what is the specific gravity of this liquid ? (12) The above bottle, when 8 grammes of a certain sand have been introduced, and the rest filled up with water, weighs altogether 30 '5 grammes; what is the specific gravity of the sand ? (13) A piece of cork weighs 10 grammes. A piece of iron is attached to it, and the two together weigh in water 20 240 ELEMENTARY MECHANICS. [EXS. XXII. grammes. The iron alone in water weighs 70 grammes. Find the specific gravity of the cork. (14) A body weighs 10 Ih. in air and 8 Ib. in water. Find its volume and specific gravity. (15) Find the density of a piece of wood from these data : weight of wood 230 grammes, weight of a piece of iron in water = 580 grammes, weight of the wood and iron together in water = 465 grammes. (16) A specific-gravity bottle, completely full of water, weighs 38-4 grammes, and when 22 '3 grammes of an insoluble solid powder have been introduced it weighs 49 '8 grammes. Calculate the density of the solid. (17) Find the specific gravity of sugar from the following data : a flask, which can just be filled with 50 grammes of alcohol, whose specific gravity is 0*8, has 20 grammes of sugar put into it, and is then filled up witli alcohol, the contents then weighing 60 grammes. (18) A cubical block of wood measures 8 inches in the side. It floats in water with four edges vertical, and with one inch above the Avater. Find the specific gravity of the wood. (19) The specific gravity of a certain solution of sugar in water is 1 *2. Find the weights of sugar and water in 100 parts of it, being given that the specific gravity of sugar is T5, and supposing that no change of volume occurs on making the solution. (20) Find the length and specific gravity of a cylinder which floats in water with 2 inches of its vertical axis out of the water, and also in a liquid of specific gravity 1*5 with 6 inches out of it. (21) Compare the apparent weights of equal masses of lead and cork, (1) in water ; (2) in air of specific gravity ff fo. [The specific gravities of lead and cork are 11 and i respectively.] (22) 3 oz. of sugar of specific gravity 1'5, are dissolved in 8 oz. of water. Find the specific gravity of the mixture (1) assuming that its volume is the sum of the volumes of its constituents, (2) assuming that the volume of the water is unchanged, (3) assuming that the volume of the solution is increased by half the bulk of the sugar. (23) Find the specific gravity of naphtha if a piece of potassium of specific gravity 0'84 and weight 20 grammes in air weighs 3j grammes in naphtha. (24) If equal weights of sugar and water result in a solution of CHAP. XIT.] RELATIVE DENSITY. 241 density 1*4, how much condensation of volume has occurred ? (25) A glass stopper weighs 2^ oz. in vacuo, 1| oz. in water, and If oz. in spirit. Find the specific gravities of glass and spirit respectively. (26) Determine the thickness of a metal wire, a piece of which, 3 metres long, weighs '24 grammes in air and '21 grammes in water. Find also the weight of a cubic centimetre of the metal. (27) A flask, which when filled with water weighs altogether 410 grammes, has 80 grammes of a solid introduced, and being then rilled up with water weighs 470 grammes. What is the volume of a kilogramme of the solid ? (28) A solid weighs 117 grammes in air, 98 in water, and 101 in another liquid. Calculate the specific gravities of the solid and the liquid. (29) A piece of opaque paraffin wax (sp. gr. *9) contains imbedded in it a sphere of glass (sp. gr. 2*5). The whole weighs 50 grammes in air and 20 grammes in water. How big is the sphere ? (30) How much silver is contained in a ' gold ' crown if it weighs 985 grammes in air and 918 in cold water, taking gold as being 19 times and silver 8 times as heavy as water? 242 ELEMENTARY MECHANICS. [SECT. 189. CHAPTER XIII. ON THE PRESSURE OF THE ATMOSPHERE, AND ON THE PROPERTIES OF GASES. (Pneumatics.) 189. Most of what we have said in the last two chapters about liquids is equally true of all fluids. Gases have the same mobility of particles, and therefore transmit pressure equally in all directions. Gases are subject to gravity, and therefore press upon all surfaces exposed to them with a pressure depending on their deptn and density ; and they exert a sustaining force on bulky bodies equal to the weight of the gas displaced by those bodies, thus causing them to lose weight, and if very light to float upwards ; thus acting just like liquids. Hence, the only part of the two preceding chapters which does not apply to gases is that which relates directly or indirectly either to the constancy of density or to the free surface of a liquid a free surface being precisely the thing which a perfect gas never has. It is infinitely expansible. This and all other peculiarities of gases as distinguished from liquids are due to the fact that their elasticity of volume is not constant or dependent on the gas itself, but depends on the pressure to which at the time the gas happens to be subject; but all the special properties of gases, qua gases, we will reserve for consideration in sect. 198 et seq.; at present we will only deal with those proper- ties which they possess in common with all fluids. CHAP. XIII.] PNEUMATICS. 243 PART I. -THE PRESSURE OF THE ATMOSPHERE. 190. Now we live immersed in an ocean of air of unknown and indefinite depth, and hence we and alf terrestrial surfaces experience its weight just as if it were an ocean of liquid ; and many phenomena of common life depend upon this pressure. Its intensity may be expressed in pounds-weight per square inch, or grammes- weight per square centimetre, or units of force per unit area ; it is not quite constant at any one place, varying with many apparently accidental and local circumstances, but its average value is 1033 grammes weight (or 981 times this number of dynes), per square centimetre, or 14 '6 Ib. weight per square inch, or roughly, a ton weight per square foot. Hence, a man's body experiences a total pressure of about 18 tons weight, for we found his average cross section (sect. 174) to be 80 square inches, which is that of a rectangle 8" x 10", whose periphery is 3 feet ; so, if the man be 6 feet high, his surface, without allowing much for irregularities, is 18 square feet. The pressure is exerted with perfect uniformity on all sides, and not only on the outside but on the inside too, so that it is not felt. The only way to make it appreciated is to destroy its uniformity by partial removal. If the pressure be removed from one side of any surface, then the other side experiences the whole uncompensated pressure of 14^ Ib. per square inch. If the air be withdrawn from any closed vessel, the outside experiences a crushing pressure, and if not very strong it will collapse. Again, if the air be removed from a vessel whose mouth is beneath the surface of a liquid, that liquid is forced up into the vessel by the atmospheric pressure on the rest of the surface, the weight of the air sustaining the weight of the liquid, and completely filling it if the vessel is not too high. The product sh, which expresses the intensity of pressure of the liquid (sect. 175) at the mouth of the vessel, must therefore be about 1033 grammes weight per square 244 ELEMENTARY MECHANICS. [SECT. 190. centimetre, if the liquid is supported by the average pressure of the air. Now, if the liquid be water, s equals 1 gramme per cubic centimetre, consequently h cannot be much greater than 1033 centimetres (or about 34 feet) ; if the vessel were taller than this, it would not be full. The atmosphere can therefore support a column of water 34 feet high, but of mercury, which is 13*6 times as heavy as water, it can only support a column 76 centimetres (about 30 inches) high. (For note that 34 x 12 = 30 x 13*6, as it happens, exactly.) Modes of Removing the Air from Vessels. 191. One way of exhausting a vessel is to drive out the air by steam, and then condense the steam. Experiment 1. Boil water in an air-tight tin canister and cork it up : remove the lamp and pour cold water over it : the uncompensated pressure outside will crush it. Experiment 2. Take a long tube closed at the top and bent as shown in fig. 91 ; fill it completely with steam, and dip its open end under mercury. As the steam condenses, the mercury is forced up to a height Jof nearly 30 inches, and the tube may then be removed from the basin of mercury and carried about. The weight of liquid in one limb of the tube Is balanced by the weight of the atmosphere in the other, which may lg ' ' be supposed to be extended to the top of the atmosphere (compare fig. 83, Chap. XI. ). A still simpler way of removing air from a tube is to fill it with a liquid. This is the way in which Torricelli originally performed the experiment and measured the pressure of the atmosphere. He filled a long tube with mercury without air-bubbles, and then inverted it with its mouth under mercury in a basin (fig. 92). On removing his finger, he saw the mercury descend till its surface was 29 or 30 inches above that of the liquid in the basin, and there come to rest after a few oscillations. CHAP. XIII.] PUMPS. 245 Above the mercury was a nearly perfect vacuum, now called a Torricellian vacuum. If any gas or vapour be introduced into this it will depress the column more or less against the force of the atmosphere. For instance, the water vapour left in the cold tube after the experiment of fig. 91 will depress the column half an inch or so. Pumps. Another mode of removing air or any fluid from a vessel is by means of an arrangement of valves which open and permit egress one way only, combined with some method of squeez- ing the fluid so as to make it move in one direc- tion or other. Such a combination is called a pump, and three kinds are shown in fig. 93. The valves in each are self-closing flaps (shown open in the figures for clearness), which will open upwards by pressure from beneath, but which only close more tightly if any pressure be exerted on them from above. (Such valves exist in the veins, and ' cause whatever flow there is to take place in one direction.) The compressing apparatus to cause motion in the fluid is in (1) an elastic bag to be alternately squeezed and relaxed by the hand such an apparatus, without valves and open only at one end, is the lung of an animal ; in (2) and (3) it is a piston fitting a cylinder which is to be pushed to and fro, or up and down ; the peculiarity in (3) being that one valve is in the piston itself. Fig. 92. Torricelli's Experiment. 246 ELEMENTARY MECHANICS. [SECT. 191. No. 1 is a pump used in surgery for producing injections or for delivering a strong jet of liquid. The heart of an animal acts on the same principle ; so does a pair of blow-bellows imperfectly, for though it has only one valve, the narrowness of the nozzle acts partially as a second one. No. 2 is a mere modification of No. 1, and is used in garden and fire engines. Both these are called force- pumps. No. 1. No. S. No. 2. Fig. 93. Pumps : with the valves displayed as if kept open by a wind from A to B. No. 3 is called a lift-pump, because it gets the fluid above the piston and then lifts it up when the piston is raised. It used also to be called a '.suction ' pump. All three arrangements evidently tend to transfer any fluid they may contain from A to B, producing an exhaus- tion in any vessel screwed on to the end A, and a con- densation in any vessel screwed on to B. Modes of Measuring the Atmospheric Pressure. 192. Aneroid Barometers. The pressure may be measured and its variations indicated by exhausting a CHAP. XIII.] BAROMETERS. 247 strong metal box with a thin and flexible (corrugated) top, supported by a spring against the weight of the at- mosphere, as shown in fig. 94. If the atmospheric pressure increases, the spring is com- pressed a little more; if the pressure decreases, the spring recovers itself a little; and so the box lid indicates variations of pressure by moving in or out, and its motions may be magnified by a rack and pinion Flgt 94> -i T . ! n i Skeleton diagram of the prin- and long index as shown. Such ciple of an aneroi(l baroiliet er, an instrument is called a c baro- but in practice the spring is out- ,,/.,, x j side the box. meter (weight-measurer), and being made without mercury this form of it is called 'aneroid.' The box must be empty, or its contents would vary in size \vith temperature, and so give spurious results. In so far as temperature affects the strength of the spring, the instrument has an error to be corrected or compensated. Mercury Barometers. The mercury column (fig. 92) is a convenient measure of the pressure of the air, and is the original form of barometer. If the pressure increases, the column is forced higher up; if it decreases, the column descends. It is found to oscillate on different days between 31 and 23 inches, at places near the sea-level ; being usually high when the atmosphere over a country is quiet and dry, and the weather therefore fine and bright or frosty; whereas, when a portion of the atmosphere is in a state of whirling motion called a cyclone, the centrifugal force of the whirl causes a depression or region of low pressure near the middle of it, and the barometric column in such places is accordingly low. These large whirls of air travel along and convey with them much steamy warmth and clouds and rain, thereby destroying the continuance of fine bright weather and breaking up spells of frost. The approach of such 248 ELEMENTARY MECHANICS. [SECT. 192. a cyclone, which, when violent, constitutes a gale or storm, is heralded by an incipient, and sometimes a rapid, fall of the barometer. Cyclonic or low barometer weather is charac- terised in Great Britain by warm, damp, muggy weather, with clouds and often rain and strong westerly winds. Anti-cyclonic or high barometer weather is characterised by calms or gentle dry winds, hot sun in summer, hard frost in winter, east winds in spring, and fogs in autumn. These facts cause a barometer to be used as a weather-glass; and a convenient form, for popular purposes, is that of fig. 91, arranged as in fig. 95, where the motion of the mercury in the short open tube is used as the indicator instead of that in the long tube, and its motion is magnified by a float counterpoised over a pulley with an index ; or else by a rack and pinion as in fig. 94. The advantage of this form is that the friction pre- vents very prompt motion, so that the accumulated changes of the last hour or two are indicated by the needle whenever you go and tap the instrument. As an accurate measurer of pressure, however, it is not of much use. The cistern form (fig. 92) is always used Fig. 95. for accuracy, and some mechanical arrangement Weather-glass. of mercmy in the cistern can either be kept constant or can be read off; for, of course, when the mercury falls in the tube it rises in the cistern, and it is the difference of levels which really measures the pressure. Sometimes the scale on which the height of the column is read is adjustable, so that, before reading, its zero can be made to coincide with the level of the mercury in the cistern; by which device the required difference of level can be read off at once. If a barometer be carried up a mountain, the mercury column must descend, because some of the column of air which formerly balanced it is left below. By this decrease of atmospheric pressure, the PRESSURE GAUGES. 249 more CHAP. XIII.] height of the mountain may be calculated. For about barometers, see Deschanel or Ganot. 193. Manometers. Columns of liquid may be used to measure pressures other than those of the atmosphere such pressure gauges are called manometers. Fig. 96 shows a gauge for measuring the pressure of the steam in a boiler over and above that of the atmosphere by the height of a column of mercury ; and the pressure may be stated as equal to so many inches or centimetres of mercury, or if very large, it may be stated as so many ' atmospheres ' every 30 inches of mercury being called one atmosphere.* Metal mano- meters are, however, preferred in practice. By ' a pressure of 76 centimetres ' on any area, then, is meant the pressure which would be pro- duced by a column of mercury 76 centimetres high with that area as base. The intensity of Flg - 90> pressure in grammes per square centimetre of a column of water is equal to its vertical height in centimetres (because 1 cubic centimetre of water weighs 1 gramme) ; or in absolute measure (dynes) its pressure is 981 times its height. That the pressure of a water column is numerically equal to its height, when expressed in gravi- tational C.G.S. units, is a convenient fact to remember. The pressure of any other liquid of specific gravity s is s times as great ; so ' 76 centimetres of mercury ' means a pres- sure of 76xl3'6x981 dynes per square centimetre.* The height of the column, of course, means the vertical height (cf. fig. 82); Fig. 97. * In the C.G.S. system of measures, a million dynes (or a megadyne) per square centimetre is conveniently called an 'atmosphere.' It is very nearly equal to 75 centimetres of mercury. Regnault unfortunately employed 76 centimetres as his standard pressure. 250 ELEMENTARY MECHANICS. [SECT. 193. hence if a manometer or barometer tube be inclined, the mercury will flow further up the tube, but so that the vertical height of its surface is the same as before (fig. 97). Modes of Raising Water. 194. The most obvious mode of raising water is to get something underneath it, and lift it up. This is the old method of a bucket and windlass. Since, however, the atmosphere can support a column of water about 34 feet high, it may be used to force water up from wells not much more than 30 feet deep. For this purpose, a tube is let down into a well, and then ex- hausted of air, either by filling it with steam and condensing it (which is nearly the oldest form of steam-engine or steam-pump, and was set up by Captain Savery at the water-works, York Build- ings, Charing Cross, and used from 1698 to 1706, and in principle is still used in the modern c puls- ometer'), or by screwing the end A of one of the pumps of fig. 93 on to the tube, and working the pump. Fig. 98 shows pump No. 3 so applied, and is a common house-pump. First the air, and then the water, is transferred from A to B, and the water finds egress at the spout. It is often required to raise water from mines several hundred yards deep. Atmospheric pressure is of course quite incompetent to effect this: the only plan is to get something under the water and lift it. Pump No. 3 is still used, only it is arranged at the bottom of the mine, within Fig. 98. House-pump. CHAP. XIII.] PUMPS. 251 20 or 30 feet of the water, and its spout is transferred higher up, so that it delivers the water at the top of the shaft. Water may be thus raised any height whatever. Such pumps are called lift-pumps, and are usually worked by engines at the top of the shaft; long rods connecting the piston of the pump with the beam of the engine. A house-pump can also be used to lift water up to a cistern on the top of the house. The piston-rod of such a lift-pump works through a water-tight stuffing-box, as in fig. 98, but the spout has a tap by which it can be closed when desired ; and a pipe leads from the upper portion, B, of the pump-barrel to the cistern. Force-pumps Nos. 1 and 2 (fig. 93) are not used to raise water from any depth, but to deliver a strong jet; and fig. 99 shows the arrange- ment in a garden- engine. The stream of water is rendered con- tinuous instead of inter- mittent, either by an elastic bag, or by an air- chamber. C is the air- chamber which con- tains air compressed by the over-supply of water, so that, if the pump stops working, ,, . , ,. n Fig. 99. Garden Force-pump. the jet continues for a few seconds, only gradually diminishing in strength as the compressed air expands. Fig. 76 showed a force-pump applied in the hydraulic press, with plungers instead of pistons. Plungers are indeed generally used in force-pumps; they act precisely like pistons of equal area, the only difference is that they fit the stuffing-box instead of the cylinder. 252 ELEMENTARY MECHANICS. [SECT. 195. 195. In any kind of lift-pump, the piston has, during its up-stroke, virtually to support a column of water reaching from the surface of the water in the well to the highest surface of water in the pipe. Calling this height h, and the area of the piston A, the pressure on it is wAJt. To work the pump, a force somewhat greater than this must therefore be applied to the piston. In force-pumps, the pressure during the up-stroke corresponds to a column of water from piston to well j and during the down-stroke to a column from the piston to the highest point reached by the water, whether it be a free jet or confined in a tube (neglecting the friction of the moving water in all cases). Mode of Lowering Water. 196. The force of gravity renders the lowering of water a very easy matter. If we have a liquid in a vessel, and wish to transfer it to another at a lower level, all that is needed is two holes in the vessel one to let the liquid out, which must be below the surface, and the other to let the air in, which is best above the surface of the liquid j if it is beneath the surface, it may act, but it will do so irregularly, letting the air in by bubbles. One hole half beneath and half above the surface will act as two holes, and this is the way one empties a jug or bottle, rotating it till its one hole occupies this position. If the hole be large, it will act as two even _. [ if wholly beneath the surface, but the Fig. 100. Pipette. J . ' now will be very irregular, llie beer in a cask with the tap open, but without a venthole, is kept in by the atmospheric pressure, unless it is fermenting and forcing itself out by means of its own gas, or unless you blow up the tap. A pipette (fig. 100) is a vessel with two CHAP. XIII.] SIPHONS. 253 holes, and the flow of liquid from it can be stopped by closing either of them with the finger. Siphon. In an open glass vessel, however, it is not convenient to bore a hole through the glass beneath the surface of the liquid, neither is it always convenient to rotate the vessel till part of its mouth is below the surface. In such cases the necessary second hole may be introduced beneath the surface as one end A of a bent tube, whose other end, B, is at a lower level say is immersed in another vessel at a lower level (fig. 101). If this tube be once exhausted of air, either by sucking liquid into it with the mouth, or by filling it at a tap before inverting it, the atmospheric pressure will after- wards keep it full of water ; and the column of water in one leg, being longer than that in the other, will overbalance it, and a steady flow from A to B will be kept up till either the water sinks below the opening A, or till the level in both vessels is the same. Such a tube is called . . , , T , , . , ,, Fig. 101. -Siphon. a 'siphon. Its shape is wholly immaterial, provided that no part of it is at a height above the surface in either vessel greater than the column of liquid which the atmosphere can support, otherwise the action will cease. So also it would cease if it were put under the receiver of an air-pump and the air exhausted.* While the air was being exhausted, the flow would go on with undiminished speed until the air pressure became too weak to sus- tain the longer of the two columns ; the liquid would soon then * It is probable that, in a perfect vacuum, a siphon of moderate height would work perfectly well, because the cohesion of water free from air is pretty strong, and might maintain the continuity of the column of liquid in spite of gravity. Under these circumstances, the cause of the flow would be exactly like that of a chain over a pulley with one end longer than the other ; and the analogy will be complete if the chain be supposed to uncoil itself from a table, and to coil itself up on the floor. 254 ELEMENTARY MECHANICS. [SECT. 196. snap at the highest point, and the longer column would fall till it was the same length as the other. As the air pressure still further diminished, the two columns would slowly sink, like barometers, until, when there was no pressure left, the level of the liquid inside and outside the tube would be the same. On readmitting the air, the action would commence again, unless either end A or B was not fully submerged. The shape of the siphon tube being immaterial, it might pass straight through the wall of the vessel from A to B (fig. 102), and such a pipe would empty the vessel to just the same extent, and at the same rate, as the tube of fig. 101 ', only it does not obviate the necessity of a hole through the side of the vessel as the tube bent over the edge does ; neither, of course, would it cease to act in a vacuum. Floating of Bodies in Air. 197. All things which displace any air (that is, which have any bulk) are pressed or buoyed up with a force equal to the weight of the air whose place they occupy (sect. 179), and so everything weighs less in air than it would in a vacuum. The true weight of a thing is its weight in vacua, and this equals its apparent weight plus the weight of an equal bulk of air. The bulkier a thing is, the more does its apparent weight differ from its true and if a very light body be also very large, it may have no apparent weight at all, but may float about in equilibrium, or even be forced upwards, like a balloon. What is called a pound of cork is therefore really more than a true pound, for it has been weighed against metal weights which are not so bulky as itself and displace much less air. A little demonstration -balance is sometimes made to hold a ball of cork and another of lead of the same apparent weight, so that they equilibrate each other in air; but if the buoyant power of the air be withdrawn by putting the whole under an air-pump, the cork will descend, showing that it is really the heavier of the two. CHAP. XIII.] BUOYANCY OF AIR. 255 A thin copper or glass sphere with a tap may be used to measure this buoyant power. When the tap is open, very little air is displaced by the sphere ; if you weigh it then, you get its true weight very nearly. But exhaust it and shut the tap. It now displaces a quantity of air, and accordingly is buoyed upwards, and will be found to be apparently lighter than before. The difference between its true and apparent weights gives the weight of an equal volume of air. In this way 1 cubic centimetre of ordinary air, when the barometer stands 76 centimetres high and the thermometer stands at zero centigrade, is found to weigh '001293 gramme. (This number '001293 is therefore the sp. gr. of air referred to water ; it is approximately equal to -g-J^-.) Another mode of quoting the same result is to say that 11'2 litres weigh about 14*4 grammes; or 1 cubic inch weighs *31 grain; or that a cubic foot weighs about an ounce and a quarter. Hence, since 1 gallon of water weighs 10 pounds, 1 pound of common air occupies about 80 gallons. A sphere of brass a yard in diameter displaces, if exhausted, rather more than half a cubic yard of air, say 14 cubic feet, which weighs l?i ounces about. If then its own weight were only a pound or so, it would ascend slowly like a balloon. But if so light as this, its walls could not be strong or regular enough to resist the pressure, and it would collapse. Such balloons are therefore impracticable. To sustain very thin walls against the air pressure, it is necessary to fill the balloon with some gas ; and hydrogen, being the lightest gas known, is always used. Hydrogen enough to fill the above sphere would weigh only 11 ounce, so it would not add very greatly to the weight, and its presence enables the walls to be of thin oil-silk instead of metal. The first balloons were filled with hot air, which occupies more room and therefore displaces more than its own weight of cold air (see Deschanel, chap, xxi., or Ganot, art. 169). EXAMPLES XXIII. (1) What is the height of the mercury harometer when the intensity of the atmospheric pressure is a megadyne 256 ELEMENTARY MECHANICS. [EXS. XXIII. per square centimetre ? (A million dynes is called a megadyne. ) (2) If a mercury barometer falls one inch, what will be the fall of a water barometer ? (3) Show that the oscillation of the column in a ' siphon ' bar- ometer, with its long and short limbs of equal cross section, is only half that of the column of a cistern barometer with an infinitely large cistern. (4) Show that the motion of the top of the mercury in a bar- ometer may be doubled by inclining the upper part of the tube at an angle of 30 to the horizon. (5) What is the total pressure inside a steam boiler when the mercury gauge (fig. 96) stands at 150 centimetres and the barometer at 75 ? (6) The piston of a lift-pump is 7 inches in diameter, and the depth of the water in the mine below the spout where the water is discharged is 533 yards. Find the least force which can raise the piston ? (7) If a rectangular mass of cork, dimensions 10x8x5 centi- metres, is counterpoised in air by 80 grammes of platinum, find the mass of the cork (neglecting the floating power of the air on the platinum). (8) A mass of wood (sp. gr. -6) is counterpoised by 105 correct grammes of iron (sp. gr. 7*5); find the mass of the wood (or its true weight in vacuo}. Ans. The volume of the iron is 14 c. c., so its apparent weight is 105- (14 x -001293); and this is equal to the apparent weight of the wood, which is x- (fit 1 x '001293), where x is the number of grammes of the wood ; hence x =105 -208. (9) A piece of metal weighs 2'4 grammes in mercury and 9 grammes in water ; what would be its weight in vacuo ? (10) A lift-pump is used to lift water from a well whose water surface is initially 10 feet below the level of the pump to a cistern 18 feet above the pump-level. The diameter of the well is 3 feet, and the internal dimensions of the cistern are 4 feet long by 3 feet broad by 2 feet high. Find the work needed to fill the cistern. The water in the well will be lowered x feet, where 2 -# = 3x4x2 _ 8_32x7_224_10 2 '' X -w 3 66 66 3 33 CHAP. XIII.] PRESSURE OP FLUIDS. 257 . . height of final lift = 28 + V + A + 2 and height of average lift = 28 + f + ^ + 1 = 30 + 1 f ^ 24 x 62 -3 (30 + f + 3 V) foot Ib. 62-3 24 1495-2 SO 44856 & 498-4 $ 498-4 A 45-3 45898 Ans. 45900 foot pounds or about 20 foot tons. (11) Determine the pressure of the atmosphere in Ib. weight per square inch, correct to one decimal place, when the bar- ometer stands at 29 inches. (12) Find the greatest height to which oil whose density is 0'9 grammes per cubic centimetre can be raised by a common ' suction ' pump, when the atmospheric pressure is a million dynes per square centimetre. (13) On a day when the barometer stands at 76 cm., find the pressure in grammes per square cm. at a point 3 metres below the surface of a pond covered with a thin film of ice. (14) A mercury barometer stands at 30 inches; find what it ought to read if it were sunk 50 feet below the surface of water. (15) Find the value in grammes weight, and in dynes, per square centimetre, of a pressure able to sustain a 75 cm. column of mercury, at centigrade, when its specific gravity is 13-596. (16) The density of mercury decreases 180 parts in a million for every degree rise of temperature, hence find the pressure corresponding to a metre column of mercury at 20 C. PART II. -ON PROPERTIES PECULIAR TO GASES. 198. A perfect fluid whose elasticity of volume (see Chap. X., sects. 154, 158) is equal -to the pressure upon Q 258 ELEMENTARY MECHANICS. [SECT. 198. it, provided the temperature is constant, is called a perfect (jas. Many actual gases namely, those called permanent gases, very nearly satisfy this definition ; and we have now to consider what properties a gas possesses in consequence of this peculiarity. First of all, gases must be very compressible : any addi- tional pressure produces a corresponding change of volume. The increase of pressure (sect. 175) is the stress; the ratio of the change of volume to the original volume is the strain (sect. 160). Let the original pressure be P, and the new pressure P' ; then the stress is P' - P. Let the original vol- ume of the gas be V, the new volume V, then the strain is V- V ^r Its elasticity, when in the compressed state, by P'- P definition (sect. 158), is y _y, ; and for gases this is now stated to equal the pressure on it when in that state namely, P'.* P'- P P' Hence ^-^ f = ^ or PV = P'V, or P:P'::V':V; or, in words, the volume of a given quantity of a perfect gas varies inversely with the pressure, other things being equal. If the pressure be doubled, the volume is halved if the pressure be halved, the volume is doubled. This is called Boyle's law, and may be verified by the bent tube of fig. 103. Its short leg is closed, its long leg open. Mercury poured down the long leg confines some air in the short one and compresses it, the whole pressure on the air * If the strain takes place very suddenly, the elasticity is greater than P, being 1*4 times P. This is because the temperature does not then remain constant heat is generated by the compression which has not time to escape. We will suppose, however, that all our compressions and expansions take place slowly enough to allow the temperature of the gas to remain without change. CHAP. XIIL] BOYLE'S LAW. 259 in the tube being that of the atmosphere plus that of the column of mercury in the tube. If the mercury stands 30 inches higher in the long leg than in the short, the original volume of the air will be found to be halved : for the original pressure it sustained was one atmosphere, and now it is two. Another 30 inches of mercury will make it shrink into one-third its original bulk, and so on. Under ordinary atmospheric pressure, 14 '4 grammes of air occupy 11 '2 litres (see sect. 197) ; but under a pressure of two atmo- spheres they shrink to 5*6 litres. The shortest statement of Boyle's law is that, ceteris paribus, PV = constant ; but remember that cetera must beparia; the tem- perature must not change, neither must the quantity (that is, mass) of gas. One gramme of hydrogen under a pressure of 76 centi- metres of mercury, and at centigrade, occupies 11-2 litres, or 11,200 cubic centimetres. Hence the value of the above constant PV for 1 gramme of hydrogen is in absolute C.G.S. units (see sects. 190 and 193): 76 x 13-6 x 981 x 11,200 = 1135 million ergs. Call this K. It is the same constant for 16 grammes V ^% of oxygen, 14 of nitrogen, 22 of carbonic anhydride, Tube. 3 and so on. It varies only with the absolute tempera- ture. For 5 grammes of hydrogen or 80 grammes oxygen, the constant is 5K ; it is, in fact, proportional to the mass of a gas, but varies for different gases with their molecular weights. A better statement of Boyle's law is that the ratio of pressure to density, , is constant ; for this is independent of everything but the nature of the gas and the temperature. If the pressure of any gas is divided by its specific weight, s or gp, the resulting constant is called the height of the homogeneous atmosphere of that gas at the given temperature (see example xxiv. 2). 260 ELEMENTARY MECHANICS. [SECT. 199. 199. The density of a gas, therefore (the mass per unit volume, see sect. 35), is directly proportional to the pressure. One consequence of this is, that as one ascends in the atmosphere, the pressure does not decrease uniformly as in the case of a liquid, but it decreases, at first at a more rapid rate, and afterwards more slowly. At a height of only three miles, for instance, the intensity of pressure is half what it is at the sea-level. For the pressure decreases not only by reason of the elevation, but also by reason of the diminution of density accompanying the decrease of pressure. Both causes combine, and the pressure diminishes upwards in what is called geometrical instead of in arith- metical progression. 200. But just as no actual liquids are perfect, so no actual gas is a perfect gas. They all deviate slightly from Boyle's law; they are probably not infinitely expansible, and are certainly not infinitely compressible, for many of them, if squeezed very much, condense into liquids ; and as they approach their condensing point, they deviate from Boyle's law a good deal, becoming more and more com- pressible. Oxygen, nitrogen, and argon have now all been liquefied in bulk, and the two latter have been frozen. Hydrogen has been momentarily liquefied as a mist, and the only gas that has so far resisted even momentary lique- faction is helium. Still, all these gases are, at ordinary pressures and temperatures, a very long way off their con- densing points, and they obey Boyle's law with considerable accuracy. They cannot, indeed, be condensed by any amount of simple squeezing; they have to be cooled enormously as well. Air-pumps. 201. Air-pumps differ in no respect from other pumps except in details of arrangement. Their peculiarity is that the vessels they are used to exhaust or to fill contain always CHAP. XIII.] AIR-PUMPS. 261 the same volume of fluid ; its density and pressure, however, are diminished or increased to any extent. Pump No. 3 (fig. 93) is generally used for exhaustion, but pump No. 2 can also be used, and it will at the same time produce condensation in any vessel screwed on to its end B. It is then called a condensing syringe. If it obtains its air from the atmosphere, the same mass of air will be injected at every stroke, and consequently the pressure in a vessel screwed on to B will increase by a fixed amount at each stroke, that is, it will increase in arithmetical progression. Fig. 104. Air-pump. Fig. 104 shows a double-barrelled air-pump with two of the No. 3 pumps arranged to exhaust a glass vessel known as the 'receiver.' At every stroke the air in the receiver expands to fill both receiver and pump-barrel, and the portion filling the latter is at the reverse stroke expelled into the atmosphere. Call the volume of the receiver V, and that of a pump- 262 ELEMENTARY MECHANICS. [SECT. 201. barrel v ; the same volume of air, v, is extracted at every stroke, but not the same mass, because its density keeps on diminishing. If the pressure of the air in the receiver to start with, is P , and after the first stroke, P x ; the product of pressure and volume being constant, we have The contents of the pump-barrel are now expelled, and the second stroke begins. During the second stroke the volume V again expands to fill the volume V + #, without the quantity of the air changing ; so, if P 2 is the pressure after the second stroke, Similarly, P 3 , the pressure after the third stroke, is given by and so on The pressure after three strokes may therefore be written V \ / V \ 2 / V \ 3 similarly, the pressure after n strokes is The pressures PoPjPgPg ...... P n decrease, therefore, in a y geometrical progression with the common ratio y -- Hence perfect exhaustion (or pressure equal zero) cannot be obtained, even with a perfect pump, without an infinite number of strokes. 202. To indicate the degree of exhaustion, a mercury gauge is commonly used, which may be simply a long tube reaching from the receiver into a cistern of mercury, some- thing like a barometer; or it may be of the form shown CHAP. XIII.] MANOMETERS. 263 separately in fig. 105, and attached to the pump at G in fig. 104. The closed limb of the U tube is completely full of mercury, and remains so till the air pressure in the little bell jar which is exhausted with the receiver gets unable to support it ; it then gradually descends as the exhaustion proceeds, and the pressure of the residual air in the receiver is measured by the difference of level between the mercury in the two limbs. 203. Compressed-air Manometers. The diminution of the volume of a gas under pressure will measure that pressure in a more compact way than the mercury gauges of sect. 193 (compare the length of the two branches of the tube, fig. 103), and a mano- meter on this principle is shown in fig. 106. 1Q5 Faraday used to measure high pressures in Air-pump his glass vessels by inserting little conical Vacuum Gauge ' glass tubes, with one end sealed, containing air and a globule of mercury (fig. 107). As the pressure of the gas Fig. 106. Fig. 107. Simple compressed-air Manometers. in which they were, increased, the globule moved up and compressed the air in the tube more and more ; and the diminution of volume measured the increase of pressure, Lord Kelvin has applied the same principle to ocean 264 ELEMENTARY MECHANICS. [sECT. 203. sounding, for, since every 34 feet of water adds another atmosphere to the pressure, if the pressure of the water be known, its depth can be calculated. A tube closed at one end is lowered into the sea, like a diving-bell, mouth downwards ; and a registering arrangement records how far the water has entered the tube, and therefore how far the air in it has been compressed. Diving-bells. Open vessels containing air, then lowered into water mouth downward, need to be heavy in order to sink. It is usual to pump more air into a diving-bell as it sinks, so as to keep the water out notwithstanding its pressure ; but if no such pumping be done, the original air will be compressed into smaller compass, the water will partly enter, and, since the displaced fluid is less, the buoyancy diminishes or the apparent weight increases as it gets more deeply immersed. On this principle little hollow glass figures, called ' Cartesian divers,' rise and sink in a liquid according as the pressure on them is diminished or increased. EXAMPLES-XXIV. (1) A barometer in a diving-bell indicates a pressure of 45 inches of mercury, the height of the barometer at the surface of the earth being (f TiT 3 -fTrr' 3 ). 2 5 <'s .'. its moment of inertia may be written = --mx 2 . = j:in -- 2"~.~ T' --- ' w ' n ch, when r =r (as is the 2 5? A 2 case in the limit), reduces to - wx ^-$--= ^ o or o EXAMPLES VIII. Pages 65, 66. (1) 9 units of acceleration ; 450 units of length. (2) 36 poundals, about equal to the weight of 1 Ib. 2 oz. (3) 2 ft. /(sec.) 2 ; 70 units. (4) 1 minute. (5) 6250 yds., i.e. about 3J miles. (6) 10 seconds. .(7) 39,240 dynes = 39| kilodynes = -fo of the weight. (8) 60 poundals, which is about the weight of 1 Ib. 14 oz. (9) 600 units. (10) 75 feet. (11) 37lb. weight. (12) About 1019J grammes weight, or nearly 2 per cent, more than the weight of a kilogramme ; 27,810 dynes will support an ounce, and 996f million dynes will support a ton. (13) 192 ft. /sec. ; between 870 and 880 miles. (14) 14 centimetres per sec. (15) 300 dynes. (16) () 37^ Ib. weight ; (b) 62 Ib. weight; (c) 50 Ib. weight. (17) Its own weight. S 290 ELEMENTARY MECHANICS. EXAMPLES IX. Pages 71-73. (1) 6 feet per second. (2) 2 feet per second. (3) 31 ft./sec. (assuming the clay to stick to the body). (4)1- (5) 7 '2 and 9 -7 ft./sec. (6) 1600 ft. /sec. (7) 1 ft./sec. nearly. (8) 11| and 21 ft./sec. (9) The velocity of the larger body will be reduced to 2 ft./sec., while the smaller body will have a reversed velocity of 7 ft. /sec. (10) 7 lb. ; reversed velocity of 5 ft./sec. (11) f (12) 607 kilogram metres, being over 95 per cent, of the initial energy. (13) f ; 5 T V ft. ; 57} ft. (14) Two cases may be drawn. In one, the direct component of the velocity of the heavier sphere, A, is changed from + 17 '32 to -11 '4 ft./sec., while its transverse velocity remains 10 ft./sec., giving a resultant 15-17 ft./sec.; and the direct component of B's velocity is changed from - 15 to +20'9, its unchanged component being 15\/3, and the resultant 33 '34 ft./sec. In the other case, the direct com- ponent of A's velocity changes from + 10 to - 22, its trans- verse component being 10\/3, and the resultant 28 ; while the direct component of B's velocity changes from - 15\/3 to +14. its transverse component being 15, and the resultant 20 '5. (15) 4J ft./sec. (nearly). (16) 500 poundals, i.e. about 15 lb. weight. (17) 230 poundals, i.e. about 7 lb. weight. (18^ The kick felt is the jerk of maximum force at the relaxation of the accumulated pressure. The speed of the bullet may have to be mainly acquired in the last inch of the barrel instead of being gradually imparted over the whole 4 feet. EXAMPLES X. Pages 82, 83. (1) 277r 2 poundals, equivalent to the weight of about 8 '4 lb. ANSWERS TO EXAMPLES. 291 (2) (Taking 7r 2 as equal to 10) 5 feet. (3) 6 inches. (4) 80 ft./sec. (5) STJ tons weight. (6) 140 Ib. (approx.). EXAMPLES XI. Pages 96-98. (1) 160 poundals, or of their usual weight. (2) At a height equal to the earth's radius. (3) In 15| seconds. (4) 981. (5) 44 f grammes weight, or 43,600 dynes. (6) 2 ft. /(sec. ) 2 ; 90 poundals, or 45 oz. weight,, (7) 6 inches-per-second per second. (8) 224 feet. (9) 4'4 seconds. (10) 20 poundals per Ib. (11) 60 poundals; 2 ft./(sec.) 2 . (12) 2ft./(sec.) 2 ; 510 poundals. (13) &g; i.e. 3'2 ft. /(sec.) 2 , or 98 centimetres/(sec.) 2 . (14) 12 feet ; 8 ft./sec. ; 146 and 138 poundals. (15) 4 Ib. (16) 12 feet; 117^ poundals. (17) 1 second. 46 ft./sec. up, or 14 ft./sec. down. Rising with a velocity of 48 ft./sec. (18) The weight of 3 tons. (19) The weight of 3& tons. (20) The weight ascends at the same rate ; and whatever the monkey does the weight does the same. EXAMPLES XII. Pages 101, 102. (1) 7| seconds. (2) 1-875 mile. (3) The tension in the string being greatest at its lowest point (because of gravity), the string is most likely to break there. The centrifugal force being equal to the weight of 100 Ib., the velocity of the stone when the string breaks must be 40 feet per second. It will start forward hori- 292 ELEMENTARY MECHANICS. zontally with this velocity and describe a portion of a parabola striking the ground after the lapse of half a second 20 feet away. (4) In 2 seconds ; 100 feet from where it would have dropped if the balloon had been stationary ; with 80 ft. /sec. vertical, and 40 ft./sec. horizontal velocity. (5) 600 feet, 225 feet. (6) 900 ft, 1800 ft. ; 3 seconds. (7) 10 seconds ; 400 feet. (8) 40V6 ft./sec. ; 45. (9) 40\/5 ft./sec. ; 25 ft.; 2| seconds. (10) 40 ft./sec. (11) 20i miles ; 15,625 feet, or nearly 3 miles. (12) 100 ft./sec. (13) Angle. Seconds. Distance in feet. Maximum elevation in feet. 2^ 3000 100 30 37i 39000 5625 45 53 45000 11250 60 65 39000 16875 -30 i 173 100 Dropped 24 100 EXAMPLES XIII. Pages 105, 106. (1) 0-999785 of a second. (2) 32-0763. (3) 216. (4) 261 ; 99-396 centimetres. (5) 87| inches. (6) 8531. (7) ?& of an inch. (8) -^; that is to say 39-14 inches, or 994 centimetres. (9) 32-0763 poundals per Ib, (10) 32-128. ANSWERS TO EXAMPLES. 293 (11) 0-3009, 0-5015, and 0'4012 sec. respectively. (12) 62i. (13) w = 4 radians per sec. ; v = S\/3 ft./sec.; t=\ir seconds. EXAMPLES XIV. Pages 109, 110. (1) 24 inches. (2) 6V3 inches (i.e. about 10 '4 inches) from the middle; 146 seconds. (3) At a distance from the hinge = of the breadth of the door. (4) 1284 ft./sec. (5) 1 seconds ; 13i inches. (6) 2-fV seconds. (7) At any point 10 inches from its centre. It will swing quickest about an axis distant 17-32 inches from the centre. (8) Length of simple pendulum %h; Mass = $m. (9) Length same as in No. (8) ; Mass = ^m. (10) Time of complete swing =1'88 sec. (11) 31b. (12) 150. (13) ^V6 seconds. (14) 1280 ft./sec. EXAMPLES XV. Pages 128-132, (1) 16V2~feet per second. (2) 2400 foot-pound als, or about 75 foot-pounds. (3) 4 ft./sec. (4) (By equating the initial and final energies) 615 '8 ft./sec. (5) 689 feet per second. (6) gfth of the weight of the projectile. (7) 400 poundals ; -$ second. (8) 93| Ibs. weight. (9) 217^ 7T 2 foot-poundals, or about 68 foot-pounds. (10) 28 million foot-poundals, or about 880 thousand foot- pounds, i.e. nearly 400 foot-tons. (11) About 15,600 foot-tons. (12) 2000 foot-poundals, or about 62 foot-pounds. 294 ELEMENTARY MECHANICS. (13) 100,000 foot-pounds per minute, or about 3 horse-power (see No. 19). (14) 13f tons weight (nearly). (15) 155 foot-pounds ; energy at lowest point, 45 foot-pounds, .-. 45-^^ = 20^ .-. x = ^, i.e. it rises U f ee t. (16) \1\\ -, 236. (17) 6*7 tons weight. (18) si7 of the weight of the truck = the weight of 1 T V cwt. j 40 ft. /sec. (19) 98^. (20) 35 T 7 T Ib. weight. (21) 314 -16 foot-pounds; 141^ British units. (22) Energy = I foot-ton; momentum = 5666 '8 British units. (23) T V of a horse-power. (24) 37 Ib. (25) 6400 feet. (26) 600 (nearly). (27) 460 (nearly). (28) 0-17 ; 3-6 seconds ; 36 ft. (29) 322 tons weight. (30) 264 foot-tons ; (a) 5 '6 Ib. weight ; (b) 7 Ib. weight. (31) 3640 foot-pounds ; 2180 foot-pounds. (32) The sliding body. (33) Energy of roller =m^A=^my 2 + ^Iw 2 =|wy 2 + ^wir 2 w 2 . [See = |wiv a . list in Sect. 43.] Energy of slider = -i >ight _1 base 15' (The general solution is fi -= (34) Energy = and is therefore the same for all the spheres. ANSWERS TO EXAMPLES. 295 (35) In the case of cylinders, v z =$ of 2gh= -- The average o velocity will be half this final velocity, hence the time occupied in descending a slope s, whose height is h, is The time taken by a sphere would be s . . the ratio of the times = ^/j^' (36) 560 foot-poundals, or about 17^ foot-pounds. (37) 140 foot-pounds. EXAMPLES XVI. Pages 145-147. (1) 10, 10V2or 14-142, 10V3orl7'32, (2) Each equals 6\/2. (3) 8^ Ib. weight (nearly). (4) The fractions of the weight in each case are V3-i . * i _L ; i . V3+i V2 ' V3 ; \/2 ; ~ V2 The best way with a single cord is to use two nails as far apart as the eyes on the picture. (5) (1) TK of 10 Ib. weight, 5 77 Ib. weight; normal pressure V* = 11-55 Ib. weight. (2) 5 Ib. weight; normal pressure = 8-66 Ib. weight. (6) ton weight. (7) (a) 200 V26 Ib. ; (b) 1000 Ib. (9) V247 = 15'7. If OC be the resultant, in the diagram, of the given forces acting on the particle at 0, and if a circle be drawn on OC as diameter, any pair of chords drawn through O at right angles to each other will be a pair of components equivalent to OC, and therefore to the given forces. Their magnitudes are given by the equation ic 2 + / 2 =247. (10) 5V57=37f units. (11) The forces can be reduced to 10^ units acting along the line of o. /q action of the greatest of the given forces, and ^- perpen- dicular to it, the resultant being VH7. (Or, they may be 296 ELEMENTARY MECHANICS. reduced to 9 units and 3 units acting respectively along the lines of action of the given forces 10 and 9.) (12) 2972. (13) (a) 2500 Ib. weight ; (b) Depends upon a new set of the sail, and the wind speed, and too many assumptions to permit a satisfactory brief answer. (14) Resultant =25 "58 units, 18'22 along AB, and 17'96 along DA. (15) 77 Ib. (nearly). (16) Resultant = 24'2 Ib. weight, its components along AB, AD being 15& and 18^ respectively. (17) f second. (18) V3=0'577. (19) In No. 18, 56V3( = 97) Ib. weight; in No. 17, 61 "94 Ik weight. EXAMPLES XVII. Pages 166-169. (1) 5 and 15, or 30 and - 10. (2) 3 inches from the middle. (3) 2 inches from the middle. (4) The diameter of the hole being |, its area is ^ of that ot the whole disc; so the centre of gravity is ^th of the radius of the disc away from its centre, on the side oppo- site the hole. (5) Magnitude of resultant, 2\/2 Ib. weight. Its line of action lies outside the square 3 inches from the point where the two larger forces act, and is perpendicular to the diagonal through that point. (6) 5 foot-pounds. (7) 50 Ib. weight. (8) 86 '6 and 100 Ib. weight. (9) A couple whose moment is 3 foot-pounds. (11) 133 Jib. weight; 166 Ib. weight. (12) Each equals 15 Ib. weight. (13) 27A and 32^ r Ib. (14) 4 feet. (15) 86f and 293J. (16) 6 - 58 inches from the opposite corner, i.e. nearly \ inch from the middle. (17) i inch from centre of disc. (18) 5 inches from the 1st side of the square, and 4^ inches from the 2nd side. ANSWERS TO EXAMPLES, 297 (19) 7 inches from the 1st side of the square, and 5 inches from the 2nd side. (20) On the diagonal joining the 3 and 6 Ib. weights, 7 inches from the latter. (22) 8 inches from the vertex. (23) 7 inches from the heavier weight. (24) 1 inch. (25) T V inch from the middle. (26) 3 inches from the loaded side. (27) 21-6 inches from the shorter bar, and 9 '6 inches from the longer. (28) 13| inches from the top, and lOf from the bottom. (29) 12-8 inches from the top, and 11 -2 from the bottom. (30) 1 '282 inch from the centre, away from the hole. (31) About 1| inch from the centre of the plate. (32) f V inch from centre of plate. (33) r y (34) f of height from the shortest side 2'06 inches from that side. (35) 2-18 inches, i.e. if of the height, from the shortest side. (36) 4 T 4 T inches from the remaining short bar. (37) lg inch from the shortest side, and 1 inch from the medium side, in each case. EXAMPLES XVIII. Pages 189-191. (4) 10 Ib. weight. (5) 5V2 Ib. weight. (7) At a height ~/o^h f the length of the rope, because then the perpendicular distance of the rope from the base of the column will be greatest, and therefore the moment of any stress in the rope about it will be a maximum. (8) 45. (9) 30. (10) 5 feet 4 inches. (11) 10 Ib. weight, at an angle of 30 below the horizontal. (13) If 21 is the length of the bar, and a is the distance of the rail from the wall, the point of the rod which rests on the rail is at a distance x from its wall-end given by x s =a?l. (14) 11-3 and 80 -8 Ib. weight. (17) 5 ft. 8 in. ; 54'7 and 41 '4 Ib. weight. 298 ELEMENTARY MECHANICS. (18) It will fall over when the slope is 3 vertical to 8 horizontal. (19) A horizontal force of (about) 7 Ib. weight applied 6 inches up. (20) 5 T 2 T inches above the ground, i.e. T 9 T of an inch below the centre of the sphere (which is the metacentre). (21) Stable. (22) 24i (by protractor, approx.). (23) 2 inches from the centre ; 1 foot-pound. Considering the circular cross-section of the cylinder which contains G (the centre of gravity), G will lie somewhere on a concentric circle of 2 inches radius. The vertical line through the point of support will cut this circle in two points ; G must be at one of these points the upper will give unstable, and the lower, stable equilibrium. (24) 1 vertical to 3 slant. In this case the vertical line through the point of support will just graze the above described gravity circle. EXAMPLES XIX. Pages 200-202. (3) 448 poundals, or a weight of 14 Ib. (4) 17 Ib. weight. (6) 1 -486 feet a second (nearly). (7) Acceleration of W=^g; acceleration of P=^ s g 2\/29 seconds. (8) f of distance between hooks from middle hook. (9) 3 tons. (10) 10^ Ib. weight ; = 1 1 7 <7, or nearly 2 ft. /(sec.) 2 . (11) 39cwt.;9'908lb. ; 1-293 Ib. (12) Pulleys; 1 Ib. each. Tensions; 7, 13, 25 Ib. weight. (13) Pulleys; 2 Ib. each. Tensions; 20, 38, 74 Ib. weight. Put: on ceiling ; 132 Ib. weight. (14) Pulleys; 5 Ib. each. Tensions; 20, 35, 65, 125 Ib. weight. (15) 4? Ib. weight. (16) Weight = 48 times the required effort. During each turn of the windlass the end of the handle (where the effort is applied) 3 x 22 moves through STT feet = feet ; and the weight is 22 22 raised ^ of this viz. = f a ^ oot - Therefore 112 turns are required to raise the weight 22 feet. ANSWERS TO EXAMPLES. 299 EXAMPLES XX. Page 210. (1) The strain is 3 parts in a thousand ; the stress is 600,000 grammes weight per square centimetre, hence the ratio of stress to strain, or Young's modulus, is 200 million grammes per square centim., or about 200,000 atmospheres linear traction per unit strain. NOTE. By an 'atmosphere' is meant a pressure of about 14'6 Ib. to the square inch, or 1 ton to the square foot, or a kilogramme per square centimetre, or 30 inches of mercury, or 34 feet of water ; or, when precise, 75 cm. of mercury at a certain temperature. (2) 200 atmospheres, or a load of two kilogrammes. (3) rb E traction. (4) W E thrust. (5) 1000 times 14 '6 Ib., or say 6 tons. (6) One-fortieth per cent. (7) Three- twentieths per cent. (8) Three parts in two thousand for about 60 atmospheres, or 000025 per atmosphere. (9) The volume elasticity is nearly the reciprocal of the last answer namely, 40,000 atmospheres; the Young's modu- lus is 60,000 atmospheres. [To complete this it may be as well to state, without proof, that the reciprocal of the rigidity of the above material would be two-tenths plus two-fortieths per cent, for the 60 atmospheres, or that the rigidity itself would be 24,000 atmospheres.] (10) 150 parts in 20,000, or about three-quarters of one per cent. (11) The average compression would be that at the depth of 1 mile, found above. Hence the rise of level would be three parts out of each 400 in the two miles namely, about 120 feet. [John Canton, 1747.] (12) 1 atmosphere. (13) 1*4 atmosphere. EXAMPLES XXI. Pages 222, 223. (1) The mechanical advantage of the lever is 60, and of the press itself 256 ; hence, the total mechanical advantage is 15,360, and the greatest weight the man can raise is 1440 tons. (2) 13,600 grammes weight on the bottom, and 6800 on each side. 300 ELEMENTARY MECHANICS. (3) 2075 grammes weight, made up as follows : (1) 125 grammes weight on the upper half of the side, due to the water, acting at a point down that half, i.e. 3J cm. from the top ; (2) a transmitted pressure of 250 grammes weight on the lower half, due to the water, and acting at the centre of that half, i.e. 7% cm. from the top ; (3) 13-6 x 125 grammes = 1700 grammes weight on the lower half, due to the mercury, acting down that side, i.e. SJ cm. from the top. These make a total pressure of 2075 grammes weight acting at 2*07 cm. from the bottom. (4) 30 ' atmospheres ' or about 440 Ib. weight per square inch. (5) TT tonnes, or 3141*6 kilogrammes weight. [The atmospheric pressure would hold them together about ten times as strongly if the interior were exhausted.] (6) 14 tons (nearly) ; 3| tons (nearly). (7) Downward pressure on base = 250 Ib. weight. Pressure on each side = 140\/5 Ib. weight, of which 140 Ib. weight is the vertical component. Weight of wedge in order to float = 280 - 250 = 30 Ib. (8) 124 tons weight, acting horizontally 8| feet from bottom of gate. (9) 6 '154 Ib. weight ; 6 inches down. (10) 3'077 Ib. weight ; 4 inches down. (11) 74f Ib. weight. (12) 1000 Ib. weight. (13) 24. (14) () or about 107 Ib. weight per sq. centimetre ; (b) 11% tons weight; J/ (20) 2 = 571f- (15) () 1543 foot-pounds; (b) & of (a) = 868 foot-pounds. EXAMPLES XXII. Pages 238-241. (1) |TT tonnes, or 41 88 '8 kilogrammes weight. (2) 1500 grammes. (3) 816 cubic centimetres. (4) 0-7. (5) 0-8421 and 1'06 ; i.e. Jf and Jf- (6) To the division 100. (7) 1-2. (8) 9-5 grammes. (9)5. ANSWERS TO EXAMPLES. 301 (10) 0-6 and 3 respectively. (11) 1-8. (12) 3-2. (13) J. (14) 55^ cubic inches (nearly) ; sp. gr. = 5. (15) . (16) 2-05 approx. (17) 1-6. (18) I (19) 50 parts of each. (20) 14 inches ; f . (21) (1) 10: -44; (2) 168:167. (22) M; (2) If; (3) If . (23) 0-7. (24) 7 : 6. (25) 2i and f . (26) 0.-113 millimetre; 8 grammes. (27) 250 cubic centimetres. (28) 6-A and & (29) Volume = 14g cubic centim.; diameter = 3*02 centim. (30) 209 & grammes. EXAMPLES XXIII. Pages 255-257. (1) 75 centimetres. (2) 13-6 inches. (5) 3 megadynes per square centimetre. (6) About 12 tons weight. (7) 80-517 grammes. (9) 9-523 grammes. (11) 14-2. (12) 11 -3 metres. (13) 1334. (14) 74 T 2 ? inches. (15) 1019-7 grammes weight, = 1,000,325 dynes. (16) 1354*7 grammes weight per square centimetre. EXAMPLES XXIV. Pages 264-266. (1) 17 feet. (3) 74 centimetres. 302 ELEMENTARY MECHANICS. (4) After one stroke, 70 centimetres; after two, 63 '63; after three strokes, 57 '9420, and so on ; each time dividing by 11 and multiplying by 10. (5) '652 ton weight per square inch. At a depth of 3366 feet. (6) (Neglecting the thickness of its wall, and taking the baro metric height as equal to 34 feet of the water), 0. (7) 15 to 17. (8) 1005 grammes weight. (9) 10. (10) It is reduced to ^ by 1 stroke, and to ^fo in 2 strokes. (11) 0-463 of its initial pressure. (12) 29 T V (13) 38 ft., or 40 ft., according as barometer is taken as 34 or 32ft. (14) 9 feet, leaving only 1 foot depth of air space. (15) 500 cubic feet. (16) 66 inches of mercury. (17) Ifi atmosphere. (18) 30 inches. (19) Mercury has risen 1 cm. ; i.e. lip is submerged 6| cm. ANSWERS TO MISCELLANEOUS EXERCISES. SET I. Pages 267, 268. [2V. . Air-resistance neglected throughout.] (1) 40,000 feet ; 100 seconds. (2) 5 seconds after the last ball is thrown, at a height of 560 feet. (3) 2 seconds. (4) 640 feet below the balloon ; velocity, 224 ft. /sec. downwards. (5) 52 ft. /sec. (6) In 10 seconds ; 240 ft/sec. (7) In |\/10 or about 0'8 second. (8) In 24 seconds, 100 feet from the top. (9) 240 poundals, or 7 lb. weight ; 1 second. (10) Vertically as seen by the passengers ; in a portion of a para- bola whose vertical and horizontal dimensions are about 9 feet and 57 feet respectively, relatively to the earth. (11) It goes up 16 feet, describing a parabola so as to be always vertically over the man's hand, to which it returns in 2 seconds, having really travelled a horizontal distance of about 150 feet. (12) 74 seconds. (13) 47cwt.; 180 feet. SET II. Page 268. (1) Magnitude = V31=5'57. Equivalent to 0, 0, 0, 0, 1, 5 in the given direction. (2) 26 along line of greatest force. (3) 19 lb. weight in the long cord ; 107 '3 in the short cord. (4) 7 '3 feet and 11 '6 feet. (5) 99-4 and 54 '6 lb. 304 ELEMENTARY MECHANICS. (6) 10 units, at 4*6 inches from first force. (7) 8* feet. (8) 7s inches from the middle of the pole. SET III. Pages 268, 269. (7) 75 centimetres. SET IV. Pages 269, 270. (3) 15*94 oz. weight ; 2 ft./sec. per second. (5) 3 inches from the side containing the first two weights, and equidistant from these weights. (7) 933,912 dynes per square cm. i.e., -934 atmosphere. (9) Horizontal pressure, 10 ; vertical, 56 ; ground resultant, pressure, about 57. (10) 576 feet ; 12 seconds ; 432 feet up. (11) 40 ft./sec. per second ; 500 feet. (12) 2-15 Ib. weight. SET V. Pages 270-272. (2) \/2 times as long. (3) 210^ feet ; 3| seconds. (4) 4*- 2 wr-rT 2 . (6) \/15 (i.e. nearly 2) seconds. (7) of the length of the rod from the axis. (8) The vertical through the weight must pass through the intersection of the lines along which the unloaded rods lie. (9) It must make an angle with the plane equal to the angle of repose. (10) 16 inches ; (11) 7; 72 grammes. (13) 3-i8 kilogrammes. (14) If p. is the coefficient of friction with the ground, and y! with the wall, and if the centre of gravity is - of its length from the ground, then, if h is the elevation of the top of the ladder, and b is the distance of its foot from the wall, 1 ANSWERS TO MISCELLANEOUS EXERCISES. 305 (15) -rAt\/3, where a is the length of an edge of the pyramid, and /* is the coefficient of friction. (The drawing cord must be fastened to an edge, to give this best chance of not up- setting. ) (16) v must be greater than \/(gr). SET VI. Pages 272, 273. (3) In 3 seconds ; 3250 feet. (5) 4 ft. /sec. per second ; 60 feet. (6) Tension = 10^?- Ib. weight ; acceleration = ^g. (8) The recoil velocity is f of the approach velocity. See also answer to IX. (13). (9) The required moment exceeds the central moment by the product of the mass into the square of the distance between the axes. (10) If is the distance of the centre of rod from the axis of sus- pension, and x the length of the equivalent simple pen- dulum, (a; - a) = iV 2 x * s a maximum, and the time of swing is therefore a minimum, when 2 = T 1 ^ 2 . (11) One five-hundredth of an inch. (12) ii. g varies apparently on account of centrifugal force, and really on account of the shape of the earth. The two effects add together. (13) iii. By equating the weight of a satellite or planet to its centrifugal force. (14) Directly as distance from centre. About 21 minutes. The motion is simply harmonic and independent of amplitude. The maximum velocity at centre is \/(gll), and g varies directly with R if density is given. Hence linear size -f velocity is constant. 1 12 (16) The rod has -th immersed, such that n+ - = -; or if water fv it S is above pivot, such that 2 = 1; otherwise the rod is vertical. SET VII. Page 274. (1) 196 feet ; 112 feet per second. (2) 2 sees. ; 2500 feet from base of tower. T 306 ELEMENTARY MECHANICS. (3) fl of 3 cwt. ; or 283^> Ib. weight. (4) 1-06 foot, 2-13 feet per sec. 22'4 Ib. weight. (5) 1664 feet per sec. (6) 24'7 revolutions per sec. (7) Speeds proportional to the chords. Times all the same. (9) 30-73 foot-second units. (10) 31,260 feet ; 44 -2 seconds ; 7812*5 feet. (11) About 80 Ib. per inch of width. 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewals only: Tel. No. 642-3405 Renewals may be made 4 days priod to date due. Renewed books are subject to immediate recall. V72-6PM9S BEC'BtO. *JUl 30 1973 REC-OLB FEBlQ 74-BPH LD21A-60m-8,'70 (N8837slO)476 A-32 General Library University of California Berkeley UNIVERSITY OF CALIFORNIA LIBRARY