AN ELEMENTARY TREATISE ON MECilANICS, EM,'BRACING TIHE THEORY OF STATICS AND DYNAMIICS, AND ITS APPLICATION TO SOLIDS AND FLUIDS. rgeparlDo for t ile lt-a'e ra'allte (9ttug in te W~ legraa Mzfbeasitp, BY AUGUSTUS W. SMITH, LL.D., PROFEESOR OF MATHEMATICS AND ASTRONOMY. NEW YORK: HARPER & B!ROTltEPI{, PUBLISHERS. Entered, according to Act of Congress, in the year one thousand eight hundred and forty-nine, by HARPER & BROTHERS, mn the Clerk's Office of the District Court of the Southern District of New York. PRE F AC E THE preparation of the present treatise was undertaken un. der the impression that an elementary work on analytical Mechanics, suited to the purposes and exigencies of the course of study in colleges, was needed. This impression is the result of long experience in teaching, and a fair trial of all known American works and reprints designed for such use. It can scarcely be necessary, at this period, to assign at length the reasons for adopting the analytical methods of investigation. Whether the object be intellectual discipline, or a knowledge of facts and principles, or both, the preference must be given to the modern analysis. It affords a wider field for the exercise of judgment, calls more fully into exercise the inventive powers, and taxes the memory less with unimportant particulais, thus developing and strengthening more of the mental faculties, and more equably by far than the geometrical methods. It is more universal in its application, shorter in practice, and far more fruitful in results. It is, indeed, the only method by which the student can advance beyond the merest rudiments of the science, without an expense of time and energy wholly disproportioned to the ends accomplished-the only method by which he can acquire a self-sustaining and a progressive power. As the hope of furnishing to the student some additional facilities for a pleasant and profitable prosecution of this branch of study was the motive for undertaking the preparation of Aiis manual, it will be proper to refer to some of the 2 Li P It EFAC E. specific objects had in view. The most formidable obstacles to the acquisition of any branch of science are generally found at the very outset. It has therefore been a specific object to introduce the subject by giving distinctness to the elementary truths, dwelling upon them till they are rendered familiar, adopting the simplest mode of investigation and proof consistent with rigor of demonstration, and avoiding all reference to the metaphysics of the science as out of place in a work designed for beginners. At every successive stage of advancement the student is required to review the ground passed over by the use of the principles learned, in the solution of examples which will require their application, and test at once his knowledge of them and his ability to apply them. Having examined in this way each division of forces as classified, more general methods are introduced, and their application illustrated by numerous problems. These methods are often employed in particular cases when others less general would be shorter; for it is a readiness in their use and a familiarity with their application that gives to the student his power over more difficult and complicated questions. To adapt the work to the exigencies of the recitation-room, the whole is divided, at the risk, perhaps, of too much apparent formality, into distinct portions or propositions, suitable for an individual exercise. In each case the object to be accomplished is distinctly proposed, or the truth to be established is briefly and clearly enunciated. The student has thus a definite object before him when called upon to recite, acquires a convenient formulary of words by which to quote and apply his arguments, and the more clearly conceives and marks hia own progress. In this institution the mathematical and experimental courses are assigned to different departments, It is not, however, for this reason alone that I have purposely abstained fion) swell PR EF A C E. iil ing the volume with diffuse verbal explanations, and the introduction of experimental illustrations. The experienced teacher will always have at command an abundance of matter of this kind to meet every emergency, and can adapt the kind and mode of illustration to the specific difficulty that'arises in the mind of his pupil. There is an objection to the' introduc tion of such matter, growing out of a tendency, in some at least, to become the'passive recipients of their mental aliment. These mental dyspeptics loathe that which costs them labor, and, satisfied: with' the stimulus of inflation, seize upon'the lighter portions, and neglect'that which alone can impart vigor to their mental constitution. Whoever caters for this class of students must share in'the responsibility of raising up a race effeminate in mind, if not in body.' The investigations are limited to forces in the same plane, except in the case of parallel forces. This limit, while it is sufficiently ample to include the'most important topics' of terrestrial mechanics, and to embrace an interesting field of celestial mechanics, is fully sufficient to occupy the time which the crowded course of study in our colleges will admit of being appropriated to it. It may, without marring the integrity of the work, be very much reduced by the omission of all those portions in which the integral calculus is employed; and still further, if desired, by omitting all that relates to the principle of virtual velocities, and its application to the mechanical powers. A copious analysis is given in the contents, convenient for frequent and rapid reviews, and suggestive of questions for examination. In this work no claim is advanced to originality. The materials have been sought and freely taken from all available sources. Nearly all the matter, in some form, is found in almost every author consulted, and credit could not be given in every case, if, in an elementary work designed as a text-book,' IV P R PREFACE. it were desirable to do so% In the portion on Statics, 1 am most indebted to the excellent introductory work of Professor Potter, of University College, London. The chapter on Couples is substantially taken from Poinsot's Elemens de Statique. The questions, generally simplified in their character, are mostly taken from Walton's Mechanical and Hydrostatical Problems, and Wrigley and Johnston's Evxamples. The works more.especially consulted are those of Poisson, 1Franccur, Gregory, Whewell, Walker, Moseley, and Jamieson. Something has been taken from each, but modified to suit the specific object kept constantly in view-the preparation of a manual which should be simple in its character, would most naturally, easily, and successfully induct the student into the elementary principles of the science, and prepare him, if so disposed, to prosecute the study further, without the necessity of beginning again and studying entirely new methods. How far I have succeeded must be left to the decision of others, especially of my co-laborers in this department of instruction. WESLEYAN UNIVERSITY, Mfiddletown, Conn., Jan., 1849. CONTENT S, &rt Pa INTRODUCTION.............................I 1. Definition of Mechanics and its Subdivisions. 2. Definition of Force and its Mechanical Effects. 3. Definition of its Intensity and Measure, its Direction and Point of Application. 4. Definition of Analytical Mechanics. 5. Definition of Concurring and Conspiring Forces. 6. Definition of Body, Rigid, Flexible, and Elastic. STATIC S. CHAPTER I. COMPOSITION AND EQUILIBRIUM OF CONCURRING FORCES IN Til, SAMIE PLANE. 3 7. Two equal and opposite Forces in Equilibriumll. 8. Two Forces inclined to each other can not Equilibrate. 9. Definition of Resultant and Components. 10. Resultant of several Conspiring Forces. 11. Resultant of two unequal opposite Forces. 12 Resultant of any number of opposite Forces. 13. Point of Application at any Point in its Direction. 14. Direction of the Resultant of several Forces. 15. Direction of the Resultant of two equal Forces. 16. Direction of the Resultant of two unequal Forces. 17. Variation of the Magnitude of the Resultant and its Components. 18. Equilibrium of three equal Forces. 19. Resultant of two equal Forces at an Angle of 1200. 20. Each of three equilibrating Forces equal to the Resultant of the other two iz Magnitude. 21. Parallelogram of Forces. 22. Triangle of Forces. 23. Resolution of Forces. 24. Polygon of Forces. 25. Representation of equilibrated Forces. 26. Graphical determination of Resultant. 27. Parallelopiped of Forces. 28. Ratios of three equilibrated Forces. 29. Expression for the Resultant of two Forces. 30. Definition of Moment of a Force-Origin of Moments. 31. Equality of the Moment of the Resultant with the Sum of the Moments of two Components. 3.2. Equality of the Moment of tl e Resultant with the Sum of the Moments of aln number of Components. 33. When the Origin of Moments IS fixed. 34. When the Forces are in Equilibrium. 35. Examples. VI CO NT NT S. CHAPTER II. PARALLEL FORCES...................,,, l~ 56. Resultant of two Parallel Forces. 37. Definition of Arms of Forces. 38. Equilibrium of two Parallel Forces by a third Force. 39. Point of Application of the Resultant. 40. When the:Forces act in opposite Directions. 41. When the Forces are Equal and Opposite. 42. Such Forces' constitute a Statical Couple. 43. Resultant of any Number of Parallel Forces, 44. Definition of Center of Parallel Forces. 45. Equality of the Moment of the Resultant with the Sum of the Moments of thse Components. 46. Definition of Moment of a Force in reference to a Plane. 47. Conditions of Equilibrium of any Number of Parallel Forces 48. Condition of Rotation. 49. Wthen in Equilibrium, each Force equal in Magnitude to the Resultant of all the others. 50. Equilibrium independent of their Direction. 51. Examples. CHAPTER III. THEORY OF COUPLES.........,...........,, 24 52. Definition of a Statical Couple. 53. Definition of the Arms of a Couple. 54. Definition of the Moment of a Couple. 55. A Couple may be turned round in its own Plane. 56. A Couple may be removed parallel to itself in its own Plane. 57. A Couple may be removed to a Parallel Plane. 58. Couples are equivalent when their Planes are Parallel and Moments are EquaL 59. Couples may be changed into others having Arms of a given Length. 60.- Definition of the Axis of a Couple. 61. Properties of an Axis. 62. Definition of the Resultant of two or more Couples. 63. Equality of the Moment of the Resultant with the Sum of the Moments of the Components. 64. Equality of the Axis of the Resultant with the Sum of the Axes of the Comn. ponents. 65. The Resultant of two Couples inclined to each other. 66. Representative of the Axis of the Resultant of two Couples. 67. Parallelogram of Couples. CHAPTER IV. ANALYTICAL STATICS IN TWO DIMENSIONS................. 31 68. Resultant of any Number of Concurring Forces. 69. Directions of the Rectangular Components involved in their Trigonometrical ValueN 70. Conditions of Equilibrium of Concurring Forces. 71. Resultant Force and Resultant Couple when the Forces do not concur. 72. Construction of the Results. 73. Equation of the Resultant. 74. Equilibrium of non-concurring Forces. 75. Equilibrium when there is a fixed Point in the Systeram 76. Equilibrium of a Point on a Plane Curve, C o N T E. N T i, Yii CONTENT.%, Vii rtI. Pap 77. (Jonditions of Equilibriunm. 78. Definition of Virtual Velocitits. 79. Principle of Virtual Velocities. 80. Principle of Virtual Velocities obtains in Concurring Forces in the same Plane. 81. Principle of Virtual Velocities obtains in non-concurring Forces in the same Plane, 92. The Converse. CHAPTER V. CENTER OF GRAVITY..4...1.......4 83. Definition of Gravity. 84. Laws of Gravity. 85. Definition of a Heavy Body. 86. Definition of the WVeight of a Body, 67. Definition of its Mass. 88. Expression for Weight. 89. Definition of Density. DO. Another Expression for Weight. 91. Relations of Masses to Volumes of the same Density. 92. Relations of Densities to Volumes of the same Mass. 93. Itelations of Densities to Masses of the same Volume. 94. Definition of Center of Gravity. 95. Connection of the Center of Gravity with the Ioctrine of Parallel Forces. 96. Definition of a Body symmetrical with respect to a Plane. 97. Position of its Center of Gravity. 98. Definition of a Body symmetrical withl respect to an Axis.'39. Position of its Center of Gravity. 100. Center of Gravity of a Body symmetrical with respect to two Axes. 101. Definition of Center of Figure. 102. Center of Gravity of any Number of heavy Particles. 103. Their Center of Gravity when their Positions are given by their Co-ordinates 104. Their Center of Gravity when they are all in the same Line. 105. Their Center of Gravity when they are Homogeneous. 106. The Center of Gravity of the WVhole and a Part -given to find that of the other Part. 107. Examples-1. Of a Straight Lne —2. Triangle-3. Parallelogram-4. Polygon5. Triangular Pyramid-6. Any Pyramid-7. Frustum of a Cone-8. Perimeter of a Triangle-9. Of a Triangle in Terms of its Co-ordinates. CONDITIONS OF EQUILIBRIUM OF BODIES FRO~M THIlE ACTION OF GRA-VITY... 54 108. When the Body has a Fixed Point in it. 109. Definition of Stable, Unstable, and Neutral Equilibrium. 110. Position of the Center of Gravity when the Equiliblium is Stable and when iUnstable. 111. Pressure on the Fixed Point. 112. Position and Pressure when there are two Fixed Points. 11:3. Position and Pressure when there are three Fixed Points. 114. Position and Pressure when a Body touches a Horizontal Plane in one Point. 115. Position and Pressure when a Body touches a Horizontal Plane in two Points. 116. Position and Pressure when a Body touches a Horizontal Plane in three' Points 117. Position and Pressure when a Body touches a Horizontal Plane in any Number of Points. 118. Measare of the Stability on a Horizontal Plane. 11.9. Case of a Body on an Inclined Plane. 120. Examples. APPLICATION OF THE INTEGRAL CALCU:US TO THE DETERMINATION OF TIHE CE.l TER OF GRAVITY........................................ M Vlii CON TENT S. Art. Paga 121. General differential Expressions for the Co-ordinates of the Center of Gravity. 122. General differential Expressions for the Center of Gravity of a Plane Curve. 123. General differential Expressions for the Center of Gravity of a Plane Area. 124. General differential Expressions for the Co-ordinates of a Surface of Revolution. 125. General differential Expressions for the Co-ordinates of a Solid of Revolution. 126. Determination of a Surface of Revolution. 127. Determination of a Solid of Revolution. 128. Examples-1. Of a Circular Arc-2. Circular Segment-3. Surface of a Spherical Segment-4. Spherical Segment. 129. Examples on the preceding Chapters. CHAPTER VI. THE MECHANICAL POWERS..................... 76 130. Classification of the Mechanical Powers.' I. THE LEVER. 131. Definition of a Lever. 132. Kinds of Lever. 133. Conditions of Equilibrium when the Forces are Parallel 134. Conditions of Equilibrium when the Forces are Inclined. 135. Conditions of Equilibrium when the Lever is Bent or Curved.'36. Conditions of Equilibrium when any Number of Forces in the same Plane act on a Lever of any Form-Examples. S II. WHEEL AND AXLE I....8............................................. 86 137. Definition of Wheel and Axle. 138. Conditions of Equilibrium when two Forces act Tangentially to the Surface of che Wheel and Axle. 139. Perpetual Lever. 140. Pressure on the Axis. 141. Conditions of Equilibrium of any Number of Forces. 142. Definition of Cogged Wheels, Crown Wheels, Beveled Wheels, Pinions, Leaver 143. Conditions of Equilibrium in Cogged Wheels. 144. Conditions of Equilibrium when the Cogs are of equal Breadth. 145. Conditions of Equilibrium in Cogged Wheels and Pinions. ~ III. THE CORD........................ 91 146. Definition of the Cord-Of Tension. 147. Conditions of Equilibrium when there are three Forces. 148. Conditions of Equilibrium when the Ends are fixed. 149. Conditions of Equilibrium when the Ends are fixed and a third Force applied to a Running Knot. 150. Conditions of Equilibrium when there is any number of Forces. 151. Relations of the Forces when in Equilibrium. 152. Relations of the Forces when thie Ends are Fixed and the Forces Parallel. 153. Relations of the Forces when the Ends are Fixed and the Forces are Weights. 154. Point of Application of the Resultant. 155. Catenary-Examples. ~ IV. THE PULLEY. 96' 156. Definition of the Pulley —Fixed and Movable. 157. Use of Fixed Pulley. 158. Equilibrium in single Movable Pulley. 159. Systems of Pulleys. 160. Equilibrium in first System. 161 Equilibrium in the second. 1C2 Equilibrium in the third —Examples. C O N'T E N'T S. X Art Pag ~ V. THE INCLINED PLANE...........................................10 163. Definition of Inclined Plane. 164. Equilibrium when the Body is sustained by a Force acting in any Direction. 165. Equilibrium when the Body is sustained by a Force parallel to the Plane, 166. Equilibrium when the Bodyis sustained by-a Force parallel tothe Base of the Plane. 167. Equilibrium when two Bodies rest on two Inclined Planes. Q VI. THE WEDGE................................................. 10J 168. Definition of the Wedge-Faces, Angle Back. 169. Conditions of Equilibrium in the Wedge. 170. Defect in the Theory. 171. Illustrative Problem. ~ VII. THE SCREW..................................105 172. Definition of the Screw. 173. Conditions of Equilibrium in the Screw. ~ VIII. BALANCES AND COMBINATIONS OF THE MECHANICAL POWERS...... 107 174. The common Balance. 175. Requisites for good Balance. 176. Conditions of Horizontality of the Beam. 177. Conditions of Sensibility. 178. Conditions of Stability. 179. Relations of the Requisites. 180. Steelyard Balance. 181. Law of Graduation. 182. Bent Lever Balance. 183. Law of Graduation. 184. Roberval's Balance. 185. Condition of Equilibrium. 186. Conditions of Equilibrium in a Combination of Levers. 187. Conditions of Equilibrium in the Endless Screw. 188. Conditions of Equilibrium in anlly Combination of the Mechanical Powers. 189. Conditions of Equilibrium in the Knee. CHAPTER VII. APPLICATION OF THE PRINCIPLE OF VIRTUAL VELOCITIES TO THE MECHANICAL POWERS.......................................................... 11 190. Preliminary Considerations. 191. Application to the Wheel and Axle. 192. Application to Toothed Wheels. 193. Application to Movable Pulley with Parallel Cords. 194. Application to the first System of Pulleys. 195. Application to the second System. 196. Application to the third System. 197. Application to the Inclined Plane. 198. Application to the Wedge. 199. Application to the Lever of any Form. 200. Application to the single Movable Pulley with inclined Cords CHAPTER VIII. FRICTION.............................. 124 201. Definition of Friction-Kinds. 202. Measurement of Friction. 203. Laws of Friction. W04. Value of the Coefficient of Friction. I CONTENTS. &rt. Pas 205. Limits of the Ratio of the Power to the Wreight on the Iiclined Plane. 206. Limits of the Ratio of the Power to the Weight in the Screw. 207. One Limit obtained directly from the other. —Exampies. 203. Examples on Chapters VI., VII., and VIII. D Y N A M I C S. INTRODUCTION.......s..s. ~a 209. In Dynamics, Time an Element. 210. Definition of Motion. 211. Definition of Absolute Motion. 212. Definition of Relative Motion. 213. Definition of Velocity-Its Measure. 214. Definition of Variable Velocity-Its Measure. 215. Definition of Relative Velocity. 216. Definition of Inertia —First Law of Motion. 217. Definition of Center of Inertia. 218. Definition of the Path of a Body. 219. Definition of Free and Constrained Motion. 220. Definition of an Impulsive Force. 221. Definition of an Incessant Force. 222. Definition of a Constant Force-Its Measure. 223. Definition of a Variable Force-Its Measure. 224. Definition of Momentum-Its AMeasure-Of Living Force. 225. Definition of a Moving Force-Its Measure. 226. The second Law of Motion. 227. The third Law of Motion. CHAPTER I. UNIFORM MOTION..................e..34.28. Point to which the Force must be applied. 229. General Equation of Uniform Motion. 230. Relation of Spaces to Velocities when the Times are Equal. 231. Relation of Spaces to the Times when the Velocities are Equal. 232. Relation of Velocities to the Times when the Spaces are Equal. 233. Measure of an Impulsive Force. 234. The Velocity resulting from the Action of several Forces. 235. Parallelogram of Velocities. 236. Rectangular Composition and Resolution of Velocities. 237. Relations of Space, Time, and Velocity of two Bodies moving in the same Straight Line. 238. Relations of Space, Time, and Velocity of two Bodies moving in the Circumference of a Circle. 239. Examples. CHAPTER II. IMPACT OF BODIES..................... 1E3 o240. Definition of Direct, Central, and Oblique Impact. 241. Definition of Elasticity-Perfect —Imperfect-Its Modulus. 242. Definition of Hard and Soft. 243. Velocity of two Inelastic Bodies after Impact. 244. Loss of Living Force in the Impact of Inelastic Bodiea. 245. Velocities of imperfectly Elastic Bodies after Impact. CONTENTS. X Art P P 246. Velocity of the ath Body in a Series of perfectly Elastic Bodies. 247. Velocity of the common Center of Gravity before and after Impact. 248. Conservation of the Motion of the Center of Gravity. 249. Definition of Angles of Incidence and Reflection. 250. Motion of an Inelastic Body after Oblique Impact on a Hard Plane. 251. Motion of an Elastic Body after Oblique Impact on a Hard Plane. 252. Direction of Motion before Impact, that a Body after Impact may pass tlrough a given Point. 253. Measure of the Modulus of Elasticity. 254. Mode of determining it-Table of Moduli. 255. Examples. CHAPTER III. MOTION FROM THE ACTION OF A CONSTANT FORCE........... 153 256. Uniformly accelerated Motion-Acquired Velocity. 257. Space in Terms of the Force and Time. 258. Space in Terms of the Force and Velocity. 2:59. Space described in the last n Seconds. 260. The Velocity and Space from the joint Action of a Projectile and Constant Force. 261. The Velocity when the Space is given. 262. Velocity lost and gained by the Action of a Constant Force when the Space is the same.,463. Scholium on Universal G-ravitation. 264. Scholium on the Numerical Value of the Force of Gravity.'f6i5. Examples. CHAPTERI IV. PROJECTILES........................ ]81 266. The Path of a Projectile is a Parabola. 267. Equation of the Path when referred to Horizontal and Vertical Axes. 268. Definition of Horizontal Range-Time of Flight-Impetus. 269. Time of Flight on a Horizontal Plane. 270. Range on a Horizontal Plane-The same for two Angles of Elevation. 271. Greatest Height. 272. Range andTime of Flight on an Inclined Plane, and Co-ordinates of Point of Impact 273. Formula for Velocity of a Ball or Shell. 274. Examples. CHAPTER V. CONSTRAINED MOTION. M I. MOTION ON INCLINED PLANES............................... leg 275. Relations of Space, Time, and Velocity. 276. Velocity down the Plane and its Height. 277. Times down Inclined Planes of the same Height. 278. Relations of Space, Time, and Velocity when projected up or down the Plane 279. Time of Descent down the Chords of a Circle. 280. Straight Line of quickest Descent from a Point within a Circle to its Circumferenceu 281. Straight Line of quickest Descent from a given Point to an Inclined Plane. 282. Motion of two Bodies suspended by a Cord over a Fixed Pulley. 283. Motion of two Bodies when the Inertia of the Pulley is considered. 8 II. MOTION IN CIRCULAR ARCS.t..................................... 17 284. Velocity lacquired down the Arc of a Circle.;85. Velocity lost in passing from one Side of a Polygon to the next Xi] CONTENTS. Art. Pa e 286. Velocity lost when the Sides are Infinite in Numler. 267. Direction and Intensity of an Impulse at each Angle, tc make a Body describe a Polygon with a uniform Velocity. 288. Direction and Intensity when the Polygon becomes a Circle. 289. Definition of Centrifugal and Centripetal Force. 290. Discussion of the Motion of a Body in a Circle by the Action of a Central Force. 291. Centrifugal Force compared with Gravity. 292. Centrifugal Force of the Earth at the Equator. 293. Ratio of Centrifugal Force of the Earth to Gravity. 294. Time of its Rotation when they are Equal. 295. Diminution of Gravity in different Latitudes by the Centrifugal Force. 296. Cause and Value of the Compression of the Earth. 297. Centrifugal Force of the Moon in its Orbit. 298. Moon retained in its Orbit by Gravitation. III. PENDULUM....................................................... le 299. Definition of a Simple Pendulum. 300. The Force by which it is urged in its Path. 301. Times of Descent to the Center of Force equal when the Force varies as the Dii. tance. 302. Expression for the Time given. 303. Discussion of the Vibrations of Pendulums. 304. Pendulum used to determine the Figure of the Earth. 305. Lengths of Pendulums as the Squares of the Number of Vibrations in a given Time 306. Mode of determining the Length of the Seconds Pendulum. 307. Determination of the Intensity of Gravity. 308. Correction of the Length of the Pendulum for a given Loss or Gain. 309. Rate at a given Height above the Earth's Surface. 310. Heights determined by the' Change of Rate. 311. Definition of Conical Pendulum. 312. Tension, Velocity, and Time determined. 313. Formula for Length of Pendulum in a given Latitude. 314. Examples. CHAPTER VI. ROTATION OF RIGID BODIES..................... 315. Angular Velocity of a Body about a Fixed Axis. 316. Definition of Moment of Inertia. 317. Moment of Inertia about any Axis compared with that about a Parallel Axis through the Center of Gravity. 318. Radius of Gyration-Principal Radius. 319. Existence of a Point at which, if all the Matter of the Body were collected, the Motion would be the same. 320. Expression for its Distance from the Axis. 321. Definition of Center of Oscillation. 322. Time of Oscillation of a Rigid Body. 323. Definition of Compound Pendulum. 324. Centers of Oscillation and Suspension convertible. 325. Length of an Equivalent Simple Pendulum. 326. Value of Principal Radius of Gyration. 327. Relations of the Centers of Gravity, of Gyration, and of Oscillation. 328. Determination of the Length of the Seconds Pendulum. 329. Examples on the Relation of the Simple to the Compound Pendulum. 330. Examples on the Determination of the Moment of Inertia. ONTENT S. Xiii ar& Page CHAPTER VII............2............. 201 331. More General Methods required. 332. Modification and Extension of Fundamental Formulse. g I. RECTILINEAR MOTION OF A FREE POINT........................... 201 333. Preliminary Considerations-Absolute Force. 334. Space in Terms of the Time when the Force is Constant. 335. General Expression for the Velocity when the Force is Variable. 336. General Expression for the. Time when the Force is Variable. 337. Velocity when the Force varies as the Distance. 338. Velocity at the Center of the Earth. 339. Time when the Force varies as the Distance. 340. Velocity when the Force varies as the Square of the Distance. 341. Velocity from an Infinite Distance. 342. Time when the Force varies inversely as the Square of the Distance. 343. Velocity and Time when the Force varies inversely as the Cubo of the loistace. Q II. CURVILINEAR MOTION OF A FREE POINT.................. 2, 344. Preliminary Considerations. 345. Motion of a Point acted on by any Number of Forces iu t.e same Plane. 346. Specific Directions for particular Cases. 347. Velocity of a Point in its Path. 348. The Velocity may be found when the Expression ia iategrable. 349. The Expression integrable when the Forces are directed to Fixed Centers, and are Functions of the Distances from those Centers. 350. Motion of a Point when the Force is directed to a Center. 351. When there is an Equable Descriptionof Areas, the Force is directed to a Center. 352. Motion of a Point from an Impulsive Force. 353. Motion of a Projectile acted upon Ly Gravity. C III. CONSTRAINED MOTION OF A POINT..................2............. l1i 354. Velocity of a Point on a given Curve. 355. Time of its Motion. 356. The Reaction of the Curve357. Motion down the Arc of a Circle by Gravity. 358. Motion down the Are of a Cycloid. 359. Tautochronism of -vibtations in a Cycloid. ~ IV. MOTION OF A POINT ACTED ON BY A CENTRAL FORCE............... 221 360. Preliminary Remarks. 361.' Expressions fot the Force in Terms of Polar Co-ordinates. 362. Equable Description of Areas. 363. Angular Velocity of the Radius Vector. 364. Veloclty of a Point in its Orbit. 365. Time of describing any Portion of its Orbit. 366. Eqtation of the Orbit. 361. Geometrical Values of Differential Expressions. 368, Kepler's Laws.:369. Accelerating Force of a Planet directed to the San. 370. Accelerating Force of a Planet varies inversely as the Square of the Distance. 371. Absolute Force the same for all the Planets. 372. Velocity of a Planet at any Point of its Orbit. 373. Determination of the Orbit when the Force varies inversely as the Squiar of the Distance. 374. Time of describing any Portion of the Orbit. 9lV C O N T E N T S. A rt. Pa 375. Value of the Abso:ute Force in Terms of the Masses-In Terms of Terrestria Gravity. 37S. Examples. HY DR O ST AT IC S. CHAPTER I................... 377. Forms of Matter. 378. Definition of a Perfect Fluid. 379. Definition of an Incompressible Fluid. 380. Definition of a Compressible Fluid. 381. Unimpaired Transmission of Pressures. 382. Surface of a Fluid at Rest. 383. Surface of a Fluid moving Horizontally. 384. Surface of a Fluid Revolving. 385. Pressure on the Horizontal Base of a Vessel. 386. Pressure on any Side of a Vessel or immersed Surface. 387. Pressure in a definite Direction. 388. Resultant Pressure on the Interior Surface of a VesseL 389. Resultant Pressure on an immersed murd. 390. Conditions of Equilibrium of an immersed Solid. 391. Definition of Plane of Flotation-Of Axis of Flotation. 392. Depth of Flotation of a Body. 393. Position of Equilibrium of a Triangular Prism. 394. Definition of Center of Pressure. 395. Formula for finding the Center of Pressure. 396. Center of Pressure of a Rectangle with one Side in the Surface of the Fluid. 397. Center of Pressure of a Triangle with the Vertex in the Surface. 398. Center of Pressure of a Triangle with a Side in the Surface. 399. Center of Pressure of a Rectangle wholly immersed. - 400. Equilibrium of Fluids of different Density in the same Vessel. 401. Equilibrium of Fluids of different Density in a bent Tube. 402. Examples. CHAPTER II. SPECIFIC GRAVITY...................,.. A 403. Definition of Specific Gravity. 404. Specific Gravity of a Body more Dense than Water. 405. Specific Gravity of a Body less Dense than Water. 406. Specific Gravity of a Liquid. 407. Weights of the Constituents of a Mechanical Mixture. 408. Hydrometer. 10. Of Constant Weight. 20. Of Constant Volume. 409. Nicholson's Hydrometer. 410. Examples, CHAPTER III. CO3IPRESSIBLE OR AERIFORM FLUIDS.................. 272;1. Tension of t Compressible Fluid found. 412. The Unit of Pressure. 413. Tension inversely as the Volume. 414. Tension directly as the Density. CONTENTS. Xv Itrt. Pa4e 415. Effect of Heat on Volume and Tension. 415. Relation of Density to Temperature and Pressure. 416. Density of the Atmosphere in Terms of the Height. 417. Barometrical Measurement of Heights. 418. Examples. HYDRODYNASIICS. 419. Velocity of a Fluid in a Tube of variable Diameter........................ 420. Velocity of a Fluid from a small Orifice in the Bottom of a Vessel. 421. Horizontal Range of a Spouting Fluid. 422. Quantity of Discharge from a small Orifice. 423. Time required for a Vessel to empty itself. 424. Vena Contracta-Coefficient of Efflux. 425. Discharge from a Rectangular Aperture. 426. Discharge from a Triangular Aperture. 427. Velocity of Efflux of an Elastic Fluid. 428. Motion of Fluids in Long Pipes. 429. General Method of determining the Discharge from small Orifices. 430. General Method of determining the Time for a Vessel to empty itself. 431. General Method of determining the Discharge from Orifices of any Form. 432. Examples. HYDROSTATIC AND HYDRAULIC INSTRUMENTS. 433. Mariotte's Flask..................................................... 434. Bramah's Press. 435. Hydrostatic Bellows. 436. Diving Bell. 437. Sea Gage. 438. Siphon. 439. Common Pump. 440. Common Pump, Condlitions of Failure. 441. Air Pump. 442. Condenser. 443. Clepsydra. ELEMENTARY 3IECHANIC S IN TROD UC T ION. 1. MIEC:HANICS is the Science which treats of the laws of Equilibrium and MI}otion. It is subdivided into Statics, Dynamics, Hydrostatics, and Hydrodynamics. Statics treats of the necessary relations in the intensities and directions of forces, in order to produce equilibrium of solid bodies. Dynamics treats of the effects of forces on solid bodies when notion is produced. Hydrostatics investigates the conditions of equilibrium ill fluid bodies. Hydrodynamics investigates the effects of forces on fluids:,vhen motion results.* 2. Force is that which produces or tends to produce motion or change of motion. The consideration of the nature of force does not belong to the present subject. Mechanics is concerned only with the effects of force as exhibited in the production of motion or rest. 3. The efect of a force depends on, 1st, its Intensity; 2d, its Direction; and, 3d, its Point of Application. The Intensity of a force may be measured, statically, by the pressure it will produce, or by the weight which will counterpoise it; dynamically, by the quantity of motion it will produce. 3y assuming for a unit of force that force which is counter poised by a known weight, the intensity or magnitude of any other force will be expressed by the numerical ratio which its counterpoise will bear to the counterpoise of the unit of force. * Theoretic Statics and Dynamics are those branches of theoretic Mechanics which cleat of the effects of forces applied to material points or particles regarded as without w-eihght or magnitude. Static and Dynamic Somatology would then embrace the appliration of theoretic Statics and Dynamics to bodies of definite form and magnitude, both solid and fluid. A 2 INT RODUCTIO N. In the same manner, by fixing on any line to represent the unit of force, any other force will be represented bT the line which bears to the linear unit the same ratio which the force in question bears to the unit of force. The Direction of a force is the line which a material point, acted on by that force, would describe were it perfectly free. The point of application of a force is that point in its line of direction on which the force acts. 4. As the magnitudes, directions, and points of lines are all determinable by the principles of Analytical Geonetry, so fbrces, of which lines are the appropriate representatives, come under the dominion of the same principles. The application of these principles to the determination of the laws of equilibrium and motion, considered as the effects of forces. constitutes analytical mechanics. 5. Forces may act on a point either by pushing or pulling it. As these are readily convertible, the one into the other, without affecting the intensity, direction, or point of application of the forces themselves, all forces will be regarded as pulling unless otherwise expressly stated. For convenience, we shall call those forces which have a common point of application concurering forces, and those which act along the same line toward the same parts, conspiring forces. 6. A body is an assemblage of material points. The material points, or elementary particles, are connected together in various ways, according to the nature of the body. A body is said to be rigid when the relative position of its particles remains unchanged by the action of forces upon it. In flexible and elastic bodies the relative positions of the particles change by the action of forces. S T A T I G S. CHAPTER I. [Y TIIE COMPOSITION AND EQUILIBRIUM OF CONCURRING FORCES.'7. PROP. Two equal forces applied to the same point in exactly opposite directions are in equilibrium. For no reason can be assigned why motion should take place in the direction of one force which will not equally ap ply to the other. 8. PRoP. Two concurrinog forces forming an angle with eat other can not be in equilibrium. If possible, let the two forces P and Q acting at A. a- d making, by their directions, the angle PAQ, be in equilibrium. Apply to the point A the force P1 equal and - other forces be applied, each equal to it, but opposite to each other, since B, by hypothesis, is invariably connected with A, one of these forces (Art. 7) will be in equilibrium with P. There will remain, therefore, a single effective force equal to P and applied at A. In thle same manner it may be transferred to any other point in its direction. 14. Pop.: The resultant of several concurring forces in one plane lies in the same plane. For if we suppose the resultant to lie out of the plane of the forces. on one side, we may always conceive a line symmetrically situated on the other side of the plane; and since no reason can be assigned why if should be in one of these lines rather than in the other, it can be in neither of them, unless we admit the absurd consequence, that it is in both; in other words, that a system of forces has two resultants. 15. Prom,. The resultant of twio eqzal concurring forces is in CONCURRIN G FORCES. 5 the direction of a line bisecting the angle formed by the comp7 nents. For no reason can be assigned why it should tend to one side rather than the other of this line. 16. CoR. When the forces are unequal, it is obvious that the direction of the resultant will make a less angle with the larger force than with the smaller, and the greater the disparity in the forces, the smaller will be this angle. 17. PRop. If all the forces of a'system, while their directions are preserved, are increased or diminished in any ratio, their resultant, without changing its direction, will be varied in the same ratio; and if the components were previously in eqzeilibrium, they will remain so in whatever ratio their intensities be varied. Let P1, P,, P3..... be any concurring forces whatever. Since (Cor., Art. 9) we can replace them by their resultant R, if we double or treble each of the components, this will only double or treble the resultant R, without changing its direction, because, in so doing, we only add to the system one or more systems precisely equal to the-first. In like manner, if we reduce the original components to one half or one third of their former intensities, the resultant will still preserve its direction, but will become one half or one third as large as before; since to double or treble these is only to double or treble their resultant, and thus reproduce the original system. If the forces are in equilibrium, by varying the magnitudes of all in the same ratio, we only add to or suppress from the system other systems already in equilibrium. 18. PROP. Three equal concurring forces, inclined at angles of 120~, will be in equilibrium. Since each force is inclined in the same P angle to the directions of the other two, any reason that can be assigned why ei- ther one will prevail, will apply with equal force to show that each of the others wvill prvn;l. P3 i-6.6sS T A T I C S. 19. Proli. Two equal concurring forces, inclined at an angle of 1200, have for their resultant a fojrce which will be represented in magnitude and direction by the diagonal of a rhombus constructed on the lines'representing the components. Let AD and AB represent the two forces P1 and P,, mak4P in0g the angle BAD-=120. To the;13X\ point A apply a force P=Pl or Po, and making, with P1 and P,, angles of x/E\ 120~. By Art. 18, these three forces A)<-. X w ill be in equilibrium. Now since P1 and Po equilibrate P, their resultant'3B must also. Therefore produce FA to C, making AC=AF; and since AC=R equilibrates P3, it will be the resultant of P1 and P2. Join DC and BC, and ABCD will be a rhombus, of which AC is the diagonal. For since CAD=-600, and AC-AD, the triangle CAD is isosceles; and since each angle is equal to 600~, it is equilateral..-. DC=AD-AB. Inthe same manner it may be shown that CB=AD, and hence the figure is a rhombus. Con. Denoting the angle BAC by a, we have AE-AB cos. a-P cos. a, and the resultant AC=R=-2P cos. a. 20. Pnop. If three forces are in equilibrium, each will be equal and opposite in direction to the resultant of the other two. For, replacing either two by their resultant, we shall have two forces in equilibrium. If they be not opposite in direction, by Art. 8, they can not be in equilibrium; and if they be not equal, by Art. 11, motion will ensue. PARALLELOGRAM OF FORCES. 21. PROP. If two forces be represenlted in magnitude and di. r'ection by the two adjacent sides of a parallelogram, the diago Ra1 will represent their' resultant in magnitude and direction. First. The direction of the diagonal is that of the resultant 1~. When the forces are equal, this is obvious from Art. 15. 2~. Let us assume, for the present, that it is also true for the trwo systems of unequal forces P and Q, and P and R; then it will be true for the forces P and Q+R. Let P and Q act CONCURRING FORCES. 7 at A in the directions AD and AB respectively, and be repre. sented in magnitude by these lines. A o. B > Ca Suppose the force R to act at B p_ (Art. 13), a point in its direction, 1? and to be represented in magnitude by BC. Complete the parallelo- D 11 gram ADFC, draw BE parallel to AD, and the diagonals AE, AF, and BF. Then P at A, in AD, and Q at A, in AB, have by hypothesis, their resultant in the direction AE, and (Art. 13) may be supposed to act at E. Replacing this resultant by its components acting in their original directions, we have a force P acting at E in the direction BE, and a force Q actin at E in the direction EF. Transfer P to B and Q to F, wit1hout changing their directions. Then P at B, in BE, and It at B, in B(C, will also, by hypothesis, have their resultant ill the direction BF, which may be supposed to act at F. We now have all the forces acting at F, and this without disturbing their effect upon the point A, supposed to be invariably 2onnected with F. Hence, if the assumption be correct, F is -. point in the direction of the resultant of the forces P and Q I-R. But the assumption is correct when Q and R are each equal to P (Art. 15). Therefore, the proposition for the direcf/on of the resultant is true for P and 2P. Again, making Q -=2P and R=P, it is true for P and 3P, and so for P and nP. Also, putting nP for P, Q=P, and R=P, it will be true for nP and 2P, and so on for mP and nP (m and n being positive integers), or for all commensurable forces. 3~. When the forces are incommensurable. Let AB and AC represent the forces. Complete A B the parallelogram, and draw the diagonal AD. AD will represent the direction of the resultant. If not, let some other line, as AE, be its direc- B tion. Divide'A.B into a number of G equal parts less than DE, and on AC c E take as many of these parts as possible. Since AC and AB are incommensurable, there will be a remainder GC less ih.an I)E. Draw GF parallel to AB, and join AF. AB and AGd ST AT IC S. representing two commensurable forces, AF will represent thlt direction of their resultant. But (Art. 16) the resultant of AB and AC will make a less angle with AC than the resultant of' AB and AG does..'. AE can not be the direction of the resultant of AB and AC; and similarly, it can be shown that no other direction than AD can be that of the resultant. Second. The diagonal will represent the magnitude of the D resultant. Let P, Q, and R.E/ ~_ ~ be three forces in equilibrium, C i/and AD, AB, AF represent AA R rI' their magnitudes respectiveB Ivy. Complete the parallelograms BD and DF, and draw the diagonals AC and AE. AC being the direction of the resultant of P and Q, must, by Art20, be in the samne straight line with AF. and is therefore parallel to DE. In the same manner, AE being the direction of the resultant of P and R, will be in the direction of BA pro duced, and therefore parallel to CD. Hence CAED is a parallelogram, and CA is equal to DE, which is equal to AF, by construction. But, by Art. 20, AF represents the magnitude of the resultant of I) and Q. Hence CA, the diagonal of the parallelogram constructed on the lines AD and AB, represents the resultant of P and Q, in magnitude as well as direction. TRIANGLE OF FORCES. "22. PRop. If three concurring forces arie in equilibrium, ana a triangle be formed by lines dirawn in their directions, the sides of the tritanle, taken in order, will represent the fBrces. Conversely, if the forces can be represented by the sides of a t itri'gle, taken in order, they will be in equilibrium. Let the forces P, Q, and R be in equilibrium, and be repre B sented by AB, AD, and AE respectJiely.:p Produce EA, draw BC parallel to AD,', R Ac and complete the parallelogram. AC' will represent the resultant of P aid Q (Art. 21); and since R equilibrates PI D and Q, it must be equal and o)pos;::e to their resu.tant..'. EA=AC. And since BC is equal sold CONCU R I NG FO R C ES. also parallel to AD, that is, has the same direction, tlie three sides of the triangle ABC taken in order represent the three forces.P, Q, and R. Conversely. If the three sides AB, BC, and CA of the triangle ABC, taken in order, represent the direction and magnitude of the three forces P, Q, and R, they will be in equilibrium. Draw from A a line parallel to BC, and from C a line parallel to AB, meeting the former in D. The resultant of the two forces P and Q, represented by AB and BC, or AB and AD, is equal to, and in the direction of the diagonal AC; that is, equal and opposite in direction to the force R. Hence P, Q, and R are in equilibrium. 2.. 23. Con. Hence a given force may be resolved into two component forces, acting in given directions. Also into two others, one of which is given in magnitude and direction. 1~. Let AB be the given force, and /c AC, AD the given directions; that is, making known angles with AB. From. B draw BF, and BE parallel to AD and' AC. Then AF and AE are the two A B components (Art. 21) acting in the given directions AC and AD. D 2~. To resolve AB into two others, one of which shall be in the direction AC, and be equal to AF. Draw FB, and complete the parallelogram. AE will be the other component. When the directions of the components are arbitrary, their valuation will be most easily effected by assuming these directions at right angles. If the angle CAD be right, then AE =AB cos. BAE, and AF=AB cos. BAF. POLYGON OF FORCES. 24. PROP. If any number of concurring forces be represented zn magnitude and direction by the sides of a polygon, taken iin order, they will be in equilibrium. Let the sides of the polygon ABCDEA represent the mag. nitudes and have the directions respectively of the forces P P2, P3, P4, and P5. These forces will be in equilibrium. STATIC S. Draw the diagonals AC and AD. By Art. 22, AC, the third P'5 side of the triangle ABC, repre~'- sents a force equivalent to the two forces AB and BC. AC is therefore the resultant of P1 and A__-__ _ =:C P,, and may be substituted for them. In the same manner, AD is equivalent to the forces AC D and CD, or to P1, Po, and P3; P \ / land AE to the forces AD an:d DE, or to tihe forces P,, P9, 39 BE and P4. That is, AE is the iesultant of the four forces P, P1, P3, and P4, and acts in the direction AE. But P5 is represented in magnitude by EA, and:acts in the direction EA. Hence P5 is equal and opposite to thle resultant of the other four, or the forces are in equilibrium. CauL. The proposition is true of forces which do not lie all irl one plane. For the proof is independent of this supposition. 25. ScHoLrIUI. When three forces are in equilibrium, any three lines, taken parallel to their directions, will form a triangle, the sides of which respectively will represent the relative magnitudes of the forces; but when there are four or more forces this will not hold, since the relation which subsists between the sides and angles of triangles does not obtain in poly gons of tore than three sides. 26. ProP. To find, graphically, the resultant of any nzmber offorces, acting at diFerent points in the same plane. Let the forces P,, P2, P3 have their points of application at A, B, and C. Producing the directions of \c the forces P1 and P2 till they meet in D, Hj construct the parallelogram AB on the G lines representing them, and the diagconal:D T1h will represent their resultant R,. Pro/ n~ ducing IR, until it meets the direction of A/ /-oP2 P3 in F, and constructing the paralleloA \:e E - gram on the lines irepresenting R, and P3, RP I tlhe diagonal will represent their resultant CONCURRING FORCES., R2, or the resultant of P1, P2, and P3 - By the same process we may find the resultant of any number of forces. If any two are parallel, we must first compound one of these with a third, whose direction is inclined to it. If all are parallel, tne case will belong to parallel forces. PARALLELOPIPED OF FORCES. 27. PROP. If three concurring forces lying in diff'e,'ent planes be represented in magnitude and direction by the three edges of a parallelopiped, then the diagonal will represent their resultant in magnitude and direction; and conversely, if the diagonal represents a force, it is equivalent to three forces represented by the edges of the parallelopiped. Let the three edges AB, AC, AD of the parallelopiped represent the three forces. Then AE, B the diagonal of the face ACED, represents the resultant of the forces AD and AC. Compounding this with the third force. represented ly AB, we have AF, the C diagonal of the parallelogram AEFB, for the resultant of AEL and AB, or of the forces AC, AD, AB. Reciprocally, the force AF is equivalent to the components AB, AE, or to the components AB, AC, and AD. i28. PROP. If three forces are in equilibrium, they are proylrtional each to the sine of the angle made by the directions of the other two. By Art. 22 and figure, P: Q: R=AB BC: CA, and by trig., =sin. BCA': sin. CAB: sin. ABC; =sin. CAD: sin. CAB: sin. ABC; or because sin. A= sin. (1800-A), =sin. DAE: sin. EAB: sin. BAD; =sin. QR: sin. PR: sin. PQ. v 29. PROP. If P and Q be two concurring forces, 0 the angle made by their directions, and R their resultant, then R'=P2+ Q2'+PQ cos. 0. [t2 -' S T A T IT'C S.; By trig. we have, from fg., Art. 22, AC2=AB2+BC2-2AB.BC cos. ABC, and cos. ABC —cos. BAD=-cos. 0. R%=P2~Q2+2PQ cos. 0. 30. DEF. The moment of a force about any point is the prod. uct of the force into the perpendicular let fall from that point on the direction of the force. The point is called the origin of moments. The moment of a force measures the tendency of the force to produce rotatory motion about a fixed point.. 31. PROP. The moment of the resultant of two forces equals the algebraic sum of the moments of the components. 1~. When the origin of moments falls without the angle mnade by the forces. A. RE F G Let AB, AC represent the two M forces P and Q, and AD the diago nal of the parallelogram constructed \ii 11H Take any point, 0. for the origin o1 moments; join OA and draw AG per \\\ "-pendicular to OA; draw 01, Omn, On respectively perpendicular to AB, AC, D AD, and BE, CF, DG perpendicular to AG; also CH parallel to AG. The triangles OIA, OmA, OnA are respectively similar to the triangles AEB, AFC, AGD. AB.O0 Whence AE: AB =01: OA, or AE= OA AC.Om AF: AC- O2z: OA, or AF= OA AD.On AG: AD=On:OA, or AG= OA But the triangles ABE and CDH neing equal, AE=CH —FG.. AE+AF=AG, or AB.OI-AC.Om=AD.On, or designating 01, Omn, and On by p, q, and r respectively, P.p+Q.q=R.r. 2~. When the origin of moments is taken within the angle CONCUR RING FORCES. 13 In -this case the moments of the forces tend to produce rota. tion in opposite directions. As- E A G r suming one direction as positive, 7 Bib to distinguish them we must regard the other as negative. Let the P positivTe direction be that of the 0 \:i c hands of a watch; then Q.Om will be positive, and P.OI negative. The proof is the same in this case, except that AG-AF-AE, and AC.Omr-AB.OI=AD.On, or Q.q-P.p=R.r.'/ 32. Con. 1. The moment of the resultant of any number of concurring forces in the same plane, is equal to the algebraic sum of the moments of the components. By compounding the resultant R with a third force, we should obtain a like result. In the same manner, the proposition may be extended to any number of forces. 33. COR. 2. If the origin of moments be a fixed point, and taken in the direction of the resultant, On will become zero. and P.OI=Q.Om, or P.p-Q.q; that is, while the fixed point O, by its resistance, counteracts the resultant force, the forces P and Q will be in equilibrium about that point, since their moments, tending to cause rotation in opposite directions, are equal. 34. COR. 3. If several forces are in equilibrium, the resultant force R is zero, and the moment of the resultant R.r-=0. Hence the moments in one direction balance those in the opposite direction, and there is no tendency to motion either of translation or rotation. 35. EXAMPLES. 1. When the component forces are P and Q, and the angle made by their directions 0; what is the magnitude of the resultant R when 0=-0 and 0=Tr i Ans. (P+Q) and (P - -Q). 14 STATICS. 2. Show that R is greatest when 0=0, least when 0-: r, and intermediate for intermediate values of 0. l 3. When P=Q and 0= 600, find R. Ancs. R=P 3. V4. When P=Q and 0=135~, find R. Ans. R=P01f2-/-V2. i/5. When the three concurring forces 3m, 4mn, and 5mn are in equilibrium, find the angle 38m,4m. Ans. 90~. 6. If P=Q and 0=120~, find R. Ans. R=P. - 7. If P=-, Q 11, and 0-30~, find the magnitude of R, an 3 of the angles P,R and Q,R. Ans. R=16.47, P,R=190~.30, Q,R=10o.30'. /'8. Apply the proof of the polygon of forces to the case ot five equal forces represented by the sides of a regular penta cron taken in order. 9-'3. A cord is tied round a pin at the fixed point A, and its two ends are drawn in different directions by the forces P and Q. Find the angle P,-=0, when the pressure on the pin is P+Q Plr. 2 2PQ-3(P' +Q9 Ans. Cos. 0= sPQ' 10. A cord, whose length is 21, is tied at the po'nts A and B mn the same horizontal line, whose distance is 2a; a smooth ring upon the cord sustains a weight w: find the force T of tension in the cord. Ans. T=- 11. Given the four concurring forces P=-1, P2=2, P3=3, P4 =4, and the angles P,P 3 =90~, P,P4=900, and P1,P2 60f. Find the magnitude of the resultant R, and its direction P,,R. Ans. R=6.889 and P,,R=102'.16' CHAPTER It. PARALLEL FORCES. / HAVING considered the subject of forces which have a common point of application, or which are reducible to this condition, we next proceed to consider forces which act on different points, connected together in some invariable manner, as in rigid bodies' and whose directions are parallel. V36. PROP. The resultant of two parallel forces, acting at the extremities of a rigid rig/it line, is parallel to the components, equal to their algebraic sum, and divides the right line, or the right line produced, into parts reciprocally proportional to the forces. 10. When the forces act in the same direction. Let the parallel forces P and P, act at A and B. At these points: apply the equal and opposite forces P' and P',; PI< / >'P these will not disturb the system. P and P' at A will have a resultant P",,<:__ and P, and P', at B, a resultant P",. The direcItions of P" and P", will p, meet in some point D, at P which we may suppose them to act. Replacing P" by P and P', and P", by P, and P', we now have, acting at D, the four forces P, P', P., and P',, of which P' and P', are equal and opposite, while P and P1 act in the direction DC, and have a resultant R=P+ P1 (Ar't. 10). To determine the point C in the line AB, where [R acts, since the sides of the triangle ACD have the directions respectively of the forces P, P', and their resultant P" (Art 22 xv we have, 16 ST ATIC S. P: P'=DC:CA, and similartly fiom triangle BCD, P'' =BC.: DC. Compounding these ratios, we have, since P=P',, P: P-=BC: CA, or P.AC==P1.BC. 2~. When the forces act in opposite directions. Let P and P, (P1>P) be two parallel forces, acting at A and B in opposite directions 1 ~f:-P i? and apply to these points, as /i/ before, the equal and opposite DP l 1)/gs,, forces P' and P'1. The resultant P" of P and P' will meet the resultant P"1 of P1 and P'1 in some point D By A / ~p: since (Art. 16) P"'1 makes a B3 C less angle with P1 than P" does with P. We shall then have, as before, the four forces P.' P, P', and P'1, acting at D, of which P' and P', are equal and opposite, while P and P, are opposite and unequal, and have a resultant R=-P1-P, acting at any point in its direction, as C in AB produced. To determine the point C, we have (Art. 22), from the triangles ACD and BCD, as in Case 10, P: P' =DC: CA and P': P =BC: CD; compounding P: P -BC:CA, or P.AC=P,.CB. 37. DEF. When AB is perpendicular to the direction of the forces, AC and BC are called the arms of the forces, and the products P.AC and P1.CB are the moments of the forces about the point C. / 38. Con. If to, the point C we apply a force R1, equal and opposite to R, the forces P, P,, and RI will be in equilibrium about that point. For the resultant R, passing through C, will be counteracted by R,, so there can be no motion of' translation. And if we draw through C a line perpendicular to the P A R ALLEL FORCES. 17 direction of the forces, calling the parts intercepted by P and P1 respectively p and px; from the similar triangles, thus formed, we should have BC: CA=p: p..'. P:P =p1:p, or P.p=P,p,. Hence the moments of P and P1, which measure the tendency to rotation in opposite directions, being equal, there can be no motion of rotation. VJ 39. PRoP. To determine the point of application of the resultant in terms of the components and distance between their points of application. 1~. When the forces act in the same direction. By Art. 36 and figure, we have P.AC=P 1.BC=P.(AB-AC), P, or AC.AB and P.AC=P,.BC=P(AB-BC), p or BC= —.AB. P+Pi 20. When the forces act in opposite directions, we have P.AC=Pl.BC=P.(AC-AB), or or AC=-'p.AB; and similarly, BC=p _p.AB. 40. Coa. 1. When the forces act in opposite directions, the resultant lies without the components and on the side of the larger. v 41. COR. 2. When the forces are equal and opposi!e, we have R=P -P=0o, or there will be no motion of translation. _ _ P Also, AC=p AB= -.AB= c. B ~~18~. STATICS. Hence an equilibrium can not be produced except by the ap4 plication of an infinitely small force at a point whose distance is infinite; that is, two equal and parallel forces, acting in opposite directions, can have no single resultant. v 42. DEF. Such forces are called a statical cozuple. Their effect is, tendency to rotatory motion only, and all tendency to rotatory motion can be referred to forces forming such couples. 143. Prop. To find the resultant of any number of parallel bforces acting at any points in the same plane. Y iz2~ P2 ll/ b A c o F -l{ G L X 10. Let the parallel forces P,, P2, P... P. all act in the same direction. Draw any two rectangular axes OX, OY, and let xi y1 be the co-ordinates of A, the point of application of P, x2 y " B, " " P x3 Y3 D, " P3 xI y? "' N, " " P." pn also, x y/ " " C, " i R, x2" y'" "6 6 E, " " R y " " 4 R, " " " R.', Draw Ac and Cb parallel to OX. The triangles ACe and CBb are similar, and give AC: BC=Cc: Bb. By Art. 36, R,=P, +P2, PARALLEL FORCES. 19 and P,: P2=BC: AC=Bb: Cc, or P,.Cc=P2.Bb, Pr. (CH- AF) =-P. (BG-CH),: (P, +P2)CH-=P1AF+P2.BG; or R, 1.y'P1y1+P2.y2 01 ~ Rl~yl 3 y'=P,y, 2 ~ Compounding R, with P3, we shall find the second result. ant, R2 =R1 +P3=P1 +P2 +P3, and R2.EL=R.CH+P3.DK, R2.EL=PIAF+P2.BG+P,DK; or R2.y"==P1,y+P2Y2 +P3Y3 By continuing the same process, we should find, ultimately, R=P, +P2 +P3 +... P, (a), Ry=Py1+P2y2 +P3y3+.... P,y,, (c). By drawing lines from A, C, B, E, D, &c., parallel to OX. we may find, similarly, R=P:xz~+P2x2+P3x3+, &c....... +Pnx, (b). 20. When some of the forces act in opposite directions. To these the negative sign must be prefixed. If R,, the resultant of'y P1 and P2, as found in the' previous case, be com- 112 pounded with P3 acting 1P1 at D, in a direction opposite to that of P1 and P2,/ by joining CD and pro- / eC ducing it (Art. 40), in the direction of the greater force, say R1, we have..x R,, the second resultant. R2=R -P3, and E being its point of application, R1.EC.=P3.ED. Drawing Ce, Ed, parallel to OX, and meeting DK in d and LE produced in e, the triangles ECe, EDd are sim;far; and EC: DE=Ee: Dd..'. R1.Ee=P3.Dd; or R1, (CH-EL)=P3.(DK-EL), ST AT I S.; or (R' -P3).EL==RR I.CH -P3.DK, or R_2.EL-=P.AF+P2BG-P3.DK, or R 2.y" Py +Py2 -P3y, and so on for any other forces acting in directions opposite to P, and P2. 3~. When the point of application of either force lies in any other angle than- OX, to its co-ordinates must be given their appropriate sign. The ordinate y2 of B, the point of application of P2, will be negative. Draw Aa,. Cb parallel to OX. 3 1 P2 iA'a 0 H G P1.CA=P2.CB. By sim. triangles, P.Ca=P2.Bb; or P1 (AF-CH) = (CH +GB), or (P +P2).CH-=P1.AF-P9 BG, or: R.y'=PyI -P2y2.. The general formulae (a), (b), (c) will apply to all' cases, by giving the proper signs to the forces and to the co-ordinates of their points of application. These formulae are more concisely written by using the Greek letter z as the sign of summation, P to represent either force, and x, y its co-ordinates; thus, R.=z(P) (~, R..=- (Pzx) (be), R.=z(P.y) (c). When, therefore, the magnitudes of the forces and the co-or PARALLE L FORCE S. 21 dinates of their points of application are given, (a) will give the magnitude of the resultant, and (b) and (c) the co-ordinates x1, y of its point of application.'/44. DEF. The point whose co-ordinates are., y is called the center of parallel forces. Its position depends on the magnitude of the forces and the co-ordinates of their points of application, but is independent of their common direction; for, by turning the forces around their respective points of application, at the same time preserving their parallelism, it will be seen that the position of the point E is not thereby affected. J45. PROp. The nomnent of the resultant of any number of parallel forces, acting in the same plane, is equal to the algebrlaic utnz of the moments of the components. Since the origin of the co-ordinate axes and their direction are arbitrary (see fisure of Art. 43), suppose the axis OX to be drawn perpendicular to the direction of the forces, and the forces produced to:intersect it. Then Pix,, P2x2, &c., will represent the moments of P1, P2 respectively, and R.Z the moment of the resultant. / 46. DEF. The moment of a force with reference to a plane Is the product of the intensity of the force by the perpendicular let fall from the point of application upon the plane. Thus Px1 is the moment of the force P1, in reference to the plane passing through OY perpendicular to OX, and Ply, is the moment of the force P', with reference to the plane through OX perpendicular to OY. /47. PRO. To determine the conditions of equilibrium of ay1?utmnbe of parallel forces. Suppose the forces in the figure (Art. 43) all turned round their points of application so as to become perpendicular to the plane of YOX. They will no longer be in the same plane, but as their intensity is not thus changed, nor their points of appii~ cation, (a'), (b'), (c') will still hold. Then, 1~. We must obviously have R=0 in (a'), or P +P2 +P3 +, &c.-=o. 2~. If this value of R be substituted in (b'), we have. - P.Px 22 -' STATICS. Therefore x will be infinite unless I.Px=O. But when =x is in finite, we have a couple (Art. 41), and, consequently, there can be no equilibrium. Hence we must also have:.Px=O, or Plx +P2x2+P3x3+, &c.=O. And in the same manner it will appear that:.Py-=-O in case of an equilibrium. But..Px and s.Py are the moments of the forces in reference to two planes parallel to their directions. Hence the necessary conditions of equilibrium are, 10. The sum of theforces must be equal to zero. 2~. The sum of their moments, with reference to two planes parallel to their directions, must each be equal to zero. 48. COR. 1. If R=0, but y.Px and S.Py are not equal to zero, there will be no motion of translation, but simply a motion of rotation. For if R1 be the resultant of all the positive forces, and x, the abscissa of its point of application, R2 the resultant of all the negative forces, and -x its abscissa, then R,,-R2z >0; or, since R =R R-R2, R, —z)0. But R1 being finite, X,- - must also be finite. That is, the points of application of R1I and R2 are not the same, and a tendency to motion of rotation around the axis of Y exists. And 2.Py<0 gives, in like manner, a tendency to motion around the axis of x. 49. CoR. 2. If the equilibrium subsist, one force, as P1, must be equal and opposite to the resultant of all the others. 50. Con. 3. If the equilibrium subsist with one direction, it will subsist whatever be the common direction of the forces. 51. EXAMPLES. 1. Two parallel forces, acting in the same direction, have their magnitudes 5 and 13, and their points of application A P A RALLEL F O RCES. 23 and B 6 ifet distant. Find the magnitude of their resultant, and its point of application C. Ans. R=18, AC=4~, BC=1 2. 2. Find the resultant, and its point of application, when the forces in the last question act in opposite directions. Ans. R=8, AC=94, BC=34. 3. If two parallel forces, P and Q, act in the same direction at A and B, and make an angle. 0with AB, find the moment of each about the point of application of their resultant. P.Q Ans. QAB sin. 0. P+Q 4. If the weights 1, 2, 3, 4, and 5 lbs. act perpendicularly to a straight line at the respective distances of 1, 2, 3, 4, and 5 feet from one extremity, required their resultant and its point of application. Solution. From equations (a) and (b) we have R=:.P=1+-2+3+4+5=15 lbs. And, taking the extremity of the line for the origin of co-ordinates, R- =15.x= Z.Px=1 X 1 +2X2+3X3+4X4+5X5=55... =33 feet. 5. Let the weights 4, -7, 8, and -3 lbs. act perpendioularly to a straight line at the points A, B, C, and D, so that AB=5 feet, BC 4 feet, and CD=2 feet; find the resultant and its point of application E. Ans. R=2 lbs., AE=2 feet. 6. Let three forces which, if concurring, would be in: equilibrium, act each in the side of the triangle which represents them in magnitude and direction. Show that they are equiv~ alent to a statical couple. CHAPTER III. THEORY OF COUPLES. / 52. DEF. A statical couple consists of two equal and parallel forces acting in opposite directions at different points of a body. - 53. DE). The arm of a couple is the perpendicular distance between the directions of the forces. V 54. DEF. The moment of a couple is the product of the arm by one of the forces. v 55. PROP. A couple may be turned round in any manner in its own plane without altering its statical efect. Let PABP, be the original couple. Suppose the arm AB P2 turned around any if ape point in it to the position ab. Apply the equal and opposite forces P3 and P4 per-. 4 pendicularly to ab at /B:~6 3 \.. a, and similarly P5 and P6 at b, and let 6~-~ each be equal to P, or P2. These forces, being in equilibriumn, will not disturb the P3 system. Thetwoequal forces P, at A, and P at a, may be regarded as acting at their point of intersection E, and will have a resultant in the direc-, tion CE bisecting, the angle P,EP. Similarly, the resultant of P2 and P6 at D will be an equal force in the direction CD. These forces,- being equal and opposite, may be removed; that is, we may remov'e fiom the system the forces P,, P2, P4, and P., and we have remaining the forces P3 and P5 at a COUPLES. 25 and b, forming the couple P3abP5, which is the same as if we had turned the original couple round the point C until its arm came to the position ab. / 56. PROP. A couple may be removed to any position in its own plane, the parallelism of its arm being preserved, without changing its statical effect. Let the arm AB of the couple P,ABP2 be removed in its own plane to the parallel po- AP2 sition ab, and let the forces P3, P4, P5, P6, each equal to the original forces, be applied perpendicularly to ab, at the extremities a and b, in oppo- C4 site pairs. Join Ab and aB. These b c lines will bisect each other in C. P1 at A and Ps atb V will have a resultant 2P1 at I' C, parallel to the original direction. Similarly, P2 at B and P4 at a have a resultant =2P, at C, opposite to the former; these will consequently balance each other, and may therefore be removed, or the forces P,, Pa, P, may be removed, and we have remaining the couple P3abP6, equivalent to the original couple, removed parallel to itself in its own plane. J 57. PROP. A couple may be removed into any other plane, par llel to the original one, without altering its statical effect, the pa-rallelism of its arm being preserved. Let the arm AB of the original couple be transferred from its own plane MN parallel to itself to ab in the parallel plane'RS.- Let forces be applied at a and b, as in the preceding proposition, each being equal to P1 or P2. Join Ab and aB. These lines will bisect each other in C. The forces P1 and P6 will have a resultant =2P, at C, and P, and P4 a result. ant =2P, at C. These resultants will be equal and opposite and may therefore be removed without disturbing the statica. effect of the system. WTe have then remaining the couple 6 S T STA T TICS -P2. P 3abP5, equivalent to the original couple, and transferred to the plane RS. 58. PROP. All statical couples are equivalent to each other uwhose planes are parallel and moments equal. Let PABP2 and QjlDEQ2 be any two couples whose Q2 planes are parallel and moments equal. If they )D are not in the same plane, PiQ i by the foregoing propoQ1 sitions let the latter be (nQ4 A transferred to the plane C of the former, and moved in that plane until the extremities D and A of their arms coincide, and DE take the positiqh AC. Apply at C, perpendicularly to AC, two equal and opposite forces Q3 and Q4, each equal to Q1. Resolve P1 into two forces P' and Q., so that P,=P'+Q,, or P,-Q~=P'. Then, by hypothesis, C O U P L E S. 27 P,.AB =Q,.DE=Q1.AC=Q,.AB.+Q1,.BC=Q. AB3+Q3.BC( and (P,-Q,)AB=P'.AB =Q3.BC, or the resultant of P' and Q3 (Art. 36) passes through B. It is also equal and opposite to P2. Hence the forces P', Q3, and P2 may be removed, and there will remain the couple Q,ACQ,, equivalent to PABP,. 59. PROP. Any statical couple may be changed into another, which shall be equivalent and have an arm of given length. Let Pp1 be the moment of any couple, of which P1 is one of the forces and p I its arm, and let p be the given arm. Find a fourth proportional P'1 to p, p,, and P,; or take p: p P1:P', which gives P',p- P,p, Hence, by Art. 58, their moments being equal, and in the same plane, the couples are equivalent. /60. DEF. The axis of a couple is a line perpendicular to the plane of the couple. If the length of the axis be taken proportional to the moment of the. couple, and drawn above its plane when the couple is positive or tends to produce rotatory motion in the direction of the hands of a watch, and below its plane, when negative or tends to produce motion in the opposite direction, then the axis will completely represent the couple in position, intensity, and sign. O 61. COR. By considering the previous propositions, it will be obvious that the axis of a couple, as thus defined and limited, may be removed parallel to itself, to any position within the body acted on by the couple. 1'/ 62. DEF. The resultant of two or more couples is one which will produce the same effect singly as the couples themselves jointly. 63. PROP. The moment of the?resultant of two or more couples in the same or parallel planes, equals the algebraic sum of the moments of the component couples. Let P1, P2, P3, &c., be the forces; p p p3, ccP3, their arms respectively. The couples may all be removed into one 28 STA 0ICS. plane (Art. 57), turned round in that plane (Art. 55), moved in it (Art. 56), and their arms changed (Art. 59) to a common arm, while their moments re. A main the same. Let p be the comrnmon arm AB; P'1 P'2, P'3, &c., the' forces; so that P'-p=Pp1,, P' 2p= -r P~p2, P'3p=P3p3 &c. Now the forces P'1, P'29 P'3 at A are equivay lent to a force P', +P' +P'3 +, &c., —+- P- + P -3 +, &C. And the forces at B are equal to the same sum. We have then for the moment of the resultant couple, (P'1 + P' +P'3 +, &c.) AB-=(P', +P'2 + P' 3 +), =Pl p P2ps p3p3+, &c..1i If either of the original couples, as P3p, tend to proi duce motion in the opposite:B pa A direction, its sign will be negD/' ative. The sum of the forces will be P' +P'2-P' -I-, &c., rs &?, and the moment of the resultant couple (P'1 +P'2 -P'3+, &c.)p=P p ~+P2P2,p,~ &c. C4. Con. The axis of the resultant couple will be equal to the algebraic sum of the axes of the component couples. ScHOL. The composition of couples in the same or parallel planes, by means of their axes, is therefore analogous to the composition of conspiring forces.,/ 65. PRop. To fi1nd te resultanzt of two couples in differend planes inclined to each other. Let each couple be turned round and moved in its own plane until its arm coincide with the common intersection of their planes, qnd let the arms be mnade coincident and of the c O UP L.E S.o 2ll same length, the moments R of the couples remaining the same. Let AB be this p common arm, and PABP, QABQ the couples. Complete the parallelograms on the lines representing B P and Q, and the diagonals will represent their resultants R, which may be substituted for them. Q /' Hence we have the resultant couple RABR equiva- R lent to the component couples PABP and QABQ.; 66. Con. The diagonal of the parallelogram, constructed oi, the lines representing the axes of the component couples, will represent the axis of a couple equivalent to them. Let OK and OL be the axes of the couples, and 0 the K angle made by the planes of the couples. Since the axes are perpendicular to these 0 planes, the angle made by the axes will be 0. We have, therefore, 0=PAQ=LOK. L By the triangle of forces, R2=P2+Q2+2PQ cos. 0..'. R.AB=AB VP2+Q2+2PQ cos. 0, P= P2.AB2+Q2.AB2+2P.ABX Q.AB cos. 0 (since OK=P.AB, &c.), = V0K2+OL'+20K.OL cos. 0, =OM. And OK, OL being respectively perpendicular to the planes of the component couples, OM will be perpendicular to the resultant couple. / 67. ScHOL. The above is analogous to the parallelogram of forces, and may be called the parallelogram of couples. 30 S T.A T I C S. If L and MI represent the axes or moments of the component couples, and G the axis or moment of their resultant, then G2=L 2+M2+2LM cos. 0. If L, M, N were the axes of three component couples, it might be shown that G, the axis of the resultant couple, would be represented in magnitude by the diagonal of the parallelopiped formed upon them. If the planes of the three couples are at right angles each to the other two, or if L, M, and N are at right angles each to the planes passing through the other two, the parallelopiped would be rectangular, and we should have, W=L" L +M2+N2. Y~ CHAPTER IV. ANALYTICAL STATICS IN TWO DIMENSIONS. IN this chapter the points of.application and the directions of the forces will be referred to co-ordinate axes at right angles to each other. 68. PROP. To find the magnitude and direction of the result. ant of any number of concurringforces in the same plane. Let P1, P2, P3, &c... P. be the n forces, and let the point 0 x at which they act be taken for the origin of co-ordinates. Let P1 make the angle a, with OX, and 0, with OY. P so a " " (f P 3 a 3 33 &cS., - &c., &c, Pn An, a n " f If OA represent the force P1, and the parallelogram OMAN be completed, OIM will be the component of P, in OX, and ON that in OY. And STATIC S. OM=X,==P1 cos. aos.= cos. Pursuing the same course with the forces P2, P3, &c., and calling the resolved parts of the forces respectively in OX, X2, X3, X4, X&c.... X,, and those in OY Y2, Y3, Y4, &c.... Y., wve shall obtain X=-P, cos. a,, Y-=P, cos. o, X2=P2 cos. a,, Y2=P, cos. 2 3, X.3=P3 cos. a3, Y3=-P3 cos. i3, &c., &c., &c., &c., Xn= P, cos. a,,, cY-P,, cos.,,. But the components in OX are equivalent to a single force, =X1,X2+X3+, &c.... X,,z.X, and those in OY to a single force, =Y +Y2+Y3+, &c... Y.=z.Y, the Greek letter L being used as the sign of summation. Hence we have Z.X=-E.P cos. a, (1) -.Y=I.P cos. j. (2) Now if R be the resultant required, and 0 the angle it makes with OX, the resolved parts of R in the axes must equal the resolved parts of the forces in the same directions..'. Rcos. 0=-Z.X, (3) R- sin. 0=-.Y. (4) Hence Tan. 0= (5) RR — (5) Also, R'. cos.1 0+R2. sin.? 0-=R= (z.X)~+(.Y)2; or R= /(z.X)2+(~.Y).2 (6) Equations (1) and (2) give the values of 2.X and B.Y, by which we obtain 0 from (5) and R from (6); or R may be obtained, numerically, more readily from the equation R cos. 0=E.X...X For R-= Z.X sec. 0. (7) cos. 0 69. ScHiOL. It is readily seen that the signs of the compo ANAL I TC AL ST AT ICS. 33 nents.along either axis are involved in the trigonomnetrical expressions of their values, (1) and (2). Assume the directions OX and OY positive, and OX', OY' negative. The components Ox1,,P 2 Oy, of P, are positive; their values, P1 cos. a1, P1 cos. A,, 1 P will also be positive, since the /,, X cosines of angles in the first x x quadrant are positive. The components of P2, which lies in the 3 angle X'OY, are Oxo, Oy, the former negative and the latter. positive. The value of Ox2= P2 cos. a2 is negative, since a,, reckoned from OX around to the left, will be in the second quadrant; and that of Oy, =P, cos. p2 positive, since /32, reckoned from'OY around to the right, will be in the fourth quadrant; or, if reckoned from OY around to the left, will be negative, and less than 90~. The components Ox3, Oy of P 3, in the angle X'OY', are both negative, and their values P3 cos. a3, P3 cos. 13 are negative, because the angles a3 and 13 are both in the third quadrant. The components Ox4,, Oy4 of P,- which lies in the angle Y'OX, are, the former positive and the latter negative. The Value of Ox4=P4 cos. a4 is positive, because a4 is in the fourth quadrant, or, if taken negatively, is less than 900; that of Oy4=P4 cos. [A is negative, because 34 is in the second quadrant. By the above mode of reckoning the angles, we always find a+P= 90~, or one the complement of the other. 70. PROP. Required the conditions of equilibrium of any nutmber of concuriing forces in the same plane. Since the forces are in equilibrium, their resultant must be equal to zero, or R-0. This gives, by (6), (B.X)2 + (Z1Y)) o But each term being a square, is essentially positive. The equation, therefore, can only be satisfied by making each term equal to zero at the same time. C. 34 S T A T I CS S. Hence.X-0, O.Y-=, (8) are the two necessary and sufficient equations of cqtuzlibrium of any number of concurring forces in the same plane; that is, the sums of the components of all the forces resolved in any two rectangular directions, must be separately equal to zero. v 71. PROP. Required the expressions for the resultant force and resultant couple of any number of forces acting at dif'erent points in the same plane. Let P,,P,, &c.... Pn be any number of forces in the same plane. Take the co-ordinate axes OX, OY in the plane of the forces, and let a,, a,, a3... a, be the angles the forces make respectively with OX, and /3,, /3, P2... P, the angles they make with OY. ~Y''.Let x,, y, be the co-ordinates OM,, Xl M, A, of A1, the point of application of the'-..~ ~ force P,, and x'2, ~,' _NT-/, A:[~ }cc~.. &e..... x,,, y~ be the co-ordinates of the points of application x, of the others respectivel. RIesolving P, in the directions of OX and OY, we shail have for the components XI=Pl cos. a,, Y,=P, cos. Ii at the point A 1. Now apply at 0, in OX, two equal and opposite forces, each equal to X,; this will not affect the system. Then,'instead of the single force X, at A,, we have X, at A,, or N, and -X1 at O, which together form a couple, and X, at 0. That is, for the force X, at Ai, we may substitute Xl at 0, and the couple X,.ON, =X,y,. In like manner, applying at O, in OY, two forces, opposite and each equal to Y,, we should have Y, at A, equivalent to Y, at 0, and the couple -Y.OM, = — Y.x,. ANALYTICA. L STATICS. 35 This last couple will be negative, since it tends to produce to. tation in the contrary direction to that of the other. By pursuing the same course with each' of the other forces we should obviously have acting at O, in OX, a sum of forces X +~X2 -_X3, &C.... X,,=.X, (o) and at 0, in OrY, 1~+Y tY3+, &c:.... -.SY, (10) and the couples - Xy1 -X2y2 +X3y 3.... X,,,, y,.Xy, (11) -Y Ix-Yx2 - Y 33 -Y,,x-=.-Yx. (12) We have now reduced the whole to a system of concurring forces and a system of couples. The couples being in the same plane, the moment of' the resultant'(Art. 63) will'be equal to the algebraic sum of the moments of the components, or the resultant axis'G will be equal to the algebraic sum of the component axes. Hence, taking the sum of (11) and (12). G= E.-Xy —.Yx=.(Xy-Yx). (13) Now if R be the resultant force acting at 0, and 0 the angle it makes with OX, we have, as before, R cos. 0=.X, R sin.' 0=-:.Y. Tan. 0 —;.X' (14) and R2- (:.X)2+ (.Y)2. (15)'? 7`2. SCIIOL. 1. The equations (13), (14), and (15) determine the magnitude of G and the magnitude and direction of R. o'0 construct these results, draw,hrougch the origin O the line OR,. making the angle 0 with OX, to - represent R. Then put G= B:.(Xy-Yz)=Rr, by which the moment of the resultant couple is changed into another, whose forces are each equal to R, and arm equal to r. Let this be moved and turned round until one of its forces acts at 0 in an 35'.. S TA TI CS. opposite d:rection to the resultant force; draw OA perpeudil ular to OR and equal to r, and AR' parallel to OR. Then R'AOR' represents the resultant couple. The two forces at O, being equal and opposite, may be removed, and we have the final resultant acting in AR', which makes the angle 0 with OX. 73. SCHOL. 2. To find the equation of this final resultant. It will be of the form y=ax+b. But a- tan. 0 — OA r G G and b=-OB o —c 0 -- -- cos. 0 cos. 0 cos. Therefore, by substitution, we have for the equation.Y G - y=. " — Z.X (16). 74. Po. i To determine the conditions of equilibrium of any nz.umber of forces acting at different points in the same plane. In order that there may be no motion of translation, we must have R=O, which gives, -as before, x. x=0, (17) and E.Y=O. (18) And, in order that there may be no motion of rotation, we must also have G=j(Xy-Yx)O-. (19) Equations (17), (18), (19) are the three necessary and suicient conditions of equilibrium.` 75. Col. When there is a fixed point in the system, if this point be taken for the origin, its resistance will destroy the effect of:the resultant force R, and the sole condition of equilibrium will then be G=O, or. z.(Xy —Yx) 0; or there must be no tendency to rotation around the fixed point. ANALY T I C A L S'rAT I C S. 37 EQUILIBRIUMI OF A POINT ON A PLANE CURVE.' 76. If a point be kept at rest on a plane curve by the action of any number of forces in the plane of the curve, the resultant must obviously be in the direction of the normal to the curve at that point, and equivalent to the pressure the curve sustains. For, if the resultant had any other direction, it might be resolved into two, one in the direction of the normal, and the other in the direction of the tangent to the curve; the former would be opposed by the reaction of the curve; the latter, being unopposed, would cause the point to move. Hence, 77. ProP. To determnize the conditions of equilibrium of as point, retained on a plane curve or line, by forces acting inz its plane. Let N be the normal force of reaction of the curve, and a the angle made by the normal with the axis of x. Also, let E.X, E.Y be the sums of the components of all the other forces resolved parallel to each axis respectively, and R their resultant. The resistance N may be considered a new force, which, together with the other forces, retains the point in equilibrium independently of the' curve. If, therefore, N be resolved in the direction of each axis, we have (8) N cos. a+1.XO-0, (a) N sin. a+Z.Y=0; (b) or, transposing, squaring, adding, and reducing, N= (Z.X)2+ (. Y)= R, or the reaction is equal to the resultant of all the other forces. Now let i be'the inclination R y to the axis of x, of the tangent to the curve through the point,, x then a=90~+i. _xR__.'. cos. a=cos. (90~-+i)= —sin. i, and sin.'a os. i. Substituting these values of cos. a and sin. a in (a) and (b), and dividing, we get ,8': STATICS. Tanl. i zy.Y or.X-4t-.Y tan. i_=O. (20 But the differential expression for the tangent of the arni dy which the tangent makes with the axis of x is. Therefore. substituting and reducing,;.X.d~cz+.Y.dy=0. (21) Whenever, the line on which the point is retained is right, (20) may be used; but if the line be a curve, (21) will, in general, be necessary, and-l must be deduced from the equation of the given curve. VIRTTUAL VELOCITIES. ~ 78. DEF. If any forces P1 and P2 act at the point A, and this point be displaced through an in-.' definitely small space Aa, and the perpendiculars a up1, ap2 be drawn from a on the directions of the forces, ipa 2 - -then Ap, and Ap- are called the viie,:al velocities of the forces P1 and P2; Apl, measured in t he direction of the force P1, is positive, and Ap2, measured in the dcirection of P2 produced, is neg'ative. a 79. The principle of virtual velocities is thus enunciated: If any number of forces be in equilibrium at one or m71nor points of a -igid body, then if this body receive an indefini.ei y small displacement, the algebraic sum of the products of each force into its virtual velocity is equal to zefro. 80. Pop. To prove the principle of virtual velocit-es for Conrzcurring forces in one plane.i Let A be the point at which the forces P,, P., PF, c. Pn act; al, aa, a3, &c.... a, the angles they make respectively with OX, and /l, 2,' 13.. ~ 3, the angles they make with OY. Let 0 be the angle which the direction of the dis. placement Aa makes with OX A N A L Y T I C A L S TA T 1 C 33 y Pi 0.. Let v1, v. v3, &c... v,, be the virtual velocities oI tle forces respec'ively. T1'hen v,-=Ap1 =Aa COS.p 1 AaAa. cos. (a, -0), =Aa(cos. a, cos. O+sin. a, sin. 0),.and Pv1=-P1.Aa(cos. a1 cos. O+sin. a, sin. 0), =Aa(cos. 0.P cos. a, +sin. 0.P, sin. a1 ) Tn like manner, for P, we should have Pov2 Aa(cos. O.P2 cos. a2 +sin. 0.P2 sin. ao), and so for the other forces; and taking their sum, we should get, remembering that sin. a=cos. 3,.Ptv-Plv~ +P V2 +P3v3+, &c.... PV,, =Aa[cos. 0 (P1 cos. a, +P2 cos. a2 +, &c......P, cos. a,,) + sin. 0 (P1 cos.3~ +P2cos.i32+,&c. P,, cos.o +, &c....., cos.,)]. But when there is an equilibrium at a point (8), P1 cos. a1 +P cos. a +P c c a3+,&c.... cos. a,,= —S.X:=O alnd P1 cos. 31 +P2 cos. 3P os. 3 +, &S c. Pn cos. A3,4=.Y-=. Hence S.Pv=O, or the principle is true when the forces allact at a point. 81. PRop. To prove the principle of virtual velocities fo7 fforces acting at diferent points in the same plane. The points are supposed to be invariably connected by rigid lines or rods without weight, which transmit the actions and reactions of the particles or points upon each other. 40 S T A:T I CS. Let A, Al, A3,&c.... An be the particles to which the forces P1, P2, P3, &c.... P, are applied, v, v,, v3... v, the virtual velocities of the forces respectively. Let raa be the action of the particle A1 upon the particle A! 7aa, " reaction of " A " Al r'aa " action of " Al' "' A3, ra3al' reaction of' A3 6 Al &c., &c., &c. Let va1a2, Va2a,, va, val3,3 Va, &c., &c., be the corresponding virtual velocities. Then valCa2=vaa, va a — v 3Ol, &c., &c., from the nature ol action and reaction. Also, ala=-ra, r-aa3 — a3l &c., &c. For, let Al RI.AiPl _A2.2 al a2 and A2 be the particles displaced to aI and a<. Draw the perpendiculars alp,, a'p2. Then, if the line aa2 is parallel to A1A2, it is obvious that Alp,=Ap2. But Apl is the virtual velocity of R1, and Alp2 of R., and they have opposite signs.'Vaa -a Vaal. If a1a2 is not parallel to A1A2, let them meet when pro--:AI P1 A2P --------------------— _ _('2 duced in some point C. Since the displacements are indefin. itelv small, the perpendiculars alp,, a2P2 coincide with circular arcs whose center is C, and Ca Cp=1, Ca2,=CDp. But Ap,=Cp -CA =Ca -CA,, and A2p =Cp -CA =Ca -CA2 =(Ca, +aa )-(CA, + A~A,), =Cal -CA1..'. Ap1 = A=p, orVaa -7Va 1 2 = 2 Let the sum of the products of all the forces PI, P2, &c into their virtual velocities, acting on the particle ANALYTICAL STATICS. 41 A, be A.(Pa1.va9), those on A2 be i.(Pa2.va2), those on A3 be.(Pa'3.va3), &c., &., those on A,, be Z.(Pa;.va). Since each particle is in equilibrium from the action of the external forces and the reactions of the others upon it, we have, by the last proposition, O=. (Pal.va,) + ra a2.va+a2 +ra a 3.va1a3+-, &c., 0-. (Pa2.va2) +ra2al.va2al + ra2(a.va.a3 +, &c., o=Z.(Pa3.va3) +raa.va3 aa.vaa2+, &., &c., &c., &c., O =.(Pa.Va) + a-,a1.Vaa +ra %a,.a.Va +, a &C. In taking the sum of the products for all the particles, tne products of the reactions into their virtual velocities will disappear, being in pairs, equal in magnitude with contrary signs; therefore we have z.(Pal.val)+z.(,.v).(,Paa)+ 3.PVa3)+, &c.... Z.(Pa.v, )-0: or, generally, when there is an equilibrium, z.(Pv) =O. 82. PROP. Conversely. If the sum of the products of the forces into their virtual velocities be equal to zero, or x.(P.V.) =0, then there will be an equilibrium. For if the forces are not in equilibrium, they will be equivalent either to a singleforce or a single couple (Art. 74). In the first case, let R be the single resultant force; then a force equal and opposite to R will reduce the system to equilibrium; let u be its virtual velocity for any displacement. Since, with this new force, there will be an equilibrium, we have, by the preceding proposition,;.(P.0) +R.u=o. But by hypothesis, z.(P.v)=O..'. Ru=0, which, being true foi all small displacements of the body, we must have R-0, om .4s 2 ASTATIC S. the body was in equilibrium from the action of the orioina. tbreces. In the second case, if the forces were equal to a resultanti couple, it would be balanced by an equal and opposite couple. Let the forces of this opposite couple be Q and Q', and their virtual velocities for any displacement be q and q' respective. ly; since- they will reduce the system to equilibrium, we have, by the preceding proposition, z.(P.v)+Qq+Q'q'=O. But z.(P.v)=O... Qq+Q'q'= O, for all displacements, which is impossible, unless Q and Q' each equal zero, since they are parallel forces, and act at different points. CHAPTER V THE CENTER OF GRAVITY. 83. DEF. EXPERIMENT shows that every particle of matter is subject to a force which attracts it in a direction perpendicular to a horizontal plane or the surface of stagnant water. In reality, the directions of the forces, acting on several particles, meet nearly in the center of the earth; but as this center is very distant, compared with the distance of any particles considered together, we may, without sensible error, regard their directions as parallel. This force is called g'ravity. 84. Experiment shows, also, that the intensity of gravity varies in different parts of the earth's surface; that it is least at the equator, and increases toward the poles in the ratio of the square of the sine of the latitude. It shows,' also, that in the same latitude the intensity varies at different points in the same vertical line; that it varies inversely as the square of the distance firom the center. But for any points in the same system, or any bodies nearly in the same place, this variation of intensity, as Well as difference of direction, may be neglected without error. IH-ence, / 85. DEF. A heavy body is an assemblage of material points. or particles, acted on by equal parallel forces in the direction of the vertical to the earth's surface. J 86). DEF. The weight of a body is the resultant of all the efforts which gravity exerts on its component particles. This resultant (Art. 43) is equal to their sum, and parallel to their common direction. ~/ 87. DEF. The mass or the quantity of matter of a body is the sum of all its component particles.; 88. COR. If W represent the weight of a body, M the mass, and g the ratio of the intensity of gravity at any place, to its 44 STATIC S. intensity at another place where it is assumed as unity, vwe shall have W=gM. (22) For the resultant of all the parallel actions of gravity on the particles of a body is equal to their sum, or the product of its intensity into the number. 89. DEF. The density of a body is the ratio of its mass to its volume or, if D represent the density, M the mass, and V the volume, M DaV; (23) in which D, M, and V represent the number of units of each kind. - 90. Con. Since W=gM, and M=VD, we have W=gVD. (24) 91. PROP. The masses of two bodies of the same density are in the direct ratio of their volumes. If AI, V, and D be the mass, volume, and density of one hody, and m, v, and d the same of the other, by (23), M=VD and mnz=vd;. M: m=VD: d; (25) or, since D=d, M: m=-V: v. 92. PRoP. The masses being equal, the densities are inversely as the volumes. Since M m=VD: vd; if M-m, VD=vd;,r D d=v: V. (26) / 93. PROP. The volumes being, equal, the masses are directly (s tihe densities. Making, in (25), V=v, we have M: mA=D d. (27) v 94. DEF. The center of grIavity of a body is the center of the parallel forces of gravity on each of its component particles. ~ 95. Con. Hence the determination of the center of gravity involves an immediate application of the doctrine of parallel forces,'and we need only refer to results already obtained for many important deductions respecting the center of gravity. CENTE R OF GRAVITY 45 1-D. Thile resultant of all the vertical efforts of gravity oii each of the elementary particles of a body passes through its center of gravity. (Arts. 44 and 94.) 2~. This resultant is parallel to' the forces (Art. 36); that is, it is vertical, and its magnitude is equal to the weight of the bodv. 3~. Whatever position we give to a body, this resultant will always pass through the center of gravity; since changing the position is equivalent to changing the direction of the forces, without changing their points of application or parallelism. 4~. A heavy body will be in equilibrium if its center of gravity be supported, whatever may be the situation of the body relative to the support, since, in this case, the resultant of the parallel forces of gravity will have a fixed point in its direction. 5~. When we wish to find the center of gravity of several bodies, we can suppose the mass of each concentrated at its center of gravity, since the weight of each is a force proportional'to its mass, and-passing vertically through its center of gravity. Hence wve have only to consider a system of heavv. points. 96. DEF. A body is said to be symmetrical with respect to a plane when the lines joining its particles, two and two, are' parallel, and bisected by the 7, plane. - Thus, let m, m' be two symmetrical particles, so that the line mm' may be bisected b in h by the plane aa'. Letting fall the perpendiculars ma, m'a', the equality of the triangles mab, m'a'b gives ma-=m'a'. Hence'the par- a mb ticles of a body,- symmetrical with respect to a plane, are situated, two and two, on opposite sides of the plane, and at equal distances froim it. 97. PRor. The center of gravity of every homogeneous body of uniform densityj,symmetrical with respect to a plane, is sitzuated in that plane. For any two particles, symmetrically placed, will be at the -46 - S T A T I C S. same distance from the plane, and their moments will be equa and have contrary signs. But all the particles, taken'two and two, are thus placed (Art. 96). Therefore the resultant of the system of forces will be in that plane, and, consequently, the center of gravity also. 98. DEF. A body is said to be symmetr'ical with respect to an azis when it is symmetrical with respect to two planes passing through that axis. 99. PROP. The center of gravity of a homogerneous body, symmetrical withi respect to an axis, is situated in that axis. By Art. 97, it must be in each plane passing through the axis, and therefore in their common intersection, or the axis itself. 100. COR. If a body is symmetrical with respect to two axes, its center of gravity will be at their intersection, since it must be in both axes. 101. DEF. This point is also called the center of figure. 102. PROP. To find the center of gravity of any number of heavy particles whose weights and positions are given. Let A, B, C, &c., be the particles, whose weights w,, w,, w3, &c., act at their respective centers of gravity, vertically downward, and therefore constitute a system of parallel for/ W2 c es.' wl and wL, have a resultD D1+ W> ant =-w,z+w, acltun -, at some v5i,+w+zi-:'3 ~ point a, such that w. Aa=w. Ba. The distance AB being t 4 i given, the distanceA a is deiiV3 termined by taking (Art. 39) w, -w,: w=AB: Aa=AB. 2 1 2 2 WI -'- W 2' Compounding the weight w, +w2 at a with another weigli't w3 acting at C, they will have a resultant w, +w2 w3 a'cting, at some point b, such that (w, +w2).ab=W3.Cb. cENfr ER OF GRAVITY. 47 First determining aC firom the triangle aCB, in which BC Ba, and the angle aBC are supposed to be known, we can de termine the distance ab by the proportion w, +wo +w: w3=Ca ab=Ca. Wi+w2+W3 By continuing the same process, we should determine the point at which the final resultant weight' acts. The point will be the same, whatever be the order in which we compound the weights. / 103. PROP. To find the center of gravity of any number oJ particles in the same plane whose positions are given by their co-ordinates. Since the weights of the particles constitute a system of parallel forces, let P1,9P, P3, &c., represent the weights of the particles which may be supposed collected in their respective centers of gravity, xl, y,, x2, yo, &c., their co-ordinates. - We shall have for the co-ordinates a, y of the center of parallel forces (A'rt. 44), or center of gravity of the whole body (Art. 94). P1x, P2x 2+P3x3+, &c._ Px - P +P2+P3+,&c. LP P, yl+P2Y2+P3y3+,&c. &.Py (28) - P +-12P;3+, &c. p " 104. Coa. 1. If the particles all lie in a straight line, this line may be taken for the axis of x, and yg =O, y =, &Y..'. &=0, and the center of gravity will be in that line. 105. COR. 2. If the particles are homogeneous, the weights of each particle (24) wvilllbe proportional to the volumes; and if vl, v2,.v3, &c., denote the volumes of the particles, and V the whole volume, we have v,x] +vx22 +-v3x3+, &c. Z.vx V qx=y v - =Z.v!(29) _.Y. +,,.Y2, + 3y3+, &c. Z.v 2 Hence the sum of all the particles, or the whole volume, multzplied by the distance of its center of gravity from a plane, is 48 S T AT'I C S.' eqriull to the sun ojf e,,,chi particle into the distance of its cenrte / gravity from the plane. f 106. Coa. 3. If the center of gravity of tile whole volumu be given, and the center of gravity of one of its parts, the center of gravity of the other is readily obtained. Let V equal the whole volume, and v, v2 the volumes of the two parts, x, y, x,, y1, 29, y2, the co-ordinates of their respective centers of gravity. Then x =v1 2 and Y= —But, by hypothesis, V, vl and xi, Y, 7,, ~, are given.'rherefore, V2- V v 1 V -v a V___-v__ and 2=,and (30) V2 2 1/ 07. We shall now proceed to apply the foregoing principles to specific cases. t Ex. 1. To find the center of gravity of a uniforim physical 4traight line. If AB be the uniform straight line, and C its middle point, C will be its center of gravity;: —--— p —- - f- for we may consider the line made up of a series of equal particles in pairs on opposite sides of C, and the weights of each pair would be equal parallel forces having their resultant at C, the middle point between them, or the resultant of all the forces will pass through C. Or, the line AB is symmetrical with respect to a plane passing'through C. The center of gravity, therefore (Art. 97), is in c this plane, and as it is also in the line AB, it must be at their intersection C. ad Ex. 2. To find the center of gravity of a thin triangular plate of uniform density. Let ABC be the triangular plate, of which the thickness is inconsidei;able. Bisect AB in D, and AC in E. A:D Join C, D, and 1B, E intersecting in G. G will be the center of gravity of the plate. CENTER OF GRAVITY. 49 Since the line CD bisects all lines drawn parallel to the base, and, consequently, divides the triangle symmetrically, the center of gravity of the triangle will be in this line. Foi the same reason, it will be in the line BE, and will therefore he at their intersection G. Join D, E. DE is parallel to BC, since it divides the sides AB and AC proportionally, and DE=-BC. From the similar triangles, DEG and BCG, we have DE: BC=DG: GC=1:2..-. 2DG=GC. Adding DG to both sides, 3DG=DG+GC=DC..'. DG=-DC. In the same way it may be shown that EG=-EB. Hence the center of gravity of a triangle is one third the distance firom the middle of either side to the opposite vertex. ExY. 3. To find the center of gravity of a palrallelogram of uniform density and thickness. Bisect the sides AB and DC in 1H and K; also the sides AD and BC'in E and F. The D D IC C plane passing through H and K will divide the parallelogram sym- E metrically, since it will bisect all G lines parallel to AB. The center of gravity will lie in this plane, and A H B will therefore lie in its intersection HK with the parallelogram For the same reason it will lie in EF, and must therefore be a' G, the common point of these lines. B Or, since each diatgonal bisects all lines drawn parallel to the other, it will be at the inter- A c section of the diagonals. i Ec. 4. To find the center of gravity of a thin, polygonal p plate, of uniforrm density and thiclcness. s ID Let ABCDEF be the polygon. Draw the lines AC, AD, AE, dividing it into triangles. 1) 50 S A.AT I S. When the polygon is given, these triangles will be knowni, and their centers of gravity, gI, g, g3, g4, may be found by Ex. 2. The mass of each triangle may be considered as a heavy particle at its center of gravity, or the weight of each, as a force acting at its center of gravity. Then (Art. 39), gl+g2 g2'=glg g G' =g 12 2 This determines the point G', the center of gravity of tle portion ABCD. In the same manner, we may find the point of application G", of the resultant of g1 +g2 acting at G', and g,3 acting at g3, and so on; the last point so determined will -e the center of gravity G of the polygon. - Ex. 5. Tofind the center of gravzty of a trian gular pyramid f uniform density. Let ABCD be the triangular pyramid. Bisect the edge BC in E, and pass a plane through E and the edge AD. This D ct s th Xd \ plane will bisect all lines in the pyramid parallel to BC, and will therefore divide it symmetrically. Bisect AC in F, and the plane through F and the edge ED will also divide the pyiamid, symmetrically; and since the center of gravity of the pyramid will be in both these planes, it must be in their intersection Df. But the pointf is the center of gravity of the face ABC (Ex 2). Hence the center of gravity of the pyramid CENTER OF GRAVIT Y. 51 lies in the line drawn from a vertex to the center of gravity of the opposite face. Take Ee=~ED, and join Ae. Since e is the center of gravity of the face BCD, the center of gravity of the pyramid will be in the line Ae. It must therefore be at the intersection G of Ae and Df. To find fG, join fe, which will be parallel to AD, since it divides the sides ED and EA of the triangle AED proportionally. Now the similar trianglesfGe and AGD, withfEe and AED, give fG: GD=fe: AD-Ee: ED=-1: 3..f. 3fG=GD. Adding fG to both members, 4fG=fG+GD= fD, or fG=4fD. Hence, the center of gravity of a triangular pyramid is one fourth the distance from the center of gravity of one face to tha opposite vertex. Ex. 6. To find the center of gravity of a pyramid whose base is any polygon. In the pyramid A BCDEF find G, the center of gravity of the polygonal base (Ex. 4), and join AG. Since AG passes through the center of gravity of the base, it wvill pass through the celnter of gravity of every section parallel to the base, and the center of gravity / of the whole pyramid will / be in AG. Join GB, GC, GD, GE, tB and GF, and conceive planes to pass through A and each of these lines, C D thus dividing the whole pyramid into as many triangular pyramids as the base has 52 S T TC S. sides. The centers of gravity of these pyramids will be ax one fourth the distances, respectively, from the centers of gravity of their triangular bases to the common vertex A. These distances being thus divided proportionally, the' points of division will all lie in the same plane parallel to the base. And since the centers of gravity of all the triangular pyramids are in this plane, the center of gravity of the whole pyramid will be in it, and, being in the line AG also, will be at their intersection g. But the plane divides all lines drawn from the vertex A to the base proportionally; therefore, the center of gravity g of the whole pyramid is one fourth the distance fiom the center of gravity of the base to the vertex. COR. Since the above principle is true, whatever be the num ber of sides of the polygon, it is true when the number becomes indefinitely great, or when the base becomes a continued closed curve, as a circle, an ellipse, &c.; or, the center of gravity of a cone, right or oblique, and on any base, is one fourth the distance from the center of gravity of the base to the vertex.'V Ex. 7. Tofind the center of gravity of a frustum of a cone or pyramid cut of by a plane parallel to the base. Let a be the length of the line drawn from the vertex of the cone, when complete, to the center of gravity of the base, a' that portion of it between the vertex and the smaller base of the firustum. Then (30) we have V2 in which -=-4-a, 1v- =a'. Now the part of the cone or pyramid cut off is similar to the whole, and similar solids are as the cubes of their homologous dimensions, or cubes of their lines similarly situated. Hence: v,=a" a", or v-v, v=a-a a". at".ad V- - -V =v and v =V-V =v a-' 3. CENTER OF GRAVITY. 53. — - _ a'3j 4-a3_-aI' _ (a+ a') (a'+ a'2) a2 +aa' al a2 Subtracting this from a, we have the distance of the center of gravity of the frustum from the center of gravity of its base equal to 3 (a + a') (a'+ a") a, 3a' a-3 _ __ a9 a a a +aa, a,2 4 4(a +aaf+a"). Ex. S. Tofind the center of gravity of the perimeter of a tryangle in terms of the co-ordinates of the angular points. 93 In the triangle ABC, let a, b, A c represent the sides respective- \ l ly opposite to the angles A, B,. C. Their centers of gravity will ]B be each at the middle point of the side; as g,, g, g30 Let x,y, be the co-ordinates of A referred to the origin O, x y " " ic B, x3y3 "( "r " C. Then the co-ordinates of g, are I-(x,+x2), g(y, +Y),: -,,,, g2,.' - (+X9), I(Y2 +Y 3)) 2\, X2 +X, -Y, g 3 (X 3 (1 )s Z(Y 3 +Y 1 ) and x, y, being the co-ordinates required, by (28), we have a(x2 +x ) +b(x, +x,) +c(x, +x) 2(a+b+c) _a(y,2+y3)+b(y1,y3) +c(y,1y,) 2(a+b+c) Ex. 9. To find the co-ordinates of the center of gravity of a triangle. Let x,y, x2y2, x3.y3 be the co-ordinates of the points A. 54 S T.T I C S. B, C respectively. Draw C AD bisecting BC in D, ~ and take AG=-AD: G is the center of gravity B of the triangle. The coordinates of D are -2(.v A. / +x3), g-(y2 +y3); and if x=ON, Y=GN, be the -x co-ordinates of G, we O:L N have ON= OL + 2(OM -- OL), GN=AL+23(DM-AL); Or t=x 1 +-~312(X2 +XX3)-X1 }=-(X1 +-2 ~-X3) and Y=y, ~1+ 2(y2 +Y3) -Y1 =-(Y1 +Y2 +Y 3) CONDITIONS OF EQUILIBRIUM OF BODIES FROM TIIE ACTION OF GRAVITY. "'108. PROP. If a body have a fixed point in it, the condition oJ equilibrium requires that the vertical line through the center of gravity shall pass through the fixed point. If the center of gravity g be in the vertical line Ag, passing through the fixed point A, the weight w of the body, being a vertical / t > | g y force acting at g, in the direction Ag, will be resisted by the reaction V ( A c, of the fixed point A. If the center of gravity be at any other point VV V / \/ 4('i g', then drawing th3 1w9 s 3 vertical line through g' and the horizontal line through A, the weight w acting at g' would have an uncompensated moment w.Am, which will not vanish until the center of gravity comes into the line Ag. Or, if g' be the center of gravity, the body will be acted upon by CENTER OF GRAVITY. 53 a couple of which the forces are, the weight of the body at g', and the reaction of the fixed point A1, and the arm Am. Therefore (19), in order to equilibrium, z (Xy — Yx) = w.Am=-0... Amnz=O0, or the point g' must be in Ag. 109. DEF. The equilibrium is said to be stable when the body, if slightly disturbed, tends to return to its original position. It is called zcnstable when, being disturbed, it tends to re move further fi om its original position; and neutral when, after being disturbed, it still remains in equilibrium. 110. PROP. 1When the equilibrium of a body containling a fixed point is stable, the center of gravity is in the lowest position it can take; when unstable, in the highest. Let A be a fixed point in the bodies M and N. g their oanier of gravity'. The center of gravity can 7i 7 only move on the surface of a sphere N c whose center is A / and radius Ag. When the body M is disturbed, its center of gravity g being removed to g', will rise through the versed sine of the are gg', and when at g', the moment w.Am will obviously tend to bring it back to its original position. The equilibrium is therefore stable, and the center of gravity the lowest possible. In the body N, the center of gravity being at g is the highest possible, and being in the vertical Ag, will be in equilibrium. When removed to g'. the moment w.Am will obviously tend to carry it further, andl the equilibrium was therefore unstable. 111. Con. 1. The pressure on the point by which a body is suspended is clearly, in the case of equilibrium, equal to tlhe weight of the body. vi 112. COR. 2. If a body is suspended fiom tzwo points, the position of equilibrium is that in which the center of gravity is in bt~: STATICS. the vertical plane passing through the two points of suspen sion, since it is then the highest or low est possible. To determine the pressures on the fixed points A and B, let GCO represent the weight of the body D\ acting vertically at G. Resolve GC( G into the two forces DG and EG acting in the directions AG and BG. These will represent the pressures on A and B. Since the directions of GC, GD, and GE, and the magnitude of GC are known, the magnitudes of DG and EG may be determined. 113. COR. 3. If a body be suspended from three fixed points not in a right line, the body is necessarily at rest. With regard to the pressures on each point, the three lines drawn from the fixed points to the center of gravity give the directions of the pressures, and the vertical is the direction of the weight, the magnitude of which is given. Hence we have, in a parallelopiped, the three sides and diagonal given in:position and one given in magnitude, to determine the magnitude of the other three.' 114. CoR. 4. If a body touch a horizontal plane in one point, it will be in equilibrium when the vertical through its center of gravity, and the perpendicular to the plane at the point of contact, coincide; for the weight will then be counteracted by the plane.', 115. CoR. 5. If the body touch the plane in two points, it will be in equilibrium when the vertical through the center of gravity, and the perpendiculars to the plane at the points; are in the same plane. Thus, if ABC be a vertical section through the two points of support P and Q. p, Q there will not be an equilibrium unless the vertical through the center of gravity G is in the same plane, or, which is the same thing, meetl CENTER OF GRAVITY. 57 the line PQ. The pressures on P and Q may be determined by the theory of parallel forces, and P: Q: wV=w: WP: PQ, P, Q, and W representing the pressures on the two points and the weight of the body respectively. If the body touch the line PQ in more than two points, the problem of the pressures is indeterminate, as the pressures may be any how distributed. /116. COR. 6. If the body touch the plane in three points, it will be in equilibrium when the vertical through the center of gravity falls within the triangle formed by joining these points. To estimate the pressures in this case, let PQR be the triangle formed by joining the three points, and W the point where the vertical through the center of gravity meets it. We must re- p solve the weight acting at W into three others parallel to it, acting at P, Q, and R. Join PW, QW, and Q [(W, and produce them to the opnosite sides. Then, by the theory of parallel forces, P: W-WK: PK=triangle WRQ: triringle PRQ, Q:.W=WM:QM= " WRP:' PRQ, and R: W-VH: RH= " WPQ: " PRQ. P;Q.: R W-W- RQ: WRP: WPQ: PQR. If the body touches the plane in more than three points, the pressures on the points are indeterminate, but their sum is equal to the weighrt of the body. 117. Con. 7. If the vertical through the center of gravity of a body on a plane meets the plane in a point within the base, the body will stand. For the resultant of the parallel forces of resistance must be within the figure formed by joining the several points of contact. If the vertical falls without the base, we have two parallel forces in contrary, but not opposite, directions, and the body wii turn over. tt3 SSTAT IC S. 118. PROP. The stability of a body is measured by the excess.f the shortest line that can be drawn from the center (f gravity/ to the perimeter of the base, above the vertical, from the center to; the horizontal plane. The stability will depend on the ex-.........:.cess of GP over GHI, since G must be elevated a distance equal to this difference, in order to turn the body over the edge of the base at P. P' I-1Con. 1. The greater the base HP the greater the stability, if the height of G remain the same; and the greater HG, the less the stability if HP remain the same. COR. 2. The stability is measured by the versed sine of the are, through which the center of gravity must move firom rest to its highest point. For GP-GH-GP-BP=AP-BP=AB - versed sine of are GA to radius PG.; 119. Pitor. If a body be placed on an inclined plazne, it uwill tescCnd when there is no resistance from friction. For the weight of the body, represented by GA, may be resolved into two others, GB and BA, one perpendicular to the plane, and the other parallel to it, of which GB, A B the one perpendicular to the plane, can alone be counteracted by the plane.. In this case, if the vertical GA fall within the base of the body, the body will slide; if it faill without, it will slide also. 120. EXAMPLES. 1. If two right cones h]ave the same base and their verticesb in the same direction, find the distance of the center of gravity of the solid contained between their two surfaces from their common base. Ans. i sum of their altitudes C E N,'T IV"o( G y 4Y/T r. 59 2. The center of gravity of a paraboloid being imll N:i X:.*iat a distance from the vertex equal to -- of the axis; finca the center of gravity of a frustum of a paraboloid from the base, a and b being the radii of the two ends, and m1 the parameter to the axis (30). a- _ 3a2b4 -+ 2b Ans. 3m(a4-b) ar 3. Two spheres, whose radii are a and b, touch each othei internally; find the distance of the center of gravity of the solid contained between the two surfaces from the point of contact. a3 +a-b+abh b3 Anzs a2+ab+b2 A4. The distance of the center of gravity of a hemisphere from its base being - the radius, find that of a hemispherical bowl lwhose internal radius is a and thickness c. 4a3 + 6a2c + 4ac2 + c' An 3a2+3ac+c2 / 5. From the result obtained in Ex. 4, find the distance of the center of gravity of a hemispherical surface from the center of the base. Ans. la. APPLICATION OF TIHE PRINCIPLES OF TIlE INTEGRAL CALCULUS TO TIlE DETERMINATION OF TIlE CENTER OF GRAVITY. V 121. By the principles of the integral calculus, when the volumes v (Art. 105) become indefinitely small, they may be re. garded as the differential elements of the body, and be repre. sented by dv. In this case formulas (29) will take the form xfdv fydvo' (31) in which x and y denote the distances of the center of gravity of dv fiom the co-ordinate axes. V 122. PROP. Required the differential expressions for the co ordinates of the center of gravity of a plane curve or line. If ds represent the differential element of the curve or line, by substituting ds for dv in (31), we have 60 ST ATICS. fxd(s - fyds - _ = ~, y_.y Y. (32) S $ If the arc is symmetrical with respect to the axis of x, the center of gravity will be in that axis (Art. 99), and Y-0. _ f':ds S s sufficient. v 123. PRor. Required the dlifferential expressions for the co. ordinates of the center of gravity of a plane area. Since the differential element of a plane area is dxdy, dv= ixdy. By substitution in (31), we have fxdxdy fyd.rdy -ffdxdy' Y fdxdy' Integrating in reference to y, we have =fydx =Y Ydx (33) If the area is symmetrical with respect to the axis of x, the center of gravity is in that line (Art. 99), and -=O. x. ydx fydx is sufficient. 124. ProP. Required the differential expressions for the coordinates of the center of gravity (f a surface of revolution around the axis of x. The center of gravity will obviously be in the axis of x, and therefore y=O; and since, for a surface of revolution, dv= 2ei/ds, the first of equations (31) become, fxyds Xfd (34) and this equation is sufficient.' 125. PRor. Required the differential expressions for the co. ordinates of the center of gravity of a solid of r evolution. In this case dv=rry2 dx. Hence, from the first of equations (31), we hav3 CENTER OF GRAVITY. 61 fy2xdx Jyd'(35) which alone is sufficient. By proper substitutions for dv in the fundan ental equations (31), we may find expressions for the co-ordinates of' the center of gravity for other forms of bodies. 2 126. PRoP. The surface generated by the revclution of a curave arouend an axis is equal to the length of the curve, multiplied by the circumference described by its center of gravity. From the second of equations (32), we have 2,.y.s=27rfy ds. Now 2,ry is the circumference of which y is tLe radius, and 27ry.s is the circumference described by the center of gravity of the curve s in its revolution round the axis of x, multiplied by the length of the curve s. But this is equal to 2r-fyds, which is the area of the surface generated by the revolution of the curve. Hence, &c., V 127. PROP. The volume generated by the revolution of a plane area around an axis is equal to the product of that area by the circumference described by its center of gravity. For, fiom the second of equations (33), we have 2frydx= 7rfy2dx. In this equation, fydx is the generating area, 2ry is the cir'cumference described by its center of gravity, and Trfy2dx is the volume generated. Hence the truth of the proposition. These last two propositions comprise the theorem of Guldin, and their application to the determination of the surfaces and volumes of bodies constitutes the Centrobaryc Method. By this method, of the three quantities, viz., the generatrix, the distance of the center of gravity from the axis, and the magnitudo generated, any two being given, the other may be determined. 62 SSTATIC S. 128. EXAMXPLES. ~ 1. Required the center of gravity of a circular arc. Let the axis of x bisect the arc MATM in A, the origin being at the center of the circle, and let MAM'=2s. From the C 1'p I x equation of the circle y'=2)-x'2, we ob.2.2 tain dx2 r2 y2rdx But ds=dx/l-Fy =Id V2 dx2 xdx2_ X (32) ~ +Xt - _N/' --- _ /7 -X'_'+C -- — F'q. S Jf Vr2_-x2 S $ W\hen y 0, 7=r, and -r. C= Hence r.-CI r.2y s-2s or, the distance of tile center of gravity of a circular arc fiom the center of the circle is a fourth proportional to the arc, the radius, and the cord of the are. If the are be a semicircle, y=?, and s=-7r' CI= —=0.63662r. 7T v 2. Required the center of gravily of a circular segmtent. Putting CP=a (Fig., Ex. 1), and taking tshe center Pbr the origin, we have y= V/2 —x'2. Hence (33), __- x'ydx _ x(r-x2) dx I (r'_ 2) 2iMP JT:'ydx. MAP MAP -MAP f l4(2MP)3 - (chord)2 M/IA'P segment If the segment is a semicircle, z-' (2 1r)' 4r x --- =.=0.42441r —=CI. 3T CE N TER OF GR A VIr Y. 63 3. Required the center of gravity of the surface of a sph&;rical segment. Taking the origin at C (ig.,, Ex. 1), the center of the gendy2 x2 cratirng, circle, we have x2 —y2=r, d —, and yds-rdx. _ ) fyxds fxdx.. (34) x —=j =jI' Integrating between x-=r and x-a-CP. _ 2- 2+). r-a Hence the center of gravity is at the middle of PA. 4. Required the ctnter of gravity of a spherical segment. Taking the origin at A, the vertex of the generating circle;, we have, for its equation, y2=2ax-zx. Y fxy'dx f(2ax-x2)xdx ___. (35) X=- 2d =. (2 -2)d Hence AG= -2k- Sa 3- ax — x 12a-4x If the segment is a hemisphere, x=a, and x={a. 5. Required the surface of a hemisphere. By the centrobarye method (Art. 126), we have The generatrix = —rr, the ordinate of its center of gr'av! v _r=- (Art. 128, Ex. 1), and the circumference described by 2r the center of gl;avity =2rr. — =4r. H- ence the surface =-rr. 4r - 2rrr. 129. EXAiMPLES ON THE PRECEDING CHAPTERS. AEx. 1. Two beams, rigidly connected at a given angle, turn on a horizontal axis through their point of union; find the po. sition of equilibrium by the action of their own weights. Let AC, BC be the beams suspended from C, and making 64 STATICS. with each oth er the angle a. 7Since C is a fixed point, the "/'11 ffionly condition of equilibrium is, that the sunml tA WL 1AJ2 of the moments about C is zero (Art. 75). Let gI, g2 be the centers of gravity of tile beams, and g1C =a, g2C= b. Also, w - weight of AC, acting at g1,and wsweight of BC, acting at g2. Draw through C the line MCN horizontally, meeting the vertical directions in which wu and w2 act, at M and N. By (19), WlCM-W2CN=O. Let BCN-0. The determin ation of 0 will fix the position of the compound beam. Since CMg1lC. cos. MCA and CN=g2C. cos. BCN, we have w1.Cg1 cos. ACM- w.CgQ2 cos. BCN=0, or w,. a. cos. (180-a0+) —w2.b. cos. 0=0. Tan b+w, a. cos. a T'.an. O= wI a. sin. a Ex. 2. When a given weight W is hung from the end of one of the beams, A (Ex. 1), find 0 in case of equilibrium. Tan. 0=w 2.b+(w i +2W)a cos. a (w, +2W)a sin. a Ex. 3. Two beams, as in Ex. 1, are suspended from one end B; find the angle 0 which the upper one makes with a horizontal line. 9 E Tan. Q(21(1 +7w2)b-w la. cos. a w a. sin. a N. B. Since the common center of gravity of the two beams is in the vertical through B, BM=Gm-Cm C _ -CG=Cm-2NB=a. cos. (a-.O )2b cos. 0. W2 1 A EXA.MPLES ON TH E PRECEDING C t APTERS. 65.. tvWb. cos. 0 —w a. cos. (a -)-1-2w 1b. Cos. 0=0. Ex. 4. Two spheres of unequal radii, but of the sane material, are placed in a hemispherical bowl; find the position they take when in equilibrium. Since the reactions of the bowl upon the spheres are i n the directions of the radii of the spheres through the points C of' contact, and since these radii produced pass through C; if C was a fixed point, A and connected with A and B B by a rigid rod without weight, the bowl might be removed without disturbing the equilibrium. The question, then, is reduced to finding the position of equilibrium oi two weights suspended from the extremities of two rigid rods without weight, and is solved like the preceding. This posi. tion will be known when 0 is known. Let R be the radius of the bowl, rl, ra2 the radii of the spheres A and B respectively, and ACB=a. Then AB=rI+r29, CA=R-rl, CB=R-r2, AC2+CB2- AB2 and cos. = 2AC.BC 2AC.BC 2(R-r1) (R-r2-) ~, which gives a. Then (19) wj.CM-w2CN=O, or r'.CMLI-? CN=O; since the weights of the spheres aie as the cubes of their radii. Substituting the values of CM-=(R -r) cos. (180-(a+0)) and CN- (R-r2) cos. 0, expanding and reducing, we get Tan. 0rn (R-r2)-nr(R,-r,) cos. a r (R-r ) sin. a J Ex. 5.5. A heavy beam rests upon a smooth peg with one end against a smooth, vertical wall; find the position cf equi. librium. E 66 STATICS. ILet ACB be the beam, resting at A against the wall ADE ~ E and upon the peg C. The center of gravity g. R'at / when there is an equilibrium, will evidently be at some point beyond C from A. Let __D \/_ FAg=a, DC=b, w= the weight C of the beam acting at g, R= pendicular to itself at A, ana R'= the reaction of the peg (IV perpendicular to the beam at C. The angle 0, which the beam makes with the horizontal direction when in equilibrium by the action of these three forces, is required. Employing (17), (18), and (19), and resolving the forces in vertical and horizontal directions, and about the point C, we have w resolved in a horizontal direction 0, R " " - R R' " =" "' sin. 0..(17).X=R-R'sin.0=0. (a) Also, w estimated vertically = w, R " i - 0, R' "( " = —R — cos. 0 ~. (18):.Y=w-R'cos. 0=O. (b) Also, the moment of w about C=w.CF= —w.(DF-DC). But DF=AH = Ag cos. 0= a cos. 0, and DC =- b... w.CF=w.(a. cos. O-b), the moment of R about C —R.CK==R.AK. tan. 0=R.b. tan. A "c R' c" = 0. (19) Y. (Xy-Yx) = w.(a. cos. 0-b)-R.b. tan. 0=- O. (c) Multiplying (a) by cos. 0 and (b) by sin. 0, and subtracting, we have R. cos. 0-w sin. 0 = 0 ~. w tan. 0. EXAMPLES ON THE PRECEDING CHAPTE.1S. GIi Substituting this value of R in (c), w(a cos. 0-b)-w.b. tan.2 0=0, or a cos. 0=b (1+ tan.2 0) =b. sec.2 0= b. —2 -.'. cos. 0-= -, and b-A C- takes place in the direction of the line Sm, m'S normal to the cogs at tho point of contact, Cm, C'm' being perpendiculars from the centers C and C', of the wheels, on that line. Taking the moments about C', when the power and weight. are in equilibrium, we have P.C'A= S.C'm', and about C, W.CB=S.Cm. Dividing the latter by the former, we have W C'A Cm AP CB C'mn Now, if the radii of the axles are equal, or C'A CB, we shall have W Cm P C'm' which gives the effect of the action of the cogged wheels alone or, since the triangles C'm'o and Cmo are similar, Cm Co W C(!m' Co P' WHEEL AND. AXLE. 89 If the direction of the line Sm' mS changes as the action passes from one cog to the succeeding, the point o will also change its position, and the relation of W to P become variable. But when the cogs are of such fornm that the normal Sm m'S at their point of contact shall always be tangent to both circles, the lines Cm and C'm' will become radii and their ratio constant, and the point o a fixed point, in which case W Co Cm R P C'o C'm' R' R and R' being the radii of the circles C and C' respectively. V/ 144. COR. When the cogs are equal in breadth, the number of cogs on C will be to the number of cogs on C' as the circumference of C to the circumference of C', or as the radius 4of C to the radius of C'. W number of cogs on the wheel of W *P number of cogs on the wheel of P' ScroIIviM. When the working sides of the cogs have the form of the involute of the circle on which they are raised, the pressure of one cog on another will always be in the direction of the common tangent to the two wheels. Thus, let IHF, KEb be the two wheels. The acting face GCH of the cog a being formed by 6 the extremity H of the flexible line Fai- as it unwinds from the circum- ference, and the acting face of b by the unwinding of the thread GE, the line FCE will always be normal to the faces of the cogs a and b at their point of contact. The circles describ- A ed with the radii AD and BD are called the pitch lines of the wheels, and will roll uniformly upon each other. 145. PROP. Required the condition of equilibrium when the action of the power is transmitted to the weight by a system of cogged wheels and pinions. o0. - S T I C 8S. Let R, R,, Ra, &c., be the radii (Pd ir the successive wheels; r, r1, r2, &c., the radii of the corresponding pinions; P, P,, P2, &c., the powers applied to the circumferences of the successive wheels. r r3ll Takling the moments about the center of each wheel, we have P.R —Pr, PR —P 1R, P2R2 P3r2, &c.; since the power applied to the circumference of the secona wheel is equal to the reaction on the first pinion. Multiplying these equations member by member, and reducing, P,_ R.R1.R,&c. P r.r.r2, &c. or, the power is to the weight as the product of the radii of tho pinions to the product of the radii of the wheels; or, as the product of the numbers expressing the leaves of each pinion to the product of the numbers expressing the cogs in each wheel. EXAMPLES. Ex. 1. If a power of 10 lbs. balance a weight of 240 lbs. on a wheel whose diameter is 4 yards, required the radius of the axle, the thickness of the ropes being neglected. Ans. r=3 inches. Ex. 2. The radius of the wheel being 2 feet, and of the axle 5 inches, and the thickness of each rope being { inch, find what power will balan.ce a weight of 130 lbs. Ans. P=28- lbs. Ex. 3. The radius of the wheel being 3 feet and of the axle 3 inches, find what weight will be supported by a power of, 120 lbs., the thickness of the rope coiled around the axle being one inch. Ans. xW= 12347.,i Ex. 4. There are two wheels on the same axle; the diameter of one is 5 feet, that of the other 4 feet, and the diameter T HE, E C O RD. 9 of the axle is 20 inches. What weight on the axle vould be supported by forces equal to 48 lbs-. and 50 lbs. on the larger and smaller wheels respectively? Ans. W=-264 lbs. ~ III. THE CORD. / 146. The cord or rope is employed as a means of communication of force. It is regarded as perfectly flexible and without weight, and transmits the action of a force applied at one extremity to any other point in it, unchanged in magnitude, so long as it is straight, or only passes over smooth obstacles without friction. The force thus transmitted is called the tension. Since the tension is the same throughout, from one extremity to the other, when employed alone, it affords no mechanical advantage; but when passed over or attached to certain fixed points, the resistance of these points may be employed advantageously. 147. PRno. Required the condition of equilibrium of a cord acted upon by three forces. Let the forces P,, P2, Pi1 P3 be applied at the extremities A and B of the \ cord ACB, and to a knot D at C. Draw any line CD in the direction of the force P,, and DN, DM parallel to CA and C CB respectively. In case of equilibrium (Art. 28), CM: CD: CNP —: P: P3= sin. DCN: sin. MCN: sin. MCD = sin a' sin. (a+a') sin. a; or, theforces are each as the sine of the angle contained between the directionzs of the other two. 92 STATICS.. 148. CoR. 1. If the cord be fixed at A and B, the reactions of the points A and B take the place of the forces P1 and P3, and are equal to the tensions:of the two parts of the cord re spectively. 149. Con. 2. If the force P2 be applied to a running knot or ring, the points A and B being fixed, the condition of equilibrium requires that the direction of P2 should bisect the angle ACB. For the point C in its motion would describe an ellipse, A and B being the foci, and the force P2 could not be in equilibrium except when normal to the curve, in which case it bisects the angle ACB. Hence,:sin. a= sin. a', and P =P3; and since P1 P2= sin. a': sin. (a+-a')= sin. a: sin. 2a=1: 2 cos. a.'. P2=2P~ cos. a. Oltherwise: since the tension of the cord is the same through. out, when the cord passes over an obstacle without friction. PI=P3; a=a' and P2 =2P1 cos. a (Art. 19). 150. PROP. Required the conditions of equilibrium when anzy?number of forces in the same plane are applied at different points of the cord. Let ABCDE be a cord to which are applied the forces P,, P2, P3, &c., at the 1P5 points A, B, C, &c., in the; directions AP,, \<.A b BP2, CP3, &c. The force P1 at A may be considered as B \ acting at B in its direction BA; and since C B, when in equilibrium. A must be so from the t action of P1, P2, anct, VP/~73 the tension of BC, the resultant of P1 and P2 must be in the direction of CB, and may be c6nsidered as acting at C. Suppose it thus applied, and let it be resolved into two, acting in the directions Cn'and Cm parallel to the original components, and equal to them. THE CORD. 93 We have thus transferred the forces P1 and i', to act at C parallel to their original directions. In the same manner, the resultant of P, P2, and P3 acting at C must be in the direction of DC, and may be applied at the point D without disturbing the equilibrium, and then replaced by PI, Pa, and P, parallel to their original directions, and so on, for any number. Hence, if all the forces be supposed to act at one point parallel to their original directions, they will be in equilibrium. The conditions of equilibrium, therefore, are the same as for any number of concurring forces (Art. 70), or, the sum of all the forces resolved, in any two rectangular directions must be zero. The form which the cord takes under the influence of the several forces is called a funicular polygon. 151. PnoP. Required the relations of the forces which, acting on a cord in one plane, keep it in equilibrium. Produce P2B to N, P3C to M, &c., and let' the. A z ABN=a, zNBC=a', N L BCM=f3, lMC&Df', t, Let t,, t2, t3, &c., be /B 2 the tensions of the several successive portions Jr2 aP4 of the cord. r3 Then (Art. 147) t,: P2: t2= sin. a': sin. (a+a') sin. a; (a) alSO, t2: P3: t= sin.: sin. (') sin.')in, (a') &~c~., &c. sin. a' sin. a From (a) we obtain t=P in (a and t =Ps a sin. (a+ a')' sin. (a+a')' sin. 3' sin.3 " (a') " t2=P3.sin (0+01)andt3=P3Sin. (0+0,) &c., &c. Equating the values of t2, t3, &c., we have sin. a sin./3 P.: -Pb) s'in. (a3~a)) sin. (SP) 94 sT AT IC S. sin./3 sin.y P 3'sin. (t3'+.') 4'sin. (y')' &c., &c., foI the relations of the forces. 152. COR. 1. If the cord be fixed at the points A and E, and the forces P2, P3, P, -&c., be parallel, we have sin. a'= sin. 3, sin.-'= sin. y, &c. Multiplying these equations by (b), (b') member bty member in their order, we obtain sin. a sin. a' sin. 1 sin.' sn. y sin. y' 2' sin. (a+a') 3 sin. (S+'p:) 4 sin. (Yd)') sin. a sin. a' sin. a sin. a' sin. (a+a') sin. a cos. a'+ sin. a' cos. a 1 1 cos. a cos. a' cot. a+ cot. a' sin. a sin. a' Pc P- P4 + cot. a+ cot. a' cot. 1+cot. 3' cot. y+cot. y A N Jg 1.53. Coa. 2. If. the cord be n ] O / fixed at A and E, and the forces P, P3, P4, &c., be weights, the B horizontal tension of each porD tion of the cord is the same.!2 ~ < For, by resolving the tension i. of each part horizontally, we p3 P4 have the horizontal tension of AB=t1. sin. ABN —t, sin. a, " "' BC=t2 sin. BCM=t2 sin. 2, &c. Substituting for t1, t2, &c., their values found above, sin. a sin. a' P2 horizontal tension of AB=P sin, aa 2 sin. (a+a') cot. a+ cot. a" sin./3 sin.' P, BC BC=P3 3sin. (P+13') -cot. 3+ cot. (.' &c., bc, which, by Cor. 1, are all equal. THIE CORD. 95 154. CoR. 3. Since the reactions of the points A and E equilibrate the resultant of all the weights, the lines AB, ED produced, must meet in same point of the vertical through the center of gravity of the system. For three forces, if in equi. librium, must meet in a point. 155. A heavy cord may be considered a funicular polygon loaded with an infinite number of small weights, and since the number of weights is infinite, the polygon will also have an infinite number of sides, or will become a curve. The curve which a heavy cord or chain of indefinitely small links will assume, when suspended from two fixed points not in the same vertical line, is called the catenary. EXAMPLES. Ex. 1. Two equal weights balance, by a cord, over any number of fixed points without friction. Required the pres.ure on each. Since the tension of the cord PABCDEP2 is the same throughout, each point is acted upon B by three forces, viz., two equal ten- n - sions on each side of it and the reac- El tion of the point, which last must be equal to the resultant of the other two. Hence, calling the angles at A, B, C, &c., a, b, c, &c., by Art. 15, and Cor.,:2 Art. 19, the pressure on A=2P, cos. ga, "'; B=2P, cos. b, &c., &c. Ex. 2. A cord of given length passes over two fixed points A and B without friction, and one extremity, to which a given weight P is attached, passes through a small ring at the other extremity C. It is required to find the tension of the coid when in equilibrium, and the length of the part CP below the ring. Since the cord passes freely over the points A and B, and through the ring C, it is of the same tension throughout, and equal to the weight P. Hence the point C is kept at rest by 06 S T ATIC S. u three equal forces, and, by Art. 18, must make A x angles of 120~ with each other. Draw the c 1 horizontal line AI); and, since A(CD=120~, P ADC and CAD are each equal to 30~. The position of A and B being given, the angle BAD must be known. Hence, in the triangle ACB we have the side AB and all the angles from which AC and BC may be determined. Then the whole length of the cord, diminished by the perimeter of the triangle, will be the distance of the weight from the ring. ~ IV. THE PULLEY. 156. The pulley is a small grooved wheel movable about an axis, and fixed in a block. The cord passes over the circumference of the wheel in the groove. The use of the pulley is to prevent the effects of friction and rigidity of the cord. The first of these it diminishes by transferring the friction from the cord and circumference of the wheel to the axle and its supports, which may be highly polished or lubricated. The effects of rigidity are diminished by turning the cord in a circular are instead of a sharp angle. The pulley is called fixed or movable, according as the block is fixed or movable. llllll lllll il l l lii!I I!l i i iiii r 157. The fixed pulley serves merely to change the direction of the forces transmitted by the cord, since, neglecting the friction of the pulley, the tension of the cord is the same in every part of it. Hence B e the power equals the weight, and the presto ure on the axis of the pulley equals their E) SUM.1 THE PULLEY. - (9 158. Piki,. Requilied the relation of the poewer tc the weizght z thy single movable pulley. The tension t of the cord, being the same pa \ throughout, is equal to the power P; also. to the pressure on the hook Q. The resultant of the two tensions in the directions CP and DQ, being equal and c opposite to the weight W, must be vertical. Let a be the angle made by the cords with this vertical, and, resolving the tensions vertically, we have 2t cos. a - 2P cos. a =W. W P.. P 2 cos. a' the same as obtained in Art. 149. COR. 1. If the weight to of the pulley be taken into account, _W+w 2 cos. a' CoR. 2. If the cords are parallel, a=0 and W+w 2 CoR. 3. If a=900, 2a-180~, or the cord becomes straight and horizontal. In this case W+w or the power must be infinite. In other words, no power can reduce the cord to a horizontal straight line while the weight is finite. 159. Of the various combinations of pulleys there are three, which we shall distinguish by the first, second, and third systems of pulleys. 160. PRoP. Required the relation of the power to the weight in the first system of pulleys. G 98 S T AT IC S. \N\\, 71,~,~,~,, The annexed figure represents this system. Neglecting friction, the tension t of the cord is the same throughout, and equal to P. The V 2 l ll ff weight W and the weight of the lower block w are sustained by the tensions of the several cords at the lower block. Hence, if n be the number of cords at this block, nt —n nP-W- -w pW+W. If the weight of the lower blockl be neglected, _W Of this system of pulleys there are various P modifications. The annexed form is the one in most common use. 161. Pior. Required the relatzon of the power to thie weig'ht tin the second system of pulleys. The annexed figure represents this system with three mova able pulleys, each pulley having its own rope. Designate by a, a2, a&, &c., the pulleys respectively in their order from the weightaW, b by wv, w?, &c., their weights, and by t,, t?, t3, &c., the tensions of their respective cords. Then, for the equilibrium of a,, we have 2t1=W+wl, or t- 2 For tne equilibrium of a., 2t2=t, +w2- 1 2 2 Wow= +2w2 or t- W+w, +2w +2 2, For the equilibrium ofr 3,2t =t2 +w W= 2 22 THE PUL L EBY. 9 W+-wI +2w2 -+2 W3 W 3 a) f 2 For the equilibrium of a,, W+w, +2w2 22w.+ -2W 4~ -... -1W, p-t,,- - 2"' Con. 1. If the weight of each pulley is the same, and equal wt W+w, (1+2+22+2+.... 2"-') 22 W w,(2"-1) 2' 2",11 — -). 2W 12 f1\ C(OR. 2. If the weights of the pulleys be neglected, w P= — or W=2".P. 4)' 162. PRoP. Required the relation of the power to the weig,; in the third system of pulleys. In this system each cord is attached to the. weight, and the number of movable pulleys is one less than the number of cords. Designating the pulleys in their order from the weight by a,, a., a3, &c., their weights respectively by w,, w2, w3, &c., and the tensions of the successive cords by tl, t2, t3, &c., we have t,=P, t2=2t + w =2P+-w1, t3=2t~ +w2- 2'P-+-2w I -w,, t4=2t 3+w3-=2P+2'w +2w, +w3. And if there be n cords, t.=2n"-1P+2"n-2w +2""tv, + &c..... 2w,,.2+-U,_1. But W=t, t2+t,+, &c.... t. =P(1 +2+2+23+, &c....2"-1) +w (I +222, &.2"-2) +w2 (1 +22,~ &C.... 2"-) -, &c. 100 STATICS. =P(2n_- 1 p+ (2n- _ 1) +~ 1 (2-2)- 1) +, &C... w,,_1(2-1). If the (n-1) pulleys are of the same weight wl, W=P(2n_- 1)+w(2n- +2 22n &c...... 2'+2+1 -n =?(2 —1)+wj-(2n1)-w~ n (P+w,) (2 — )-Wln. If the weights of the pulleys be neglected, Y=P(2 -_ 1). EXAMPLES. Ex. 1. At what angle must the cords of a single movable pulley be inclined in order that P may equal W? Ex. 2. In the first system of pulleys, if there be 10 cords at the lower block, what power will support a weight of 1000 lbs.? Ex. 3. In the second system of pulleys, if 1 lb. support a weight of 128 lbs., required the number of pulleys supposed without weight. Ex. 4. In the third system of 6 pulleys, each weighing 1 lb., find what weight will be supported by a power of 12 lbs eil k ij Ex. 5. Find the ratio of the power to the weight in the annexed modification of the second system of pulleys. wK ~ V. THE INCLINED PLANE. 163. The Inclinedplane, as a mechanical power, is supposed perfectly hard and smooth, unless friction be considered. It assists in sustaining a heavy body by its reaction. This reaction, however, being normal to the plane, can not entirely counteract the weight of the body, which acts vertically down. THE INCLINED PLANE. 101 ward Some other force must therefore be made to act upon the body, in order that it may be sustained. 164. PROP. Required the conditions of equilibrium of a body sustained by any force on an inclined plane. Let AB be a section of an inclined plane, of which AB is the length, BC the eP height, and AC the R - base. Let i be the in- clination of the plane to the horizon, s the angle a made by the direction of the power P with the plane AB, W= the weight of the body a, and R= the reaction of.A\ the plane. The body is kept at rest by the action of P, W, and R. Resolving the forces' parallel and perpendicular to the plane, we have P cos. E- W sin. i= 0, (a) R +P sin. E-W cos. i-O. (b) From (a) we obtain WM cos. E (c) i Sill. i () W sin. i sin. e From (b), R=W cos. i-P s.n. e —W cos. IW- -S. - wc. Cos. (i+ c) e COS. 8 cos. E W/ cos. e or (3d) I cos. (i+) ( the same relations as obtained in Ex. 24, Art. 129. 165. Con. 1. If the force P act parallel to the plane, E=0 and (c) becomes W 1 AB -P sin. i BC' or, the power is to the weight as the height of the plane to iti ten, th. 1.02 STATICS. W 1 AB From (d) we get R-cos. i= AC' or, the reaction of the plane is to the weight as the base to tht length. 166. COR. 2. If the power act parallel to the base of the plane, s —i, and (c) becomes W cos. i AC P sin. i BC' or, the power is to the weight as the height to the base. W cos. i AC From (d), 1 AB' or, the reaction of the plane is to the weight as the length to the base. 167. PROP. Required the conditions of equilibrium of two bodies resting on two inclined planes having a common summit, the bodies being connected by a cord passing over a pulley at the summit. 1B Let W and W1 be the 4 v \tt' weights of the bodies, and C*,~ a i, i1 the inclinations of the planes. If t be the tension of the cord, we have for equilibrium on the plane AB A;D D C (Art. 165), t=W. sin. i, on the plane BC t=W1 sin. i1... W sin. i=WI sin. il, BD. BD or. BC' or or, the weights are proportional to the lengths of the plane3 (,n which they rest respectively. T 1H1 W E D G X, 103 EXAMPLES. Ex. 1. What force acting parallel to the base of the plant is necessary to support a weight of 50 lbs. on a plane inclined at an angle of 150 to the horizon? Ex. 2. If the weight, power, and reaction of the plane are respectively as the numbers 25, 16, and 10, find the inclination of the plane, and the inclination of the power's direction to the plane. ~ VI. TIE WEDGE. 168. The wedge is a triangular prism whose perpendicular section is an isosceles triangle. The dihedral angle formed by the two equal rectangular faces, is called the angle of the wedge. The other rectangular face is called the back. It is used to separate the parts of bodies, by introducing the angle of the wedge between them by a power applied perpendicularly to the back. The equal rectangular faces are regarded as perfectly smooth, in which case the only effective part of the resistance must be perpendicular to these faces. 169. PROP. To determine the conditions of equilibrium in thI wedo e. Let ABC be a section of the wedge perpendicular to the angle or edge A. Draw AD bisecting _D B C the angle, and let BAD=CAD=a. Let 2P be the power applied to the back BC of the wedge, which must be in equilibrium with the pressures R on the two faces AB and AC. If an equilibrium exist, the forces 2P, R, and R must meet at some point in AD (Art. 74). Resolving the forces vertically, A or in the direction AD, we have 104 STATICS. 2R11. sin. BAD-2P- 0, or P=R. sin. a, P BD 3DI - the back of the wedge or s' 1:1. ~ —... ri B BB 3A./ thce of the wedge where I- the length of the edge or breadth of the face. 170. In the foregoing investigation of the theory of equilibrium in the wedge, we have omitted the consideration of the friction, and have supposed the power to be a pressure; whereas, in practice, the wedge is kept at rest by friction alone, and the power arises from-percussion. The following problem will serve to elucidate the theoretical view here taken of the wedge. 171. PROB. A heavy beam is attached, by a hinge at one end, to a smooth, horizontal plane, while the other rests on the smooth face of a semi-wedge. Required the horizontal force necessary to keep the wedge from moving. Ad> /. C Let DE be the beam and BAC the,.wedge, \ LBAC a, LADE =, K, YE"'b/cz( the length of the _.,[ / ac6 beam, g the center of 0) >;B A gravity, and Dg=a. The wedge is kept in equilibrium by the pressure of the beam upon it at E, and the horizontal force P acting upon it at some point H. The beam is kept in equilibrium by its weight w acting at g and the reaction R of the face of the wedge at E. Taking the moments about D for the equilibrium of the beam, we avoid expressions involving the unknown thrust R' and have w.Dg. cos. ADE-R.DE. sin. DER= O, or w.a. cos. 3-R.L. cos. (a-3)=O. a cos. 3. R=w. Icos. (a-0)' But the principles of the wedge give P=R. sin. a. THE SCR EW. 1o0 a sin. a cos. 2 I cos. (a-a)' By an examination of this value of P, it will be se en that the power necessary to keep the wedge from moving will diminish as the wedge advances beneath the beam. ~ VII. THE SCREW. 172. If we divide the rectangle ABCD into equal parts by the lines mn, m'n', &c., parallel to AB, and draw the parallel diagonals of the rectangles thus 77/ formed, and if we suppose the n7: 72'-, whole rectangle to be wrapped,e' 7 /' round the surface of a cylinder, ";'- t/ the perimeter of whose base is equal to AB and altitude to BC, c, the diagonals of the rectangles will trace on the surface of the cylinder a continuous curve, which is called the helix. If a projecting thread or rib be attached to the cylinder upon this curve, we have the screw, sometimes called the externa2 screw. Similarly, if we take a hollow cylinder of exactly the same radius as the solid one, and generate a groove in the same curve, we have the internal screw or nut. The screw works in the nut, either of which may be fixed and the other movable. 173. PRop. To determine the conditions of equilibrium in ti,.e screw. F:E D From the construction of the screw, it appears that the thread Q B i-f it is an inclined plane, of 1i' 3 a which the base is the circum-.k C ference of the cylinder and the height the distance between the threads. The force is generally applied perpendicularly to the end of a level inserted into the G06 STATICS. cylinder, and in the plane perpendicular to tne axis of the aylinder. The power P thus applied, in turning the sciew round, produces a pressure on the threads of the screw in the direction of the axis of the cylinder. In case of equilibrium, let the counterpoise of this pressure be W. Let ABC be the inclined plane formed by unwrapping one revolution of a thread, and let w —th part of W, be supported at a, q being the same part of Q, the force applied at the circumference of the cylinder. Put FD = a, ED the radius of the cylinder -r and angle BAC=-a. Then, Art. 133, FD_ a 2ra (a).ED Y-'P. r P. 27r' By Art. 166, q= I. AC Q W distance of two threads n n circumference of the cylinder' and the same holds for each of the other portions of W at the other points of the plane. Therefore, we have distance of two threads circumference of the cylinder' 27ra circumference described by P By (a), -.27rr circumference of cylinder W circumference described by P P distance of two threads' or, the power is to the weight as the distance of two threads is to the circumference described by the power in one revolution of the screw. It will be seen that the ratio of P to W is independent of the radius of the cylinder. EXAMPLES. Ex. 1. What force must be exerted to sustain a ton weight on a screw, the thread of which makes 150 turns in the height of 1 foot, the length of the arm being 6 feet? Ex. 2. Find the weight that can be sustained by a powce; B A LA NC:ES, E T C. -10 of -1 lb. acting at the distance of 3 yards from the axis of the screw, the distance between the threads being 1 inch. Ex. 3. What must be the length of a lever at whose extremity a force of 1 lb. will support a weight of 1000 lbs. on a screw, whose threads are 4 inch apart? V TIII. BALANCES AND COMBINATIONS OF THE MECHANLCAL POWERS. 174. The common balance, in its best form, is a bent level, in which the weight of the lever must be taken into consideration. In the annexed figure the points A and B, from which the scale-pans and weights are suspended, are called the points of suspen- B sion; C is the fulcrum, being the lower edge of a prismatic rod of A steel projecting on each side of the beam; when the balance is in use, these edges on each side of the beam, as at C, rest on hard surfaces, so that the beam turns freely about C as a fulcrum. 175. The requisites of a good balance are, 10. That the beam rest in a horizontal position when loaded with equal weights. 20. That the balance possess great sensibility. 30~. That it possess great stability. 176. PROP. To determine the conditions that the beam rest in a horizontalposition when loaded with equal weights. Supposing the beam horizontal, in case of equilibrium, if we neglect the weight of the beam, the moments of the weights must be equal, and, therefore, the arms must be equal (Art. 133). 108 s T A T I C S. But taking into consideration the weight of the beam, its ceit ter of gravity must be in the vertical through C, the center of motion (Art. 110), and, in order to this, the beam must be symmetrical on opposite sides of the fulcrum (Art. 97). The line AB, joining the points of suspension, is obviously bisected by the vertical through C and the center of gravity of the beam, and the point of intersection, for reasons which will appear, should be below C. 177. PROP. To determine the conditions that the balance may possess great sensibility. Let C, A, and B be the fulcrum and points of suspension, as in the preceding figure, and 3c Cda _ join Cg, the centers of mo- Cdct N_ tion and of gravity. Cg is D g perpendicular to AB and biV sects it, if the beam is constructed in accordance with the first requisite. Let M C E N be a horizontal line A P Q through C, meeting AB in E making with it an angle equal to 0. The sensibility is measured by the amount of deflection 0 of the line AB from a horizontal position by a given small difference P-Q of the weights. Draw the vertical lines Dd and ga, and put AD=BD-=a, CD=d, Cg=h, and weight of the beam =w. Now MN is bisected in d, and Md=a cos. 0, Cd=d. sin. 0, and Ca=h. sin. 0. If the system is in equilibrium, the moments about C give P.CMI-Q.CN-w.Ca=0, or P.(Md-Cd)-Q(Nd+Cd)-w.Ca=O, or (P-Q)a. cos. 0.-(P+Q)d sin. O-w.h. sin. 0=0, or (P-Q)a- (P+Q).d+w.h tan. 0=0. tan. O= (P-Q).a (P+q)d+wv.h Hence the angle 0, and, therefore, the sensibility, is increased for given values of P and Q, by increasing the lengths of the B AL A NCES ETC. 109 arms (a), by diminishLing the weight of the beam (to), or by diminishing the distances of the fulcrum from the center of gravzty of the beam (h) and from the line joining the points of suspension (d). 178. PROP. To determine the conditions that the balance may possess great stability. If the balance be loaded with equal weights and disturbed firom its position of equilibrium, the rapidity with which it returns to that position is a measure of its stability. But this rapidity of return to a horizontal position will depend upon the moment which urges it back. But this moment is, since P-Q, P.CN-P.CM+vw.Ca, or P.2.Cd+w.Ca, or (2P.d+w.h) sin. 0. Hence, for given values of P and 0 the stability is greater, as d, h, and w are increased. 179. CoR. Hence, by increasing the stability, we diminish the sensibility, but the sensibility may be increased by increasing the length of the arms, without affecting the stability. For commercial purposes, when expedition is required, ana the material weighed is not of great value, sensibility is sacrificed to stability; but for philosophical purposes great sensibility is required, and stability is of little comparative importance. TIlE STEELYARD BALANCE. 180. The steelyard balance, or Roman steelyard, is a level of the first kind, with unequal arms. The body C W to be weighed is hung -OB at the shorter arm A; and a given constant weight P is moved along the other arm till it balances W; then the weight of XW is knowin from the place of the counterpoise P. 110 ST AT I CS. 181. PROP. To determine the law according to which tih longer arm of the steelyard must be graduated. Let G be the center of gravity of the beam AB and w its weight. Put OA=p, OC=2p, OG=g, and taking moments about the fulcrum O, in case of equilibrium we have Wp-wg-Pp =0. wg Wp... p+ p p=* Taking Ox, a fourth proportional to P, w, and g, we have wg Wp + —p=CO+Ox=Cx= —. Hence the distance from x to the counterpoise P varies as the weight; and if the weights be in arithmetical progression, the distances xl, x2. x3, &c., will also be in arithmetical progression. THE BENT LEVER BALANCE........ 182. This balance is represented by the annexed figure, where ABC is the bent lever turning about a pivot at B. A scale (E) hangs from A, and at C an index points to some division on the graduated arc GCF. 183. PROP. To determnine the principle of graduation in the bent lever balance. Let g be the center of gravity of the beam at which its weight w acts. The weight of the scale (E) and the wv;ght (W) of a body placed in it will act vertically through A. Let the horizontal line through B meet the vertical lines thrr ugh g and A respectively in D and K. Then the moments roonet B, in equilibrium, give w.DB- (E +W)BK O. (a) BALANCES, Er C. 11 As greater weights are put into the scale E, the point A approaches more nearly the vertical through B fiom the bent form of the beam, or BK diminishes, while BD increases. Suppose the point A to be at K when the scale is unloaded. and let, in this position of the beam, the angle DBgz=0. When a weight is put into the scale the point A will descend through some angle q, and the arm Bg will rise through the same angle. In this new position the angle DBg will become 0-~. Let B1g=a and BA=b; then DB=a cos. (0-p), and BK=b cos. ~. These substituted in (a), give w.a. cos. (0 —)-E.b. cos. ~-W.b. cos. ~=0, or w.a. cos. 0 cos. b+w.a sin. 0 sin. O-E.b cos. p-W.b. cos. 0=0, or w.a. cos. 0+w.a sin. 0 tan. ~-E.b - W.b=O. W.b E.b- w.a cos. 0 *tan. w.a. sin. 0+ w.a. sin. 0 Hence, tan. ~ varies as W, and the limb GF must be divided into arcs whose tangents are in arithmetical progression. Practically, the limb may be graduated from the positions of the index at C for a succession of weights put into E. This instrument possesses great stability. ROBERVAL S BALANCE. 184. This instrument is of greater interest from its paradoxical appearance than from its use as a machine for weighing bodies. Its discussion affords an interesting application of the doctrine of couples. It consists of an upright stem upon a heavy base A, L c a - - with equal cross-beams II turning about pivots at aa ( _ ] and b. These cross-beams b are connected by pivots at c, d, e, and f, with two oth- L A er equal pieces in the form 112 STAT' CS. of a T. The weights are suspended frornm he horizontal armi of the latter pieces. 185. PRor. In Roberval's balance, equal weights balance at all listances from the upright stem. P2 r3 Let the letters in the annexed figure indicate the same parts as Q2 -i d2s in the former. B 271. c Let equal and opposite forces P, and P2. e b b<~ each equal to P act in QJ1.:B ec, and, similarly, let P3 and P4 act in df. l P >1. / 4 These forces PI, P2, P3, and P4 do not dis-,urb the equilibrium. Now P at B, as in Art. 71, is equivalent to P, at e and the couple P, Bh, P., and, similarly, P at C is equivalent to P4 at f and the couple P, Ck, P3; and since P, at e balances P4 at f, we have only the two couples to dispose of. Now for the couple P, Bh, P2, Arits. 58 and 59, we may substitute a couple Q1, ec, Q., in its own plane and of equal moment, in which the forces Q, and Q2, acting in the directions of the cross-beams cd and ef (which always remain parallel to each other as they turn on the pivots a and b), are destroyed by the resistance of the pivots a and b. Similarly, the couple P, Ck, P3 may be replaced by the equivalent couple Rl,fd, R2, in which R. and R2 are destroyed by the resistance of b and a. These new couples, therefore, do not disturb the equilibrium, and the original forces P at B and P at C must be in equilibrium. If the beams cad and ebf be moved round the pivots into any oblique position, the same reasoning would apply, and the equilibrium still subsist. CoR. Unequal weights can not balance from whatever points suspended BALAN CES, ETC. 11I 186. Piop. To determine the ratio of thel power to the weight in a combination of levers. Let the power P act at A, A and the weight W at B"'. _ The first three levers are of W' wB C the second kind, and the last A' one of the first kind, the ful- WA crums being at C, C', C", and Iw/ C"'. Let BA', B'A", B"A"' N" c" B' be rigid rods connecting the'evers, and let the action of the first lever on the rod BA' be W', which becomes the " power acting on the second, WV" and W"' the weights to the second and third respectively, and powers to the third and fourth. P CB W' C'B' WV"' C"B"' "'B" T Vhen WCA, W -. and W;WVY-CA9 W/-CRAM9 WA /V-C/A//9 ttA -CRO/AK @ Taking the continued product of these equations member,y member, we have P CB x C'B' x C"B" x C"'B" W -CAX C'A' x C"A" x C"'A"" or, the ratio of the power to the weight in the combination is equal to the product of the ratios of the power to the weight in each lever. 187 PROP. To determine the ratio of the power to the weight in the endless screw. This machine is a combination of the screw and wheel and axle. Let P be the power applied to the handie of the winch, W' the pressure of the screw on the teeth of the wheel, and WT the weight suspended from the axle of the wheel. Then II 141 STATICS. P distance between the threads of the screw'W' circumference described by the power' W' radius of axle and W = radius of' wheel' P distance between two threads of screw'W circumference described by the power radius of axle Xradius of wheel' or, the ratio of the power to the weight in the endless screw zs equal to the product of the ratios of the power to the weight in the screw and in the wheel and axle. 188. Pino. To determine the ratio of the power to the weight zn any combination of the mechanicalpowers. Let P= the power for the whole combination, W" — " weight " " "s W' =" " 4 to the 1st in the series and power to the 2d, W_= 6" " " 2d " " " 3d, &c., &c., &c. Let a,= the ratio of the power to the weight in the 1st, a = - " " 2" " " 2d, a3= c " " 3d, &c. P W' W" Wn-1 Then =a =a W,=a3. =a,,; and, taking the product, P W, —l.a.a3 *a..... Hence the ratio of the power to the weight in any combination of elementary machines is equal to the product of the rctios of the power to the weight in each of the simple machines. 189. PROP. To determine the ratio of the power to the pressure in the combination of levers called the knee. This combination of levers is used with advantage where very great pressure is required to act through only a very small space, as in coining money, in punching holes through thick plates of iron, in the printing-press, &c The lever A I nBALA NCES. ETC. 115 turns about a firmly fixed pivot A at A, and is connected by another pivot at B to the rod BC, whose extremity C produces the pressure on the obstacle at E. N Let the power P act horizon- B1 tally at some point F in the lever AB, ANC be a vertical line meeting the direction of P in N, and DE a horizontal plane, on L which, at E, is the substance D e subject to pressure. Let R= the reaction of the rod BC in the direction of its length, AMI, DL perpendiculars upon its direction from A and D, and W the vertical resistance of the substance at E. Taking the moments about A and D in equilibrium, we have P.AN-R.AM, and W.DE-R.DL, P AMXDE or W -AN X DL' WVhen BC becomes nearly vertical, DL becomes nearly equal to DE, and AN to AF, while AM becomes very small. P AM In this situation, W-AF nearly, so that P is a very smiall fraction of W. CHAPTER VII. APPLICATION OF THE PRINCIPLE OF VIRTUAL VEIOCITIES TO T1aI MECHANICAL POWERS. 190. IN Arts. 80 and 81, it is shown that the principle of virtual velocities obtains for all cases of equilibrium of a free body under the actions of any number of external forces in the same plane. In the mechanical powers, the parts by which the actions of the forces are transmitted being rigid or inextensible, the forces may be considered as acting in the same plane, and the internal reactions and tensions will not enter the fundamental equation v.P.v=O. Also, the virtual velocities of the supporting parts will in general be zero for the possible displacements of the system. In some of the mechanical powers, the principle applies to all possible displacements, however great, since they must be in the direction of the forces. This is true in the wheel and axle, toothed wheels, pulleys with parallel cords, the inclined plane, the wedge, and screw. In the lever, and pulleys with inclined cords, the displacements must be taken indeniitel/Y small. 191. PRor. The principle ('j /~ d a, B virtual velocities obtains in ti, c n / wheel and axle in equilibrium. The forces which act on lh:~ AS\ wheel and axle are the bpowei P, the weight NW, and the reacP tion R of the steps which support each end, C, of the pivot about which it turns, and, in conP' Yw' sequence of the rigidity of the PRINCIPLE OF VI RTUAL'ELOCITIE r, ETC. 1"17 system, they may be considered as acting in the same plane. Also, the wheel and axle receiving a displacement turning about C, the virtual velocity of R equals 0. Let A and B be the points at which the cords left the wheel and the axle respectively before displacement; A', B' afterward. Then W ascends through the space WW'= are BB', and P descends through PP'-arc AA'. PP' is the virtual velocity (Art. 78) of P, and positive; WW' is the virtual velocity cf W, and negative. Hence (Art. 81), P.PP'- W.WWV' 0,.or P. are AA'-W. are BB'=O, or P. AC. L A'CA- W.BC. L BCB'-0. W AC P. AC- W. BC = o, or W= P - -BC' the condition of equilibiium found in Art. 138. 192. PROP. Thie principle of virtual velocities obtains in a pair of toothed wheels. Let the circles in the annexed figure represent the pitchlines of the wheels (Art. 144), and D 1, C' D 2 the points which were in contact in the line CC' before displacement. Since the pitch- lines roll on each oth- er without slipping, P are DD, arc DD2, T~t and P's displacement =PP'=AC'. D C'D=AC'. D, CD are D.D W's displacement =WW -CB.LzDCDCB. DBy the principle of virtual velocities, P.PP'- W.WW =0, 'ils ST AT IC S.'Care DD C arc D D or P.AC'. =0 W.C. ~C'D CD -09 AC' CB or P.DC' W.-D=o. W AC' CD W CD'" P-BC'C'D; and wvhen AC'-=BC, -'' as found in Art. 143. 193. Pror. The principle of'virtual velocities obtains in the single movable pulley with parallel cords. If the pulley A be raised to A we shall have AA'=WW'-=PP', P since each of the cords passing round the pulley A must be short. ened by a length =WW'. WW', the virtual velocity of W, is negative..P. P.PP'-W.WW1'=o, A or P.PP'-W. 4PP' =0. Hence -=2, as found in Art. 158. V''P 194. PRoP. The principle of virtual velocities obtains in the first system of pulleys. In the figure of Art. 160 we see that, if WV be raised through a space WWV', each of the n cords at the lower block will be shortened by the same quantity, or that P will descend through a space n.WW'. Hence the equation of virtual velocities P.PP'-W.WW'=O becomes P.n.V W'-W.WWV'=C r -=n, as in Art. 160. PRINCIPLE JF VIRTUAL VELOCITIES, ETC. 119 195. PROP. The principle cf virtual velocities obtains in the second system of pulleys. Referring to the figure in Art. 161, we see that, if P descends through the space PP', PIP the pulley an would rise through a space 2-, PPP &c, &c., PP/ a2 2n-1l PP, 6" al or the weight W " And the equation of virtual velocities P.PP'-W.WWV'=u be.comes PP, P. PP, - W..- O W ~or — =2n, the same as in Art. 161, Cor. 2. 196. PROP. The principle of virtual velocities obtains in the third system of pulleys. Referring to figure of Art. 162, and designating the pulleys as in that article, we see that, if W be raised a space =WW', each cord will be shortened by a space equal to WW'. The highest movable pulley an-_ will descend a distance =WW'. The next pulley a,,_ will descend a distance =2.WW' by the descent of a1_, and a distance WW' by the elevation of W, on will descend on the whole (2+I)WW'. Similarly, the pulley a,, will descend through t2(2+1)+1 }W.W'=(22+2+1)WWI. Proceeding in the same way, we find that pulley a(.,,_, or a,3 will descend through the space 12G S T ATICS. (2 -+ 2w-5+, &c.... 2+ 1)WW, and a through the space (2-3+2"-+, &c.... 2+ I)WW, and a, " " (2"-n 2"+n-3, &c.... 2+~l)WW', and P will descend through twice the last found space by the descent of the pulley al, and through the space WW' by the elevation of the weight; or PP'=WW/'. 2(2n"-+2.3+, &c... 2+1)+1j =WW'.(2n"-l2"-t —-, &c... 22+2+1) =WW' (2"- 1). The equation of virtual velocities is P.PP'-W.VW'= 0, which becomes P.WW'(2n- 1)-W.WWV=O, W or -=2" —1 as in Art. 162. 197. PROP. Tile principle of virtutal velocities obtains in the inclined plane., -P Let the force P make any angle s with the plane, a= LBAC, a the first position of the body whose weight is W, a' the posi a!~k ~ ZWb tion of it after displacement. A \:aC,~ C Drawing the perpendiculars av, a'u, we have -a'v=-a:/n' cos. S= the virtual velocity of w' l T P, and au=aa'. sin. a= the virtual velocity of W. By the equation of virtual velocities, P.a'v-W.au=O, or P.aa'. cos. E-W.aa'. sin. a= 0. W sin. E Hence, P sin. a' as found in Art. 164. 198. PROP The principle of virtual velocities obtains in the wedge. PRINCIPLE OF V IRTUAL VELOCITIES, ETC. 121 Let 2P be the whole power, R r? and R the pressures perpendicular to the faces of the wedge ABC, which produce equilibrium. Let a the wedge be displaced to the posi-:B'.' a' /..r tion A'B'C'. The displacement of tne point of application of P is aa'= AA'; that of b, the point of application of R, is bb'=Am, a perpendicular from A on A'B', and ARmz=AA' I 1E BAC sin.. The equation of virtual velocities is P.aa'-R.bb'== O, BAC or P. AA'- R.AA'. sin. = 0. P BAC' sin. - 2 as found in Art. 169. 199. PROP. The p-inciple of virtual velocitzes obtains in a lev,ri of any for-m. Let ACB be the lever be- c fore displacement, A'C'B' 13 its position afterward. A From A' draw A'v perpendicular to AP, and firom B',' A' B'u perpendicular to BQ produced. Av is the vir- Pi tual velocity of P, and B P Q that of Q. Now when the displacements are indefinitely small, the circular arcs AA', BB' become straight lines, and Av=AA' cos. A'Av=AC./LACA'. cos. (PAC-90~) =AC. sin. PAC. ACA'; Bu=BC. sin. QBC. LBCB', and Z ACA' = Z BCB'. The equation of virtual velocities is 122 S. TATIC S. P.Av- Q.Bu=O, or P.AC. sin PAC-Q.BC. sin. QBC-O. P BC. sin. QBC -Q AC. sin. PAC' as found in Art. 134. 200. PROP. The principle of virtual velocities obtains in thi single movable pulley with cords inclined. B c Let A be the point where, the cords produced would meet at the first position of the pulley, when P and W are the positions of the power and weight. Let P be displaced to -iii~ T 7 P P', when the weight is XAi.- 8 raised to W', or the point )(W f p, of meeting of the cord is raised to A'. Draw the Wy circular arcs A'm, A'n with centers B and C. When the displacement is indefinitely small. the arcs A'm, A'n become straight lines, and Am=AA' cos. BAA'=An, BAC PP'=Am+An=2AA' cos. —, WW'=AA'. The equation of virtual velocities is P.PP'-W.WW'w=o, which becomes BAC P.2AA'. cos. 2C W.AA'=0, W BAC or -p-=2cos. 2' as found in Art. 158. SCHOLIUM. In the preceding propositions, the expressior P.PP'=W.WW', PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 12~ WW' P orPP -- W' explains the principle that, "in using any machine, what we gain in power we lose in time." For, in order that W may be moved through any given space, we must have the space moved through by the power P, increased in the same ratio tnat P is diminished. CHAPTER VIII. F I I C T IO N. 201. TnIE surfaces on which bodies pressed have hitherto been regarded as perfectly smooth, so that they offered no resistance to motion parallel to themselves, their only reaction being perpendicular. When rough surfaces are in contact, the motion, or tendency to motion, parallel to the surfaces, is affected by the roughness, and the effect is called Friction. Friction may be divided into two kinds: sliding friction, when one rough surface slides on another, and r olling friction, when one rolls on the other. The former only will be consid-.red here, under the term Statical Friction. 202. The laws of friction are determined by experiment. If the body A rest upon the perfectly smooth [A ----- plane BC, the smallest C possible force applied to it will cause it to move. But if the body or the plane, or both, be rough, a force within certain limits of magnitude may be applied to it without causing motion. The greatest force which can be so applied to the body in the direction of the plane will measure the friction. Let W be this force, acting by a cord over a pulley on the body A, F being the opposing force of friction. Then F=XV. 203. The following laws of friction are deduced by this or some similar process. 1~. The friction of the same body, or a body of the same material, wh/en the weight is the same, is independent of the extent FRICT I N. 125 of the surfaces in contact, except in extreme cases, whele the weight is very great compared with the surfaces in contact. Thus the friction of the body A will be the same whichever side rest on the plane, or whatever be the form within the excepted limits. 2~. The friction is-proportional to the pressure on the plane, or the reaction of the plane, within moderate limits. If other weights, as m, be placed on A, W or F will vary as the whole reaction R of the plane. Con. The friction is therefore some function of this pressure, and we may represent it by /tR. M, is called the coefficient of F friction, and is equal to R, or the ratio of the friction to the reaction of the plane. 204. PRoP. The coefficient of friction between tiwo given substances is equal to the tangent of the inclination of the plane formed of one of the substances, when the body formed of the other is about to slide down it. Let the inclination a of the plane AC be increased till the body a is just on the point of sliding down it. The body a will then be in equilibrium c from the normal reaction of \C the plane Rt, the friction tR A acting up the plane, and its weight W acting vertically downward. -w Resolving parallel and perpendicular to the plane, we have pR —W. sin. a=O, R-W cos. a=0. Eliminating R, we have t1 cos. a-sin. a=O, or t= tan. a. 205. PIOP. To determine the limits of the ratio of P to W on an inclined plane, when friction acts up or down the.plane. 126 ST ATI CS.. -) - pLet the power P make. ar 13.. 4 *ItR angle E with the plane whose ij'~ < inclination is a, and W the weight of the body. 1~. To determine the greatest value P can have without causing the body to move up the plane. In this case, the friction 1-R, opposing the motion up the plane, will act down it, and, resolving parallel and per pendicular to the plane, we have P. cos. E-/uR-W sin. a-O, (a) P. sin. e+ R- W cos. a=O. (b) Multiplying (b) by [t and adding to (a), we have P (cos. e+~t sin. s)-W. (sin. a+t* cos. a)= 0, W. (sin. a +s cos. a) or P cos. E+t*t sin. e 20. To determine the least value P can have without causing the body to move down the plane.' In this case the friction will oppose the descent, and will therefore act up the plane. Hence equations (a) and (b) become P. cos. s+bR-W sin. a=O, (a') P. sin. E+ R-Wcos. a=O. (b') Multiplying (b') by t* and subtracting, we find W. (sin. a-It cos. a) cos. E-ft sin. e No motion will take place while the value of P is between these two, which are its limits. These two values of P may be combined so as to take the form WV cos. E-t, sin. E P sin. a-t*- cos. a' in which the upper sign is to be taken when friction acts down the plane, and the lower when the friction acts up the plane. 206. Pnop. 7'o determine the limits of the ratio of P to W in the screw, when. friction acts assisting the power or the weight. FRICTIO N. 127 Proceeding as In rt. p E D 173, let ABC be the inclined plane formed R by unwrapping one revolution of the P thread, the angle BAC q =a; let W= the Rk whole weight sustain- -_ ed by the screw, w= that part of it supported at a, Q= the whole force acting at the circumference of the cylinder, r=ED the radius of the cylinder, a=FD the lever at which the power acts, and q that part of Q which supports w at a. Then Q=P-. The forces which are in equilibrium at a are the weight w, the reaction R, the horizontal force q, and the friction FjR acting up or down the plane. Resolving parallel and perpendicular to the plane, we have q. cos. a-t=iR-w. sin. a=O, (a) q sin. a- R+w cos. a=0. (b) Multiplying (b) by /, and adding and subtracting, we have q(cos. a~(4- sin. a)-w(sin. a:fL cos. a)=O, q Q sin. a:F:t cos. a or w- W-cos. a-+- sin. a P r sin. aiy, cos. a W a'cos. a-+- sin. a' The two values of this expression give the limits required, P and W may have any intermediate value. 207. COR. From the two preceding propositions it will appear that, when we have obtained one of the limits, the other may be had by simply changing the sign of y in the former. [L2S3 ZS T STATICS. EXAMPLES. Ex. 1. A uniform straight beam rests on a rough cylinder of given radius; required the greatest weight that can be suspended from one end of the beam without causing it to slide off. Let g be the center of gravity of the bC eam BC, whose length is 2d, and B3 =dsince the beam is uniform, w= its k S / vA weight, and W the weight suspended A' from B. Before the weight W was suspended from the beam the point g' o / must have been at A. Let A' be the point of contact with the cylinder when the beam is on the point of sliding off, a the angle it then makes with the hol'wT 5 ('} rizon, and r= the radius of the cylinder. Resolving parallel and perpendicular to the beam, we have 1tR- w. sin. a- W sin. a O, (a) R-w cos. a-W cos. a=0; (b) whence -= tan. a. Taking the moments about A', we have IW.BA' cos. a-w.A'g. cos. a=O, or W. (Bg-A'g) -w.A'g=O. But A'g= are A'A- radius X angle AOA'=ir.a..W. W.(d-?ra)-w.raz=O, w.ra w.r tan.- t o r W-, tan_ the weight required. d-ra d-r tan.- In Ex. 2. A ladder rests with one end on a rough horizontal plane, and the other on a rough vertical wall; given I= its length, d= the distance of its center of gravity from its lower end, /t and /t'= the coefficients of friction on the horizontal and vertical planes respectively; required its inclination 0 when on the point of sliding down. Let AB be the ladder and the fores acting upon it, as represented in the figure. HXAMPLES ON CHAPTERS V. VIIl. AND VIII. 129 Resolving horizontally and rertically, we have 11.B R' —tR=0o, (a) R+j!'R'-w=0. (b) Multiplying (a) by /', and subtracting firom (b), we find w = R(I +') and from (a), R'=/tR. Taking moments about A, we get w.d. cos. 0-R'.l. sin. 0-t'/R'.1. cos. 0=0, or tan.= R'l Substituting the values of w and R', tan. 0= d(1.-(I+L') —IpL' If the center of gravity of the ladder be at the middle point, i=2d, and tan. 0= 208. EXAMPLES ON CHAPTERS VI., VII., AND VIII. Ex. 1. A beam 30 feet long balances on a prop ~ of its length from the thicker end; but when a weight of 10 lbs. is suspended from the other end, the prop must be moved 2 feet toward it to maintain the equilibrium. Required the weight of the beam. Ex. 2. The forces P and Q act at arms a and b respectively of a straight lever, which rests on a fixed point to which it is not attached. When P and Q make angles a and / with the lever, required the conditions of equilibrium. Ex. 3. A uniform beam is sustained by three persons, one at one, end, and the other two by a hand-spike placed at some I I0 O ST ATICS. point beneath it. At what point must the hand-spike be placed that each person may sustain one third of the weight? Ex. 4. A Roman steelyard, whose weight is 10 lbs., has its center of gravity 2 inches from the fulcrum, and the weight to be determined is supported by a pan placed at a distance of 3 inches on the other side. Find the respective distances from the fulcrum at which the constant weight of 5 lbs. must be placed, in order to balance 10, 20, 30, &c., lbs. placed successively in the pan. Ex. 5. Find the ratio of the power to the press,'~e ill the comnmon vice. Ex. 6. Find the ratio of the power to the pressure in the Screw by the principle of virtual velocities. 7z. 7. An isosceles triangle, whose base is to one of its equal sides as 1: /7, is placed with its base on an inclined plane; and it is found that, when the body begins to slide, it also begins to roll over. Find the coefficient of friction. Ex. 8. A ladder rests against a vertical wall, to whl-ich it is inclined at an angle of 450; the coefficients of fiiction of thes wall and of the horizontal plane being respectively. and }, and the center of gravity of the ladder being at its middle round. A man whose weight is half tihe weight of the ladder ascends it. Find to what height he will go before the ladder begins to slide. Ex. 9. In a uniform lever of the second kind, whlic1h weighs 2 ounces per inch, required the length of the lever, in order th'atthe power may be the least possible when in equilibrium with a weight of 48 ounces placed at a distance of three inches firom the fulcrumn. Ex. 10. Two given weights, P and Q, are suspended from two given points in the circumference of the wheel, a being the angle made by the radii drawn to the points of suspensioA. Requii'ed the angle 0 which the lower radius makes with the vertical when the weights cause the greatest pressure on the axle. D YIN A M11 I C S. INTRODUCTION.-DEFINITIONS. 209. IN Statics we have investigated the relations of the in. tensities and directions of forces necessary to produce equilibrium, this result being entirely independent of the time during which the forces act In Dynamics, forces are regarded as producing motion or change of motion in bodies, and these effects'must obviously depend on the duration of the action of the forces. In dynamics, therefore, time becomes an element in our investigatiops. 210. MOotion is the transit of a material point or body friom one position to another in space.'211. The absolute motion of a body is its transit from one fixed point in space to another. i212. Relative motion is a change of distance firom a point which is itself in motion. All motions are relative in any practical view which we can take of them, since we have no means of determining the absolute rest of any point in space.:'213. The velocity of a body is its r-ate of motion. It may be uniform or variable. The velocity of a body is uniform when it passes over equal spaces in equal times, and is measured by the spacepassed over in the unit of time. Let v be the velocity, or space passed over in one second, then the space described in two seconds will be 2v, in three seconds 3v, and so on; and if s be the space described from the commencement'of motion, and t the number of seconds also reclioned fi'om the commencement of motion. then 132 D Y N AM I C. s-vt. LI.] The units of time and space are arbitrary. It is usual to take one second for the unit of time, and one foot for the unit of space. When no mention is made of different units, these will be understood.'/214. Variable velocity is that which continually increases or decreases, so as to be the same at no two successive instants. To find a measure of the velocity of a point or body so moving, let us assume, at first, that the velocity which the body has at the end of t1 seconds is uniform from t1 seconds to to seconds, a very small interval, the space passed over at the end of t, seconds being s5, and the space passed over at the end of t2 seconds being s2. Then, using the symbol A to signify finite difference, by [I.] we have s -Sl As v = 2 —t, -At' If, however, there is no time, however small, during which the velocity is uniform, then the smaller we take At, and consequently As, the more nearly t. [II.] That is, the velocity, when variable, is measured by the limit of the ratio of the space described to the time of describing it..i 215. Relative velocity is the velocity with which two bodies approach or recede from each other.' 216. Matter at any given moment must be in one of the two states, motion or rest. The inertia of matter is the entire absence of power in itself to change this state. It implies equally a disability, when in motion. to change its rate or its direction. Hence A body, when not acted on by any external forces, if at rest, will remain so, or, if in motion, will continue to move in a straig1M line and with a uniform velocity. This is called the first law of motion. 217. It is a consequence of the inertia of matter, that when a force is applied to a body to move it, each of its pa.rticles opposes a resistance to motion in directions parallel but oppo. DEFI IT IONS. 133 site to the direction of the applied force. The center of these parallel forces (Art. 44) of resistance is called the center of inertia of the body. It is the same point which, in statics, was called the center of gravity of the body in reference to the force which was there supposed to act on the body. The sum of these parallel resistances, or their resultant (Art. 43), is obviously proportional to the number of particles in the body or to the whole mass. Hence the inertia of a body is a surer test of the quantity of matter or mass of a body than its weight is; for the latter (Arts. 84 and 88) varies by a change of position on the earth, while the former is always the same. -'218. The path of a body is the line, straight or curved, which its, center of inertia describes when it passes from one point to another in space. v219. A body is said to be free or move freely when its path depends on the action of the impressed forces only. Its motion is'said to be constrained when its path is limited to a given line, straight or curved, or limited to a given surface. 2S20. An impulsive force is one which acts instantaneouslv flnd without sensible duration. t' 221. An incessantforce is one which acts without intermission. If a material point move from rest by the action of an incessant force, its rate of motion or velocity must continually increase. The amount of this increase, or the increment of the velocity in the unit of time, will obviously be greater or less as the intensity of the force is greater or less. This increment of the velocity in one second is therefore the measure of the intensity of the force. L' 222. A constant force is an incessant force whose intensity is at all times the same. If a material point move by the action of a constant force, the increments of the velocity in each successive unit of time must all be equal, and each increment will be a measure of the force. If, therefore, we put, for the increment of velocity, or the velocity generated by the force in one second, 9 will represent the force. The increments being all equal, the velocity generated in two seconds will ha, 134 DYNAM' CS A M C 2~, in three seconds 3~, and so on. Hence, if be the vel}cit) generated in t seconds, v=~t. [III.] 223. A variable force is an incessant force whose intensity either increases or decreases, so as to be the same at no two successive instants. To find an expression for a variable force, let us assume it to be constant from the end of the time t,, when the velocity is v;, to the end of the time to, when the velocity is v.' Then, by [III.], we have Vt -Vt A 2 1 /t If, however, there be no interval during which the force is constant, then the smaller we take At, and consequently Av the more nearly will Av a.d [Iv.] At That is, a variable force is measured by the limit of the ratio of the velocity caused by it to the time of causing it. 224. The momentum of a body is its quantity of motion, ana is measured by the product of the mass of the body by its velocity. For the motion of a single particle is its velocity, and the motion of any number of particles, having the same velocity, is obviously as much greater as the number of particles is greater.'Hence the whole motion is equal to the whole number of particles in the body, or its mass,' multiplied by their common velocity. Or, if Q be the quantity of motion of a body, M its mass, and V its velocity, Q-MVy. [V.] The mass, multiplied by the square of the velocity, is called the vis viva, or the living force of a body. / 225. In estimating the effects of incessant forces, we have considered only the acceleration or velocity which each force, will produce when acting on a free material point or a unit d'f mass.. When so measured, they are called accelerating forces. If the mass moved differs fiom that which we have called the unit of mass, and it is taken into consideration in estimating DEFIN IT IONS. 135 the effects of the forces, they are then called moving forces. Let 0 be the acceleration, or velocity generated in a unit of time, M the number of units of mass; then %, the moving force, will be measured by the quantity of motion generated in tile unit of time, or [VI.] q) Hence (b — or the accelerating force, is equal to the moving force divided by the mass. v 226. If a. body already in motion be acted on by a force in the direction in which the body moves, the superadded motion is just the same as that which would have been produced in the body if at rest when the force began to act. If the force act in a direction different from that in which the body moves. the new motion produced by the force, estimated in the direction of the force, will be the same as if the body had been at rest. If the force act in a direction opposite to that in which the body moves, the motion destroyed in the body is equal to that which the force would produce in the body if at rest when the. force began to act. These facts are consequences of the inertia of inatter, tandl will receive additional illustration in the sequel. They are embodied in the following enunciation, called the second law of motion: All motion or change of motions in a body is propor'tional to lhe force impressed and in the direction of that force. p'227. When one body impinges on another at rest or in motion, the quantity of motion, or momentum of the two bodies after impact, is the same as before impact. For matter being incapable of originating motion, can not add to the motion of other matter, or take from it except by imparting its own motion. Hence whatever motion the second body receives in the direction of the striking body, just so much must be lost by the striking body. This fact is usually enunciated as follbows, and is called tile third law of motion.. Action and reaction are equal and in opposite, directions. CHAPTER I. UNIFORM MOTION. 228. IN considering the effect of an impulsive force, we shall suppose the force applied, at the center of inertia (Art. 217). When so applied the parallel forces of resistance of all the particles situated on opposite sides of the direction of the force will balance each other, and the body will not rotate, but all its particles will describe parallel lines with a common velocity. 229. PRoP. To find the general equation of uniforr motion. p-. Let OB be the path of the body (Art. 218), v its velocity (Art. 213), and t the time of its motion in seconds. O Let 0 be the origin, or point from which we estimate the A4 successive positions of the body, s the distance of the body from 0 at the end of the time t, and OA=s1 its distance from the origin O at the commencement of the time. When the body is at B, we have [I.] AB= vt. But OB=OA+AB, Bor s=s, +vt, (36) which is the general equation of uniform motion. If the body, instead of receding from the origin 0, approach it, then v will be negative; or if the body begin to move from A', then s, will be negative. COR. 1. From (36) we obtain s-s, AB t t 9 or the velocity of a body is equal to the constant ratio of the space described to the time of describing it. CoR. 2. If we estimate the position of the body from the UNIFORM M OTION. 137 point where the body is when t=0, or suppose the space and time to commence together, then s,=0, and s=vt. (37) 230. Ptor. If two bodies move during the same time, their velocities will be proportional to the spaces described by them respectively. Taking the points of departure for the origin of spaces, we have from (37) s=vt and s'-=v't'..'.: s'=vt: v't'. And, since t=t', s: s'=v: v'. 231. PRoP. If the velocities of two bodies are equal, the spaces described are proportional to the times. As before, s: s'=vt: v't'; and if v=v', s: s'= t:t'. 232. PRoP. If the spaces described by two bodies are equal. their velocities are reciprocally proportional to the times. For, since s: s'-Vt'v't', if s-=s', vt-v't'..'. V'=t:' t. 233. PROP. An impulsive force is measurled by thie momentum it cadn produce, in any mass. If v be the velocity produced by the force F, in a body containing M units of mass, by [V.] the momentum will be Mv. Now if the mass M move from rest by the action of the force, all the motion it receives is the effect of the force. And, admitting the principle that effects are proportional to their causes, if X be the constant ratio of the force to the momentum, then F, =?..Mv. But the unit of force being arbitrary, we may assume it to be A. Hence, putting F for the number of units of force, or the ratio of F1 to i, we have F= — 1=Mv. (38) COR. If M=1, F,=-).v, in which A is constant. Hence the force isproportional to the velocity produced in the unit of mass. 133 D Y NAMICS. SC-IOL. Since we know nothing of the nature of forces, we can not determine their effects a priori. The foregoing proposition ought, therefore, to be regarded as depending ultimately on observation and experiment. The fact that on the earth, which is subject to the double motion of rotation and translation, forces are found to produce precisely the same effects as if the earth were at rest, is a confirmation of its truth; also that a pendulum performs its vibrations in the same time, whatever be the direction of these vibrations in reference to an east -aid west line. 234. If any number of forces act upon a body in tile snamle direction, the velocity imparted to the body will be equal to the sum of the velocities imparted-by each. Let F, F', F"... be any number of forces acting upon a body, v, i', v"... the velocities imparted by each respectively, p the resultant force, and, u the velocity due to b. Then F:=2v, F'=Xv', F"=2vv", &c. Since the forces are conspiring, their resultant will be equal to their sum... F+F'+F"+ &c., =05-:(v+v'+v" — &c.). Lut P =)X. HIenc, u v' - v'~ 4 — &c.: 235. PnoP. If the adjacent sides of a parallelogram rep2resent in magnitude and direction the velocities which two forces, by their separate action, would respectively produce in a body, the diagonal of the lparallelogram will represent the actual velocity produced by their joint action. Let AB and AC represent the velocities which the forces D P and Q respectively impart to a body placed at A, or, in other words, the spaces over which the body. would pass in a unit of time by the separate ac" tion of the forces, then AD _Bf. will represent the velocity produced by their resultant, or the space ovei UNIFORM MOTION. 139 which the body will pass in the unit of time by the joint action of the forces. We may assume AB to represent the force P; then, by Art. 233, Cor., P: Q=AB: AC, and AC will represent the force Q. Also, Art. 21, AD will be the resultant R of P and Q.' Let x be the velocity due to the resultant R, then (Art. 233, Cor.) P: R:=AB: x. But P: R=AB: AD... x=-AD, and AD being the direction of the resultant, the velocity due to it will have the same direction. The two preceding propositions illustrate the second law of motion (Art. 226). a/ 236. PnoP. The velocity of a body being given, tofind the component velocities in any directions at right angles to each other; and two component velocities at right angles to each other being given, to find the resultant velocity. Let AD represent the velocity v of the body in direction and magnitude, D AX and AY the rectangular directions in which the components are required, and 0 the angle which AD makes with AX. Completing the rectangle, AB will represent the velocity a in AX, and AC the velocity b in AY. A B x Then AB-=AD. cos. 0, AC-AD. sin. 0, or a=v. cos. 0, b=v. sin. 0. (39) Again, if a and b represent the given component velocities in AX and AY, to find the resultant velocity v we have v= V/a2+b, (40) and tan. 0=- (41) as required. 140 D Y N A M I CS. Con. Hence velocities may be compounded and resolved like forces in statics. 237. Pnop. Two bodies, A and B, describe the same path ithil the velocities v and v', and at the commencement of their motion are at a distance a from each other: tofind the time t when they will be at a distance b from each other and the position of each at the end of that time. Taking the position of A when t=- for the origin of spaces, the equations of their motions are, by (37) and (36), s=vt, (a) and s'=a+v't'. (b) By the conditions of the question, t=t'. Also s-s'=b or s'-s=b,.. s-s-'=ib. Subtracting (b) from (a), and putting t=t', we have s-s'==vt-a-v't= ~ —b. *~~ t-, (42) Also, from (a) s=vt=v. a, (43) a''b va-4v'b and from (b) s'=a+v't=a+v'.,. (44) CoR. 1. If the bodies move in opposite directions, v' will be negative, and acb a=+-b vaFv'b i,- t S-v. S' I'v' -v'9 v, + v v +v v'. CoR. 2. If the time when the bodies are together be required. then b=0, and ac v aZ _v a t=-, S =S, S —Cl= V — - 1)1 V 238. PROP. WVhen two bodies A and B move in the circurN ference of a circle with uniform velocities, to determine the circumstances of their motion. Let v and v' be their velocities, c the circumference of the circle, and a the distance apart at the commencement of the time. Then, putting b=O in (42), we have U N I F ORM MO T I ON. 141 a t =-v = the time of their first meeting. a+c 2 s - " second " V-V t a+2c " third " a+(n-1)c t-' " " nth "*? -— _ Vt a+ (n-1)c v- -v also, V a+(n- -)c = -pace described by A, and s-av'a+v(n-1)c B. and s-a -- = "i so B. v-v1 ('on. The interval between two successive conjunctions Ls t2 -t I=t1'= (45) V__Vt" 239. ExAMIPLEs. Ex. 1. If an iron rod have one end against the sun and the other resting on the earth, the distance of the sun from the earth being 95,125,000 miles, in what time will a blow applied to the end on the earth be felt by the sun, the velocity of an impulse in iron being 11,865 feet per second? Ans. 490 days. i Ex. 2. When the earth is in that part of its orbit nearest to Jupiter, an eclipse of one of Jupiter's satellites is seen 16 minutes 36 seconds sooner than it would be if the earth were in that part of its orbit most remote from Jupiter. The radius of the earth's orbit being 95,125,000 miles, what is the velocity of light? Ex. 3. The star 61 Cygni is ascertained to be 56,319,996,600,000 miles distant from us; were its light suddenly extinguished, in what time would the intelligence reach us, the velocity of light being 191,000 miles? i Ex. 4. Suppose 964 tons of ice to be floating directly to the east at sunrise on the 21st of March, with a velocity of 12 feel 142 D Y N AMi 1 C S. per minute, how many grains of light from the sun would be sufficient to stop it? Ex. 5. A train of cars moving with a velocity of 20 miles an hour, had been gone three hours, when a locomotive was dispatched in pursuit, with a velocity of 25 miles an hour; in what time did the latter overtake the former? Ex. 6. Had the trains in Ex. 5 started together and moved in opposite directions around the earth, 24,840 miles, in whalt time would they meet? Ex. 7. Suppose it to be 12 minutes past noon by a clock, in how long a time will the hour and minute hands of the clock be together? Ex. 8. The daily motion of Mercury in his orbit is 40.09239; that of Venus 1~.60216; that of the earth 00.98563: what are the intervals between the epochs at which Mercury and Venus respectively will be in the same direction from the sun as the earth? Ex. 9. A man being caught in a shower in which the rain fell vertically, ran with a velocity of 12 feet per second. He found that the drops struck him in the face, and estimated that the apparent direction of the drops made an angle of 10~ with the vertical line. What was the v.locity of the drops? Ans. 68 feet. iEx. 10. When the path of the earth in its orbit is perpen dicular to a line drawn from a star to the earth, the path of the light from the star appears to make an angle of 20tt,445 with the perpendicular to the path of the earth. The velocity of the earth being 68,180 miles per hour, what is the velocity of light? CHAPTER II. IMIIPACT OF BODIES. 240. DEF. When two bodies in motion impinge, if their centers of inertia move in the same straight line perpendicular to a plane tangent to the bodies at their point of contact, the impact is said to be direct and central. If the straight line described by the center of inertia of one )f the bodies is not perpendicular to the tangent plane, the im)act is said to be oblique. In the cases discussed the bodies will be supposed, spherical and of uniform density. 241. DEF. When the bodies impinge, they exert a mutual but varying pressure during the interval between contact and separation, an interval of time which is generally very short. and we suppose them to suffer a degree of compression, by which, during a portion of this interval, their centers will approach each other, and during the remaining portion will recede by the action of an internal force tending to restore them to their original form. The force urging the approach of their centers is called the force of compression; the opposing force causing them to separate again is called the force of restitution or elasticity. The ratio of the force of restitution to that of compression is called the modulus of elasticity. When this ratio is unity, or the force of restitution is equal to that of compression, the bodies are peifectly elastic; when it is zero, or the force of restitution is nothing, they are inelastic. If the value of the ratio is intermediate between zero and unity the bodies are imperfectly elastic. 242. DEF. If the bodies suffer no compression, they are called hard; if, when compressed, they exert no force to recover their original form, they are called soft. There are no known bodies either perfectly elastic or per. 144 DY N A MI CS. fectly inelastic, but these states may be considered as limits to the various degrees of elasticity presented in nature. 24.3. PRoP. To determine the velocity of two inelastic bodies after direct impact. Since the bodies are inelastic, the force of restitution is zero, and the bodies will move on together with a common velocity. 10. Let m, and m, be the masses or bodies moving in the M2 m same direction with the velocities vL -1 and v2 (v,>v2), and v their common velocity after impact. If F and F' are the forces which impress on the bodies their respective velocities, then (38) F=m vI, and F'=m v2, and their resultant, F+F'=mv1 ~+m2v2. After impact the bodies move on together as one mass, and its momentum (m, +m2)v must be a measure of the force F+F'. (m +mZ2)v=-m v +m2ve2, or miv- (46) ni, +7 m2 2~. If the bodies move in opposite directions, the resultant of the forces will be F-F'. F. - F'=mv, -m2v =(m, +m2)v, mn v, -m, v or V= 2 (47) m,.+m2 The same result will be obtained by changing the sign of vo in (46). Hence the velocity of two inelastic bodies after impact is equal to the algebraic sum of their momenta before impact, divided by the sum of their masses. CoR. 1. If two bodies move in opposite directions, with velocz-., ties reciprocally proportional to their masses, they will rest aftei' impact. For mI1: m2 =V' s: v gives mn v,=m2 v2, which, substituted in (47), gves vo0. IMPACT OF B ODIES. 145 Con 2. if n is at rest before ihr act, v2-=0, and mvv1 and if, at the same time, the masses are equal, 2- Cor. 3. If the masses are equal, and move in the same di rection, 13 -2 fLw opposite directions, v! -v2 244. PtoP. In the impact of inelastic bodies there is a loss oJ living force, and this loss is equal to the sum of the living forces Jdue to the velocities lost and gained by the bodies respectively. For the livings force before impact =m,v~+mv' (Art. 224), and " " " " after " =(mI +m2)v2..'. the loss of living force by impact is?n,11 + mgv3 - (in +Fm) Vr2 m1V2 +m'v, -2 (ml + mn) v2+ (,in1 +m)v' (46), = +- (,,2 v +,2) v + (ml, m) +m2) -m(v —v)2+m2(v — v)2, which is necessarily positive, and in which v, — v is the velocity lost by mI, and v-v, the velocity gained by mi. From this proposition it appears that in machinery made of inelastic materials all abrupt changes of motion are attended with a loss of living force, by which loss the efficiency of the r-iachinery is impaired. 245. Prop. To find the velocities (f two imperfectly elastic bodies after direct impact. Let mi and m2 be the masses, vl and v2 their velocities betore impact, and v', v' their velocities after impact. The bodies being elastic, will suffer compression. Let v be their common velocity at the instant of greatest compression or when the distance between their centers is least. Then the velocity lost by m, at this instant wil be v,-v. K 146 D Y N A M I C. Let E be the modulus of elasticity, or the ratio of the force of restitution to that of compression. Since these forces are pioportional to the velocities they generate or destroy in the same mass, the velocity destroyed in m, by the force of restitution will be e(v, -v). Hence the whole velocity lost by Inn will ibe'1-v+E(V1 -V)=(1 q-) (V1 -). This, subtracted from the velocity of m, before impact, will give its velocity after impact, or v'- =v, -(1 +E) (v 1-t-)-=v-e(v — v). (a) In like manner, the velocity gained by m, during compression will be v-v,, and the velocity gained by the force of restitution e(v-v2 ). Hence the whole velocity gained by m, will be (1 +e),v-v,). This, added to the velocity before impact, will give the velocity after impact, or v'=v -. (1+0 ( —V2o)= ~+,~(6- )'). (b) Substituting in (a) and (b) the value of v (46), and redlucingr, V, v1 +m2 nz, (v 1 (v v) v- I I -m- + 2 (48) I +lm2v?11 )nz v,-,:L' f'q' (49) -- M +?M2 m, + 7n, As in Art. 243, if the bodies move in opposite directions, we must change the sign of v~, or, if one of them be at rest before impact, make v =O. Also, if we put e=1, the results will be those for perfectly elastic bodies, or make -=O, the result will be that for inelastic bodies. Con. 1. If the bodies are perfectly elastic, their relative velocities before and after impact are the same. For, making =l in (a) and (b), and subtracting the latter from the former, we have,,l-v'-v v-v,. (50) COR. 2. In the impact of bodies no motion is lost. For, multiplying (48) by ml, and (49) by m2, adding and reducing, we have m,lv' +m v'=m, 1m2v, (51) IMPACT OF BO DIES. 147 in which the first member is the sumh of their momenta tfter impact, and the second member the sum of their momenta before impact. CoR. 3. If the bodies are perfectly elastic and equal, they will interchange velocities by impact. For. making m =nzm and E-1 in (48) and (49), we have 1-2V1 ( +V2)-+-(V1-V2)=v. V22-(V-1.+V.)+21(-')=V I Cot. 4. The velocity which one body communicates to an other at rest, when perfectly elastic, is equal to twice the velocity of the former divided by one plus the ratio of the masses. Making in (49) m, =rm,, v =O, and e1=, we obtain 2v -2 r+ l 246. Pior. WYhen, in a series of n perfectly elastic bodies whose masses are in geometrical progression, the first impinges directly against the second at rest, the second against the third, and so on, to find the velocity of the nth body. Let m, rm, r2m.... 2.-' m be the bodies, and v the velocity of m. By Cor. 4, Art. 245, 2'0 the velocity of the second will be r+ 1 2v 1 2'v: 6 fu" r thtird 2 (+r " " fourth', 1 " nth " Con. 1. The momentum of the nth body is. 7) n1- 1 2)"CoR. 2. If the bodies are equal, r= 1, and the velocity of the last equals v, the velocity of the first. If all are in contact ex cept the first before impact, all except the last will remain iri contact after -impact, and the last will move off with the velocity of the first. 14E1 DYNAMICS. 247. Por. The motion of the common centei of Lg cavity Uf two bodies after direct impact is the same as before imzact. Let v be the velocity of the common center of gravity of the bodies ml and m2, moving in the same direction with velocities v1 and v2 respectively before impact, and:5' the velocity of their common center of gravity after impact. Let xl, x.,, - be the distances respectively of the centers of gravity of m, and m2, and of their common center of gravity from any fixed point in their line of motion at any instant; x'1, x2, x' the same quantities after an interval t, so that (36) X1 =x- +vt, X = X2 +V2 t x' =z Vt. By (29), Art. 105, (n 1 +m72)' =m. Ix+m2X2, (a) (n,. + m m ) XI'=n2, ~'~ +m 2x. (b) Substituting in (b) the values of x', x', x', above, and subtract ing (a) from the result, we have V= +mv: I5 r 1+2 V2(c) which is the same as the velocity of two inelastic bodies after impact, and therefore equal to the velocity of their common center of gravity after impact, since the masses move on together. If the bodies are elastic, and v'1, vt' are their velocities after impact, then X1 =X +1t, 2 2 ~V2t a' =x -V't; from which we deduce, as before, M VI'+m.Vt = m +m2 (d) for the velocity of the common center of gravity of elastic bodies after impact. Now the denominators of (c) and (d) are the same, and by (51) the numerators are elual; IMPACT OF BODIES. 149 or he velocity of the common center of gravity is unchanged by impact. 248. SCHOL. This proposition is only a particular case of a general principle in Mechanics, denominated the conservation of the motion of the center of gravity. The principle consists in this, that the mutual action of the several bodies or parts of a system upon each other can produce no change in the motion of the center of gravity of the entire system. 249. DEF. If a body impinge on a surface, the angle which its path, before impact, makes with the perpendicular to the surface at the point of impact is called the angle of incidence, and the angle which its path, after impact, makes with the same perpendicular is called the angle of reflection. 250. PRoP. To determine the motion of a smooth inelaslic body after oblique impact upon a smooth,' hard, and fixed plane. Let the body?m impinge on the plane N AC at B, with the velocity v, making ~7, the angle of incidence mBN=-O. The component of the velocity v parallel to the plane (39) is v. sin. 0; A \3 and this velocity will not be changed by impact, since the body and plane are smooth. The component of v perpendicular to the plane, viz., v. cos. 0, will be destroyed by the plane, and, since the body and plane are inelastic, there will be no vertical velocity after impact. Hence the body will slide along the plane with the velocity v. sin. 0. 251. PRor. To determine the motion of ani elastic body after oblique impact upon a smooth, hard, andfixed plane. Let the body m impinge on the plane,, RS at B, with the velocity v, making the 6 angle of incidence PBN=0. Let PB rep- N resent the velocity v before impact. Draw M Q PR and PN perpendicular and parallel to the plane RS. PN=RB-v. sin.0 is the 7I, S component of the velocity parallel to the plane, and is not affected by the impact. PR=NB=v. cos. 0, the component of ql'e velocity perpendicular to the plane, will be destroyed by' ID5Y N A I IIC S. the plane. But the body being elastic, the force of restitution will give it a velocity e.v. cos. 0 in the direction EN. Take BM=-.v. cos. 0, BS=v. sin. 0, and, completing the parallelogram, draw BQ. BQnm' is the direction, and BQ the velocity of the body after impact. Now BQ=2-v/2- BM,12+MQ2 = 2e.v2.V coS.2 0+ 2 Sin.2 0'. BQ=v' =v.2 Cos.2 0+sin.2 0; MQ PN tan. 0 o, n. MB e.NB 8 COR. If e=1, 0=0', and v-v', or if the body be perfectly elastic, the angle of incidence equals the angle of reflection, and the velocity is the same after as before impact. If E=0, tan. 0'=-, and 0'=90~. 252. PRoP. To determine the direction in which a body of given elasticity must be projected, in order that after refection from a given plane it may pass through a given point. Let MB be the given plane, A the A. N D- position of the body before projection, D the point through whi'ch it is required to pass after reflection, and e the modulus of elasticity of the body, /B supposed known. Draw AMC perpendicular to the plane, and take AM': MC= 1: Draw C D cutting the plane in B, and join A B. AB is the direction in which the body must be projected, and AB, BD will be its path. Since s.AM=AIC, MB MB e.AM=MC' tan. MAB or -tan. MCB; tan. 0 or tn -, as in Art. 251. IMP A CT O F BOD I ES. 151 253. PROP. The modulus of elasticity is equal to the ratio oj the relative velocities of the bodies after and before impact. For, eliminating v from (a) and (b) (Art. 245), we have 254. ScIHoL. Bodies suspended by fine cords, and made tc oscillate like a pendulum, acquire velocities at the lowest point proportional to the chords of the arcs of descent, and will rise through arcs whose chords are proportional to the velocities impressed upon them at the lowest point of the arcs. Also, if the arcs be small, the times of descent will be equal, so that bodies descending through arcs of different lengths will impinge at the 6~ 6 lowest points. If, therefore, two spher- 2 o 2 3 ical bodies of the same material be made to descend, in the manner described, through arcs of given length, and the arcs through which they rise after impact be measured, the velocities v,,v2, v',, v will be known, and these, substituted in (52), will give E the modulus of elasticity. The following table exhibits the rezults of experiments, perfect elasticity being unity: Substances. Moduli. Substances. Moduli..l,~ so...M.. Glass........ 0.94 Bell-metal...... 0.67 Hard-baked clay... 9.89 Cork.........65 Ivory.........81 Brass........41 Limestone.79 Lead.........20 Steel, hardened..79 Clay, just malleable by the 17 Cast iron.......73 handl... Steel, soft.......67 255. EXAMPLES. Ex. 1. Two inelastic bodies, weighing 12 lbs. and 7 lbs. Iespectively, move in the same direction with velocities of 8 feet and 5 feet in a second. Find the common velocity after impact; also the velocity lost by one, and that gained by the other. Ex. 2. A mass m,; with a velocity 11, impinges on m2 moving in the opposite direction with a velocity of 5; by impact m, 152 D Y N A M I C S. loses one third of its momentum. What are the relative magnitudes of m, and mn? Ex. 3. mi, weighing 8 lbs., impinges on m,, weighing 5 lbs., and moving in ml's direction with a velocity of 9; by impact m2's velocity is trebled. What was mn,'s velocity before impact? Eix. 4. Two bodies m, and m2 are moving in the same straight line with velocities v, and v2. Find the velocity of each after impact when 6nz,=5-z2, v,-7, 4v, 5v-=0, and E-2 Ex. 5. Two bodies m, and mnz are perfectly elastic and move in opposite directions; ml istreble of ma, but m2's velocity is double that of ml. Determine their motions after impact. Ex. 6. There is a row of perfectly elastic bodies in geomet. rical progression whose common ratio is 3; the first impinges on the second, which transmits its velocity to the third, and so on; the last body moves off with G the velocity of the first. What was the number of bodies? Ex. 7. n(,(=3m,) impinges on m2 at rest; m,'s velocity after impact is - of its velocity before impact. Required the value of E, the modulus of elasticity. Ex. 8. Two bodies mn and m, whose elasticity is -, mo- ing in opposite directions with velocities 25 and 16 respectively, impinge directly upon each other. Find the distance between them 4. seconds after impact. Ex. 9. At what angle must a body whose elasticity is ~ be incident on a perfectly hard plane, that the angle made by its path before and after impact may be a right angle.? Ex. 10. A ball whose elasticity is c, projected from a given point in the circumference of a circle, after two reflections from the interior of the circle, returns to the same point. Eiquired the angle 0 made by the direction of projection withe tte radius at the given point. 2 Ans. Tan. 0= Vd _L..... CHAPTER III. )MOTION FROM THE ACTION OF A CONSTANT FORCE. 256. BY the definition of a constant force (Art. 222), the velocities generated in equal successive intervals of time by the action of the force are all equal, and the increment of velocity in a unit of time is a measure of the force. Hence the velocity is the same at no two successive instants, and, if the body or point move from rest, will increase uniformly, or be uniformly accelerated. By the velocity acquired in t seconds is meant the space over which the body would pass in the second next succeeding the t seconds if the velocity should remain the same during this second as it was at the end of the t seconds; or the space described by the body in the interval between t seconds and t+1 seconds, if, at the end of t seconds, the force should cease to act. Puttingf for the force, and v for the velocity acquired in the time t, we have, as before, v-=ft. (53) J 257. PRoP. To find the space in terms of the force and time when a body moves from rest by the action of a constantforce. Let f be the force, and s the whole space described in the time t, and let t be divided into n equal parts, each =-. The n intervals reckoning from the commencement of the time, will be t 2t 3t 4t (n-1)t nt nnnn n n By (53), the velocities at the end of these intervals will be t 2t 3t 4t (n-l)t nt n n n n n n 154 DYN AM IC S. If now the body moved unijformly during each interval z with the velocity it had at the beginning or end of this interval, the spaces described during the intervals respectively, would be equal to the product of this uniform velocity by the interval (37), and the whole space described would be equal to the sum of these partial spaces. If, therefore, the body moved uniformly during each interval t - with the velocity it had at the beginning of the interval, we should have t2 2t2 3tJ (n — I) t O+f. +-f+ &i. fl t2 -f.-?1 +2+3+, &c.,... (n-1)}, t2 n(n-l) 12 t2 Bu if the 2body 2move (am ) But if the body moved uniformly during each interval -- with the velocity it had at the end of the interval, we should have 2t: + A f.3.- + &c......2.. 3. (n -1)t2 nt t2.-~2(l +2+3+, &c.,.. n), t2 n(~1 l) t2 t2 Since the velocity is uniform during no sensible interval, thk true value of s will lie between the two quantities (a) and (b), however small each interval may, be, or however large n may be. But when n becomes indefinitely large the last terms in (a) and (b) vanish, and (a)=(b). ACTION OF A CONSTANT FORCE. 155 s.-. s-1ft and s oc t. (54) Herice the space described from rest by a body from the action of a constant force is equal to half. the product of theforce by the square of the number of seconds, and the spaces vary as the squares of the times. C 258. PROP. To determine the space in ternms of the force and velocity; also in terms of the time and velocity. 10. Eliminating t from (53) and (54), we have S_ 0f, ors QCv. (55) 2". Eliminatingf from (53) and (54), we obtain sIvvt, or s vt. (56) Cor. I'The space described in any time by a body moving trom rest by the action of a constant force, is half that it would describe in the same time if it moved uniformly with the acquired velocity. For the space s, described in the time t, with a uniform velocity v, is, by (37), s 1 =vt, which, compared with (56), gives / 59. PROP. To find the space described by a body in the last n seconds of its motion. The space described in the time t (54) is s =ft2 (a) The space described in the time (t-n) is sS.2 f(t-2 n) (2) Subtracting (b) from (a), we have, for the space s described in the last n seconds, = s1-s -= f(2n!t-n2). (57) COR. If the space described in the m seconds next preceding the last n seconds is required, we have, for the space s3, described in the time t- (n+m) seconds, s — f(t-n-m), which, subtracted firom (b), gives for the space s required, S= S 2 — Z = ILA~ml 2mn-m'). (58) 156 Y N A MI C S. / V 260. PnoP. A body being projected with a given velocity in the direction in which a constant force acts, to find the velocity of the body at the end of a given time, and the space described in that time. By Art. 234, the velocity due to the joint action of the impulsive and constant forces will be equal to the sum or difference of the velocities-due to each, according as they act in the same or opposite directions. If, therefore, v, be the velocity of projection, the whole velocity v, at the end of the time t, will be V=V1~ft. (59) In the same manner, the space due to the joint action of the forces will be equal to the sum or difference of the spaces due to each, or s/ =v -- ftl. (60) 261. PnoP. A body being projected with a given velocity in the direction in which a constant force acts, to find its velocity when it has passed through a given space. Let s be the given space, and h the space through which the body must pass to acquire the velocity v, by the action of the constant force. Then (55) 2= 2flh and for the space (hIs) v =2f(h/s), =2 fih I2fs, -=vli2fs, (61) the signs to be taken as in the last proposition. i 262. PRoP. lVhen a body is projected in a direction opposite to that in which a constant force acts, the velocity acquired in returning to the point of departure is equal to the velocity o( projection. If v, be the velocity of' projection in the direction AB, fromr (61) we have v 21 2f (a).3,-= A CT ION OF A CONSTANT FORC E. 15 Now the actual velocity v of the body is continual- B.y diminished by the action of the force f, and when the body has reached its greatest distance from A, as B, v=O. Her ce (a) gives S =f, or vl= V2fs. But the velocity v acquired in moving from rest through BA=s is, by (55), V-. v-v. C Goa. 1. The velocity of the body at any given dis- l tance from A is the same in going and returning. For in the expression (a), since v, and 2f are constanlt, v is always the same for the same value of s=AC. If v>v:, s is negative, or on the opposite side of A from B. CtoR. 2. The whole time of flight T= 2v' Making v=O in (59), we have t= -' when the body is at B. fJ But, in returning, it acquires the velocity v,=. ft. hence the time of return is t= —, f and the whole time 2t=T=2 >a 263. SCHOL. 1. There is no known instance in nature of a force which is constant. The law of Universal Gravitation is, that every particle of matter attracts every other particle with a force which varies directly as the mass of the attracting particle, and inversely as the square of the distance. A sphere of uniform density, or one whose density is the same at equal distances from the center, attracts a body exterior to it as if the matter of the sphere were collected at its center, and with a force varying inversely as the square of the distance of the body fi'om the center, but a body or particle in the interior with a force varying directly as its distance from the center. Regarding the earth as a sphere, this is the law of the earth's 158 DYNAMIC S. attraction for bodies exterior to it; but for all small distances above the surface the intensity of gravity may be considered constant, since at a distance of one mile above the surface the actual diminution of gravity is only or about the 2000th part of that at the surface; a variation too small to affect sensibly the circumstances of the motion of a falling body com puted on the hypothesis that the force suffers no variation at all. In reality, the force by which a body is drawn toward the earth is equal to the sum of the attractions of each for the other; but when the mass of the body is inconsiderable with regard to the mass of the earth, the effect of the former is insensible, and the accelerations of all bodies of moderate size are the same. Within these limits, then, gravity may be taken as a constant or uniformly accelerating force, whatever be the mass. 264. SCIOL. 2. Representing the intensity of gravity at the surface of the earth, as before, by g, we have, from (54), s-=-Igt2, and if t=1, s —=g, or g=2s from which it appears that the acceleration is equal to twice the space described in the unit of time. It has been found by experiment that the space through which a body falls freely in one second in the latitude of New York is equal 16.0799 feet, or 16-'- feet nearly. Hence g-32.1598 feet, or 32- feet nearly. Hence, also, all the relations between the space, time, and velocity due to the action of a constant force are true in relation to the action of gravity near the earth's surface. Collecting these, and substituting g for f, we have the following relations, suppos'ng no resistance from the air: s —gt —-2=- t, (54') 2s r= gt -= V-2gs, (53') V 2s /2s g.V g! v 2s vu ACTION OF A CONSTANT FORCE. 159 Also, when a body is projected with a velocity vl vertically upward or downward, (260) and (261), s=-v t+ —gt,2 (60') v2=- v +2gs. (61') 265. EXAMPLES. /Ex. 1. A body has been falling 11 seconds. Find the space described and the veloctiy acquired. By (54'), s=-gt"-16 —X 121 =1946 - feet. By (53'), v= gt =321 X 11=- 353 5 feet. i Ex. 2. Find the time in which a falling body would acquire a velocity of 500 feet, and the height friom which it must fall. v 500 By (56'), t= - = —=15.544 seconds. g 32. v2 250000 By (54'), s=- = 3886 feet nearly, s2-g 64kor s=.ltv=- X 15.544 X 500= 3886 feet, as before. 7Ex. 3. What is the velocity acquired by a body in falling 450 feet? and if the body weigh 10 tons, what is the momentum acquired? Ans. v =170.14 feet. M=3811136 lbs. Ex. 4. A body had fallen through a height equal to one quarter of a mile. What was the space described by it in the last second? Ans. s=275 feet. Ex. 5. A body had been falling 15 seconds. Compare the spaces described in the seventh and last seconds. Ex. 6. A body had been falling 12.5 seconds. What was the space described in the last second but 5 of its fall? Ex. 7. The space described by a body in the fifth second of its falll was to the space described in the last second but 4, as 1 to 6. What was the whole space described? Ans. s=15958.69 feet. Ex. 8. A body is projected vertically downward with a ve 160 D Y NAMICS locity of 100 feet. What is its velocity at the end of 5 sec, onds? Ex. O. A body is projected vertically upward with a velocity of 100 feet. Find its velocity at the end of 5 seconds, and its position at the end of 8 seconds. Ans. v=- 60 feet. s= -- 229IL feet. Ex. 10. A body is dropped into a well and is heard to strike the bottom in 4 seconds. What is the depth of the well, the velocity of sound being 1130 feet? Ans. 231 feet. Ex. 11. A body is thrown vertically upward with a velocity v1. Find the time at which it is at a given height h in itc ascent. If t be the time required, (60) gives h=v t —}gt; whence t2-t 2- +2 g g or t= v~ Vv,-2g/i The lower sign gives the time when the body is at the height h in its ascent, and the upper in its descent. Ex. 12. A body is projected vertically upward, and the interval between the times of its passing a point whose height is h in its ascent and descent is 2t. Find the velocity v of prolection, and the whole time T of its motion. Ex. 13. A body whose elasticity is E is projected vertically upward to the height Ih above a hard plane, to which it returns, and from which it rebounds till its motion is destroyed. What's the whole space described by the body? CHAPTER IV. PROJECTILES. IN. the preceding chapter we have discussed the motion of a body by the joint action of a projectile force and the foceo of gravity, when these forces were coincident or opposite in direction. We now proceed to determine the circumstances of the motion of a body, when the direction of the projectile force is other than vertically upward or downward, supposing, as heretofore, no resistance from the air. 266. PnoP. The path of a body, moving under the joint influence of a projectile force, and the force of gravity considered as a constant, accelerating force, is a parabola. Let v be the velocity of pro- y jection from the point A, in the direction ANY, and t the time In which the body will describe AN=y, with the uniform velocity v, if gravity do not act. Let / IP AIM=x be the space through which the body will fall in the time t by the action of gravity. A Completing the parallelogram MN, the actual place of the body at the end of the time t is M P. By (37), AN-=y=vt; and by (54),AM=NP==x= gt2. Elim- x inating t from these two equat:ons, we get 2va y =-X. (a) If h be the space due to the velo':ity v, or the space through L 162 D Y N AIM I C S. which the body must fall to acquire the velocity of projectior. (55), v2=2gh, which, substituted in (a), gives y2= 4hx, (62) the equation of a parabola referred to the oblique axes AX, AY; and, since 4h is the parameter to the diameter through A, h is the distance from A, the point of projection to the focus F, or to the directrix. CoR. The velocity of the body at any point of its path is that which the body would acquire in falling vertically from the directrix to that point. For if the body were projected from any point of its path in the direction, and with the velocity it has at that point, it would obviously describe the same path. Therefore the velocity at that point must be equal to that due to one fourth the parameter to the diameter at that point, which is the distance from that point to the directiix or to the focus.' 267. PROP. To ind the equation to the path of a projectile whuen referred to horizontal and Vertical co-ordinate axes. Let v be the velocity of pro~yi / jection in the direction AN, N / which makes with AX the angle of elevation NAX=a, APB the path of the body, AM=x, PM=y, the co-ordinates of the point P, t the time in which the body describes the are AP, and produce MP to meet the direc. tlon of projection in N. Then AN =vt, NP=-gt2, and MN=vt sin. a. Also, AM-=x=vt. cos. a, (a) and PM =y=vt sin. a-Igt'. (b) Eliminating t from (a) and (b), we obtain g yZx tan. a —x 2a cos.2 a' and, substituting the value of v2=2gh, PROJECTILE S. 163 x2 y —x tan. a- 4h cos.2 (61) 268. DEF. The horizontal range of a projectile is the distance AB from the point of projection to the point where it strikes the horizontal plane in its descent. The' time of fight is the time occupied in describing APB. The height through which the body must fall to acquire the velocity of projection is called the impetus. / 269. PRop. To find the time of flight of a projectile on a hori~. zontalplane. At the points A and B the ordinate y= —O. This value of y, substituted in (b), Art. 267, gives vt sin. a —2gta= 0. 2v sin. a. t=0, and t= --. (65) The former value of t applies to the point A, and the latter to the point B, which is, therefore, the thime of flight required. Or, since v sin. a is the vertical component of the velocity, or the velocity of projection estimated vertically, if this value of the velocity be substituted for v, in (60), and we make s-=O we get, as before, O=vt sin. a —-gt2, 2v sin. a t =0, and t=- J 270. Pnop. To find the range of a projectile on a horizontal plane. Put y=0 in (64), and we have O-x tan. a- a 4h cos.' a; from which we obtain for the point A, x=0O, for the point B, x=4h sin. a cos. a, or AB=R=2h sin. 2a, (66) the horizontal range required. Con. 1. The horizontal range is greatest when a=45~. For in this case 2a=1900, and sin. 2a=1..-. x=~h. I 64 D Y N A M IxC S. or tile greatest horizontal range is equal to txwice the height due to the velocity of projection, or twice the distance from the. point of projection to the focus of the trajectory. Coi. 2. The range is the same for any two angles of elevation, the difference- between which and 450 is the same, or for (45-4-0). For sin. (90~+20) sin. (90~0-20), or sin. 2(450+0)=sin. 2(45~-0). If, therefore, we'put either 450+0 or 45~-0 for a in (66), the value of R remains the [ P // same. Thus, if AB be the range of a projectile when /! / P \ the angle of elevation is 45~, and APB its path, the range AB' will be that due to the A\ B elevation 450-0, for which A]3- B B the path is AP'B', and to 450+0, for which the path is AP"B'. CoaR. 3. When the velocity of projection is given, and we know the range R due to the elevation a, we can readily find the range R' due to any other elevation a'. For R — 2h sin. 2a, and R'= 2h sin. 2a'. sin. 2a'' R' —.R. (67)'sin. 2a COR. 4. Since the horizontal velocity of a projectile is urnform, the range is equal to the horizontal component of the velocity into the time of flight, or 2v sin. a 2v2 sin. a. cos. a. R=v. cos. aX - =2h sin. 2a, as before. g g / 271. PnoP. Tofind the greatest height which a projectile attains. The greatest height H is evidently the value of the ordinate at the middle of AB, or when the'time is one half the time of v si n. a flight." Putting t= - in (b), Art. 267, we have g P.P I. E.CTIE S. 1 65 vt sin., a vy sin.' a 2Xv sin?2 a 11=, 2 2 - g g g -h sin.' a. (68)' 272. PRoP. Tofind the co-ordinates of the point wh6re a projectile will strike an inclined plane passing through the point of projection, the range 3n the inclined plclie, and the time of fight. Let y=x tan. / be the equation y of the line AC, which is the intersection of the inclined plane, with the vertical plane of the path of the body./ C Substituting this value of y in (64), we obtain, after reduction, A 1 for the abscissa of the point C, 4h cos. a. sin.(a —i) cos.t (a) By substituting in (64) for x, its value y cot. 3, and reducing, we get for the ordinate of C, 4h. cos. a. sin. d. sin. (a —3) COS.2f (b) To find AC=R', multiply (a) by sec.: —, and we. Cos. [3 have,AC=Ri=x sec. j34h. cos. a. sin. (a —) (69) If the inclined plane cut the path of the projectile If the inclined plane cut the path of the projectile below the axis AX, 3 will be negative. The time of flight is equal to the time of describing the abscissa AM with the horizontal component of the velocity. Hence (37), Jividing (a) by v cos. a., we get 4h sin. (a —3) V. Cos. P 2v sin. (a —p) g. -Cos. 3(70) = g. cos. sO 166 DY N AM IC S. 273. SCHOL. It has been found by experiment, that if w be the weight of a ball or shell, p the weight of the gunpowder used in discharging the ball or shell from a mortar, and v the velocity generated by the powder, v=- 1600-o feet. (71) 274. EXAMPLES. Ex. 1. A body is projected at an angle of elevation of 15" with a velocity of 60 feet.- Find the horizontal range, the greatest altitude, and time of flight. v2 3600 From (54'), h-= 4 — 55.96. From (66), R=2h sin. 2a=-2h. sin. 30~ —-2.h. =h=-55.96 From (68), H-= h sin.2 a. Log. sin. a= 9.4129962 "' " - 9.4129962 "' hI = 1.7478777. H=3.7486.... H=.0.5738701 2v sin. a From (65), T- g.. Log. 2v= 2.0791812 Log. sin. a=- 9.412996'2 (a.C. " g= 8.4925940.. T=-0.96554..... "6 T-= 9.9847714 Ex. 2. A body is projected at an angle of elevation of 450 and descends to the horizon at a distance of 500 feet from the point of projection. Required the velocity of projection, the greatest altitude, and time of flight. Ans. v = 126.82 feet H=125. " T =5.58 seconds. Ex. 3. The horizontal range of a projectile is 1000 feet al4>; the time of flight is 15 seconds. Required the angle of ele,)'a. tion, velocity of projection, and greatest altitude. Ans. a =74~.33/09/. v =250.29 feet. H=904.69 " PRO J E CT ILE S. 167.Ex. 4. If a body be projected at an angle of elevation of 60, with a velocity of 850 feet, find the parameter to the axis of the parabola described, and the co-ordinates of the focus. Ans. p-11230.57 feet. x- 9725.67 " y= 5615.28 " Ex. 5. Find the velocity and angle of elevation of a ball tlia it may be 100 feet above the ground at the distance of one quarter of a mile, and may strike the ground at the distance of one mile. Ans. a= 5.46'.04".6. v= 921.566 feet. Ex. 6. What must be the angle of elevation of a body in order that the horizontal range may be equal to three times the greatest altitude? What, that the range may be equal to tihe altitude? Ex. 7. A body is projected at an angle of elevation of 600 with a velocity of 150 feet. Find the co-ordinates of its position, its direction, and velocity at the end of 5 seconds. Ex. 8. A body is projected from the top of a tower 200 feet high, at an angle of elevation of 60~, with a velocity of 50 feet. Find the range on the horizontal plane passing through the foot of the tower, and the time of flight. Ex. 9. A body, projected in a direction making an angle of 30~ with a plane whose inclination to the horizon is 450, fell upon the plane at the distance of 250 feet from the point of projection, which is also in the inclined plane. Required the velocity of projection, and the time of flight. Ex. 10. The heights of the ridge and eaves of a house are 40 feet and 32 feet respectively, and the roof is inclined at 30~ to the horizon. Find where a sphere rolling down the roof from the ridge will strike the ground, and also the time of descent from the eaves. Ex. 11. How much powder will throw an eight-inch shell, weighing 48 lbs., 1500 yards on an inclined plane, the angle or 168 D Y NA M I'C S. elevation of the plane being 28~.45', and that of the'mortal being 480.30'? Ex. 12. Find the velocity and angle of elevation that a pro Jectile may pass through two points whose co-ordinates arc x=300 feet, y=60 feet, x'=400 feet, and y'=40 feet. Also find the horizontal range, greatest altitude, and time of flight. CHAPTER V. CONSTRAIN E D MOTION. ~ I. MOTION ON INCLINED PLANES. 275. PnoP. To determine the relations of the space, time, and velocity when a body descends by the action of gravity down an inclined plane. Let the body fall from C down C the inclined plane CA, whose inclination is a, M be the position of the body at any time t from rest at C, CM=s, and v= the velocity at A/ D M. g The force of gravity g, acting in the direction Mg, may be resolved into two forces, onef=g sin. a acting in the direction CA, the other g cos. a acting perpendicularly to CA, and wholly ineffective in producing motion; the body is, therefore, urged down the inclined plane by the constant accelerating force f=g. sin. a. Ift therefore, this value off be substituted in (53), we have v =gt sin. a, (72) in (54), we have s -=gt2 sin. a, (73) in (55), we have v2==2gs sin. a, (74) from which all the circumstances of the motion may be determined. 276. PRoP. The velocity acquired by a body in falling down an inclined plane is equal to that acquired in falling freely through the height of the plane. If s=:AC, the length of the plane, and h=BC, the height, by (74), v2-2g.s. sin. a =2g.CB. 1 70 D Y N AMICS.. v='2gh, which, by (53'), is the velocity due to h, the height of the plane. 277. PROP. The times of descent down different inclined planes of the same height vary as the lengths of the planes. By (73), s=-'t2 sin. a, BC or'C- 1A =-gt AC' V 2.. t-=AC B.C x AC, when BC is constant. 278. PROP. To find the relations of the space, time, and veloc dty when a body is projected down or up an inclined plane. Substituting for f, its value, g sin. a, on an inclined plane, in (59), we have v =v:-gt sin. a, (75) in (60), we have s =vt~t4-gt2 sin. a, (76) in (61), we have v =v' -42gs sin. a, (77) which give all the circumstances of the motion of the body. 279. PROP. The times of descent down all the chords of a circle in a vertical plane, drawn from either extremity of a vertical diameter, are the same, and equal to that down the vertical diameter. B Let AB be the vertical diameter, BD and AD any chords drawn from its extremities, and DC perpendicular to AB. To find the time of descent down BD E we have, from (73), BDz=1-gt2 sin. BDC 2BC 2BD2 g. 2B 2.BA g CONSTRAIN ED MO IT DN. 17] /2.BA or =V': which, by (56'), equals the time down the diameter in the same manner, 12CA DA- A from which we obtain, as before, ~ /2.BA t= /-. 280. PROP. To find the straight line of quzckest descent from a given point within a vertical circle to its circumference. Let P be the given point.: Draw the A vertical radius CA, join AP and produce it to meet the circumference at B';' PB will be the line' required.... Join BC and draw POD parallel to- C E AC.'- Since' BC AC,'BO-PO. - With B n O as a' center,'and'radius PO;, describe' the circle PBD, which' will be tagrient to the circle C at B.. Now the- time down PB will equal the time down the vertl cal diameter PD, and the time down every other chord drawn from P (Art. 279).' But any other chord from P, produced to the circumference of C, will be partly without the circumference of 0, and, therefore, the time down it will be greater than the time down PB, which is therefore the line of quickest descent to the circumference of C. 281. PROP. To find the straight line of quickest descent from a given point to a given inclined plane.: Let A be the given point, AG a vertical line passing through this point, and AB a perpendicular to the inclined plane BG. The line AC, bisecting the angle BAG, will be the line required.' Draw CE parallel to BA, and therefore perpendicular to BG. Since EAC=CAB G 172 DY N A I C S. =ECA, the triangle EAC is isosceles. With E as a center, and EA or EC as a radius, describe the circle ACD, to which BCG will be a tangent. Now the time down AC will be equal to the time down AD, and to the time down every other chord drawn from A. But any other chord of the circle must be produced to meet the plane BG. Therefore the time down any other line, drawn from A to the plane, will be greater than the time down AC, which must be the line of quickest descent required. 282. PROP. Two bodies are suspendedfrom the extremities oJ a cord passing over afixed pulley: to determine the circumstances. of their motion. A Let P and Q be the weights of the bodies, of which P Q P is the greater. By (22) their masses are - and respectively. Neglecting the rigidity of the cord, the inertia and'friction of the pulley, if P=Q, they p will counterpoise, and no motion will ensue; but P>Q, P will descend, and Q rise through equal spaces by a force equal to the difference P-Q of y their weights. But this being a moving force (Art. 225), is equal to the accelerating force into the mass moved, =f(PcQ), where f represents the accelerating force. Hence P-Q=fP+Q P-Q This being a constant accelerating force, by substituting its value in (53), (54), and (55), we have expressions for the space, time, and velocity. 283. SCHOL. If the inertia I of the pulley be taken into con-,' sideration, Ig is an additional force to be overcome by the differenc:e of the weights, and in this case P-Q f=g'p+Q+Ig CONSTRAINED MOTION. 173 This is the formula for Atwood's machine, an instrument foA illustrating the laws of falling bodies. The friction of the pulley is reduced by friction wheels, and the rigidity of the cord by employing a fine flexible thread. By making the difference between P and Q small, the motion is made so slow as to render the time, space, and velocity easily determinable. ~ II. MOTION IN CIRCULAR ARnS. 284. PnoP. lVlzen a body descends by the action of gravity down any smooth arc of a circle in a vertical plane, the velocity at the lowest point is proportional to the length of the chord of the arc. Let a body descend firom the point D (Fig., Art. 279) down the are DA. Since the reaction of the curve is always perpendicular to the path of the body, it can neither accelerate nor retard the motion of the body. Its velocity, therefore, at A will be that which it would acquire in falling through the vertical height CA of the are. Hence, by (55). v= -2g CA'AD2 — 2gAB' v-=ADA acAD. When the arcs are small, the velocities are nearly proportional to the arcs, the principle referred to it t Art. 254. 285. PROP. When a body is constrained to describe the sides of a polygon successively, to findthe velocity lost in passingfromZ one side to the succeeding one. Let AB and BD be two adjacent sides of the polygon, w the angle made by their A C directions, and v the velocity of the body at the point B. In passing from AB to D BD some of the velocity will necessarily be lost. Resolving the velocity in the direction BD, we have, for the velocity in BD, 174 - DYN AMIC S. V COS. W. This, subtracted from the primitive velocity v, will give for the velocity lost, V =-'V cos. W =v(1 —cos. w) — v versin. w v sin.' w 2-versin. w 286. COR. If the sides of the polygon be increased in num ber, the angle w diminishes; and when the polygon becomes a circle, w, and consequently its sine, becomes indefinitely small. In this case, versin. w, small in comparison with 2, may be rejected, and V vu =_ sin. w. But if sin. w is infinitely small, sin.2 w is infinitely smaller, and the velocity lost v = 0. When, therefore, a point or body is constrained to describe a curve, no velocity is lost by the reaction of the curve. 287. PROP. If a material point move through one side of a regular polygon with a uniform velocity, to find the direction and intensity of the impulse which must be given to the material point at each angle of the polygon in order that it may describe the entire polygon with the same uniform velocity. A, Let ABC... be the polvnon. and let, / D the material point describe AB with the.velocity v in the unit of time. If, when E at B, no other force act upon the point, it will describe BD=AB in the same time. Join DC, and draw BN equal and parallel to DC. If, when at B, the point- receive an impulse in the direction BN, such as to cause it to describe BN in the same time that it would describe BD, the point will describe the diagona BC, the succeeding side of the polygon. CONSTRAINED MOI 0 T I 0 N. 17b in the same time. But since BC=BD, the velocity in BC is the same as that in AB, and the point will describe the two adjacent sides with the same uniform velocity. Now, since the triangle BCD is isosceles, BDC=BCD=CBN. But ABN=BDC;.'. ABN=CBN, and BN bisects the angle ABC. Hence the direction of the impulse will pass through the center of the circumscribing circle, and, if similar impulses be applied at each of the angles of the polygon, the point will describe the entire polygon, with the original velocity unchanged. To find the magnitude of the component force BN=f, let r be the radius of the circle; and, since BN is perpendicular to the chord AC, BM-=,2 2' BC2 and f=BN=BC But BC=AB is the space described in the unit of time, and is therefore represented by v. Hence -f= (78) or, the intensity of the impulse is equal to the square of the velocity in the polygon divided by the radius of the circumscribing circle. 288. Con. The force which must continually urge a material point toward the center of a circle, in order that it may describe the circumference with a uniform velocity, is equal to the square qf the velocity divided by the radius. Since the reasoning in the proposition is independent of the number of sides in the polygon, the sides of the polygon may be increased in number, and the fiequency of the impulse increased in the same ratio, without affecting the relation of the intensity of the impulse to the velocity in the polygon. But when the number of sides becomes infinite, or the polygon becomes a circle, the impulses will no longer be successive, but an incessant action of the same intensity. The force is, therefore, a constant accelerating force, and 175 DDY N A MI I C S. f=-. (79) r If the mass of the body be taken into consideration, the moving force (Art. 225) will be my2 F mf= r. (80) It is obviously immaterial whether the body be retained in ts path by the reaction of a smooth curve, or by an inextensible cord without weight, by which it is connected with the center of the circle. F will be a, measure of the resistance of the curve in the one case, and of the tension of the cord in the other. 289. DEF. The force which constantly urges a body toward the center of its circular path is called a centripetalforce. The tendency which the body has to recede from the center, in consequence of its inertia, or the resistance which it offers to a deflection from a rectilinear path, the resistance being estimated in the direction of the radius, is called a centrifugalforce. 290. PROP. To discuss the circumstances of the motion of a body constrained to move in a circle by the action of a central force. 1o. When the masses are equal, by (79), V2 f=7. if r be constant, f Oc v Jr, when a body moves in the circumference of a circle by means of a cord fixed at the center, the tension of the cord, or the centrifugalforce, will vary as the square of the velocity. 20. If v be constant, 1 or, if equal bodies describe different circles with the same velocity, the centrifugalforces will be inversely as the radii of the circles. CONSTRAINED MOTION. 177 30. Let T e the periodic time, or time of one revolution. Then (37) vT=2-rr, or v - - (79), =fr-. 47r2. r * f= T9 (2 ) T; or, the centrifugalfolrce varies directly as the radius of the circle, and inversely as the square of the periodic time. 4~. If T2 G r", by substitution in (81), r 1 f rI r'; or, when the squares of the periodic times are as the cubes oi the distances from the centel, the centrifugal force will be inversely as the square of the distance. 5~. If o be the angle subtended by the are described by the body in the unit of time, o is called the angular velocity. Assuming for the angular unit the angle whose arce is equal in length to the radius, v=ro. Substituting this value of v in (79), f=-'w2; (82) or, the centrifugalforce varies as the product of the radius and the square of the angular velocity. 6~. If in each of the above cases the masses be not the same, the centrifugal forces will vary directly as the masses in addition to the other causes of variation. The foregoing principles admit of a very simple and satisiactory illustration by means of an instrument called the whirling table. 291. Pnor. To find the relation of the centrifugalforce in a circle to the force of gravity. Let h be the height due to the velocity v which the body has in the circle. By (53') v'=2gh. This value of v', substi tuted in (79), gives M 178 DYNA MICS. 2gh f _2 or --- g r THence t'ne centrifugalforce is to the force of gr avity as twice the height due to the velocity in the circle is to the radius of the circle. Con. If f=g, r=2h, or, in a circle whose radius is equal to twice the height due to the velocity in the circle, the centrifugal force is equal to the force of gravity. 292. Prao. To find the centrifugalforce at the equator. 47r2R By (81), f= T The equatorial radius R of the earth is 3962.6 miles =20,922,528 feet. 7r=3.1415926. And, since the earth revolves on its axis in 0.997269 of a day, T=0.997269X86400-=86164s. These values, substituted in the above, give f=0.111255 feet. 293. SCHOL. Since the force of gravity g at the equator has been found to be 32.08954 feet, if G be the force of gravity on the supposition that the earth does not revolve on its axis, then g=G-f, or G=g +f, =32.08954+0.111255=32.200795,.. f 0.111255 1 and. —U 3)0735=28 nearly8 (83) G-32.200795 289 near]v or the centrifugal force at the equator is 289 the force of gravity. 294. PROP. To find the time in which the earth must revolve on its axis, in order that the centrifugal force at the equator may equal the force of gravity. Let T' be the required time, and f' the corresponding cen trifugal force. From (79) we have CONSTRAINED IMOTION. 179 R R' f'f'=' But in this case R-R', andf' is to become equal to G. or T 2=- TT T2 ~~or CG 289 and T' 7-T nearly. 17 Hence the earth must revolve in I-th of its present period. 17 295. PROP. The centrjfugal force diminishes gravity at diferent places on the earth's surface in the 2ratio of the square of the cosine of the latitude. Regarding the earth as a sphere, from which it differs but very little, let EF be z A the axis, AC=R the equatorial radius, B B any point whose latitude is the arc AB, ~d7 measured by the angle ACB=-, and A BG=R' the radius of the parallel of latitude passing through B. Let the centrifugal force at B, which acts in the direction GB, be represented by BD. By (81), BD= Resolving this in the direction of the vertical CZ, opposite to that in which gravity acts, and calling this componentsf, we have f'=-Bb= T Cos.~. But R'==R cos. b,.'. f'- TM COS.'~, (84) tn which the coefficient of cos.2 q' is constant for all latitudes. Hence f' O cos.2 q, 180 I Y N A M C S. f' being the diminutic(1 of gravity by the action of tile centrif ugal force. The latitude of Middletown being 41~.33'.10", the centrifugal forcef', in a. vertical direction at this place, will be found to) be f'=0.062305. If this be added to the observed gravity g'=32.16208, we shall nave for the whole gravity, undiminished by the centrifugal fOrce, G g'=+' —f' =32.224385. 296. Con. Resolving the centrifugal force at B in a direc tion perpendicular to the radius CB, we have f' — d= T- cos. ~ sin., 27rR in. 2 If, therefore, the matter of the earth were susceptible of yielding to this component of the centrifugal force, it would necessarily cause the earth to deviate from a spherical form. The fluid portions of the surface are therefore urged toward the equatorial regions, thereby increasing the equatorial diameter and diminishing the polar. If the solid portion did not also partake of the same general form as the fluid, we should expect to find vast equatorial oceans and polar continents, with polar mountains far exceeding the equatorial in height. But the actual distribution of the waters and mountains on the surface of the earth is widely different. The amount of deviation fiom a spherical form, which the earth must take from the ac tion of the forces developed by its motion, will depend in part on the law of variation of its density from the surface to the center. By measurement the equatorial diameter is found to exceed the polar about 26 miles. The ratio of the difference of the equatorial and polar radii to the equatorial radius, called the compression or ellipticity of the earth, is about 297. PROP. To find'he centrifugcl force of th1 moon in its orbit. CONSTRAINED MO TION. ]81 Let R be the radius of the earth, nR the mean radius of the moon's orbit, and P the periodic time of the moon in seconds. Then the velocity of the moon will be 2rrnR P This, substituted in (79), gives, for the centrifugal force, 4rr2nR Of=- p2* (a) If the earth were an exact homogeneous sphere, the intensity of gravity would be the same at all points on the surface. Being a spheroid, differing little from a sphere, the theory of attraction of spheroids shows that it is necessary to use the intensity of gravity at a latitude ~, of which the sine is the V-; whence b=35~.15'.52". Supposing the compression of the earth to be 0.00324, in the latitude ~, R=20897947 feet. To find n, conceive two lines drawn from the center of the moon, one to the center of the earth, and the other tangent to the earth at the equator. The angle made by these lines at the center of the moon, and subtended by the equatorial radius of the earth, called the moon's equatorial hoirizontalparallax, is found by astronomical observations to be at a mean 57'.1". This angle, when subtended by the radius of the earth in the RP 1 latitude of 35~.15'.52", is irz 56'.57".33. Now = sin. 7r=-, from which we find n —60.3612. P-23605S5s. These values, substituted in (a), give f= 0.00894. 298. CoR. The moon is retained in its orbit by the gravity of the earth. For, since the intensity of gravity is inversely as the square of the distance from the earth's center, its intensity g' at the moon may be found by the proportion 1 1 g g'-=: L2R2; whence' n' The intensity of gravity in the latitude of 35~.15'.52", un 182 D Y N AM IC S. diminished by the centrifugal force of the earth in its diurna, motion, is g= 32.24538. Substituting for n and g their values, we find g'= 0.00885. The difference between f and g' is less than one ten thousandth of a foot, a difference which may be attributed to errors of observation. ~ II1. PENDULUM. 299. DEF. A simple pendulum is a material point suspended by a right line, void of weight, and oscillating about a fixed point by the force of gravity. The path of the point is the arc of a vertical circle, of which the fixed extremity of the line is the center. 300. PRoP. To find the force by which the pendulum is utrged in the direction of its path. Let MI be the material point suspended from C by the line CM. When removed from its'vertical direction CM to the inclined position CM', gravity will cause the point to descend and describe the are MI'M. Resolving the force of gravity g in the directions M/'B and M'T, parallel and perpenB T dicular to the radius CM', the component g cos. 0, in the direction of CM', will be counteracted by the fixed point C. The component g sin. 0, in the direction of the tangent M'T, will be the only effective force to produce motion. Hence the accelerative force is f=g:siin. 0. Let CM=I, the length of the pendulum and the arc M' M-.s the semi-are of the vibration; then s=110... f=g. sin.8 CONSTRAINED MOTION. 183 NIow f x be any arc of a circle, it is shown in trigonometry tUat X3 x x sin. x l - - -1- 4, &c. in 1.2.3 1.2.3.4.5 Hence f=g - 1 &C. e 1.2.3.13J 1.2.3.4.5.1 &. But when the arc s is small compared with 1, the cube ani( higher powers of I may be neglected, and f=-s (85) CZ S, or the force varies as the distance from the lowest point measured on the arc. 301. PROP. When a body is urged toward a fixedpoint by a force varying directly as its distance from that point, the times of descent to that point from all distances will be the same. For the space described will obviously depend on the intensity of the force and the time of its action, or will be measured by the product of the force by some function of the time. Hence s=f.~b(t). Since s varies asf, let s=nf, n being the constant ratio of the space to the force; then, by substitution, n=0(t) i or the function of the time is the same whatever be the dis tance, and, in the case of the pendulum, n=-. (85). 302. ScHOL. It will be shown hereafter that the time of de scent to the center of force, or the point toward which the 7r 1 force is directed, is always equal to /- when / is the accelerating force. In the case of the pendulum, tl=-, and we shall have, for the time of descent to the lowest point, or the time of a semi-vibration. 184 D Y N A MI C S. When the material point has reached its lowest pos tion, its momentum will cause it to rise on the other side of the vertical line; and, since its velocity at the lowest point will be that due to the vertical height fallen through, and the force will require the same time to destroy the velocity that was required to generate it, the whole time of a vibration will be T = -. (86) As the circumstances of the material point at the end of the time T are the same as at the commencement, it will then perform another vibration in the same time, and so continuing, all its vibrations will be zsochronal. It should be recollected that these results are obtained on the supposition that the arcs of vibration are small. 303. PROP. To discuss the circumstances of the vibration oJ different pendulums. 10. Since in (86) 7r is constant, T 9 V/g or, the time of vibration of a pendulum varies directly as the square root of the length, and inversely as the square root of the accelerating force. 20. If g be constant, which is the case in the same latitude, and at the same elevation above the surface of the earth. T G a/1; or, the time will vary as the square root of the length. 30~. If I be constant, or the length of the pendulum remain the same, then Tc-_; Vg or, the time of a vibration will vary inversely as the square roo, of the intensity of gravity. CONSTRAINED MOTION. 185 4'. If T be constant, since gT2= r2, lacg; or, the lengths of pendulums vibrating in the same time vary as the accelerating force. 304. Con. Hence the force of gravity in different latitudes, and at different elevations above the surface of the earth, will vary as the length of the pendulum vibrating seconds. If, therefore, the length of the seconds pendulum be ascertained at various places on the earth, we shall have the relative intensities of gravity at those places. Since these intensities of gravity at different places are dependent t upon the figure of the earth, they will serve to determine it. The pendulum, therefore, becomes an instrument for ascertaining the form of the earth. 305. PROP. The lengths of pendulums at the same place are inversely as the square of the number of vibrations performed by each in the same time. Let N and N' be the number of vibrations performed by each respectively in the time H. Then the duration of a vibration by each will be H H T=- and T'=,. But T: T'='/: /' = j, Hence 1: I'=N': N2. Also, I= N2. (87) 306. SCI-OL. To determine experimentally the length of the seconds pendulum, or a pendulum which will oscillate 86400 times in a mean solar day, let a clock, the length I of whose pendulum is required, be regulated to vibrate seconds. Suspend in front of the clock a pendulum of known length l'(1'<1), and observe the exact second when the two simultaneously commence a vibration. As the vibration of 1' will be more rapid than that of 1, count the number of returns to coincidence of vibration in a period of fiSe or six hours; and, finally, iote 186 DYNAMICS the exact second of the simultaneous termination of a vibra tion. If now to the whole number of seconds N shown by the clock, there be added twice the number of coincidences of vibration, we shall have the number N' of vibrations of 1'. These values, substituted in (87), will give the'ength I of the seconds pendulum. The length of the seconds pendulum in the latitude of New York has been determined to be equal to 39. 10168 = 3%25847. 307. PROP. To find the value of g, the measure of the intensity of gravity. From (86), we deduce 7Tr Making T=l, and using the length of the seconds pendulum above, we find, for the value of g in the latitude of New York, g=3855n.9183=32t. 1598. 308. PROP. To find the correction in the length of a pendulumn which gains or loses a known number of seconds in a day. Let I be the length of the seconds pendulum, n=86400, y — the number of seconds gained or lost in a day, and x= the corresponding correction in the length. By Art. 305, and recollecting that a diminution of length corresponds to an increase in the number of vibrations, n2 (n+y)-2=I-X:,'from which we deduce 2nvy+y2 X n2+2ny+y2' or, rejecting y2, as small in comparison with 2ny, 2y 2y + n.1: -.Ate (88) lo y Hence, divide the le ngth of the seco:,ds pendulum by one plu. CONSTRAINED MOTIO N. 18 the ratio of the number of seconds in a day to twice the gain or loss, and the quotient will be the correction in the length. x will have the same sign as y. 309. PROP. To determine the rate of clock when carried to a given height above the surface of the earth. Let N=-86400, N'= the number of vibrations in a day when the clock is carried to the height h above the surface of the earth, and r= the radius of the earth. The length of the pendulum remaining the same, 1 1 1 1 I 1 * But r2: (r+h)2.N N'=r+h: r, rN and N' rN hN N-N' — ~ N (89) N-N', the loss in a day, is the rate of the clock. Hence, di'vide the number of seconds in a day by one plus the ratio of the radius of the earth to the height, and the quotient will be the rate. The quantities r and h must be in the same denomination. 310. COR. If the loss in a day by a seconds pendulum be determined by observation, we may find the height to which it is carried, for (89) gives hr(N-N') -= N' (90) 311. DEF. A body suspended by a cord and performing revolutions in a horizontal circle is called a conical pendulum. 312. PROP. To determine the motion of a conical pendulum. Let 1= the length of the cord: AP, fixed at A and attached to the body at P, PC=r the radius of the circle which the' body I88 ~DY N AM I CS. A describes with the uniform velocity v, m-= the mass of the body, and the angle PAC=O. S:nce Am. the body is in the same circumstances at each point of its path, the forces acting upon it must be in equilibrium. These are the ten"......', sion t of the cord in the direction PA, the centrifugal force f= - in the direction CP, and the w-eight of the body mg downward. Resolving horizontally and vertically, we have my2 tsin. 0- -=O. (a) t cos. 0-mg= O. (b) From (b), we have the tension in the cord. mg cos. 0' Eliminating t from (a) and (b), gr'sin. 0 cos. 0 sin.' 0 g Cos. 0' The time t of performing one revolution is 2rrr t=I cos. 0 ACo =27/r 313. Dr. Bowditch, in the second volume of his transiatlon' of the 3I6canique Celeste, gives the following formula for comrn puting the length I of the seconds pendulum in any latitude p l= 2 + o sin.2 ~, in which.2=39.01307 in. and w=0.,20644 in. CONSTRAINED MOTION 189 314. EXAMPLES. Ex. 1. The length of an inclined plane is 400 feet, its heigllt 250 feet: a body falls from rest from the top of the plane. What space will it fall through in 3- seconds? in what time will:t fall through 300 feet? and what velocity will it have when it arrives within 50 feet of the bottom of the plane? Ex. 2. The angle of elevation of a plane is 250.30'. A body, in falling from the top to the bottom, acquires a velocity of 450 feet. What is the length of the plane? Ex. 3. The length of a plane is 240 feet, and its elevation is 36~. Determine that portion of it, equal to its height, which a body, in falling down the plane, describes in the same time it would fall freely through the height. Ex. 4. A body descending vertically draws an equal body, 25 feet in 2- seconds, up a plane inclined 30~ to the horizon by means of a string passing over a pulley at the top of the plane. Determine the force of gravity. Ex. 5. The time of descent of a body down an inclined plane is thrice that down its vertical height. What is the inclination of the plane to the horizon? Ex. 6. If a body be projected down a plane inclined at 30~ to the horizon with a velocity equal to 3ths of that due to the vertical height of the plane, compare the time of descent down the plane with that of falling through its height. Ex. 7. A given weight P draws another given weight WV up an inclined plane of known height and length, by means of a string parallel to the plane. When and where must P cease to act that W may just reach the top of the plane? Ex. 8. Divide a given inclined plane into three parts such that the times of descent down them successively may be equal. Ex. 9. At the instant that a body begins to descend down a given inclined plane from the top, another body is projected upward fiom the bottom of the plaine with a velocity equal tc 190 1 Y- N A MI CS. that acquired in falling down a similar inclined plane n time. its length. Where will they meet'? Ex. 10. A ball having descended to the lowest point of a circle through an arc whose chord is a, impels an equal ball up an arc whose chord is b. Find E tile modulus of elasticity of the balls. Ex. 11. Determine that point in the hypothenuse of a right angled triangle whose base is parallel to the horizon, from which the time of a body's descent in a straight line to the right angle may be the least possible. Ex. 12. Two bodies fall fiom two given points in the same vertical line down two straight lines to any point of a curve in the same time, all the lines being in the same vertical plane. Find the equation of the curve. Ex. 13. 193 oz. is so distributed at the extremities of a cord passing over a pulley, that the more loaded end will descend through 3 inches in one second. What is the weight at each end of the cord? Ex. 14. If an inelastic body be constrained to move on the interior of a regular hexagon, describing the first side with a uniform velocity in one second, find the time of describing the last side. Ex. 15. A stone is whirled round horizontally by a string 2 yards long. What is the time of one revolution, when the tension of the string is 4 times the weight of the stone? Ex. 16. What is the length of a pendulum which oscillates twice in one second? Ex. 17. A seconds pendulum, carried to the top of a mountain, lost 48.6 seconds in a day. What was the height of the mountain? Ex. 18. The length of a pendulum vibrating sidereal seconds, being 38.926 inches, what is the length of the sidereal day? How much must it be lengthened that it may measure mean solar time? CHAPTER VI. ROTATION OF R IGID BOD)IES. 315. Pnop. When a rigid body, containing a fixed axzs, is acted upon by a given force in a plane perpendicular to thai zaxis, to determine its motion. Let the annexed figure be a section of the body by a plane perpendicular to the axis at c C, and the given force F act at B, in the direction AB, at a given distance from C. If v be the velocity which the force F can impart to a mass MI when firec, F=Mv. \ 9 When the force acts on the body, each particle is constrained to move in a circle whose center is in the axis through C. If o be the angle made by the body from the action of the force in a unit of time, the linear velocity of any particle m will be rt, and its momentum f=mro. The moment of this force, in reference to the axis through C, is?f=mr2o. The moment of any other particle will have the same form, and hence?f+rf'+, &c., =mr2W+m'r'2w, &c., or X.rf= Z.mr'w =t mr2, since w is the same for each particle. But the moment of F is rF=Mv.CB, and, as the former is a measure of the effect of the latter, &OZ.mr2= Mv.CB. MV.CB W 2.=. r2' (91) Thus, if the body consisted of two particles m and m' at the distances r and 7' from the axis, and the force F would impar' to m, when free, the velocity v. then the angular velocity 192 D Y N A I I CS. mvr mr72 +-nm l/2 and tlie linear velocity c f m will be mv,2 mr2 +m/' / If the two particles were equal, and at equal distances fromn the axis, then mr-2-m'r'2, and V If'm=m', and r'-2r, Con. If the body were in motion before the force acted on it, then the change of angular velocity will be given by (91). 316. DEr. The moment of inertia of a rigid body about an axis is the sum of the products of the mass of each particle by the square of the distance of that particle fiom the axis. Thus, if m be the mass of a particle of a rigid body, r its distance from a fixed straight line,:Z.mr' is the moment of inertia of the body about that line. Thedefinition given above is to be regarded as a verbal enunciation of this analytical expression, which has required nomenclature by the frequency of its occurrence in dynamical investigations. 317. PnoP. The moment of inertia of a body about any fixed axis exceeds its moment of inertia about a parallel axis passing the center of gravity, by the product of the mass into the square of the distance between the axes. cfA Let mn be a particle of the body, and let the plane mCG, passing through m, cut the two axes at C and G. Through the axis at G pass a plane perpendicular to CG, intersectBI / G AF ing the plane of the figure in AB. Let, i mC=r, mG=r,,andCG=h. DrawmFperD pendicular and mD parallel to AB. Then r 2= 2+hI2h. GD, according as the angle mGC shall be obtuse or acute. Hence ROTATION OF RIGID BODI ES. 193 mr" z= Zlm2 lf/ ~i 2m h. GD, anld ~ -.?r2.r + 1h2Ym~2hn.m.GD.!Now, since the plane through the axis at G passes through the center of gravity, by (29),,.m.GD= z.m.Fm=O. * 2.mr'==zXmr 2+hA/m.'. mrl+ -h', M being the whole mass of the body. 318. SCHOL. Assuming k, such that MkA2=:.mr-, we have?Y,,a =:M(k2 +h2). M/k' is the moment of inertia of a body about an axis through,he center of gravity, and M(kI+h') is the moment of inertia about a parallel axis at a distance h fiom the former. The moment of inertia about any axis will, therefore, be easily determined when the moment of inertia about an axis through the center of gravity is known. Since k2= M' the determination of k will, in general. require the aid of the integral calc ulus. The length v/k2+h2 is called the radius of gyration about the axis considered, and, similarly, k is called the radius of gyration about an axis through the center of gravity. Since h is the least value of Vk2"+h', it is sometimes called the principal radius of gyration. 319. PROP. If a body oscillate about a fixed horizontal axis not passing through its center of gravity, there is a point in the right line, drawn from the center of gravity perpendicular to the axis, whose motion is the same as it would be if the whole mass were collected at that point and allowed to vibrate as a pendulum about the fixed axis. Let the horizontal axis be perpendicular to Rc AB the plane of the figure at C, G be the center of gravity, CA be horizontal, and GA vertical. Then, if M be the mass of the body, and v the velocity which gravity can impart to the body G if free in each instant of time, the moment of 0 N 194 DVY N A MI CS. gravity will be Mv.CA. By Art. 315, Cor., tne change of an. gular velocity produced in each instant of time is Myv.CA 6).mr2 Produce CG to some point O and draw OB vertically, meeting CA in B. Now, if the whole mass of the body were collected at 0, the moment of inertia of the body would be M.CO2, and the change of angular velocity in the same instant of time would be Mv.CB M.CO' But the position of the point O being arbitrary, CO, and thercfore CB, will vary at pleasure. We may, therefore, put y.mr2: M.CO2=CA CB=CG: CO, (a) which gives CA CB 2.mr2 M.C0O'9 or to —; that is, the change of angular velocity is the same in the two cases, and, therefore, the motion from rest will be the same. q20. ConR. The point 0 may be found from (a), which gives CO= M.CG M.h h =" lh' (92) 321. DEF. The point O is called the center of oscillation of the body with'respect to the axis through C. It is thus defined: when a rigid body moves about a fixed horizontal axis under the action of gravity, in the straight line drawn through the center of gravity perpendicular to the axis, a point can be found such that, if the mass of the body were Collected there and hung by a thread from the axis, the angular motion of the point would, under the same initial circumstances, be the same with that of the body, and this point is called the center of oscillation of the body with respect to the axis. R O ATIC N ) F RIGID BODIES. 195 822. Cor. Herbee, when a body makes small oscillations about a fixed horizontal axis, it is only necessary t) calculate the value of the expression h+ =-1, and the time of a small oscillation will be 7rV/'. (86). 323. DrEF. A body of any form suspended firom a fixed axis, about which it oscillates by the force of gravity, is called a compound pendulum. 324. Prnop. Tae centers of oscillation and suspension are convertible. Let h' be the distance of the center of oscillation from the center of gravity, when h is the distance of the axis from the center of gravity. Then, by (92), CO -Ah X —, -- 2and and h=-; (93) so that if h' be the distance firom the center of gravity to the axis, h will be the distance from the center of gravity to the center of oscillation, and a body will oscillate in the same time about an axis, through the center of oscillation, as it oscillates about the original axis, the extent of vibration being the same. 325. COR. 1. If in a straight line through the center ofgraN ity, perpendicular to the axis of motion, points be taken at distances h and h' from the center of gravity, and on opposite sides of its then the length of the equivalent simple pendulum is h+h'l; so that the time of vibration about the axes through each of these two points is the same, and the length of the equivalent simple pendulum is in each case the distance between the two axes. Each of these points is the center of oscillation in reference to, the other as a center of suspension. 326. Con. 2. From (93), we have hh' = kA; or, the principal radius of gyration s a mean is ean proportional tw 196 - u Y N AM I CS. tween the distances of the centers of osciliation7 an! stusyJensIon from the center of gravity. I,2 327. Cor. 3. Since h: k +h2 —= -Vk2+-: h+J —= CO, we in. fer that the distances of the centers of gravity, of gyration, and of oscillation from the axis of motion, are continued proportionals. 328. SCHOL. The convertibility of the centers of oscillation and suspension was employed by Captain Kater in finding the length of a simple pendulum vibrating seconds, and consequently the force of gravity at the place of observation. A bar of brass one inch and a half wide and one eighth of an inch thick was pierced by two holes, through which triangular wedges of steel, called knife edges, were inserted, so that the pendulum could vibrate on the edge of either of these as an axis, resting on two fixed horizontal plates of agate, between which the bar was suspended. The axes were about 39 inches apart. VWeights were attached to the bar and rendered capable of small motions by screws, by which means the position of the center of gravity of the bar could be changed. These weights were so adjusted by trial that the time of a small vibration through an angle of about 1~ was the same when either knife edge was the axis, so that each gave the center of oscillation belonging to the other. The distance of the knife edges was obtained by placing the pendulum so that the edges were viewed by two fixed microscopes, each furnished with a micrometer of ascertained value, and afterward placing a scale of known accuracy in a similar position under the same microscopes. This method, combined with that referred to in Art. 306, served to determine the length of the seconds pendulum, and thence the torce of gravity. 329. The relation of the simple to the compound pendulum will be illustrated by one or two examples in which the prin, cipal radius of gyration is supposed to have been previously determined. Ex. 1. A material straight line vibrates about an axis perpendicular to its length: required the length of the isochronal. simple pendulum. ROTATION OF RIGID BODIES. 197 ILet 2a be the length of the line, and h the distance of the point of suspension from its center of gravity, which is its middle point. The radius of gyration of a straight line about an axis through its center of gravity, perpendicular to its length, is ~. Since the radius of gyration is a mean proportional between the distances of the centers of oscillation and of sus. pension from the center of gravity (93),.2 aa2 l'- and (Art. 325), I= _h 3h' 10. Let the point of suspension be at the extremity of tile line, in which case h-a. Then h'=1a, and la I +-a= 4.2a, that is, the center of oscillation is two thirds of its length below the axis of motion. 2~. Let h== a; then and 1- 2.2a, the same as before. Hence the time of a small vibration is the same, whether the line be suspended from one extremity, or from a point one third of its length from the extremity. This also illustrates the convertibility of the centers of oscillation and of suspension. 30. If /z-laz and I — 2 a=- 5.2a 6.2a; or, when the center of suspension is three eighths of its length from one end, the center of oscillation is one sixth of its length below the other end. 40~ If h=10, h' —, and -= A; or, when the center of suspension is at the center of gravlty, the leng'th of the equivalent simple pendulum is infinite. and 1 9,)8 D Y N A M I C 5. therefore the time of one vibration is infinite. This is obvious, also, from the fact that, when the line is suspended from the center of gravity, no motion can result fiom the action of gravity. Ex. 2. A sphere being made to oscillate about a given axis; required the length of the equivalent simple penduluin. Let r be the radius of the sphere, and h the distance of the axis of motion from its center. The principal radius of gyration of a sphere is r V. 2 Hence 5h' 3L. Let h=r, or the spnere vibrate about an axis tangent to its surfate. Then and I=r+ -r; or, the center of oscillation is two fifths of the radius distant friom the center of the sphere. 2~. If h=10r, h'l=2 25' and l=10rI —-. 30. If h=Ir, 1'=2r, and 1=2 +; or, the center of oscillation is without the sphere and at a dis. tance r from the surface. MOMENT' OF INERTIA. 330. Since (Art. 316) the expression for the moment of in ertia of a system consisting of a finite number of points is.mrw2' /n being the portion of the mass which is at the distance r fironr the axis of rotation; when the number of points becomes indefinite, the expression will evidently become fr'dM, dM hbe. ingy an element of the mass at the distance r. Hence (Art. 318), Mk2=fr12dM. (a) ROTATION OF RIGID B BDIES. 199 We shall now illustrate the method of determining Mka in a few simple cases. Ex. 1. To find the moment of inertia of a straight line re volving about an axis perpendicular to it at any point of it. Let the axis be at a distance a from one end and b from the other, and let r be any distance from the axis; since the thickness and density of the line are supposed to be uniform, er.}n nay ~re taken =1.. M a+b and dlM-=dr. Hence (a) (a tb)l2 =fir~ dr= + c, and the integ'al being taken between the limits -a and +b). a2 3+b3 3(a+b). If a=b, or the axis be at the center of gravity, we have, for the principal radius of gyration, a Ma2 2 3 3 and 3k~-2=- = whatever be the thickness and density of the line. Ex. 2. To find the moment of inertia of the circumference of a circle about an axis perpendicular to its plane through its center. In this case M-=27r, and since all the points are at the same distance from the axis, r is constant... Mk=fr2dM=-rfdMI=27rr3, and — r. Ex. 3. To find the moment of inertia of a circle about an axis through the center perpendicular to its plane. Putting a= the radius of the circle, its area will be 7ra2', and at a distance r from the axis the area will be Trr2. Hence, in this case, dM=g2rrdr and Mk2=7ra22=2grfr'dr=-r"'4= a'. when r-=a, and V~ _2. k/ = 2-, /2 200 - DYNA M IC S. Ex. 4. To find the moment of inertia of the circumferenco of-n circle about a diameter. Taking the proposed diameter for the axis of x, the distance of any point (x, y) from the axis of rotation is y. From the equation of the circle, y2=a2-xz, we find a /dx2+dy2=d aM= —dx.'. r dlM=y2dM =aydx, and Mk:=2Wake= afydzx=ax area of circle-=7-a'. Ex. 5. To1 find the moment of inertia of a sphere about a diameter. Let the axis of rotation be the axis of x, and conceive the sphere to be generatsd by a circle of variable radius, y, whose center moves along ti:e axis. By Ex. 3 the moment of this generating circle is ~rTy'. Hence the whole moment will be the sum of the moments of the gene';ating circle in all its posi tions..M..M2]=wrfy4dx. But y"=a-x2..'. Mk2=- rrf(a2 —x2)dx, and this integral, taken between the limits +a and -(a, is Mk2= 8A r5ra But M=P ra3... k=a i/ CHAPTER VII. 331. THE methods employed in the preceding chapters t; determine the motion of a body are partial in their application and limited to the simplest cases. When the force is variable, or the motion of a body is due to the action of several forces, varying, in direction as well as in intensity, with the varying position of the body, some more general method is required. Wc now proceed to show, to a limited extent, how the circumstances of the motion of a body or point, in such cases, may be determined. In order to this the fundamental formula w-ill need some modification and extension. / 332. By Art. 214, variable velocity at any instant is measured by the limit of the ratio of the space to the time, that is, from the nature of the differential coefficient, it is the differential coefficient of the space regarded as a function of the time Hence dst In like manner, it appears from Art. 223, that the force, when variable, is measured at any instant by the differential coeffi. rient of the velocity regarded as a function of the time. Henco dv [VIII.] From [VII.], by differentiation, we obtain dv d's dt dt~ d2s..d. [Ix.l From [VII.] and [VIII.] we obtain, by eliminating dt, vdv=l-ads. [X.j 202 D Y N A MI S. ~ I. RECTIINEAR MOTION OF A FREE POINT. 333. If a fiee point at rest is acted on by a single force, or forces whose resultant is equivalent to a single force, the mo. t1!n must be entirely in the direction of that force, and may at once be determined from [VII.] and [VIII.]. But in order that we may integrate these expressions, the law of variation of the force must be given as a function, either of the space, time, or velocity; and as the intensity of a variable force would depend on the position of the body, the force is natu-'rally and generally given as a function of the distance of the body from some fixed point. In order that we may obtain an exact expression for the value of the force at any variable distance, its value at some given distance must be known. This given distance is most conveniently assumed as the unit of distance, and if /t be the intensity of the force at this unit of distance, it is usually called the absolute force. If, then, the force be given to vary as the n"v power of the distance x from the fixed point, we have,e=1L;x". (94) We shall first consider the case of a constant force, for the purpose of showing with what facility the calculus enables us to determine the relations of the time, space, and velocity, although these relations have already been deduced in chapter III.' 334. PROP.'To determine the space described by a point acted on by a constant force in terms of the time, the velocity of the point in the direction of the force at the commencement of the time being given. The velocity due to the action of the force in the time t is, by [III.], v'=pt, where r represents the velocity generated by the force in the unit of time. If v, be the given velocity, or the velocity when the time commences, we have, for the whole velocity, as in (59), v=v,1~t. RECTILINEAR M.OTLON O.F A FREE POINT. 20: Substituting this value of v in [VII.], we have ds=v dt16tdt, and, integrating, s= t2 -+- -c; in which c is an arbitrary constant depending on the position of the point when t=O. If the space be reckoned from the position of the point when t=0, then c=O, and s=v,t~bt2. (60). If the point move from rest, v, =0, and s t. (54). All the other relations are readily deduced from these. 335. PROP. To determine generally the velocity of a particle moving in a right line to or from a fixed point by the action of a variableforce. By [X.], vdv-5=ds; or, measuring the line on the axis of x, vdv = Obdx. Hence, by integration, v2=2fPdx+ c; (95) and if the force varies as some function of the distance x, the integration may in general be effected. If the motion is toward the fixed point, dx will be negative; if from it, positive. To determine the constant, we must know the velocity at some given point. v 336. PROP. To determine generally the time of the motion. From [VII.] we have at once t=z-f+c, (96) and since v is determined by (95) in terms of x, the integra. tion may in general be effected. To determine c, we must know the position of the particle at some given time. 204 DY NA MI CS.' 337. PRoP. To determine the velocity of a particle attracted to a fixed point by a force varying directly as the distance of the particle from that point. Assuming the axis of x in the direction of the motion, andi the origin at the center of force, by (94) we have' -Jtx; the sign being negative, because the force which is directed to the origin tends to diminish the distance x. Substituting this value of O in (95), and integrating, we get V2 C -tX2. If a be the distance of the particle from the origin when the motion commences, when x=a, v=O. Hence and v)2=U(a2-x2), 97) from which it appears that x can not pass the limits ~-a. 338. COR. Since the force by which a material point, at liberty to move along a perforation from the surface to the center of the earth, varies directly as the distance from the center, (97) will serve to determine the velocity of the particle at any distance x from the center. In this case, if a=r, the radius of the earth, and g the force of gravity at the surface, ft will be found from the proportion 1': 1=g: M. ~g Hence v2' ==(r2 -x2). 4s this velocity must be spent before the particle stops, if we make v=O, we get x-=r, or the particle will go to the opposite point of the earth's surface. 339. PROP. To find the time when the force varies directly as tne distance. dx From [VII., we have dt=1 dx'~U ~a — X2. intrate ts, le t a sin 0, whece a- =a To integrate this, let x=a sin. 0, whence an-x2=a2 COS.'O. RECTILINEA R MOTION OF A FREE PO IN T. 205 1 a.d sin. 0 ~*. =: V'/ a cos. 0 - dO, 1 and t= _. +c. Hence 0= V't(t-c), and x=a. sin. xVf(t —c). Suppose the time to commence when the pai'icle is at t"ne origin, or x=O; then 0 = -.sin.'l6.C, or c=0.,'. t=.0. (a) If x=a, sin. 0=1, and 0= r,, or -r,, &c., 1 1 and tr- or 27 —= &C., = i VT - firom the commencement of the motion. If x=-a, sin. 0=-1, and 0=- 70r or r. &., 1 1 and t= 3r- or 7r-, &c. Hence the particle moves firom +a to -a, while t changcs from 1r-to 3- or the time of one vibration is T= _1, (98) and this is true whether a be large or small. Compare Art. 302. CoR. 1. When a particle is attracted to a fixed point by a force varying directly as the distance, the time of descent to that point will be the same for all distances. COR. 2. The time of descent to the center of the earth. ill which case P=9' is s06 DY N A.-M I C S. = /f 2~ g = 21'.5".8. CoR. 3. Let S be the center of force, AS=a the distance of the particle at the commencecment of motion, P the position of the particle at any instant, and t the time of describing PS-x. S~ B On AS describe a quadrant, and draw PQ perpendicular to AS. Then x=a. sin. QSB... QSB=0 and QB=a.0. 13y (97), v= 2. a2-~x- V/j.PQ C P Q, 1 1 QB QB and by (a), t= _ =- -QB - QB. VI /E.L* a vel. at SQB When the particle is at A, are BQ becomes BA=-'ra, and -T'= 2 Vft 340. Poro. To determine the velocity when a particle is at. tracted by a force varying inversely as the square of the distance. In this case (94), s= /x-t, dnd [X], vdv= —z-x-t dx. Integrating, v2= 2x-' +c. If v=O when x=-a, c= - 2/ta-'..'. v2t=2(x-1-' —a-')=21a. (99) 341. Con. Since the intensity of gravity above thle earth's surface varies directly as the mass of the bo7dy, and inversely as the square of its distance from the center, (99) will give the velocity of a body falling from any height to the surface. When the body arrives at the surface x-=r, and 1 1 2 since l: g:= ~ - y, =grF. Hence a-r v2=2gr. a Ifa=-c, v= V2gr=6.9428 miles, RECTILINEAR MOTION OF A FREE i UINT. 20)''iom which it appears that the velocity can nevei equal seven miles, and if a body be projected upward with the above velocity, supposing no resistance from the air, it would never r1o turn. If we make x=0 in (99), v — o 0; or the velocity at the center would be infinite if the same law of force continued. 342. PROP. To find the time when the force varies in cerselt as the square of the distance. We have, from [VII.], -dx -dx dt= - By adding ladx to the numerator, and then subtracting thle same quantity from it, we have t-(a)/ -ad/x-xdx dx 12x =a (a -2 -ver sin.-' —+c } When t=O, x=-a, and versin.-'~=rr. C=. 7T ()t= ~ - (acx-x) +-(,r- versin.-'- (100) COR. 1. If x-O, we have f a 2,7:2z \ ) SQ- =SC./_ SCQ- veasin.-. 2 a or the square of the time of falling, to the center varies as the cube of the distance. COR. 2. On AS=a describe a semicircle, and d(Iaw PQ pespendicular to AS. Then PQ- Vax —xe, arc AQS=-7;a, arc S;Q-SC. SCQ= — versin. -. 2 a 208 DYN AM ICS. A R Hence the time through AP is, by (100), s If SQ be produced to meet in IT a tangent at A SA.P a~a xa~-zxa-3x P by (99), ( PQ)+ are AQ- are When P coincides with S, AR is infinite. 343. PROP. To find thle velocity and time welen theforce varir nversel as tIfe cSube pro ducedf t toe distance. Integrating, 2PS x - x But when(x=, v=;.. c- AR-, or v= v/~t. — e (101) -dxnverse as the cube of the distance.-xd and [II.], vdtv - a2dx. Integrating, t=. vx'2- t+ c. But hen X-=a, t v=O; c C= a-, and t. Va(x —. (102) ax2 the time to he center of force, Cca. (See Fig2., Art. 331) XI0ASQ/ /a- 2 =. tfi x-O, t=(-4, Q CURVILINEAR 1MOTION OF N FREE POINT. 209 ~ Ii. CURVILINEAR MIOTION OF A FrEE POINT. 344. If several forces in different directions in the same plane act continually on a material point, it will have a resulting motion at any instant, whose direction and rate must be determined by the relations between the directions and intensities of the forces which act upon it. The motion may or may not be rectilinear, and, in order to investigate the circumstances of the resulting motion generally, we shall employ the method of resolution to two rectangular axes, as in Statics, using X and Y, instead of x.X and I.Y, to denote the sums of the resolved forces. 345. Pror. To determnzine, generally, the zmotion of a point acted upon by any number of forces in the same plane. Resolve the forces in the direction of two rectangular axes, and let X and Y represent the sums of the resolved forces in each axis, x and y being the co-ordinates of the position of the point. The point may be regarded as acted upon by the two forces X and Y, independently of each other. Hence, by [IX.], d x d2y dt dtO' (103) If ds represent any small element of the path of the point, by LVII.], -t /will represent the velocity of the point in its path. Hence, if a and i3 represent the angles which ds makes with the axes, the velocities in the direction of the axes (39) are ds c'lr ds (104) C s. o s. *co s13-. (104) dt (at' dt O dt' Such are the general equations by which the motion cf a point in a plane may be determined. 346. If the velocity is given, we must, by resoiution, obtain the velocity in each axis, and thence, by differentiation, the force in each axis, and, by composition, the resultant force may be determined. If the forces are given, we may, by integration, find the ve 0 210 DY N A I C s. locity in each axis, and thence, by composition, the velocity oi the point in its Fath. If the path is required, each of equations (103) must undergo two integrations, at each of which operations a constant must be introduced. The constants introducecl at the first integration will depend on the velocity of the point when the time commences, or at some given time; those introduced at the second integration will depend on the position of the point. We thus have two'equations involving the co-ordinates x and y and the time t, and, by elimination of t, an equation between the co-ordinates will be obtained, vwhich is the equation of the path of the point. 347. PRor. To determine the velocity of a point in its path. The velocity may be found as above indicated, but the following is the preferable method. Multiply the first of equations (103) by 2dx, the second by 2rgdy, and, adding the results, we have 2dcxd x + 2dyd y — (XYdx+ ~dy). dti The first member of this equation is the differential of 2 c 2~ d ds2 dt2 lt2 Therefore, integrating, we find v = 2f(Xdxl Ycdy) + c. (105) 348. Con. If this expression Xdx+Ydy is integrable, the x elocity may be found, provided we can correct the integral, or know the velocity at some given point. Thus, if the expression is a differential of the co-ordinates of the position of the point, so that its integral is a function of these co-ordinates, or vo- =2f(x, y) + c, and if v, is the velocity at the point whose ce-ordinates are. and b, so that V2=2f(a, b)+c; then V- V= 2f(x, y)- 2f(, b). Hence it appears that the velocity acquircd by the particle CIRVILIN AnR MOTION OP A FREE POI T. 21 1 in passing from one point (a, b) to any other (x, y) is the same; uwhatever be the curve described between these points, since the change of velocity is independent of the co-ordinates of any intermediate point. 349.. PROP. T/e expression Xdx+Ydy is always integrable whenever the forces ar'e directed to ixed centers, and their intensity is afunTction of the distancefrom those centers. Let F, F1... be two forces directed to fixed centers, a, b, a,, b be the co-ordinates of the centers, x" y.... be the co-ordinates of any position of th:e particle, 7',9 r,1.. be the distances of the particle from the centers, " a,, a1,, be the angles which r and r, make witlh the axes. Resolving F parallel to each axis, we have for the components F. cos. a, F cos. 3. x-a y-b But cos. a=- and cos. P= Hence the components are F x-a Fy-b r r The other forces may be resolved similarly. Therefore the expression But'2 - (x- a)'+ (y-b), and, by differentiation, dr- X -dx+Y dy. Similarly, d, x-a dx+ 1dl..'. Xdx+Ydy=Fdr+F,dr,. knd s'nce, by hypothesis, each f- -ce F;i a function 9f the dis. 212 DYN AM IC S. tance r, each of the terms of the second member of the equa. tion is an exact differential, and therefore the first is al'so. In the same manner, the reasoning may be extended to any number of centers of force. 350. PRoP. To determine the motion of a point acted upon by a force directed to a fixed center. Assuming the fixed point as the origin of co-ordinates, let F be the force, whether a single one or the resultant of several, and r the distance of any position of the particle from the center, called the radius vector. Multiply the first of equations (103) by y, and the secona by x, and sultract the former from the latter. This gives xdy-y(l =zYX -X. (106) dt2 Resolving F in the direction of the axis, we y/;have X-=F- Y=-F S "- I \ X 1' b, Aultiplaini the first by y, and teie second bear xs and talking the difference, we get xY-yX=O. Therefore the first member of (106) is zero, and since it is the differential of xd, by integration we obtain fxd2Y-yd 2x S — xdy —ydx J ~dt dt or xd.y-ydx= clt. (107) Let o be the angle made by r with the axis of x; then x=1 cos., (a) y=r sin. {); (b) also, dx- cos. odr —r. sin. o).dw, (c) dy= sin. wdr+ r cos. o.dw. (d) Multiplying (c) by (b), and (d) by (a), and subtracting, we get xdy-ydx = r-'dw, = cdt.'d =c. ( OS) dt) CURV [LINEAR MOTION OF A FREE POINT.'213 But l-2r2dw is the differential of the area of a plane curve referred to polar co-ordinates. Hence the ratio of the element of the area to the elem'ent of the time is constant, and the area described by the radius vector in any time is proportional to the time. By integrating Ir'do= —}cdt, we have for the area described in any time A=-ct, (109) and if t=1, we have c=2A, or the constant c is twice the area described in the unit of time. 351. PRoP. Conversely, if a material point describe by its radius vector around a fixed center areas proportional to the times, the force is directed to that point. When the radius vector describes areas proportional to thI times, we have (107) xdyy-ydp dt c Differeitiating, we get xd'y —ydx0 dt2 (106) xY-yX=0, and X: Y=x: y. Hence the forces X and Y, which are parallel respectively to the co-ordinates x and y of the point P, are proportional to these co-ordinates, and the resultant PR of X and Y must take the direction of PS, which is the hypothenuse of the right-angled triangle PMS, in which PMi=y and SM=x. As an application of the general equations of motion of a firee point (103), we may take the following cases. 352. PRor. To determine the motion of a point moving from the action of an impulsive force. Since, after the action of the impulse, the point is abandoned to itself, there are no accelerating forces acting on the point, and we have (103) -d12 =X=O - dY O —, 214 -- r D h- I) A,M; C N. dx di IntegratinS,: t=v; (a) v, and v being constants, added to complete the integral. Compounding these velocities (Art.' 236), we have for the resultant velocity dX +dy's,,i 2(I dt" dt ='' The, velocity is therefore uniform. Integrating equations (a), x=v1 t-SI, y-v' or-soS (b) tlhe constants s, and s2 being the co-ordinates of the point whien the time commences. Eliminating t from equations (b), we have yz-x + v2 (s, -s ), tile equation of a straight line. Hence the path of the point is a straight line and the motion uniform, which accords with the first law of motion. 353. PROP. To detelrmine the motion of a projectile acted upon' by gravity regarded as a constant force. In this case, taking the axis of x horizontal and that of y veitical, (103) give d2x d2y_ d — =Xo, Wd = -- fMultiplying by dt, and integrating, d= vi, -=v -gr t AMultiplying by dt, and integrating again, we have x=v 1t+s, y= v2t —gt2 + s (a) But if the point of projection be at the origin of co-ordinates, and t be reckoned from the commencement of motion, s, anra s. will each be zero. Putting v =av,, s,=O, so=O, and eiimi? nating t from the equations (a), we get y=a -x 2 Compare this with (63). CONSTRAINED nlOTION OF A POINT. 215 ~ III. CONSTRAINED MOTION OF A POINT. 351. PRor. To determrine the velocity of a poizt moviewg oiz a given czlrve. If a point is constrained to mnove on a curve, the reaction of he curve will be a new force normal to the curve, and may be resolved in the direction of the axes and combined with the other components, as in Statics, Art. 77. Let N be this normal force, a and p the angles which the normal mnakes with the axes, and X and YT the sums of the components of all the other forces in the direction of tile axes. Resolvincg N in the same directions, we shall have from (103) d:2Y Yd -X- N cos. a dcl~~~t,42 ~t dy But cos. a= sin. Ptm= od supposing,, ds tle motion in the direction tP, and cos. B= sin. a- cos. Ptm..=. dt ds Multiplying (a) by 2dx, and (b) by 2dy, and adding 2d.xd'x +2dyd2y = 2 (clx - +Ydy). dt1 The first member is the differential of dx.- 2 + d4 /= 2 ds2 dt2 dt' Whence, integrating,,v= 2f(Xcldx+Yy)+-. (110) COR. 1. Since this result is exactly the same as that obta'nedl for unres'rained motion (105), we may conclude, 1216 D' ~ N AD M IC S. 1~. That, if no accelerating force act an tile point, or X=G and Y-0, its velocity will remain constant. and be not at all retarded by the action of the curve. See PArt. 284. 2~. If any accelerating forces do act on the point, the veloc ity is independent of the curve on which it is constrained to move. See Art. 286. Con. 2. If the accelerating forces are all parallel, they may be assumed parallel to the axis of x, and in this case (110) be comes v"=2fXdx.+c. (111) CoR. 3. If in this last case the force, or resultant of the forces, is constant, and equal tof, we have v2=+~2fxc, where the upper or lower sign is to be used according as the force tends to increase or diminish x. If the distance of the point firom the origin, xza, when v —O. co=2fa, and v =2f(a-x); (112) and since only the ordinates on the axis of x are involved, tile velocity of the point on the curve depends not on the curve described, but on the difference of the ordinates on the axis of x. 355. PnoP. Tojind the time of motion (f a point on a givenr curve. ds In all cases [VII.], dt=-. Hence, when the nature of the curve and the velocity at any point of it is known, the value of v being found from (110) an( substituted in that of dt, the time may be found by integration COR. If the forces act in parallel lines, and their resultant it constant, by (112) dt=- (113) V2f(a-x) 356. PnoP. To find the reaction of the curve. Multiplying (a), Art. 354, by /, and (b) by -, anc subtract inD the latter firom the former, we get CONSTRAINED MOTLuN OF A POINT. 217 N=dx y~ _ X y/ /,- - d:~dd/ N ds -X l s dsdt2 ds Eliminating dt by the equation V=dt, NYd —X dy dyd2 2x-dxxd 2y = ds Xcs ces' But if p be the radius of the osculating circle, ds3 P-dyd2x-dxd2y' ds ds p =Y cos. cos. a cos.+p The first two terms of this value of N are equal and oppo, site to the forces X and Y resolved in the direction of the nor. mal. They give, therefore, the pressure on the curve due tc the action of these forces, while the other term - is the reacp tion of the curve or pressure due to the motion. Otherwise. If we suppose no accelerating forces to act on the point, then X and Y are each equal to zero in (a) and (b), A1t. 354. Hence N2( )=d2d N2(d ) +(d Ny); ds' dt2 ds or, eliminating dt by the equation rct=-, N=,V (d2 X)2+(d2y)2 ds2 But the coefficient of v2 is the reciprocal of the radius of curvature when s is the independent variable..'. ~ —— _ (114) Compare this with (79). As an application of the formulle )f constrained motion, we may take the following:ases: 218 D Y N A AI C S. 357. PROP. To determine the motion of a btody descendizng by? the force of gravity down the arc of a vertical circle. 10. To find the velocity. Formula (112) applies to this case, and if hI=SA be the height from which the J0= A body at P descends, and x any distance from the lowest point on S A, the axis of x, we havk S v= /V2g(h-x). When x=0, we have for the velocity at the lowest point, v= v 12gh, which is the same as that due to the vertical height ih. 20. To find the time, wxe have from (113) ds dt _ ds t/ 2g, (h - x) The equation of the circle is y2 2ax-x2. d (a -.x)2dx2 Hence dy"- (a-:x But ds2 —d+l y2oz2e( (l- x) ( ) a2Clx2 2ax-x"" adx'. ds= a/ 2ax- x2 This, taken negative, because the are is a decreasing lurc tion of the time, and substituted in the above, gives a (Ix dt= / 2g V (h/-x) (2axz- ) a dx v/ 2,g-V(hX -X2) (2a-x) a (2a-x) —;dx v2g /Ix —x2 -~~( X ). V-x2 CONSTRAIN El) MOTION OF A POINT'. 219 E xpanding (1 — ) by the binomial theorem, we have 1 a x,+ 1 13 X ) 2+35/x( t- - {1+, +'4 +',+ 2 g/ Vh X- 2 4 4 6xo. &c.}. -x"dx Thus the terms to be integrated are of the form __ Vhix-x" the exponents n being natural numbers beginning with 0. Performing the operations and taking the integral between the limits x=h and x=O, we find 1+a ( l)2 ( ) 1 ( 2( 2 ( 2 3 52 (1 2 3 If h be small, the first term will give an approximate value for the time, viz., a/ Otherwise. If in (a) x be rejected as small in comparisor, with 2a, tl.e equation reduces to dx dt — / g V.'/hx —x2,/a. 2x t= --.versin.- — +c. But t=0 when x —h; /d and t=,/- (rr —versin.- ), and when x=0, t= -,/ -. (See (86)). 358. PaPop. To determine the motion of a body descendi-ng( oy gravity down the inverted arc of a cycloid whose?Sase is horizontal and axis vertical. The cycloid being inverted, let the origin be at the lowest 220 D N A MI CS. point, the axis of z being vertical, and that of y horizontal Let the axis AB-2a, AM=x, PM-y, and AP=s. 3B The velocity, as in the preceding propo, ~S,7 ~'sition, is (h —), A ih being the abscissa of the point K fiom which the boay begins to descend. The differential equation of the curve, referred to the vertex is the-origin, is d (2ax-x)dx V/2ax —x2 But ds2=dy2+d 2-( - (2a -x+)dx 2axZ2+)d.ds. =dxs- X. Substitutingc this value of ds with its proper sign in [VII.], we find /a dx dt= -- \g'/h- x Integrating, /- versin.-' _), and therefore for the time of descent to the lowest point, where / xr /4a t = t-= V' V (115) Con. The time t being independent of the abscissa h, will be the same, from whatever point in the curve the body begins to descend; in this respect differing from the motion in circulax arcs. On account of this remarkable property the cycloi(l is called tautochronal 359. ScnloL. The property of tautochronism which attaches to the cycloid was supposed to give peculiar advantages to a pendulum made to oscillate in this curve. The method of accomplishing this is naturally suggested by the property of the evolutes; and as an evolute of a cycloid is another equal cy, ENTRAL F Ot C Ei. 221 cloid, w. have only to take two equal semicycloids and place the extremities of their bases contiguous in the same horizontal line as in the figure. If we then suspend a body at the point S, I C common to the two semicycloids, by a flexible string equal in length to either semicycloid, or, which is the same thing, to twice A the axis, and make it oscillate between the two, it will generiat. the involute, or another equal cycloid. The constant change, however, in the center of motion arising from the contact of the string with the two curves or checks, which changes also the relative velocity of the different parts of the vibrating body, renders this contrivance, although beautiful in theory, yet useless in practice; independently of the difficulty of obtaining a string sufficiently flexible, and of ensuring accuracy in the plates. Vibrations, therefore, in small circular arcs, which are at the same time also most natural, have been adhered to in practice, and it has been shown (Art. 302 and 300) that, as long as the arc is small, these vibrations have all the advantages of the vibrations in cycloidal arcs. ~ IV. MOTION OF A POINT ACTED UPON BY A CENTRAL FORCE 360. The case of a point revolving in an orbit by the action of a force tending to a fixed center is of sufficient impoitance to justify a distinct discussion, especially as the formulm are susceptible of considerable simplification, when the force is given as a function of the distance from the center. The fundamental equations d2x d2y X d12 Y= dt2, are components of the force in two rectangular directions But when the force is directed to or from a fixed center, and is given in functions of the distance of the particle from that center, it is most natural and convenient to introduce polar co. ordinates. 222 D Y N A MI CS. 361. Pnor. To transform the expressions fo the forcefromn rzctangular to polar co-ordinates. Suppose the origin to be at S, the center of force, and let r= —SP, the radius vector or distance of the particle, and,o=PSX, the angle Sj ~i —-\~;made by r with axis of x. Then X:-F cos., Y= - F sin. o, the signs being negative because the force is supposed to be attractive, or act toward the center in opposition to X and Y. Multiplying the first by sin. r), and the second by cos. o, and subtracting, we have X sin. &(-Y cos. (j=0. (a) Similarly, multiplying the first by cos. o, and the second bv sin. ro, and adding, we get X cos. w+Y sin. o= —F. (b) But x and y being the co-ordinates of the position of the particle X —r COS. 6, y=r Sin. w. Differentiating each twice, we obtain d x=d 2r. cos. ~W-2dr.d&.sin. o-)-r do2.cos. w- rd'o.sin. o, d2y=-dr. sin. o+d-2dr.dco.cos. wo-rdo2.sin. o+rd'.cos. 6o. By dividing these expressions by dt2, the values of X and Y will be obtained (103), and if we substitute these values in (a) and (b), we get dr dA d2o d2id. 3+2..(1 ~1 ) d'ct dwt2 d2rl c102 t-r. tF=0, (117) which are two polar equations to determine the motion. 362. Con. 1. The radius vector of a point revolving about a center offorce describes equal areas in equal times. For, multiplying (116) by rdt, we obtain dW d%2o r'2d 2 rdr. -dt 2 = 0 =.d dt t. -- dt CENTRAL FORCES. 223 Integrating, dt = (118) ind fr&dwc=ct +c,. But 1fr2dwo is tle area described by the radius vector in the time t, and it varies as t. 363. Con. 2. The angular velocity of the radi'us vector varies invesely as the square of the distance of the particle. c.w c From (118) we obtain t (119) But is the angular velocity of r, and it varies inversely as 2. 364. PROP. To determzne the velocity of the particle in its orbit. Multiply (116) by rdo, and (117) by dr, and add the results. This gives d 2r dwo2 2 dr. - + rdr. d-+ rd'Fdr= f. dt2 dt2 dti2 or d7. (d r dt- ) + 2Fdr= 0. Therefore, integrating, we have it2. +2fFdr=c'. (120) 2ds2 dr2 + r12dro2 But v2=ds2=dr2+rdc2. (See Art. 367, 40) v2 =c' — 2fFdr. (121) CoR. Since the value of v depends only on the distance ot the particle from the center, the velocity of the particle will be the same at any two points equally distant from the center, and the velocity acquired in passing from one point to another will be the same, whatever be the path or curve between them, if the law of force remain the same. 365. PRor. Tofind the time of describing any portion of tie orbit. 224 Y N AMIC S. By (119), dt2 -r4 which, substituted in (120), gives d1.2 C2 j+ r++2fFdm =c'. (122) In this expression we have the relation of r to t, independent of the orbit. 366. PROP. Tofind the equation of the orbit. r4d6w 2 Substituting in (122) for dit its value, obtained fior 0r~-, obaief'C (118), we get C2 ( d + fdr-c'; (123) r4dw-2 or, to give the equation a more convenient form, put r=-, fronm U du which we find dr- -. These values of r and dr, substituted in (123), give (du2 du. C- (d 2+u') - 2f F 2-c'. (124) When the law of force is given, we can obtain from this expression, by integration, the relation bet-ween u and 6, or the polar equation of the orbit; or, if the relation between u and c. is given, the law of force may be determined. 367. SCHOL. A familiarity with the geometrical representatives of the terms and factors in the foregoing formulae will conduce to facility in their application. In the annexed figure let the Y p center of force be at 8, the origin of co-ordinates, SP=r the radius vector of the particle at P, P'P=dci T X x an element of its path coinciding A S with the tangent PT, (o=PST the angle made by the radius vector with the axis of x, P'SP=(7w the angle described by the radius vector in the indefinitely CENTRAL FORCES. 225 small time dt, and nzP =dr the increment or decrement of the radius vector in the same time. Let the are Pm be described with S as a center, and radius SP, and the are nn' with the radius Sn-=1. Then, 10. Since Sn: SP=nn': Pm or 1: r-dow: Pm, Pm = rdw. 20. The area of the sector SPmn is SP.Pm A= = ='d. 2 2 30. If the point P' be indefinitely near to P, PP' and Pm may be considered straight lines, and the triangle PmP' a plane zectilinear triangle, right-angled at m. Hence P'P2=pm2+Plm, ds2-r= 2do2 + dr2. 4~. If Sp —p be a perpendicular on the tangent, when the triangle PmP' is in its nascent or evanescent state, the angle PP'm=pPS and the triangle PP'm is similar to pPS. Hence r2dr' PP': Pn=PS: Sp, or ds: rdw=: p. P=ds and [VII.], r'do r~dw P vdt.. vp= V dt — by (118), twice the area described in the unit of time.?'4d63 r4doa Also, P2 cs d=2~r dw2 5~. Since Sp: pP: PS=Pm: mP': P'P=rd: dr: ds, Sp Pm rd&o -=sin. P, SP PP' ds Pp P'm dr P P=-= d-cos. p, pS mP rdo dnd Pp- = -=tan. P. PP nzPI dr 6~. If the force F be resolved in the di ection of the curvo, wre shall have d s dr =-F cos. P=-F P 226 D Y N A AI I C S. Mlultiplying both sides by 2ds, and integrating, we have ds2 dts=v2-c'-2fFdr, as before (121). By reference to (117), we see that the first term has the form of the differential expression [IX.] of a force in the direcrdo2 r'dW2 trdo tion of r; the second term dt But -- is the arc of a circle of radius r, divided by the time. It is, therefore, the linear velocity v, of a body in a circle. HIence dt2 -d -tf' d22,, the centrifugal force in a circle... (117) becomes dtr=f-F, F being the force by which the point describes any orbit, and f the centrifugal force in a circle whose radius is r, the disd'r tance of the point at the instant. dt- the difference of these forces at any time, is that by which the radius vector is increased or diminished, and is called the paracentricforce. In d2r dr like manner, the integral of -or is the velocity of ap proach or recession from the center, and is called the pa'a centric velocity. The paracentric force is the difference lie tween the centrifugal and centripetal forces. * rCO2 1?r4dw2 C2 CoR. 1. Since d't" r'4 dt2 rj' f=. (125) COR. 2. The quantity by which c' is multiplied in (123) is, hy (40), equal to -. Hence C-~ -+2 Fdr =c'. Differentiating, 1p beinrg variable, 2c d Fdp p CENTRAL FORCES. 227 Hence F= c (126) an expression most useful in finding the law of force by whict a given orbit may be described. ds2 drg 2+r2d6-2 r4dwo'2 c' COR. 3. Since v'2- dr& dt (112 p2dt2p c2 Fpdr pdr pdi and (126),. —-2.' -2F.'_. (127) p dp 2 dp dp pdr p But p2dprl- chord of curvature through the pole, and (if I' be constant) is (55) the space through which a body must fail to acquire the velocity v. Hence the velocity of a body at any point of its orbit is equal to that acquired by falling through one fourth the chord of curvature through the pole. Application of the preceding formulm to the motions of the planets. 368. Observation has established three facts respecting the motions of the planets, which, firom their discoverer, are called Kepler's Laws. 10~. The areas described by/ the radius vector' o/'a p)labnet (art proportional to the times. 20. The orbit of a planet is lan ellipse of' which the center of the sun is one of the foci. 30. The squares of the times of revolution of' the different planets are proportional to the cubes of their mean distances from the sun, or the semi-major axes of their orbits. These laws relate only to the center of inertia of each planet, and conclusions from them must be limited to the motion of' that point; in other words, to the motion of translation of the planets. 369. PROP. The accelerating' force by whlich a planet describes its orbit is directed to the center of the suIn. For, by the first of Kepler's laws, the areas described by the radius vector are proportional to the times, and when this is the case, by Art. 351, the force must be directed to the point about which this equable description of areas takes place. 228 D Y NAM IC S. 370. PRop. The force by which a planet describes its orbtn varies inversely as the square of its distance from the center. By the second law the orbit is an ellipse. The equation of an ellipse referred to the focus as a pole is. a(1 -e2) 71- - a (I (128) 1+ +e cos. (-0)' (12 in which a is the semi-major axis, called the mean distance, anti e the eccentricity. The angle 0 is an angle 3.. made by a fixed line, through the focus with the major axis, reckoned from the lower apsis, or least distance, and called the longitude of the perihelion, the lower apsis being called the perihelion, and the higher the aphelion. The angle o)-O, which expresses the angular distance from the perihelion, is called the true anomaly. Putting r=- in the above equation, we get 1 +e cos. (o —0) a(1-ee') Differentiating in reference to o, du e.sin. (t —0) do a(1 —e') Differentiating, again, d2u e cos. (w-0) do2 a(1-e) (b) Adding (b) to (a), d2u 1 + -a (IC —- c) Differentiating (124), d2u F -M-uHence F 2 c 1 (129) a(1-e2) a(1-e')'.' The coefficient of - being constant, the force varies inverse tv as the square of the distance. CENTRAL FORCES. 229 If r=-1, F= a(1 -e)' Hence tile coefficient of - is the intensity of the force at the unit of distance, or the absolute force. 371. PROP. The intensity of the force is the same for all the planets at the same distance. The quantity c (109) is twice the area described in the unit of time. If, therefore, T be the time of one revolution of a planet, cT will be twice the area of the ellipse described by the radius vector. But the area of an ellipse is 7rab=-c=a / 1-e2.. cT=2ra' V 1-e', C2 4rr2a3 and a(l-e2) T- Similarly, if a', e', c', T' express the same quantities in any other orbit, we shall have c/2 4'2a"3 a'(1-e'2) T'" By Kepler's third law, T2: T/2a'. a", c2 C/2 a'1 -e-) a'(1 -e"2) Hence the force does not vary from one planet to another except in consequence of change of distance. 372. PROP. To find the velocity of a planet at any point in its orbit. By (121), va2=c'-2fFdr. Substituting in this the value of F (129), we have - 2c' dr Integrating, v =c' +jC 1 To determine c', e must know the velocity at soe gi'r To determine c', we must know the velocity at some given 230 D Y N A MI CS. distance, or at what distance and with what velocity the planet was originally projected into space. Let v, be this primitive velocity, and r, the corresponding distance. 2c2 1 Then -' qC a(1-e2)'r, 2c' I a(I — e)'?-,' ~. V =-V-{a (I -e)(. r ). (130) Otherwise. From Art. 367, Co?'. 3. we have C2 V =and (Analytic Geometry) b 2r a2 (1-e2) P 2a-7r- 2a-r 2 2a —r.'. = ) r (131) a'"(1- e2)' r. COR. 1. Hence the velocity is greatest when r is least or al perihelion, and least at aphelion, where r is the greatest. COR. 2. If a body move in a circle whose radius is r, v, bo ing its velocity, the central force is V2 C2 1,F=v =' (l_)..~ (129), if the force is the same as that which retains the planet in its orbit c2 2a-ra c2 1 V 2(1 -e2)' r a(l- e2)' ==2a-r: a. That is, the square of the velocity in the ellipse is to the square of the velocity in the circle as the distance of the planet from the unoccupied focus is to the semi-major axis. CoR. 3. If r1 = perihelion distance, and r, = aphelion disi ance c2 2a-r - C2 r then, if in (131) r,, v _e r- - (1-' ( r= 2 Tc2 ca-r c, (c) 9 V -a=(1 -2 e2)' r.2 — a2(1 —e2) r2'. I ~r=a, C 2== a2(l1. - e2) CENTRAL FORCES. 231 or, the velocity at the extremity of the minor axis is a mean proportional between the velocities at the perihelion and aphelion. 373. PncP. To determine the orbit which a body will describe when theforce varies inversely as the square of t.he distalnce. If,l be the force at the unit of distance, then F=-=Z tu. dui.., FF =ffidu=ftu. This, substituted in (124), gives 2 (du2 _j_ U) - hU - C which is the differential equation of the orbit. To integrate this, we find c c2dllt C'+2/u- c"2 it Put u=z+,, and b"- (c'c'+ft), and we have C; C4 dz:r, integrating, =O —-cos.-b.'. z- b cos. (o-O); or, replaciilg the value of z, u= d-b Cos. (C)-O). Hence 2 bc_, (13S-) 1+ — cos. ((o-0) which is the equation of a conic section referned.o thle focus. Hence the orbit which a body will describe when the force varies inversely as the square of the distance is a conic section. To determine the constants, and therefore the dimensions and form of the orbit, some other circumstances must be given '232 D Y N A MI C S. Suppose w~ know v, the initial velocity or velocity of projection into space, p the distance of the planet from the sun, and b, the angle made by the direction of projection with the distance. pdw By Art. 367, 5~, sin. ds' p2Odw ds pdw Hence (118), c= dt -dt ds -p.. sin. Vb. (a) By (121), v =c'-2fFdr=c' + 2C. = l- 1 — (be Comparing (132) with (128), we see that C2 1w2C 1 a(1-e2)=- and e= - (C'C2 +L2)2 a=- It- -lip (C),'. a —— c, (c) c' 2,au —v2' andcc2 Pv Sin.' lb(v1 — ) ) and e2- + 1-. (d) COR. From (c) it appears that the semi-major axis a depenlds only on the distance p and the velocity of projection v,, and is independent of the angle of projection b. H-ence, in whatever direction the body is projected, the major axis of the orbit will be the same. The orbit will be an ellipse, hyperbola, or parabola, according as the value of e is less than, greater than, or equal to unity, or (d) according as P;s negative, positive, or equal to zero. Or, since (99), v'= (1-xA if a-ac, v=2. Therefore X X when a body is attracted from an infinite distance to a centei of force which varies inversely as the square of the distance the square of the velocity at any distance p is CENTRAL FORCES. 233 v2 / 2 and v2- 21- 0. P P Hence tne orbit will be an ellipse, parabola, or hyperbola, ac. cording as the velocity of projection is less than, equal to, or greater than, that acquired from an infinite distance. 374. PROP. To find the time of describing any portion of the orbit. Substituting the value of F= - in (122), we get dr2 c2 22t dt2 I r r The values of the constants found in terms of the elements of the orbit in the preceding proposition are c': -- and c-=1-/a(1-e2). a Substituting these values, and multiplying by -, we obtain a r'2dr2 2 2 dt2 ~+(a-r) 2- ae'- 0.O Hence dt= ( t e'V -(a-r) In order to integrate this expression, let (a-r)=aez, then (a ) (1- ez)dz It V1_ 2 -_Z23' - - V_ 1-dz " Integrating, t- (- ) cos.-'z —e(1 -zz) +k. When z=-1, or r-a-ae, and the body is at the nearer apsis, then t =:k= the time at the perihelion. Hence the time through any portion of the are firomn lhc perihelion is 234 DYN AM IC S. If -- i, =r-a+ae, and the body is at the further apsis, in which case t-t, is the time of half a revolution, and — ti, a — ( Also, if T be the periodic time, or the square of the periodic time varies as the cube of the nean distance. 375. ScIOL. The quantity t, which enters into the results of the preceding investigations is the value of the accelerating force at the unit of distance from the center of force. Now the attractive forces of the sun and planets vary directly as their masses, and if M be the number of units of mass of the sun, and m the scame of any planet, and if we assume for the unit of force the attraction of a unit of mass at a unit of distance, M will express the attractive force of the sun at the unit of distance, and m that of any planet; and the whole force by which they will tend to approach each other, or the whole force which the sun, regarded as fixed, exerts on the planet at the unit of distance is M+m=-e; and for any other distance 7r AMl+m_ t -2 - 2 The intensity of the solar and planetary attractions may be expressed in terms of terrestrial gravity. For this, let r? be the radius of the earth, and ml its mass, and since the forces are directly as the masses, and inversely as the squares of the distances, we have m M Mlr2 r-'=g' 9r g, the fourth term being the attractive force of the sun at any dis, tance r. Similarly, the attractive force of the planet is g; M+m r.2 t hence their joint influence is g, in whic 1 I CENTRAL FORCES. 235 —.g is that which was assumed above for the unit of force, OI the attraction of the unit of mass at the unit of distance. 376. EXAIMPLES. Ex. 1. Required the velocity and time when a material particle is attracted to a fixed point by a force varying inversely as the square root of the distance. Ans. v= 2pb(a2-x-)2 1 1 1 1.1 2 t= 3[z 1o' +2a-) (a0-X~)". Ex. 2. Find the velocity and periodic time of a body revolvingc in a circle at a distance of n radii fiom the earth's center. Ans. v= - \n/.. Find the least elocity th a body must be Ex. 3. Find the least velocity with which a body must be projected from the moon in the direction of a line joining the center of the earth and moon, so that it may reach the earth. Ex. 4. Given the velocity, distance, and direction of projection, when the force is attractive and varies as the distance, to find the orbit. Ex. 5. A body is projected in a direction which makes an angle of 60~ with the distance, with a velocity which is to the velocity from infinity as I 3: the force varying inversely as the square of the distance. Find the major-axis, the position of the apsis, the eccentricity, and the periodic time. Ex. 6. A body is projected at a given distance from the center of force with a given velocity, and in a direction perpendicular to the distance, when the force is repulsive and varies inversely as the cube of the distance. Find the path of the body. IHYDROSTATICS. CHAPTER I. 377. MIATTER exists in the various states of solid, fluid, and aeriform. In solid bodies the homogeneous integrant particles cohere firmly, and do not admit of interchange of position. The homogeneous integrant constituents of fluids possess less cohesion, and change their position among each other on the application of a moderate force. Fluids differ from each other in the degree of cohesion of their constituent parts and the facility with which they will yield to the action of an impressed force. 378. A pe7fectfluid is one in which the cohesion of the constituent particles is so feeble that the least force is capable of efiecting a separation and causing motion among each other in all directions. Perfect fluids are the only ones which will be made the subject of investigation. Fluids are divided into two classes, incomprcssible or liquid. and compressible or aeriform. 379. Incompressiblefluids are those which retain the same volume under a variable pressure. They may be made to take an infinite variety in form, which form they will retain when acted upon by no external pressure nor accelerating force. Of this class water may be regarded as the type. 380. Compressible fluids, of which atmospheric air is the type, change at once in form or volume with every variation' of pressure which they sustain, and return again to the same form and volume when the same circumstances of pressure and temperature are restored. EQUALITY OF PRESSURE. 237 PRINCIPLE OF EQUALITY OF PRESSUEE. 381. PltoP. A force or pressure applied to a given poortion of the suface of a fluid is transmitted, undiminished in intensity, to every other equal portion of the surface. A force impressed on a solid is effective only in the direction of its action, and is sustained by an equal force impressed in an opposite direction. Applied to a fluid the force is effective in every direction, and can be sustained only by forces appiied to every point in the surface of the fluid. Let the annexed figure represent a section of a vess6l filled with a fluid without weight. In this A B case, supposing the press-'"\,,x tiure of the atmnosphere re-, moved also, the surface of the fluid would be subject to no pressure, and would'.; retain its form if the sides of the vessel were removed. P4 P Let an aperture of a giv- -E 3 en size, as one square inch, be made in either side, AB, and a piston be accurately fitted to it, and supposed to move without fiiction. Now, since the particles are without friction, a force P1 of one pound, applied to this piston, will act upon the stratum of particles in contact with its base, and this stratum upon the next, and so on. The force will therefore be transmitted to the opposite side, EF, and will cause motion in the equal piston PI, unless counterpoised by an opposite force of one pound. If this counterpoise be applied, then, since the particles move freely among each other without resistance from friction or cohesion, the particles in the direction of the applied forces will act on those lying without, and will communicate motion to them, unless resisted. A motion will therefore be communicated to any equal piston PI', unless sustained lby a 238 HYDROSTATICS. pressure of one pound. Hence both pistons P', and P" will sustain the same pressure by the action of P1. Further, if the pistons P'1 and P': were contiguous, or one of them twice the size of P,, then it is obvious that a force of two pounds would be necessary to preserve the equilibrium; or, if the base of the piston P4 be four times that of P1, then a force of four pounds would be necessary to keep P4 at rest; and if the base of one piston be n times that of the other, the pressure on the former will be n times that on the latter, or a force of one pound will produce a pressure of n pounds, where n may be the ratio less one, of the whole surface of the vessel to the base of the piston. COR. 1. The pressures P, and P~ on any two portions A, and A, of the surface will be proportional to their areas. or P,, A, OAnI_ (113 ) Also, the normal pressure p on a unit of surface will be P, P. - - ---- =P. (135) COR. 2. Every stratum of particles in the interior, of the same dimensions as the base of the piston, wherever situated and however inclined, is subject to a pressure equal to that applied to the piston; and since the fluid is without weight, every particle presses and is pressed equally in every direction. COR. 3. If all the pistons except P and P' be firmly fixed, and P be forced in, since the fluid is incompressible, whatever fluid is displaced by P will be forced into the tube F'. If h and h' be the spaces through which the pistons::nve, a and a' the areas of their bases, then i a= h'a'. But P and P' bping the equilibrating pressures on these areas, a P a'-P; hence IP= h'P', or hP-h'P'= O, or the principle of virtual velocities applies to fuids in eqtuilibriunl. SURFACES OF EQUILIBRIUM. 239 SURFACES OF EQUILIBRIUM. 382. PROP. The free surface of a fluid subjected to the actzon of an accelerating force is, when in equilibrium, perpendicular to the direction of the force. A liquid void of gravity will take and retain any form im pressed upon it; but when subject to the action of an accelerating force, a containing vessel will be necessary in order to preserve a coherent mass. Let ABCD be the section of a vessel containing a fluid subject to the ac- B tion of gravity, the base DC being hori- n zontal. If any portion of the free surface', as abc, have a direction not perpendicular to the direction of gravity, D, C this force may be resolved into two components, one of which is parallel to this surface; and since the particles are free to move in that direction as down an inclined plane, they will yield to this component. But whenever every portion of the surface is perpendicular to the direction of gravity this forcewill have no component in the direction: of the surface, and every portion of it will be urged vertically downward with equal intensity. CoR. Hence the surfaces of fluids at rest, and acted on only by gravity, are horizontal. But since the directions of gravity, acting on particles remote from each other, are convergent to the center of the earth nearly, the surfaces of large masses of fluid are not plane, but curved, and conform to the general figure of the earth. 383. PorP. If a vessel containingr a fluid be moved horizontally with a constant accelerating force, the surface will take the position of an inclined plane. Let the vessel ABCD containing a fluid be moved horizon. tally with a unifbrmly accelerating force P. Then any element, k, of the surface, whose weight is w, and mass - (22), g will be urged by its weight wtv in the direction kg, and by its 240 HYDR OS TATIC S. A_:B inertia Pw in the direction khi, and - therefore by their resultant in the p direction ho. As the same is true of every element of the surface, the D ~-Al;& zga C whole surface (Art. 382) will be perpendicular to ko. Let the angle of inclination to the horizon HIM-J-I. Then since HIM-ok-g, P-P tan. olkg-tan. I=- gzo,384. Piop. If a vessel containing a fluid be made to revolve uniformly about a vertical axis, thle surface of the fluid will take the form of a paraboloid of revolution. A x 3B Let the vessel ABCD revolve uni formly about the vertical axis XX,. NX Any element k of the surface will be 1 X p Turged horizontally by a centrifugal force directed from the axis, and vertically downward by gravity. Let o be the D I f 2 C angular velocity of the fluid, w the weight x, of the element k, and y its distance kM3( from the axis. Then (82) we have, for the centrifugal force P of this element, w Being urged vertically downward by its weight w, the surface of the element will be perpendicular to the resultant kR=R of these two forces. If this resultant be produced to meet the axis in N, we have z Rkg= LkNM, Rg kM and kg MN' w r g y w -MN' PRESSURES ON IMMERSED SURFACES. 241.'. the subnormal MN — --- a constant, a property of the parabola. HIence the surface is a paraboloid of revolution. NORMAL PRESSURES ON IMMERSED SURFACES385. PROP. The pressure on the horizontal base of a vessel containing an incompressible fluid is proportional to its depth below the suiface of the fluid, and independent of the form of the vessel. Let HI-O be the fiee surface of the fluid in the vessel ABCD, whose base DC is horizon- A\ 1 tal, and suppose the fluid divided into hori- zontal strata of small but equal thickness. Let a1, a2, a3, &c., denote the successive.strata, B~/ \C A0, A1, A2, the units of surface in each, lO,, P1, P2, the whole pressure on each, P0, P0, p, P the pressure on a unit of each, a the common thickness of the strata, and p the density of the fluid. Since the thickness of each stratum is supposed indefinitely small, the upper and lower surfaces of each may be regarded as equal. Now the weight of a is w I gpaA, (24), " 4 " 6" "a2 is W2 gpaA,, "c ". " 6 aa,, is wv =gpaA,,_1. The pressure on aI is P0, on a unit of a1 is A1 -PO (135), on a~ is P,=Po+w1-=P0+gpaAo, P1 Po gpaA, on a unit of aa is A- I ~A A+ A or p-=p p+gpa, on a3 is P=P +w,2=P1+gpaA1, P2 PI gpaA,n a unit of a. is A orpv2=p, gp. a a=p+gp.2a, A2 A, A P P, tru a,,+ is P" = P__ +w _ = Pn_'gpaA_,, 242 I YD R O S T A' CS.P. P.,_ gpaA,_onl a unit of (1,+1 is A-+ A —,,= +gp.na. But na=h is the depth of a,,+l below the surface of the flLId,, and if the upper surface of a,,,+ represent the surface of the base, A the number of units in the base, then the pressure p on a unit of the base will be P =Po 0+gph, and the whole pressure on the base P=Ap=Ap0 +gphA= P0+gphA h. (136) COR. 1. Since hA is equal to a prism whose base is A and hleight h, and gphA is its weight, if we disregard Po, which may represent the pressure of the atmosphere on the surface of the fluid, we have, for the pressure of the fluid on the base of the vessel, the weight of. a column of the fluid whose base is'lhat of the vessel, and height the height of the surface of the fluid above the base. It is obviously immaterial whether the surface pressed is that of the base of the vessel or a horizontal surface of an immersed solid. COR. 2. Since a cubic foot of' water weighs 1000 oz.-62.:j lbs., we have, for the pressure on the base of any vessel containing water, P-62.5 hA lbs., (137) where h is the height in feet of the surface of the water above the base, and A the number of square feet in the base. Coa. 3. The pressure on every portion of a. horizontal stra tum of the fluid will be the same, and since this pressure is transmitted equally in every direction, the pressure on every element of the sides of the vessel having the same depth will be equal to that on the surface of the stratum. If the sides of' the vessel are inclined, this pressure will be the normal press ure on the sides at that depth. Let the annexed figures, A, B, C. D, represent vessels of different forms and capacities with equal bases. Then the pressures on the base of each will be equal when filled to a common height, and the points a, b, c, e, g, h, k, 1, m, having a common depth, will be equally pressed normally to the surfaces. In B the horizontal surface df of the vessel will experience the same pressure verticallx PRESSURES. ON IMMERSED SURFACES. 24M3 A B C D Cd ~ t If upward as that to which the stratum ftg of the fluid is subject. iMoreover, the stratum fg being in equilibrium, must be pressed equally upward and downward. The downward pressure is equal to the area of the section fg into the depth h of the section below the surface of the fluid in the vertical branch. But the upward pressure is due to the pressure of the fluid in the inclined branch; and since the area of the section fg Is common to the measures of both pressures, the vertical heights of the fluid in both branches are necessarily equal. 386. Pnop. The normal pressure on any plane surface inclined to the horizon is proportional to the depth of its centcor of gravity below the surface of the fluid. Let k be any indefinitely small element of the immersed sui face, x its depth, and p the pressure it sustains. Py the preceding proposition, p =gpxh; and since the pressure is the same in every direction, p will be the pressure normal to this element, whatever be its position or inclination. The expression for the normal pressure on every other element will be of the same form. Hence the whole normal pressure P will be P-= gpxlk-gpXI:t. Let z be the depth of the center of gravity of the immersed surface below the surface of the fluid, and A its area. Then (29), A=r-Zxk;. P. -P=gpAx m t. (138) CoR. Since gpAx is the weight of a prism of the fluid whose base is A and altitude 2, the normal pressure on any immersed surface inclined to the horizon, or on any side of the containing vessel, is equal to the weight of a fluid prism iwhose basce 244 HYDROST ATICS. is tihe surface pressed, and height the depth of its center o0 gravity beloWv the fluid surface, or, if the fluid be water, P=62.5 A Ibs. (139) 387. PnoP. To find the pressure on any immersed suiface or.side of a vessel in a given direction. D ~ Let P be the normal pressure on /P2 p / the inclined side, CDFE=A of the C ( vessel CBE, and let it be required A\ \ "p1 Fto determine the horizontal and vertical pressures on A. Let a be!~A;"-S 7 _ the angle which the side A makes'A, 2 with a vertical plane, and t its in-'E; clination to the horizon. Then, aince PAP=-a and PAP2-P, resolving P horizontally and veltically, we have P,=P cos. a=gpzA cos. a, and P2 = P cos. = —gpxA cos. 3. But CDHG-A, is the projection of the side A on a vertical plane, and EFEIG=A2 is the projection of the same side on a horizontal plane. Therefore, A-=A cos. a and Ao-A cos. P, and P1=gpA,x and P2=gpA2X. Hence the horizontal pressure on A is equal to the weight of a prismatic volume of the fluid whose base is the verticai projection of A and height the depth of its center of gravity; and vertical pressure, a volume whose base is the horizontal projection, and height the depth of the center of gravity of A. And generally the pressure on any immersed surface in a given direction is equal to the weight of a prismatic column of the fuid whose base is the projection of the surface on a plane perpendicular to the given direction, and height the depth of the center of gravity of the surface below the surface of the fluid. 388. PizoP. Tofind the resultant pressure of a fluid on the ic. terior su7face of the containing vessel. Let ABCD be a section of a vessel of any form filled with a fluid. When the sides of the vessel are curved, the norrna. rRESSURES O N- I1MMERSED SURFACES. 2415 pressures on the elements of the surface will not be parallel. But a, A B a2 we may resolve the normal pressure on each element into vertical and horizontal components, and each horizontal component into } two others parallel to two rectan- k - i gular axes. We shall then have: to find the resultants of three sets. of parallel forces, and, finally, the resultant of these three resultants. We may, however, determine the resultant more simply, as follows: 10. The resultant of the horizontal pressures. Let HK be the intersection of a vertical plane by the plane ABCD produced, and let k be the vertical projection of any element k1 on this plane. Now the normal pressure on the element kl of depth al1A,=h is hkl, and the horizontal pressure (Art. 387) is hk. The horizontal pressure on each element being of the same form, the whole horizontal pressure perpendicular to HK of the surface convex toward HK will be:.hk. But to every element k, of the surface convex toward HK there is a corresponding element ko of the surface concave toward HK, the projection k of which is the same. Hence the horizontal pressure on lc2 is -hk, and the pressure on the whole surface concave toward HK is — 2ihk. The re sultant, therefore, of all the horizontal pressures perpendicular to KH is.hk —/.hk=O. Since the same is true on whatever side of the vessel the plane HK be drawn, the resultant of all the horizontal pressures will be zero. 2~. The resultant of the vertical pressures. The pressure on the element k, vertically downward is the weight of a filament of the fluid whose base is ik, and height ablk,. The pressure on k' vertically upward is equal to the weight of a filament of the fluid whose base is k' and height a,k'. The resultant of the pressures on the corresponding elements k' and k, is therefore equal to the weight of the filament cf fluid I'lc,. In the same manner, the resultant of the vertical pressures on any two corresponding elements will be .246 H YDROS T ATICS. the weight of the filament of fluid whose bases are these eles inents, The resultant of all the vertical pressures will then obviously be the sum of the weights of these filaments, or the weight of the fluid. hence the resultant of all the partial pressures on the interior surface of the vessel is equal to the weight of the fluid, directed vertically downward, and will pass through the center of all the parallel vertical pressures (Art. 44), or the center of gravity of the fluid (Art. 94). Con. Since the horizontal pressures balance, there will be no resultant pressure to cause motion in the vessel horizontally. If, however, an aperture be made on one side of t-he vessel, the pressure on this portion of the vessel will be removed, and the > pressure on the corresponding opposite portion will tend to produce motion in that direction, and if the vessel be free to move, motion wvill actually ensue. 389. PRoP. To find the resullant pressure on any solid imrnersed in a fluid, and its point of application. 1~. The horizontal pressuie on any element of the immersed solid is the same as that on the vertical projection of the element 1 (Art. 387), and the entire pressure on each of the opposite surfaces of the solid will be equal to that on their vertical projections. x; L o But the projections of any two op)posite surfaces on the same vertical plane are equal. Hence the horizontal pressures on any two opposite surfaces are equal, and the resultant of the horizontal components of the pressur ea at every direction is zero. 2~. The vertical downward pressure p' on any element k of the upper surface is equal to the weight of a filament of the fluid ak' whose base is k' and altitude ak'=h', or if k be the horizontal transverse section of the filament through the Cen PRESSURES ON I MMERSED SURFACES. 24 ter of gravity of k', p'=gph'k. But the sum of all the filamenta resting on the upper surface of the solid is equal to the volume V' of the fluid vertically above the solid, and the whole downward pressure P' is the weight of this volume, or P'I-.p'=gpZ.h'k=-gpV'. In the same manner, the vertical upward pressure p1 on tihle corresponding element kI of.the lower surface is equal to the weight of a filament ak, of fluid whose base is k, and height ak:-ht, or p,=gphlk. But the sum of all these filaments is the volume V1 of the solid and fluid above the upper surface, and the whole upward pressure P, is the weight of this volutyme of fluid, or P =p1,=gpl.h k-=gpV1. Now h,-h'=1h is the length of the filament k'k, of the solie P1-pI=p its weight, and V1-V'-V is the volume of the solid. Therefore, the difference P of the upward and downward pressure is P=P1-P'- zIp I-.],' —'., and P=gp2.h k-gp..h'kl =gph/l- -gpV;.. P=gpY=-p, (140) or the resultant pressure is vertically upward, and equal to the weight of a mass of the fluid of the same volume as the solid. 3~. To find the point of application of this resultant. Lei the distance of the point of application of pihk from the vertical plane HK be x, and the distance of the point of application of the resultant P be x. Then, since the moment of the resultant equals the sum of the moments of the components, P-:=.px, _ X.px I.px and a (141) But (29) the point thus determined is the center of gravity of the displaced fluid. Hence the resultant of all the pressures on the immersed solid is equal to the weight of the displaced fluid, acts vertically upward, and its point of appication is the center of gravity of the displaced fluid. !248 H Y D R O S T A T I C S. 390. PROP. To find thle conditions of eqttzlibrzum if an znz mersed solid. The conditions of equilibrium involve a considera tion of the weight of the body. Let V be its volume, and a its density. Its weight will be gaV. The body will be urged downward by a force equal to gaV applied at its center of gravity, and by Art. 389 it is urged upward by a force equal to gpV, since the volumes of the solid and displaced fluid are equal. In order to equilibrium; then, 10. These forces must be equal; and, 2~. Their lines of direction coincident. The first condition gives gaV=gpY, or =-p; that is, their densities must be the same. Con. 1. If a be not equal to p, the body will ascend or descend by a force equal to g(p-o)V, according as p-a is positive or negative. If p>a, the body will rise to the surface, and be but partially submerged. Let V, be the displaced fluid, or the part of the solid immersed when the equilibrium is restored, Then gpY~ =fgV, (142) and V: V,=p:; or the whole volume of the solid is to the part immersed as the density of the fluid is to the density of the solid. CoR. 2. If the centers of gravity of the solid and displaced fluid be not in the same vertical line, the body will be acted upon by two parallel forces in opposite directions, and will cause the body to turn round. The point of application of the resultant of these forces may be found by Art. 29. 391. DEF. The section of a floating body made by a plane coincident with the surface of the fluid is called the plane of flotation. The line joining the centers of gravity of the solid and of the displaced fluid is called the axis o filotation. DEPTII OF FLOTATION. 392. PROP. To find the depth of fJotatioi whenl the volum, and density of the body are known. DEPTH OF FLOTATION. 24(J Let V be the volume of the body, a its density, V, the volume of the displaced fluid, and p the density of the fluid. Then (142) goY=gpV,, and V1 =-V. P Now whenever V1 can be determined in terms of the depth of flotation x, this expression will suffice to determine x. If the solid be a right cylinder whose axis a is vertical, and the radius of whose base is r, we have V1 =7rr2x, and V=7rr2a; x. x —a. p If the body be a right cone with the axis vertical and vertex downward, let r be the radius of the base, and a the altitude a Then the radius of the plane of flotation is Lx. Hence 2 and V= ir'2r'a; x=a / -. If the vertex of the cone be upward, X=a 1 —. 393. PRoP. To find the positions of equilibrium of a rightf triangular prism when the axis is horizontal. Whenever a body can be conceived to be generated by the motion of a plane surface perpendicular to itself, andt the body floats with its axis horizontal, it is evident that, in whatever position it be turned, its center of gravity and that of the part immersed will lie in the same vertical section; and, further, that the center of gravity of the body will be at the center of gravity of the section, and the center of gravity of the immersed part at the center of gravity of its section. Also, the ratio of the part immersed to the whole mass is the same as 250 II YDROS TATICS. the i atio of their sections. We can therefore limit the investi gation to the positions of equilibrium of one of the generating sections. A Let ABC, Fig. 1, be a generating section when only *one angle is imBmi\ /7fl ( mersed, and Fig'. 2 when two angles are immersed. M7~\ /\1 LetAB-=c,AC=b,BC=a, B \ AQ-x,AP=y,andlet abe 91M AIic the ratio of the density of A ~ the solid to that of the fluid. Now from (142) we have, in F ig. 1, AQP Ixy sin. A xy ABC ~bc sin. A bc.'. xy=-a.bc. a) In Fnig. 2, BQPC ABC-AQP bc-xy ABC ABC bc.. x-y=bc(l —o). (b) Since (b) is derived directly fiom (a), by changing a into 1 —a, we may examine the case of Fig. 1, and deduce that of Fig. 2 from the result, by changing a into 1 -a. Bisect BC in M, and QP in m. Join AMI and Am. Take MM' —IMA and mm'-' -mA. MI' and m' are the centers of gravity of ABC and AQP. Join Mm and M'm'. It is obvious that the center of gravity of BQPC is in the line M'm' produced. Now the second condition of equilibrium requires that M'nz' should be vertical, and since Mm is parallel to M'm', Mm is perpendicular to QP and MQ=MP. Let AM/=h, MIAQ-O, and MAP=0. Then MQ2 0=2 Ir+h-~2_ lx cos. 0. and MP2=,y2 +12 -2hy cos. q. Hence T"' —? —2hx cos. 0+2hy cos. pr0. (c) Substituting in (c) the value of y from (a), and reducilrg, X4-2h cos. 0.x"+2obch cos. ~.. -ab2c2 —0, (d) DEPTH OF FLOTATION. 251 and changing a into 1-a, we have for Fig. 2, X4-2h cos. O.x2+2(1 — a)bch cos. (.x- (1 -a)2b2c2-0. (e) The values of x deduced from (cl) and (e), and the correspond ing values of y from (a) and (b), will give the positions of equi librium required. Now, since the degree of the equation is even and the absoute term is negative, there are at least two possible roots, one positive and the other negative. The other two roots may be real or imaginary. If real, Descartes' rule of signs indicates that three wvill be positive and one negative. I-Ience, since the values of x and y, which are applicable to the question, are necessarily positive, equations (d) and (e) indicate no more than three positions of equilibrium. The values of x must also be less than b, and when subtituted in (a) and (b) give for y values less than c. Let us take the case of an equilateral triangle. Then 0~=-30, a=b-c, and h-b cos. 0=c cos. p. These values, substituted in (d), give xC4 ax'3 a3 3a3_ 2a4=09 0or (X2 -aa2) (X2 -ax+aa') —. (f) This equation is satisfied by putting.~ _ ffa2 = 0; whence x - a/ o, the negative value of x being inapplicable. Since a<1, we have xe real, we must have 16a<9 or 16o=9. 352 hYDROSTATICS. 9 9 9 or a can not be greater than 1But if v/9-16a>1, then x>a, which is inconsistent with tlw supposition that only one angle is immersed. The greatest value, therefore, which V9-16a can have is unity, and in this case the second angle will lie in the surface of the fluid. Putting v 9-16oa= 1, we get I= for the least value a can have when one angle only is im rnersed. The limits of a for this case are. therefore, 1 9 - and 2 16' If in (g) we rcange a into 1-a, we get x=x3l+ v'9-16(1-a)} =-3+4-3V 16o-7. (h) From which it may be shown that the limits of the value o-, for this case are 7 1 16 and -. Since equations (d) and (e) may have three real positive roots each, there may be three positions of equilibrium for every single angle immersed, and three for every two angles immersed, and therefore eighteen in the whole. But in the particular form considered above these are not all possible, 1. except — =, in which case x-a or- y-a; that is, either the angle B or C lies in the surface of the fluid. This would render six of the positions pertaining to Fig. 1 the same as six of those pertaining to /Fig. 2, making but twelve really different 1 O Mones. IfIa> and a<.9 there will be nine positions with f2 16! CENTER OF PRESSURE. 253 one angle immersed and three with two immersed; if a-2 7 7 and a>- ( nine of the latter and three of the former; if a<9 or o>-16, there will be only three of each. CENrTER OF PRESSURE. 394. DEr. The center of pressure in any immersed surface is the point of application of the resultant of all the pressures upon it. It is therefore that point in an immersed surface or side of a vessel containing a fluid, to which, if a force equal and opposite to the resultant of all the pressures upon it be applied, this force would keep the surface at rest. 395. PROP. To find the center of pressure on any imnmersea plane sulface. Let the immersed surface hbe the inclined side AE of the vessel ABE supposed to L o be filled with a fluid. Let x and y be the co-ordinates of.A\ | any element k referred to the rectangular axes AY and AX, and hk=h the vertical distance of the element below the surface of the fluid. x Then the normal pressure p on k will be p =gphk. The pressure on each element will be of the same form, and their sum or resultant R will be R=z.p. T he moment of each partial pressure, in reference to the plane through AY perpendicular to AX (Art. 46), will be px, and the moment of each, in reference to the plane through AX p)erpendicular to AY, will be py. tHence, if.7 and. are thle 254 H Y D R O ST A T I CS. co-ordinates of the point of application C of the resultant PI, we have (Art. 43) R-y_.p, R. —rpx, R —-v.py. If 0 be the inclination of the side of the vessel or irmmnersed surface to the surface of the fluid, wSe have h=x sin. 0. HIence p=gpxk sin. 0, and R —. gpxk sin. O=gp sin. O.I.xK, R-= Y.gpx2k- sin. O-gp sin. O.2.x2k, R —.gpyxk sin. O=gp sin. O.-.xyk. 2.x"k z.1xk' (143) _.xyk and Y.xk' (144) When the upper boundary is below the surface of the fluid and at a distance a from it, then, since the pressures p are limited to the immersed surface, h=(ad-x) sin. 0, and we shall have' (ax + x2) k X (a+x)k (145) Y (a+x)yk (14) Further, if the axis of x is so taken that it will bisect everx horizontal line of the immersed surface, the pressures on oppo site sides of this axis will obviously be equal, and the center of pressure will lie in this axis or y=0. It will be observed that the numerator of (148)) is the mo ment of inertia of the surface, and that the denoriniator is the statical moment. Hence the denominator is equal to the area of the surface multiplied by the depth of its center of gravity (29). Hence, if A be the area and Yr, this depth,.x2L.:.x- Ak (147):.xyk = Ax l (148) 396. PROP. To find the center of presszere of a rectang'uta; CENTER OF PRESSURE. 255 suface vertically zmmersed, and hlaving' one side in thie stqfacd of tle fluid. Let YY, be the line of intersection of the immersed surface ABCD with -L A - B Y the surface of the fluid, and conceive ABCD divided into n rectangles by horizontal lines, n being a very lairge number. Let AB —b and AD-h=, D -- C and draw OX bisecting the rectangle. x The center of pressure is obviously in OX and y=0. To find:, we have, since the height of each small rectangle is, for the area of each,, bh/ k -. The distances of these rectangles fiom the surfaice are h 2h 3h -,-9 -, &c.; n n n h/2 22/12 32h2 their squares, -2,, &C. bh/3 bh3.22 bhl'.32 Hence z.zwk= n3 + +, &c.., n3 n3 nn bn, (I2+2"+32+...... n2). But when 1 is very large, the sum of the rmth powers of the natural numbers 1, 2, 3, &c., to n is..x2kC= —. — l-b/.3 ~u. S.X2; 3 3 Also, A, =bh.h=h- b/2. b/3 2 or the center of pressure is two thirds the height of the rect. angle below the surface of the fluid. 397. PROP. To find tile center of pressure ithen the imnmersea 256 I Y DR O ST A T IC S. swuface is a triangle, hayv.ng one side horizontal and the oppost,-: vertex in the suwface. A_ F B Let BDC be the triangle, I its height, and b its base, and suppose the triangle divided as before into n horizontal divisions. 90 ~ W.CD, the embankment will be overturned. Since the embankment is uniform throughout its length, as also the pressure on it, we may determine the stability by taking one foot in length. Now W=1 X 14X6X X 120=5040 lbs., and P=1X 14 X 7 X 62.5=6125 lbs., OC=BM-M=X14 and CD- X6=4... W.CD=20160, and P.CO=28583-.. The latter being greater than the former, the embankment will be overturned. Ex. 17. A wall of masonry, a section of which is a rectangle, is 10 feet high, 3 feet thick, and each cubic foot weighs 100 lbs. -Find the greatest height of water it wili sustain without being overturned. Ex. 18. If the height of the wall be 8 feet, its thickness C feet, and each cubic foot weigh 180 lbs., find whether it wiiL st'ind or fall when the water is on a level with the top. CHAPTER II. SPECIFIC GRAVITY. 403. DEF. The speciflc gravity of a body is the ratio of the weight of the body to the weight of all equal volume of some other body taken as the standard of comparison, and whose specific gravity, therefore, is taken as the unit. Water is generally employed as the standard of comparison for solids and liquids, and atmospheric air for aeriform fluids. COR. If v, p, a, and w be the volume, density, specific gravity, and weight respectively of one body, and vI, p,, a,, w, the same of another body; then, since v=v,, a w g.p.v p (150) a, w1 g.p v, P, or the ratio of the specific gravities of two bodies is equal to that of their densities. 404. PROP. To find the specific gravity of a body more dense than water. If the body be immersed in water it will descend (Cor., Art. 390). Let w be the absolute weight of the body, and w, its weight or the force with which it will descend when immersed in water. Then the loss of weight in water is w —w,, and this is equal to the upward pressure of the water, or to the weight of a volume of water equal to that of the solid. Hence w absolute weight w-w, loss of weight The absolute weight as well as the loss of weight is ascertained by the hydrostatic balance, which differs firom the ordinary balance only in having a hook appended beneath one of the scale pans, to which -the body may be suspended by a fine thread and allowed to sink in a vessel of water beneath it. The body is first placed in the scale pan and counterpoised 1v r4 HYDROSTA T I CS a weight w, and then suspended to the hook, and when im,nersed in the water its counterpoise w, again determined. 405. PROP. To find t,7e specific gravity c)f a body less dense than water. Since the body A is less dense than water, it will not descend in the water by its own gravity. Let a more dense body B be attached to it, and call the compound body C. Let w -absolute weight of A, W' " " B, and w' —=its weight in water WI = " " "' C, and w — " " " 6 Then w'-Iwl =loss of weight of B, W'- -u:// (' " " " C, and (w"-w'y)-(w'-w',)=1oss of weight of A, the upward pressure of the water on A, and therefore equal to the weight of a volume of water the same as that of A. But and, by substitution, (w"- w'/) - (' -W'l)-w'+w-w'/-w'* w', =w+w' - w' w'W =w'-w' (152) Hence, add to the absolute weight of the body the difference of the weights of the more dense and compound bodies in water, and divide the absolute weight of the body by the sum. 406. PaoP. To find the specific gravity of a liquid. Let a body whose weight is w be weighed both in water and in the liquid, the weight in the former being w,, and in the latter w2. Then w-w1 and w-zw, are the weights of equal volumes of the two liquids. Hence W-W2 -w2; (153) W-W1 or, if an empty bottle whose weight is to, weighs when filled with water wl, and when flIed with the liquid Iov, then C= W W (154) SPE CIFIC GRAV ITY. 265 407. Pror. To find tlhe weighfs of the constituents in a me. chanical composition when the sp]ecific gravities of the compound and constituents are known. Let w, w1, w2 be the weights of the compound and constituents respectively. a, al, 62 their respective specific gravities, v, v1, v2 their volumes. In all merely mechanical combinations, v=v1 +v2 (a), and ze-wl +w2 (b). W WI Wo But v —, 9v V L, and v2. gP gPl gP2 W WI W2 (a~) + P P1 Pa:r, since their densities are as their specific gravities (150), W U)1 Wo a a1 a2 Substituting in succession the values of zow and w2, obtained from (b), we have Z I, - I I I ((72 r /I 7 tU,=w( —-). ( —-)=( 2 )).su, (155) /1 12): W (156) 408. Hydrometer. Instruments for determining the specific gravity of fluids are called Hydrometers or Areometers. They are made of glass, brass, &c., and are of two kinds: one in which the weight is constant, the other in which the volume is constant. In its simplest form the areometer consists of a hollow globe, one of whose,diameters is prolonged in a flat 01 cylindrical stem of uniform size, and to the other extremity is attached a smaller globe loaded with mercury or shot, that it may float in a vertical position. 1~. Areometer of a constant weight. The annexed figure may represent this form of the instru ment. To graduate the stein, suppose it to sink in distilled water, at a given temperature. to a point s, and let the distance sr be divided into any number of equal parts, and continued z.;~(..YIIYDROSTATICS. A. upward. from's. When immersed in another fluid. suppose it to sink to t, distant from s, x divisions. Let V be the volume of the portions immersed in water, v the volume included between'any two divisions of the stem, and a the specific gravity of the second fluid, Then p and p, being the densities of this fluid and C'r water, the weight of the water displaced by the in[rI strument will be gp,V, the weight of the other fluid displaced will be gp(V —vx), and, since each is equal to the weight of the instrument, gP 1 v=gp(V-v x). V p V v pa, ATV-mv. v *(a) v-x I. be previously known, and x be observed, we can de...'mine the value of V a - X. v 1 -a Substituting in (a) this value of -, and putting for x, 1, 2, A; &C., the corresponding values of a for each division of the scale will be known. These may be.marked on the instrument, or the divisions may be numbered, and a table of their values formed to accompany, the instrument. COR. From (a) we have: V., v a Giving to a a small increment da, the corresponding increment of x is x'-x=dx= —.-.da; that is, for any small increment of the specific gravity of the fluid the corresponding change in the depth of immersion of the instrument varies as -.- which may be considered a meas. re f the ssceptibiity of the inst ent. tre of the susceptibility of the instrument. SPECIFIC GR A:\ IT i. 2V0 2". Areometer of a constant volume. The principal obstacle to the use of the areometer of a constant weight is the inconvenience and difficulty of calculating and marking against the different.divisions of the stem of each instrument a different scale of specific gravity, and constructing the stem of that perfectly uniform thickness which is necessary to the accuracy of the observations. To obviate these difficulties,- Fahrenheit conceived the idea of sinking the instrument always to the same depth by means of weights to be placed in a cup at the end of the stem. Let W be the weight of the instrument, w and w the weights respectively necessary to'sink it to the same point a of the stem in watei-, and in the fluid whose,i' specific gravity a is required, V the constant volume of the portion immersed, p1 and p the densities of the water and the fluid. Then gpY=W+w and gpIV=W+w. / \ p. W+w p1 W+w1 CoR. Differentiating, we get dw=(W+wl)da=gp Vda; that is, for any given small variation do in the specific gravity of the fluid, the variation of w is as V, or the susceptibility of the instrument is as the volume of the portion immersed. 409. Nicholson's Hydrometer. This is a modification of the preceding, to adapt it to the determination c of the specific gravity of solids as well as liquids. For this purpose, a metallic basket f f is attached to the lower extremity, in which t t the body may be placed, and its weight in water ascertained. The basket'admits of V v reversal, so that the body may be retained under:water when specifically lighter. Let w be the weightlin C necessary to sink the instrulnent tof. Replacing w b L the body, let w, be the weight which must 266 H Y DR OS TAT 1 CS. be added to sink the instrument to the same depthf. RemoovTin the body to the basket beneath, let w2 be the weight in C requisite to sink the instrunment a third time tof. Calling W the weight of the instrument, W, that of the water displaced by it when immersed to f, V the volume of the body, p its density, that of water being p,. Then W+w-W1, (a) gpV+W+w =W1, (b) and gpV+ W+w, =W + gp 1 Y. (c) Subtracting (a) from (b), and (b) from (c), gpV-w-w1, gpV=w2 -W1. p W-W] p w-w, P W2 -W1 410. EXAMPLES. Ez. 1. A piece of wood weighs 12 lbs., and when annexed to 22 lbs. of lead, and immersed in water, the whole weighs 8 lbs. The specific gravity of lead being 11, required that of the wood. The specific gravity of lead being 11, if v be its volume,. 11v=22 or v-2. But this is the volume of water displaced by the lead, and the specific gravity of water being 1, its weight will be 2. Therefore the loss of weight in water is 2 lbs., and the actual weight in water 20 lbs. Hence, Art. 405, w=12, w11=8, and w' -20. w 12 1 w'w, -w/' 12 +20-8 2' Ex. 2. A diamond ring weighs 691 grains, and when weighed in water 642 grains. The specific gravity of gold being 164. and that of diamond 3-, what is the weight of the diamond? Let v, v' be the volumes of the gold and diamond respective: ly, and A the weight of a unit of volume of water. Then vA is the weight of a volume of water equal in bulk to the gold, and v30vX the weight of the gold. In the same manner, Wv'2 is the weight of the diamond. Hence SPEC I F IC GRAVITY. 269 692-= 1323 V+ 7v, (a) and 64-= V33x+7v'X-(V+V')2a. (b) Subtracting (b) from (a), we have 5=Xv+Av'. from (a), 69~= 33(5 — v')+7,v', 139= 165-264v', v' =1, laiv'=3}, or the weight of the diamond is three and a half grains. Ex. 3. A body A weighs 10 grains in water, and a body B weighs 14 grains in air, and A and B connected together weigh 7 grains in water. The specific gravity of air being.0013, required the specific gravity of B, and the number of grains of water equal to it in bulk. Let 2', 2A" be the number of grains in the volumes of water equal too the volumes of A and B respectively, and o', a" their specific gravities. Then, by the conditions of the question, ('-1)X'- 10, (a) (a"-.0013)X"= 14, (b) (61- 1)x,+ (,,- 1)x,,"= 7..) From (a), (b), and (c), we have 14(a"- 1) 10+- =7, o" —.0013 14(1- c")= 3 (o"-.0013).. "=.8237. Hence, also, from (b), 14 14 A"=....17.023 grains., o/ —.0013.8224 Ex. 4. When 73 parts by weight of sulphuric acid, the spe cific gravity of which is 1.8485, are mixed with 27 parts of water, the resulting dilute acid has a specific gravity equal to 1.6321. Required the amount of condensation which takes place by the mixture. Let X' be the number of parts by weight in a quantity of water equal in volume to that of the sulphuric acid, and V!' in a quantity of water equal in volume to that of the mixture. Then 1.84852'=73, or 2' =39.4915, 270 HYDR OS TAT IC S. and 1.6321A"= —73+27=1 00 or "-=61.2707. Now the condensation of the mixture will be expressed by the ratio of the diminution of the volume of water and acid, when mixed, to their united volume before mixture. Hence condensation= 2'+27 5.2208 4 —5=0.0785. 66.4915 Ex. 5. Two fluids, the volumes of which are v and v', and specific gravities a and a', on being mixed, contract -th part of the sum of their volumes by mutual penetration. Required the specific gravity of the mixture. Let a" be the specific gravity of the mixture; then the surrl of their weight before and after mixture will be equal. Hence (vf+v'a'); =(I- (v+v')G"1, n va+v'a' which gives.a" lEx. 6. A body whose weight in a vacuum is 73.29 grains loses 24.43 grains by immersion in water. Required its sreeific gravity. Ans. a=3. LEx. 7. Required the specific gravity of a body which weighs i;5 grains in a vacuum, and 44 grains in water. Ar7S. a=3.0952. Elx. 8. An areorneter sinks to a certain depth in a fluid whose specific gravity is.8, and, when loaded with 60 grains it sinks to the same depth in water. What is the weight of the instrument? Ans. w=240 grains. Ex. 9. A compound of gold and silver, weighing w=10 lbs. has a specific gravity a=14, that of gold being a'=19.3, and that of silver being "all10.5. Required the weight w' and w' of the gold and silver in the compouno. Ans. w'-5.483, w"=4.517 CHAPTER II!. COMPRESSIBLE OR AERIFORM FLUIDS. 411. PROP. To find tle tension of the atmosphere cr;te, compressible fluid. Let a glass tube AB, open at one end and A closed at the other, be filled with mercury. Retain the mercury in the tube by the press- L ure of the finger, let it be inverted and the open end immersed beneath the surface of the mercury in the vessel CDFE. It will now be found that a column of mercury, as BL, will occupy a portion of the tube, while the remaining portion AL will be void. This column of mercury is sustained in the tube by the pressure of the atmosphere on the surface of the mercury in the cistern CF, and as there is no pressure on the mercury at L, the equilibrium is due to the equality of pressures of the atmosphere and mercury on the common base B (Art. 401). If we now suppose the cistern covered, so as to sepal ate the air without fromI that within the cistern, the pressure of the ex- i-t. - ternal air can not be communicated to that E _' within, and. the mercurial column must be sustained by the expan sive force or tension of the inclosed air, and be a measure of it. If the tension of any other elastic fluid inclosed in a vessel be required, let the tube I from the cistern CEFD be fitted to an aperture in the vessel, and a communication be thus established between thei inclosed fluid and the mercury in the cistern. The mercury in the tube will then rise or fall till Can equilibrium takes place between the expansive force of the fluid 272 HY DR OS A T I CS. and the weight of the mercurial column. The height at which tne mercury stands in the tube above the surface of mercury in the cistern is ascertained by a graduated scale attached to the tube. 412, ScHIoL. The mean pressure of the atmosphere at or near the level of the sea is generally employed in mechanics as the unit of pressure, and other expansive forces are corn pared with this and expressed in atmospheres. It has been ascertained by the barometer that the mean pressure of the atmosphere, at a temperature of 50~, is equivalent to a column of mercury 30 inches in height; or, the specific gravity of mercury being 13.598, to a column of water 13.598X2.5 feet=34 feet; or, the specific gravity of air, at a temperature of 50~, being 810' to a column of air of uniform density, 27,540 feet =5.2 miles in height. The tension is also measured by the pressure of the atmosphere on a unit of surface. Now 30 cubic inches of mercury is equal to 13.598X30 cubic inches of water=407.94X1728 cubic feet of water=0.23607X 1000 oz.14.75 lbs. on the square inch. The instrument above described involves the essential parts of a barometer, a more particular description of which belongs to Physics. Generally, instruments employed to determine the tension of elastic fluids are called manometers. 413. PROP. To show that the tension of an aeriform fJuid is inversely as its volume. Let AB be a vertical tube which communicates with the cylindrical vessel DEO, 11,, closed at the top, and mercury be carefully introduced into the tube, so as to fill the horizontal part H1O,BC, and leave the air in DO, of the same density as the exterior air. The inclosed air will then have a tensionof one atmosphere. Take DH,=-IDH,1, DHE3= DH1, &c., and draw the lines H20OR,, H303R3, &c. Now if mercury be poured into the tube AB until it stands at H202 in the cylinder, it will be found to stand in the tube at the height COMPRESSIBLE OR AERIOR I FLUO I DS. 2738 L2 R- 30 inches above H, 02; or, when the air A in the cylinder is reduced to half its volume,:ts tension or expansive force is two atmospheres. When the mercury rises in the cylinder to the height 11t03, it will be found to stand at the height L3R3=60 inches above t130,; or, when the air is reduced to one third of its volume, it has a tension of three atmospheres, or, general- L2 Iy, the tension is inversely as the volume. The tlruth of the law has been tested experimentally as far as 26 atmospheres, and for all firactions D E of an atmosphere. If V and V' be the volumes of a given mass 1........3 0lo 3 of air, p and p' the corresponding tensions or I2 ---- -- o2 — i2 pressures on a unit of surface, then -------- 0 R......... I p,-V or pv='v'. (15,) p [....I This law is called the'law of 3Mariotte, firom the discoverer 414. Con. Let p and p' be the densities of any mass of ait corresponding to the volumes V and V'. Then, since the density is inversely as the volume, P V' P (158) or the elastic force is directly as the density. 415. Pnop. To estimate the efect of ]heat on the vollume and tension of atmnosyheric air. When air is inclosed in a vessel and heat is applied, its elastic force is increased, as may be shown by the method indicated;n Art. 411, and by the same method the increase in the tension for a given increase of temperature may be ascertained. Experiment indicates that the tension of a given volume of dry air increases, by being heated from the freezing to the boiling point, 0.367 of its original value, and therefore, if the Y:onsion remains the same, the volume will increase 36.7 pel cent. Let v be the increment of volume, the tension remain-,ug the same, for one degree of Fahrenheit's ihermometer; ther S 274 I Y D R OS T A T IC S. v=-0.002039, and for an increase of t degrrees of temperature, tlhe increase of volume will be vt=0.002039t. If now V0 be the original volume of a given mass of air, and it be heated t, degrees, the tension remaining the same, the new volume V1 will be V,-l=(1 -vt,)Vo; qnd when heated t2 degrees the corresponding vAlume will be V o=(l+vto)Vo. v 1 +vt 1 +0.002309t, V2 1 -vt2 1 +0.002309t, (15) But the densities p1 and P2 are inversely as the,Alumes. p, V2o 1~vt2 Hence p -V I+vt2 2 V i +vtI If, also, a change take place in the tensions at the same time, letp 0be the tension at 320,p1 the tension at 32~+t,, and P2 that at 3g~+t2. Then, since the tension is inversely as the volLum-e, V,(= (1 — vt 1)-V0 and V, = (1;t) 0V0. P1 PI2 V1 (l+vt1)p~ _(l+Vt2)p V2 (1 + vt2)p1';and I +,tP1 (l+ vt1)p, (vt,)b 60 p +Vt2)'P2 (160) in which b and b2 represent the measures of the tensions p,.ind p2, or the corresponding heights of the barometer. 416. PROP. To find the density of the air at different tempe'a(tures and under different pressures. By accurate experiments the weight of a cubic foot of air at a temperature of 32~, when the barometer stands at 30 inches, is found to be p=0.08112 lbs. avoirdupois. Hence, for the temperature 32~ + t~. P 0.08112 lb ( 1 +vt 1 +0.002039t (161) If, also, the barometer, instead of b=30 inches, should stan'd iat some other height, as b1 inches, the density will be express. ed by p b, 0.08112 b, 0.002704 lbs. 12) P =+ b +. 4t' +0 lbs. (162) 1 +vt' b 1 + 0.00f204t'30 I + 0.00204t C OMI P It E S S. I P I E O 1 A E It I V O t M F L U I D S. 275 Whenever the elasticity of the air is expressed by the pressure p= 14.75 lbs. on the square inch, instead of the baronmetriic height b, the density for any other tension pl will be p - pp 0.08112 p, 0.005'5 lbs. (1) l+vt p 1 +0.00204t'14.75 1 +0.00204t The density of steam is five eighths of the density of atmos. pheric air for the same temperature and tension. Therefore we have, for steam, 0.00344p 1 lbs. 1 +- 0.00204t 417. Prnor. To determine the height corres]ponding to a given density of the atmosphere, and, conversely, the density in terms of the height. Since the density of the atmosphere at the surface of the earth is due to the pressure of the superincumbent portions 1 it, the density must decrease as the height increases. Let HOBA be a vertical column of air whose base AB is one square foot. Conceive it divided 0 into portions of equal weights w, and heights x 0, x 1, x2, &c., beginning at AB, so small that the density of each may be regarded as uniform. Then x 0x1 +x, &c... x,,=:.x=h is the height of the nth stratum. Let p0 be the tension of the lowest stratum, and p,, that of the nth, p0 and p,, their densities respectively. Then the weight of ---- 1he nth stratum is w-1.x.P9, — p'. (1 58)..p0 Po 71W Po P... (a) But the tension of the (n- )th stratum is p,-tw, and tihe height, therefore, p0 w P0'P, -w+W 27{, Hu D R OSTATICS. Po W In like manner, flp 2= _ p,_.2w, Po w on-t p oplnw Po Po since pn+nw=p o. Now, in the exponential series, 1.2 1.2.3 when y is very small, we have ey=l~y, or y-yl.e=l.(1 +y) =2.30258L. (1 +y), (b) in which I denotes the Naperian logarithm, and L the corntrnu logarithm of a number. w If in (b), for y, we put -, we get p,, from (a), we have p0 x, = — l. (p,, + W) - L1,. Po Now, substituting p,+w successively for p,,, we obtain xn = — 1. (P(,+ w)-.. l, Po p0_= /(p,, + 3tv) 1. + 2) Po: t-=X —1-Z= p (p,, +3twn) -.(p, +(2-w ) w, Po By taking the sum of these equations, the te':ms in thle' brackets will all cancel, except l.(p,i+-nw)=l.p, in the last and l.p,, in the first, and we shall have 2.== h = P ~ I.p -I 0.p } l= P ~. (164) Pa Po P,, COMPRESSI BLE OR A E, I O F L UIDS. 271 To find p,, when h is given, we have Po poh - O 9h 2n P0 pch or P ePo. Pn psh Hence pZ=p o.e ~, (165) where e-2.71828, the base of the Naperian system of loga, rithms. 418. SCHOL. Formula (164) may be adapted to the deterreination of heights by the barometer. To this end let b0 and b,, be the heights of the barometrical columns at the lower and P 0 b0 uppsr stations respectively. Then, since -bo P b A=2.3025 P0.L ~-. (a) Po b,, The value of h_ may be determined from the consideratiol PO that p 0 expresses the weight or pressure of a column of the ate mosphere on a unit of surface as one square foot, and P_ must Po express the height of this column on the supposition that its density is uniform. Now a cubic foot of air at 320 weighs 0.08112 lbs., and a cubic foot of water at the same temperature weighs 62.37917 lbs., and therefore the specific gravity of 62.37917 water referred to air is -.08-l-2=769. Hence the height of a homogeneous atmosphere at a temperature of 32~ is 34X7 76:26146 feet..'. 2.30258P o=2.30258 X 26146=60204 feet: Po It is here assumed, however, that the lower station is at or near the level of the sea, and no account is taken of the variation of gravity at different elevations. From numerous observations made at different elevations above the sea, and at known diffierences of heig(ht. this coefficient is found to be 60345 feel 278 riYD.OSTAT CS. at a temperature of 320. But the actual temperature of the all at both the lower and upper stations will, in general, c(ffer from the standard temperature of the formula, and, since the density of air varies uniformly with the temperature, we may use the mean of the temperatures of the air at the two stations. Let to and t, be the indications of' Fahrenheit's thermometer; then the mean temperature will be ~-(t0 +t,), and the deviation t from the standard will be t- (to + to)- 3' The expansion of dry air is 0.00204 for a change of 1~; but when the atmosphere contains vapor, it is found, by comparing the rates of expansion of vapor and of dry air, and assuming a certain mean humidity for the air, that the rate is expressed by 0.00222. Incorporating this correction in (a), and using the coefficient determined by observation, we have hl=60345(1 +.00222t)L.~. (166) It is further obvious that a change in the length of the mercurial column will be produced by a change of temperature of the mercury. Let ro and b%, be the temperatures of the mercury, as shown by a thermometer attached to the cistern of the barometer; then, since mercury expands at the rate of 0.0001 for each degree, b,,-b,(l +.0001) (, 0 —*), where bi, is the observed height of the barometer at the upper station. Using this value of b,,, the difference of elevation between two stations or the height of a mountain may be determined with considerable accuracy. In the determination of the constant coefficient, the variation of gravity at different elevations is allowed for in the assumption that this coefficient is that which belongs to the mean height above the sea at which observations are usually made, and to the latitude of 450. When the latitude differs from this. it will be necessary to multiply the result by (1 +.002837 cos. 2~), ~t being the latitude at which the observations are made. C U MI P R E S S I B LE OR A E RIFO RM FLU I DS. 279 419. EXAMIPLES. Ex. 1. A cylindrical tube 40 inches long is half filled with mercury, and then inverted in a vessel of mercury. How high will the mercury stand in the tube, the pressure of the external air being equivalent to 30 inches? Let I be the length of the tube, a the length of the portion occupied by the air before it was inverted, h the height of mercury due to the pressure of the external air, x the height of the mercury after the tube is inverted, and h' the column of mercury equivalent to the tension of the air in the tube. Then (157), ha=h'(l —). But I=h'-hx or h'=-h-x..-. ha-(h-x)(I-x), whence we get x= I (l+-) ~ / (l —x)2 +4alh. Using the data of the question, 1=40, a=20, and h=30, we gei x —10 or 60. The first value is that which pertains to the specific question. Ex. 2. A tube 30 inches long, closed at one end and open at the other, was caused to descend in the sea with the open end downward until the inclosed air occupied only one inch of the tube. How far did it descend? Ex. 3. A spherical air-bubble having risen from a depth of 1000 feet in water, was one inch in diameter when it ieached the surface. What was its diameter at the bottom? Ex. 4. Required the equation of the curve described by the extremities of a horizontal diameter of the air-bubble of Ex. 3, supposing its center to move in a vertical line. Ex. 5. What number of degrees must a given volume of air be heated to double its elasticity? Ex. 6. The following barometrical observations were made at the White Rocks, on the bank of the Connecticut River some two miles below the city of Middletown: Parometer. Det. Ther. At. Ther. At the base, bo =30.09 ill. t0 =83.0,0- 84.5 On the summit, b, = 29.65 in. t,, = 85.0 -, = 83.5 280 H YDROSTAT I CS. Hence t — (t0 + t,) - 32 = 520 and 1 +.00222t1 1, 1 1544. Also,, 0-r,,= 1~ and bb,,= b(1+.0001X 1)-29.653 in. 30.09.h=. =60345X 1.11544. X L29 30.09 log. 1.4784222 29.653 a.c." 8.5279314 0.0063536 log. 7.803015D 1.11544 " 0.0474462 60345. " 4.7806413 427.67 feet " 2.6311074. Ex. 7. The following observations were made by Humboldt at the Mountain of Quindiu, Newv Grenada, in lat. 5~: Barometer. Det. Ther. At. Ther. At the level of the Pacific, 30.036 in. 77~.54 770.54 On the summit, 20.0713 65~.75 680.00 Required the height of the mountain. Ans. 11500 feet nearlyr. HYD R ODY N AM3I C S. 420. PR. The velocity of a fluid in a tube of variable dauimeter, kept constantlyfull, is in different transverse sections in.versely as the areas of the sections. Since the tube is supposed constantly full, and the fluid incompressible, the same quantity of fluid must pass through every section in a unit of time. But admitting the fluid to have the same velocity in every part of the same section, the quantity which flows through any section in a. unit of time will be the product of the area of the section by the velocity. If;, therefore, k and k' be the areas of any two sections, and v and v' the velocities at each respectively, kv= -k'v', and v: v'-=k': k. (167) In reality, the velocity is diminished by the sides of the tube, and is therefore in any section least near the sides of the tube and greatest near the central portions. 421. PROP. The velocity with which a fluid issues from a small orifice in the bottonz of a vessel kept constantly full, is equal to that acquired by a body falling freely through the height of the fluid above the orifice. Let VF represent a very small orifice in A L M B the bottom of the vessel ABCD, filled with a fluid to the level of AB, GF a stratum whose thickness FH=-h' is indefinitely small, and FM=h the whole height of the column vertically above the orifice. If now the stratum GF fall by its own D weight through iHF=h', the velocity will be E C 282 II Y D ROD Y N A MI CS. v-= v'2gh'. But if the stratum be urged by its own weight and the weight of the column LH, calling the force in this case g', the velocity u' will be v' — /'2gllh'. But the forces g and g' are as the weights of the collimns GF and LF, or as their heights th' and h. HIence g h gh - or g" — Substituting this value of g' in that of v'. we have v'= /2gh, (168) which is the velocity of a body filling freely through the height h. CoR. If the orifice be made in the side of the vessel, and a tube be inserted so as to direct the current obliquely, horizoztally, or vertically upward, the velocity of efflux will be the same, since the pressure of fluids is the same in every direc. tioln. In the first case, its path will be a parabola whose equation 1s (64). In the second case, the angle of elevation a-O, which reduces the equation to x2=-4hy, the equation of a parabola whose axis is vertical and origin of co-ordinates at the vertex or orifice. In the last case, if all obstructions are removed, the fluid will rise to the height of the surface of the fluid in the vessel. 422. PROP. To determine the horizontal distance to which a fluid will spout from an orlfice in the vertical side of a vessel. Let the vessel ABCD be filled to the level AB. If the fluid issue horizontally from the orifice 0, the equation of its path.is x2=4hy, in which h=OB, the height of the fluid above the orifice, is the impetus or height due to the velocity. To determine the range, let y=OC=a. Hence DISC H ARGE i 0UM SMALL ORIFICES. 283 x=CP=2 Vhy=2 V/ha=2.OM, or the horizontal distance is A 3 equal to twice the ordinate at the orifice, in a semicircle whose diameter is the vertical distance o0 from the surface of the fluid to the horizontal plane. CoR. When the orifice is at the middle of BC, the range is a maximum, and at equal dis- D tances above and below the middle the range will be the same. 423. Pnop. To find the quantity of fluid discharged in G given time from a small orifice in the bottom of a vessel when the fuid is maintained at the same constant height. Let h be the constant height of the fluid in the vessel, t the given time, k the area of the orifice, and Q the quantity. Then Q-=kvt=kt 12gh. (169) If the orifice be circular, and r its radius, then k=Trr', and Q=wr V 2g.trt2 h/1, =25.195.tr2 i/h. The time being in seconds, and r and h in feet, Q will be.n cubic feet. If the weight W be required, - = 62.5Q. CoR. 1. The time required for the efflux of a given quantity is Q V2 V 2gh CoR, 2. Since t and g are constant in (169), Q ct k Vh. Hence the quantity discharged in the same time from orifices differing in size and distance from the surface, varies as tho size of the orifice and square root of its depth. If the orifices are the same, Q oc h. In order, therefore, that the discharge from one orifice may be twice that from an. other, its depth must be four times as great. 284 HYDRODYNAMIC S. 424. ProP. To determine the time in which a cylindrical ves. sel filled with water will empty itself by a small orifice in its base. The velocity of efflux at any instant is v= V/2gh, h being the height of the fluid in the vessel. But since the velocities v and v', at the orifice and at any transverse section of the vessel, are inversely as their areas, k and k' (167), k k v'=-vr —-V2gh GctVh. k' ki/ The surface of the fluid, therefore, in the vessel descends with a velocity proportional to the square root of the space over which it must pass, and the circumstances of its motion are precisely the same as those of a body projected vertically upward. But the velocity of a body projected upward is such that if it continued uniformly it would move over twice the space through which it must move in the same lime to acquire or expend that velocity. Hence the time of descent of the surface of the fluid to the base is twice that required for the descent of the same superficial stratum when the vessel is kept full. In the latter case, if Q be the whole quantity of water fn the vessel (169), tk V/2gh Therefore, in the former, 2q t 2 (170) h- v2gh If r' be the radius of a tranverse section of the vessel, r that of the orifice, Q=rzrr'Th and k —,rr2. 2r2_.'. t= ~h. r2 V2g 425. SCuOL. The preceding deductions are founded on thile hypothesis that the fluid particles descend in straight lines to the orifice, and ahl issue with a velocity due to the height of the fluid surface. Experiment shows that this is true only uof D I SC II A R G E FROM SMALL ORIFICES. 28 5 those vertically above the center of the orifice, that those situated about the central line of particles, take a curvilinear course as they approach the orifice, being turned inward toward this line or spirally around it, and this deviation from a vertical rectilinear path is the greater the more remote they are from the central line. This deviation will necessarily occasion a diminution of vertical velocity, and therefore a diminution of total discharge. The oblique direction of the exterior particles gives to the vein of issuing fluid, when the orifice is circular, the form nearly of a conic frustum, whose larger base is the area of the orifice. This diminution of a section of the issuing fluid is called the contraction of the vein. and the vein itsellf from the orifice to the smallest section, the vena contracta, or contracted vein. The results of most experiments agree in making the length of the contracted vein, when the orifice is circular and horizontal, equal to the radius of the orifice. There is a greater discrepancy in the results of experiments for determining the ratio of the diameters of the two ends of the contracted vein. When water flows through orifices in thin plates it is found to be about 0.8. The ratio of the areas of the two ends, called the coefficient of contraction, is therefore 0.64. The actual discharge is found by experiment to differ sl!ight. ly from the theoretical for other causes. The ratio of the former to the latter is found to be about 0.97, and is called the coefficient of velocity. The product of the coefficients of velocity and contraction, called the coefficient of efflux, is 0.62. Hence, for the actual velocity of discharge through orifices in thin plates, we have v =0.62v=0.62 /V2gh, and for the quantity in a unit of time (169), Q=kv 0 =O.62k V'2gh. When the orifice or pipe through which the discharge is lnaae has the length and form of the vena contracts, the velocity will be v 1-0.97v=0.97 v 2g/i, and the quantity Q-=lv,= 0.97k V2gh, ic being the smallest section of the contracted veili 280 HY DR OD Y N A Al IC S. By the use of cylindrical or conical adjutages, the quantity of the discharge is increased. More seems to be gained by the adhesion of the fluid particles to the sides of the tube, in preventing the contraction of the vein, than is lost by the friction. The discharge is greater when the adjutage is conical rind the larger end is the discharging orifice. 426. PRoP. To determinze the quantity of water' which wil/ flow from a rectangular aperture. 1~. When one side of the aperture coincides with the surface of the water. BE FT Let h be its height and b its breadth, and conceive the aperture to be divided horizonLI K tally into a very large number n of equal divisions so narrow that the velocity of the G I —-'HI fluid in every part of each may be regarded as the same. The larger n is, the more nearly will the hypothesis be satisfied. The depth of these successive divisions below the surface will be h 2h 3h -, -,, &C. 7n 7 n The velocities in each will be / 2h 3h 2: 2'' 2gV ac.; h and since the areas of these divisions are each b-, the quanti. ties discharged by each in a unit of time will be (169) bit h bh / 2h a 3/h -V 2g-.,I 2g~.-i-V: 2.-7 &c., nv n 71 n7 n and their sum, or the whole quantity Q discharg ed, will be bh-Vgh 1 1V Q=-'bh VPg(12 + 32 +3 +, &c... n2). noon 1 n V' But 1+2"+322+, &c....?= —- =-2. +It1 DISCHARGE FROM RECTILINEAR APERTURES. 281... Q= Vbh2gh= -b V2gh/. bh Or. if v= be the mean velocity. v= 2 /2gh. 20. When the upper side of the rectangular aperture is not coincident with the surface of the fluid. Let its depth EL=h1l, and the distance EG=h, as before. Then the quantity which issues from the aperture LH will be equal to the quantity which would issue from EH, diminished by the quantity which would issue from EK, or 3_ 2 3 3 Q=-3b'2ghl3 - -3b V2ghlz= -b x/2g(h'-h1); and if Q=b(Alh-h 2)v, 3 3 ~h2 _h. v=' v- 2g. h 427. PROP. To determine the quantity of water which uwili flow from a triangular aperture. 10. When the vertex of the triangle is in the surface. Let the base GH=b, and the height HF=h, h and let the triangle be divided into n parts, as before, of equal but very small heights. The altitudes of these elements are each -, and the G lengths, beginning at the vertex of the trianb 2b 3b gle, are 3n' n &c. Their areas, regarding them as parallelograms, which we may do, since n is indefinitely large, are bh 2bh 3bh *.-.-, &c. 7nIn n7In n n n Hence the discharges through each are bh A h 2bh / 2h 3bh/ 3h n V2gh wo dis c ha2grg &wl b and the whole discharge will be 288 tH Y D R O D Y N A M I C S. bh V23gh k Q= — -(1".2 2 +3 2+, &c.... 2 n 2Vn q 1/ Also, = b- V/2 h. 20. When the base of the triangle is in the surface. In this case the quantity which will flow through EFG wi! equal the discharge through EFGH minus that through FGH or Q-= bh V/2gh- -2bh V 2ghAih= V4 b 2g,11 and v= - A/V 22g1h. E K IL,I IF Con. If the aperture be a trapezoid, a3 EFGH, whose upper base EF-=b lies in the surface, lower base HG=b2, and altitude KH=h, we may find the discharge by regarding the aperture as made up of the o rectangle KHGM, and the two triangles EKH and AMGF. Hence Q- bo bh V/2gh+ -4 (b -- b) V2gh= 1 (2b1+ 3b2) hV2gh. Also, through the triangle VHOG, whose base HG=b,, and depth of vertex LO=hl, the quantity of the discharge will be Q-A-ob 1t1 V 2gh-, I (2b, +3b )hV 2gh. In a similar manner, rectilinear orifices of other forms mayl be divided into triangles, trapeziums, &C., and the discharge determined. 42S. Pior. To determine the velocity with which an elastit fluid will issuefromn a small orfjice into an unlimited void wheA urged by its own weight. By reference to Art. 421, it mwill be seen that the result there obtained is independent of the density of the fluid, and wiN therefore be true of all fluids, whatever be their density. For the forces g and g' are as the weights of the stratum at the orifice, and of the column vertically above it; that is, g = g=w: w'=gh ~ gh'= h: h'. MOTION OF FLU DS IN LONG PIPES. 289 Consequently, v=- V2gh' expresses the velocity with which every fluid of uniform dens ity will issue from a small orifice. COR. The velocity with which air will rush into a vacuunn is that which a heavy body would acquire in falling from the height of a homogeneous atmosphere. The height of a homogeneous atmosphere (Art. 418), when the temperature is 32~ and the barometer stands at 30 in., is h'-26146 feet, which gives v 1297 feet. But for a temperature of 60~ the height h, will be (159) hl =h'( l+vt), in which t-60~-32~=28~, and the value of v for a mean state of humidity of the air will be 0.00222 instead of 0.002039 (Art. 412)..v. v'= 2gh = V/2gh'(1 +0.00222 X 28) — / / X 32 1 X 26146 X (1+ 0.00222 X 28) - 1 33) feet MOTION OF FLUIDS IN LONG PIPES. 429. The subject of the conveyance of water in pipes is one of considerable practical importance. But the motion of fluids in long pipes is so much affected by adhesion and friction in the interior, by the resistance occasioned by curvature, and by the disengagement of air, which remaining stationary when the pipes are laid along a level surface, or collecting in the higher portions of them, where they are curved, opposes the flow of the fluid, that theoretical results are of little practical value; besides, investigations conducted on hypotheses involving all the causes which affect the motion of the fluid are too difficult for an elementary work. The, experiments of Bossut, in 1779, are those which are usually relied on for information on this subject. Water was allowed to flow through pipes whose diameters were 13 in. and 2 in., and lengths from 30 to 180 feet. They were chiefly of tin, and inserted in the side of a reservoir filled to a constant height, either one foot or two feet above the center of the pipe The following principles were established: 1~. The discharges in given times with pipes of the same T C 9 H HY D R O D Y N A M I C S. length, and with the same head of water, are proportional to the squares of the diameters. 2~. For pipes of different lengths and of the same diameter, the discharge was inversely proportional to the square roots of the lengths. To determine the supply which may be expected foml a pipe of given dimensions, it was found that a pipe 30 feet long and 1 inch in diameter would discharge at its extremity about one half of that which would issue from a simple orifice, or short pipe of the same diameter. Couplet, in 1730, found that a pipe of stone or iron, 600 fathoms in length and 12 inches in diameter, with a head of 12 feet of water, discharged -lth, and a pipe of the same diameter and 2340 fathoms in length, with a head of 20 feet, discharged l-th of that which would have been obtained from a simple orifice To determine the quantity which flows through a section of a natural stream, it is usual to measure the breadth and the depth at different points of a transverse section, and find the area of the section. The velocity at various points of the sece tion is measured by the hydraulic quadrant, or rheometer of some kind. The mean velocity, multiplied by the area, gives the quantity which flows through the section in a second. 430. PROP. To detelrmine generally the velocity with which a fluid will issue from an or/iice of any size in t/e bottom of ft cylindrical vessel.,A We shall adopt the usual hypothesis of a division of the fluid into thin lamina, and thai in their descent their parallelism is preserved. P P' Let the distance of the lamina pp'qq' fionm q q, the surface A be x, and K the area of its surface. Thenl if dlx be its thickness and p its density, pKdx will be its mass, and gpKd.xr its weight, or the force with which it will' c descend if free. Now the resistance with, a which it meets in its descent is the difference. I of pressures on its lower and upper surfaces. ) B 1E Let p represent the pressure on a unit of t!t, GENERAL M ET 1O D S. 291 upper surface by the water above it; then pfdp will be the upward pressure on a unit of the lower surface, or the resistance of the water below it, and the difference — dp, multiplied by K, will be the resistance experienced by the whole lnmin;a Hence the moving force will be gpKdx-KdIp, and the acceleration, [VI.] and [IX.], -gpKdx- Kdp gpdx - d dp (a)d pKdx pdrx drt' Let v be the velocity of discharge at B, k the area of a section of the issuing fluid, and dt the element of time in which the surface K descends a distance equal to dcl. Then (16`7) we have Kdt x=kvdt or dx. (k) dt =K' (a) By differentiation, regarding dt as constant, we get d2x kdv dt2 Kdt' /dv _gpdx - ap Kdt- pdx pkdv dx or gpdx —dp- Ki. dt' Hence (b) gpzdxd-r-Ip=PK-.vdt. k2 2 Integrating, gpx-p=p. -+ C. Let now x-h=AB and k=K; then 2 gPh-p==-P C; and subtracting the preceding from this, we have g(h —~,=) =2 (1 - - i), Or~ U=\/2g (k-x) 2 H Y D R O D Y N AM I C S. If k be very small, v=- V2g(h-x) nearly; and if the vfssel bf. kept constantly full, x=0 and v- V2gh (168). If k=-K, v= x, or the velocity must be infinite. From whichl we infer that a section of the issuing fluid can never be equal to a section of the vessel. If a cylindrical tube be vertical and filled with a fluid, tile portion of the fluid at the lower extremity, being urged by the pressure of all above it, will necessarily have a greater velocity than those portions which are higher, and therefore (167) a section of the issuing fluid is necessarily less than a section of the tube. 431. P}iop. To determine tlhe time inz which a cylindrical ves s'e will exhaust itself by a s.!all or/i~ce in the base. Since k is very small, we have v= V2g(h-x) for the velocity at the end of the time t when the surface of the fluid has descended through the space x. Let this velocity be supposed constant during the indefinitely small time dc. Then the quantity discharged in this time will be dQ =hdt V 2g(h- x) -= Kdc, dJzl beinog the descent of the surface in the time dt. K eHence d(t __.(h — )`d.-c. k V2g 2K _ Integrating,. t=- _.(h — )+C. If x:0, t= 0, and C — Vh. 2K - - -v' k V2g and when:c; —=h, or the whole is discharged, 2KVh 2Kh 2Q t= = = (170). (173) k V 2g k V 2gh 1 ~ 2gh The time is therefore twice as great as that which is required to discharge the same quantity when the vessel is kept constantly full. E x A MI P L E S. 293 432. PWope. To determine the quantity which will issue fowrrj an aperture of any size and form in the side of a prismawb: ressel. Let the Figi., Art. 430, represent one face of the vessel, -an( DCE the aperture. The element of the area of the orifice will now be ydx..-. dq=tydx V2g (h, +x:), in which h,=AC, xzCm, and y-=zn. This, integrated between the limits x= O, and x.= -/., = B;vill give the quantity discharged in the time t. If the aperture be rectangular, y=a constant —b. Then dQ —bt f /2g(h 1 +x)dc, 3 and Q=2-bt V 2g(h 1 +x) + C, and, between the limits above named,.3 3 - 3lit V' 2g- (l -2- h ). (174) If the orifice extend to the top of the vessel, then hI -0, ano Q 2= ibt V2g..h- 2 _h2lt v' 2gi, Q or 3 which is the velocity due to the depth of the center of pressure of the aperture below the surface of the fluid. In the foregoing the vessel is supposed to be kept constantly full, or the surface of the fluid to be very large compared with the aperture. 433. EN.XAI PL ES. Ex. 1. A vessel, formed by the revolution of a semi-cubical parabola about its axis, which is vertical, is filled with a fluid till the radius of its surface is equal to its distance from the vertex; to find the time in which the fluid will be dischalreed through a small hole at the vertex. The equation of the semi-cubical parabola is ayr-a.bo Hence (167) kvdt=- Kdx, or k / 2g,xdl= -'ry2ed,= -.XdX a 294 I~ Y' R O D YN AM I CS. 7T 5.'. d= -.dx. ka V'2g Takingr tile integral between the limnits x=0 and == we 11al~e 7t= o rr2a2 ka V2g " a7k — /g Ecx. 2. To find the time in which a paraboloid of revolution whose altitude is h and parameter p, full of fluid, will empty itself through a small orifice at its vertex, its axis being vertical. Ans. t- h 3kV /2g Ex. 3. A conical vessel, the radius of whose base is r and altitude h, is filled with a fluid. Required the time in which the surface of the fluid will descend through half its altitude, the orifice being at the vertex and the axis vertical. I.S ~rv,~~- ( - 1) Ans. t- t I. 20kg2 Ex. 4. Find the times in which a fluid contained in a vessel, formed by the revolution of the curve whose equation is y4=a'x about the axis of x, will descend through equal distances h supposing a small orifice at the vertex, and the axis vertical. A ns. Each t= a2 Ex. 5. The horizontal section of a cylindrical vessel is 10u square inches, its altitude is 36 inches, and it has an orifice whose section is one tenth of a square inch. In what time, if' filled with a fluid, will it empty itself, allowing for the contrac tion of the vein? Ans. t= 1136t.5 HYDROSTATIC AND HYDRAULIC INSTRIMNII1N S MARIOTTE'S FLASK. 433. Tins piece of apparatus was devised to illustrate severai successive positions of equilibrium of air and water, and variations in the velocity of dis-D charge. In the annexed section of this flask, A, B, C are apertures so small that when open and water is issuing the air can not enter, and M the reverse. The apertures at first are all 0........ closed, the vessel is filled with water, and the A 1 tube DF is inserted at MI and filled with water to the height D. B=. 1~. Let the aperture B be opened. F If ht be the height of a column of water equivalent to one atmosphere, and x represent, in every c case, the difference of level between the surface of the water in the tube and the open orifice, then the pressure of the air inward at B on a unit of surface will be h, while the pressure outward on a unit of surface will be that due to the pressure of the air on the surface at D, together with the pressure of the column of water DE-x, or the outward pressure at B is h+x. Therefore the pressure outward is greater.than the pressure inward by the difference h+x —h=x, and the water will issue with a decreasing velocity due to x, or u= V2gx. This discharge will continue till x vanishes and the water in the tube descends to E, on a level with B, when the equilibrium is restored. 20. Let B be now closed and A be opened. Now the pressure downward on the surface of the water in the tube at E is h. The pressure upward on the same surface is that due to the pressire of the air at A, and to the weight 2~96 I Y D ROSTATIC AND HYDRAULIC IN ST RUENTS. of a column of water equal in height to AB-HE=-x. or the upward pressure on E is h+x. The resultant upward pressure on E is therefore h+x-h-x, and the surface of the water in the tube will rise till x vanishes, or till it arrives at H, when the equilibrium will again be restored. In the mean time, however, the air has entered at A and risen to the upper part of the vessel, where it occupies the space above ho. Since an equilibrium now exists, the expansive force of the air above ho, together with a column of water equal in height to LH, equilibrates the pressure Ih of the air on the water in the tube at H. If h' be the height of a column of water equivalent to the tension of the air above ho, ana yx-LH represent the difference of level of the surfaces of the fluid in the tube and flask, we have h'+y — h or h' —h-y. 3D. Let A and B be closed and C opened. The pressure at C inward is I, outward is h+:x, representing by x, as before, the vertical distance of the surface of the water in the tube from the open orifice C. Therefore the resultant of the pressures at C is h+x-h-x, and the water will flow from C with a velocity due to the height x. As the surilce of the fluid in the tube descends, x diminishes, and the velocity of discharge also, till the surface in the tube arrives at F, at which time x becomes equal to the height of F above C. In this position of the fluid, the equation h-'=h-y still obtains, y being now the distance LF. But an equilibrium not being established, air will enter the vessel at F and rise to ho. This operation will continue till y-o and h'-h. During this entrance of the air, and the descent of ho to F, the value of x has remained unchanged, and therefore the velocity of discharge has remained constant. After this the value of x will diminish, and the velocity of discharge also, till the water in the flask descends to the level of C. BRAMAHI'S HYDROSTATIC PRESS. 434. In the annexed section of this instrument L and Hi are vertical cylindrical cavities, in a solid mass of metal or other B RAA 1 An H IIYYDROSTATIC PRESS. 297 strong material. The diameter of H is con- K B c siderably less than that C of L, and they communicate through an aperture N, in which is a valve opening into L. A is a strong Ipiston or solidl cylinder of iron D A fitting closely to the [.] surface of the hollow F cylinder, and movable o n it. IH is a'piston _ G similarly applied to the Dther cavity H, and movable by means of a lever KP, whose fulcrum is at P. At 0 is a valve opening upward, and below it the cylinder H is continued downward to a reservoir of water. The lever KI being raised, the water ascends, as in the common pump, from the reservoir G into the cavity H. The lever being then pressed down, the valve O closes, and the water is forced through the valve N into L, and, acting on the piston A, communicates a pressure to any substance, as books, included between its upper surface and a strong cross bar BC firmly connected with the solid cylinder DE. When the water from H is forced into L, its reaction closes N, and the same operation is repeated. The pressure on the base of A (134) is to the force impressed by H as the area of the base of A is to that of H. Let R and r be the radii of the pistons A and H, L and l the longer and shorter arms of the lever KP, p the power ap. plied at K, and P the resulting pressure on thie base of A Then the force p', applied to the piston H., will l e, pL P?rR2 But (134), 2, 298 HYDROSTATIC AND IIYDRAULIC INSTRUMENTS. R2 L R2 This press seems to present the simplest and most effectii.e of all contrivances for increasing human power. L R lfp=100 lbs., = 10, and — =10, the P= 100,000 lbs HYDROSTATIC BELLOWS. 435. This iistrument presents an illustration of what is termed the Hydrostatic Paradox: that fluids, however unequal in quantity, may be made to equilibrate. It consists of two circular boardcs, A and B, firmly connected by ta cylindrical coating of leather or other flexible material. CM is a tube communicating with the lower portion of the cylinder. Water being poured into the tube CM, the boards A and B will separate, B will rise, B and a weight WV, very large compared with r that of the water in CM, may be supported on B. The fluid in CM which counterpoises W is that above P, the level of B. Let k be the area of a horizontal section of the tube, K that of a section of the cylinder, or that of B, and p the pressure of the fluid NP above P on k. p k Then (134) pk Tf a be the specific gravity of the weight W, V its volume, and NP=h, then gPlhk k gpV K' ph V If V' be the whole volume of the fluid in the instrument, and 1i the height of the cylinder AB, then DIVING BELL. 299 V'=H(K+k)+hk=lH(K+I+k)-d V'IK-oVk K(IK+k) Let now the quantity of fluid be increased by v, ard the co'. responding increment AH in the height of W will be A K(V'+v)-oYk KV'-oVk v K(K+k) K(K+k) K+k' DIVING BELL. 436. The diving bell is commonly a hollow cylinder o: paiallelopiped, one end of which is closed. It is immersed with the open end downward, weights being added to sink and keep it steady in its descent. As the vessel descends the fluid continually exercises a greater pressure on the contained air, condenses it, and occupies a greater portion of the vessel. If the form be as above supposed, to find the height x of that portion of the bell free from water, when the top is sunk to the depth H, let h be the height of a homogeneous atmosphere, a the height of the bell, p' the density of the external air, p thC t of the air in the bell, and p, that of water. Then the pressu e p, on a unit of surface of air in the bell, will be p-=gjhp'+(H+z)p, }. But the unit of pressure arising from the elasticity of the inI rvr al air is gph..'. gIhp'+(H+x)p1 }=gph. The quantity of air in the bell being constant, and a hon vn. tal section of it the same throughout (155) and (156), a px-p'a, or p=p'.Z... hp'-tIp1 +xp1 ==p'a-, or P hx- Hxa+x=p —ah, and a being the specific gravity of air, ohx+IHx+x'=Cah. ~L — i T- I-nh t-~/1/ T LIrr~.~ 300 ItYDROSTATIC AND HYDRAULIC INSTRUMENTS. SEA GAGE. 437. The vessel AB is perforated with holes, and has with. in it, fixed in a vertical position, a glass tube, having one end closed and the other immersed in a cup of mercury. A is a hollow sphere whose buoyancy is sufficient to raise the instrument when the weight W suspended at the bottom is detached. The instrument is allowed to sink in ~pi I the water whose depth is to be determined, and by a mechanical contrivance the weight is detached when it strikes the bottom, so that the gage xM,. will ascend by the buoyancy of the ball. The ilB'J height MP to which the mercury has risen in the tube is marked on the interior of the tube, by the adhesion of oil or other viscid substance placed on. its surface. Let h be the height of the barometer at the surface, 1 the length of the tube above the surface of the mercury in the cup, x the depth of the water, and MP-=h. If ha be the height of a column of mercury which would be sustained by the elastici ty of the air in NP, then (155) i (i-hl)-=zl or hi 1-h Now the elasticity of the air in NP+ the weight of the col unin MP= the pressure of the atmosphere on the surface of the fluid+ the weight of a column of water extending to the bottom: or, p being the density of mercury, and p, that of thie water, ph2 +ph =phf+p x..=. — P (2+h, -h) =aho +o(h, -h) Pi hl - a/, +a(h 1-) = lh, —o(h,- h). 1- + in which a is the ratio of the specific Fravity of mercury to.hat of the water of the ocean. SIP HO N. dOI SIPHON. 439. The siphon is an instrument for transfe ring fluids firom one vessel V to another V' in which the surface of the fluid is lower. It consistt of a bent tube with one branch longer than the other. Suppose this tube to be filled with the fluid from the vessel V, and to have its extremities immersed in the fluids in the two vessels. Let CB=a be the difference of level of the fluids in the vessels, p, the density of the fluid, o the density of the surrounding air, and k the area of a section of the tube. The press- ure on the surface of the fluid in V' without _ v the tube by the atmosphere will be greater than that on the surface of the fluid in V by gpak. This excess of pressure acting at O upward will urge the fluid from V' to V. At the same time, the pressure exerted by the liquid in the siphon at the level of the fluid in V will be greater than that at the level of the fluid in V by gp1Iak. This excess of pressure will urge the fluid from V to V'. Hence the resultant pressure is equal to gak(p -p), and in the direction DAO. The fluid will therefore flow from V to V', and, since p is small, it will move with a force nearly equivalent to a column of the fluid whose height is a and base k. It is not necessary that the longer branch should be immersed in the fluid of V'. If h be the height of a column of the fluid in equilibrium withl the atmosphere, and AB=x, then the upward pressure on k in the shorter branch, at the level of the fluid, is that due to h1, while the downward pressure is that due to x. Therefore the fluid is urged in the direction DAO, at A, by a force represented by h-x. As long as x4ah, there will be two points. If a= 10, 1- 36, and h1= 32, x=20 and 16. AIR PUMP. 305 AIR PUMP. 442. B and B' are cylinders or barrels, commonly of the same size, in which pistolls P and P', with valves V and V' opening upward, are movable by means of rack-work, the one ascend- p ing as the other descends. At the lower extremity of B B the barrels there are apertures with valves opening P' upward, and communicating by the pipe ac with the a receiver R, from which the air is to be exhausted. The ascent of the piston P tends to produce a vacuum in B, the valve V being closed by the pressure of the external air. The air in R, by its elastic force, opens the valve v and fills the barrel B. When P descends, the elasticity of the air in B closes v and opens V, througTh which the contents of B escape into the external air. The action of both pistons is manifestly the same, and thus for each descent of either piston a volume of air equal to that of either barrel is expelled from the machine. Let V be the volume of the receiver pipes and one barrel, V, the capacity of either barrel, and p,, the density of the air in the machine after the nth stroke. Now the quantity of air in the machine after the nth stroke is p,,V, and by the (n+1)th stroke the volume VI, or quantity p,,V,, is expelled. There will remain then in the machine after the (?n+l)th stroke the quantity p,,V-p,,V,, and this being difftsed through the space V, we have, by (155) and (156), p,,+1 — =P -p,,V, I=p,(V - V ) or p,,+,=p,,(1. If p, be the initial density of the -'o, by making n equal to 0, 1, 2, &c., successively, we have U :306 HYDROSTATIC AND HYDRAULIC INSTRUMENTs. p,,-PO (1 - VV, P2 Pl ( T) =p o(1-T>) It is obvious that p, can never become zero as long as n is finite, and therefore, even theoretically, perfect exhaustion is impossible. CONDENSER. 443. If in the preceding figure the valves V, V', v, r', were made to open downward, it would represent a condenser. By each descent of the piston a volume of air equal to that of tle barrel will be forced into R, and will by its reaction close the valves v and v', and be retained there. To find the density of the air in the receiver after n strokes of the piston, let pn and p,,+H be the densities of the air in R after the nth and (n+l)th strokes, V the volume of the receiver R and the pipe ac, V1 that of either barrel, and p0 the density of the external air. Then pl V= p,,V +p'V 1 ~Pn. P,+l=Pt+P-o V' If n be made successively equal to 0, 1, 2, 3, &c., we have P2- =P +Po0 - V=p(1 +2i) nV ( ~ pX=po(l+ V CL E PSYDR A. 307 CLEPSYDR A. 444. The clepsydra, or water clock, is a contrivance for measuring time by the descent of the surface of a fluid which flows through a small aperture in the base of the vessel containing it. Suppose the vessel a prism. It is required to determine what scale must be marked on its side, that the coincidence of the descending surface with the successive lines of division may mark equal successive intervals of time. Let a be the altitude of the prism, x the distance of the surface from the base of the vessel at the end of the time t, from the beginningc of the motion. Equation (172) gives, using x instead of h-x, Ik V'2ga k 2 x a- t t2k 22. Let /x and /t be corresponding increments of x and t; then k V2ga k 2 x+/Ax=a — Kg (t +t) +-, (t At). Subtracting the preceding equation from this, and we have k2o (K 2a Ax — I-2 -K -k/ -- t — At At. The time t is in seconds, and, to determine the divisions corresponding to successive minutes of time, put At=60", and give to t the successive values 0, 60", 120", &c. In a vessel of the form supposed in Ex. 4, Art. 433, the vertical distance of the successive divisions will all be equal, or the surface will descend equal distances in equal times. THE END.