miiHi E ii^i[,T„?,^?^! F °^'A. SAN 3 1822 02128 6786 cmcncE ft ENGINEERING LIBRARY 0075-6 ^SoF CALIFORNIA. SAN DIEGO ^lTjOLLA. CALIFORNIA 92093 s&-£ V" I, : i . cms / % k * AN ELEMENTARY TREATISE MECHANICS, EMBRACING THE THEORY OF STATICS AND DYNAMICS, AND ITS APPLICATION TO SOLIDS AND FLUIDS. ^veparcij for tlje £intier=£va*iuate ©ouvsc in tije S&cskjjau iHmbcvsftg. BY AUGUSTUS W. SMITH, LL.D., NEW YORK: HARPER & BROTHERS, PUBLISHERS. 18 68. 1°[ *3l I S Entered, according to Act of Congress, in the year one thousand eight hundred and forty-nine, by Harper & Brothers, ia the Clerk's Office of the District Court of the Southern District of New York. PREFACE, The preparation of the present treatise was undertaken un- der the impression that an elementary work on analytical Me- chanics, 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 in- vestigation. 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 particulars, thus developing and strengthening more of the mental faculties, and more equably by far than the geometric- al 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 en- ergy 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 prepara- tion of this manual, it will be proper,to refer to some of the 2 PREFACE. 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 consist- ent with rigor of demonstration, and avoiding all reference to the metaphysics of the science as out of place in a work de- signed for beginners. At every successive stage of advance- ment the student is required to review the ground passed over by the use of the principles learned, in the solution of exam- ples 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 accomplish- ed is distinctly proposed, or the truth to be established is briefly and clearly enunciated. The student has thus a defin- ite 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 his 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 from swell- PREFACE. ill ing the volume with diffuse verbal explanations, and the intro- duction of experimental illustrations. The experienced teach- er 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 ter- restrial mechanics, and to embrace an interesting field of celes- tial 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, conven- ient for frequent and rapid reviews, and suggestive of ques- tions for examination. In this work no claim is advanced to originality. The ma- terials have been sought and freely taken from all available sources. Nearly all the matter, in some form, is found in al- most every author consulted, and credit could not be given in every case, if, in an elementary work designed as a text-book, IV PREFACE. it were desirable to do so. In the portion on Statics, I am most indebted to the excellent introductory work of Profess- or Potter, of University College, London. The chapter on Couples is substantially taken from Poinsot's Elemens de Sta- tique. The questions, generally simplified in their, character, are mostly taken from Walton's Mechanical and Hydrostatical Problems, and Wrigley and Johnston's Examples. The works more especially consulted are those of Poisson, Francaeur, Gregory, Whewell, Walker, Moseley, and Jamieson. Some- thing has been taken from each, but modified to suit the spe- cific 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, Middletown, Conn., Jan., 1849. I^H1 CONTENTS. Irt. t*& INTRODUCTION 1 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. STATICS. CHAPTER I. COMPOSITION AND EQUILIBRIUM OF CONCURRING FORCES IN THE SAME PLANE. 3 7. Two equal and opposite Forces in Equilibrium. 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. 1C. 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 120°. 20. Each of three equilibrating Forces equal to the Resultant of the other two in 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 the Resultant with the Sum of the Momenta ol any member of Components. 33. When the Origin of Moments is fixed. 34. When the Forces are in Equilibrium. 35. Examples. VI CONTENTS. Art. Pag, CHAPTER II. PARALLEL FORCES 15 'J6. Resultant of two Parallel Forces. 37. Definition of Amis 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 the Components. 40. 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. When 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 Amis of a given Length. GO. 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 Com- 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 Values 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 System, 76. Equilibrium of a Point on a Plane Curve. CONTENTS. Vll trt Page 77. Conditions of Equilibrium. 78. Definition of Virtual Velocitius. 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. L»2. The Converse. CHAPTER V. CENTER OF GRAVITY 43 83. Definition of Gravity. 84. Laws of Gravity. 85. Definition of a Heavy Body. 66. Definition of the Weight of a Body. 67. Definition of its Mass. 88. Expression for Weight. 89. Definition of Density. 00. 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. Relations of Densities to Masses of the same Volume. 94. Definition of Center of Gravity. 95. Connection of the Center of Gravity with the Doctrine 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 with respect to an Axis. 99. 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 Whole and a Part given to find that of the other Part. 107. Examples — 1. Of a Straight Line — 2. Triangle— 3. Parallelogram — 4. Polygon — 5. 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 FROM THE ACTION OF GRAVITY... 54 108. When the Body has a Fixed Point in it. 109. Definition of Stable, Unstable, and Neutral Equilibrium. 110. Position of the Centerof Gravity when the Equilibrium is Stable aud when Unstable. 111. Pressure on the Fixed Point. 112. Position and Pressure when there are two Fixed Points. 113. 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. Measure of the Stability on a Horizontal Plane. 119. Case of a Body on an Inclined Plane. 120. Examples. APPLICATION OF THE INTEGRAL CALCULUS TO THE DETERMINATION OF THE CEN. TER OF GRAVITY 53 Vlil CONTENTS. Art P'S* 121. General differential Expressions for the Co-ordimites 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 7fe 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. 136. Conditions of Equilibrium when any Number of Forces in the same Plane act on a Lever of any Form — Examples. §11. WHEEL AND AXLE 86 137. Definition of Wheel and Axle. 138. Conditions of Equilibrium when two Forces act Tangentially to the Surface of the 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, Piuions, 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 the 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 •-■• 88 156. 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. If2. Equilibrium in the third — Examples. CONTENTS. IX Art P"gl § V. THE INCLINED PLANE 108 163. 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 Body is sustained by a Force parallel to the Base of the Plane. 167. Equilibrium when two Bodies rest on two Inclined Planes. § VI. THE WEDGE 103 168. 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 103 172. 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 any 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 116 190. 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 134 201. Definition of Friction— Kinds. 202. Measurement of Friction. 203. Laws of Friction. S04. Value of the Coefficient of Friction. * CONTENTS. Art. Pif. 205. Limits of the Ratio of the Power to the Weight ou the Inclined Plane. 206. Limits of the Ratio of the Power to the Weight in the Screw. 207. One Limit obtained directly from the other. — Examples. 203. Examples on Chapters VI., VII., and VIII. DYNAMICS. INTRODUCTION *.. 181 209. 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 Measure — 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 j 3fi 228. 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 143 210. 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 Bodies. 245. Velocities of imperfectly Elastic Bodies after Impact. CONTENTS. X Art. Pa^t 246. Velocity of the nth 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 through 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. 259. 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. {(63. Scholium on Universal Gravitation. 264. Scholium on the Numerical Value of the Force of Gravity. °.65. Examples. ' CHAPTER IV. P ROJ E CT I L E 3 161 266. 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 and Time 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. $ I. MOTION ON INCLINED PLANES 169 275. 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 Circumference. 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. § II. MOTION IN CIRCULAR ARCS 173 284. Velocity acquired down the Arc of a Circle. 285. Velocity lost in passing from one Side of a Polygon to the next \ XU CONTENTS. Art. Page 286. Velocity lost when the Sides are Infinite in Numler. 267. Direction and Intensity of an Impulse at each Angle, to 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 183 299. 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 Dia- 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 19i 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. CONTENTS. Xll] Art Pag« CHAPTER VII 201 331. More General Methods required. 332. Modification and Extension of Fundamental Formulae. § I. RECTILINEAR MOTION OF A FREE POINT 202 333. Preliminary Considerations — Absolute Force. 334. Spac« 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 Cube of tie I»ista«ce. § II. CURVILINEAR MOTION OF A FREE POINT 2i« 344. Preliminary Considerations. 345. Motion of a Point acted on by any Number of Forces in £.e name Plane. 346. Specific Directions for particular Cases. 347. Velocity of a Point in its Path. 348. The Velocity may be found when the Expression id i-ategrable. 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 Description of Areas, the Force is directed to a Center. 352. Motion of a Point from an Impulsive Force. 353. Motion of a Projectile acted upon hy Gravity. §111. CONSTRAINED MOTION O? A POINT 215. 354. Velocity of a Point on a giveu Curve. 355. Time of its Motion. 356. The Reaction of the Curve- 357. Motion down the Arc of a Circle by Gravity. 358. Motion down the Are of a Cycloid. 359. Tautochronism of Vibrations in a Cycloid. § IV. MOTION Of A POINT ACTED ON BY A CENTRAL FORCE 221 360. Preliminary Remarks. 361. Expressions fof the Force in Terms of Polar Co-ordinates. 362. Equable Description of Areas. 363. Angular Velocity of the Radius Vector. 364. Velocity of a Point in its Orbit. 365. Time of describing any Portion of its Orbit. 366. Equation of the Orbit. 367. Geometrical Values of Differential Expressions. 368. Kepler's Laws. 369. Accelerating Force of a Planet directed to the Sun. 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 S(pnare of th© Distance. 374. Time of describing any Portion of the Orbit. <1V CONTENTS. Art. Pagl 375. Value of the Absolute Force in Terms of the Masses— In Terms of Terrestria. Gravity. 373, Examples. HYDROSTATICS. CHAPTER 1 836 377. 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 Solid. 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 - ... 36« 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. 1°. Of Constant Weight. 2°. Of Constant Volume. 409. Nicholson's Hydrometer. 410. Examples. t CHAPTER III. COMPRESSIBLE OR AERIFORM FLUIDS 279 ■ill. Tension of a Compressible Fluid found. 412. The Unit of Pressure. 413. Tension inversely as the Volume. 414. Tension directly as the Density. CONTENTS. XX irt. Pag , 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. HYDRODYNAMICS. 419. Velocity of a Fluid in a Tube of variable Diameter 22) 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 885 434. Bramah's Press. 435. Hydrostatic Bellows. 436. Diving Bell. 437. Sea Gage. 438. Siphon. 439. Common Pump. 440. Common Pump, Conditions of Failure. 441. Air Pump. 442. Condenser. 443. Clepsydra. ELEMENTARY MECHANICS, INTRODUCTION. 1. Mechanics is the Science which treats of the laws of Equilibrium and Motion. It is subdivided into Statics, Dy- namics, 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 motion is produced. Hydrostatics investigates the conditions of equilibrium in fluid bodies. Hydrodynamics investigates the effects of forces on fluids when 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 effect 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 counter- poise it ; dynamically, by the quantity of motion it will produce. By 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 arc those branches of theoretic Mechanics which treat of the effects of forces applied to material points or particles regarded as without weight or magnitude. Static and Dynamic Somatology would then embrace the appli- cation of theoretic Statics and Dynamics to bodies of definite form and magnitude, both (solid and fluid. A 2 INTRODUCTION. In the saute manner, by fixing on any line to represent the unit of force, any other force will be represented bv 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 ol direction on which the force acts. 4. As the magnitudes, directions, and points of lines are all determinable by the principles of Analytical Geometry, so forces, of which lines are the appropriate representatives, come under the dominion of the same principles. The appli- cation 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, with- out 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 concurring forces, and those which act along the same line toward the same parts, con- spiring forces. 6. A body is an assemblage of material points. The material points, or elementary particles, are connected together in vari- ous 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. STATICS. CHAPTER I. ON THE COMPOSITION AND EQUILIBRIUM OF CONCURRING FORCES. 7. Proi\ Two equal forces applied to the same point in ex- actly 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. Prot. Two concurring forces forming an angle with each other can not be in equilibrium. If possible, let the two forces P and Q, acting at A, and making, by their directions, the an- p" gle PAQ, be in equilibrium. Apply to the point A the force P,, equal and -< ^ > opposite to Q. Since, by hypothe- Pl Q sis, the forces P and Q, are in equilibrium, the point A will move in the direction of P,. But P, and Q, being equal and opposite {Art. 7), will be in equilibrium, and hence the point A must move in the direction of P. Therefore the point A must move in two directions at once, which is absurd. - 9. Definition. The resultant of two or more forces is a force which singly will produce the same mechanical effect as the forces themselves jointly. The original forces are called components. Cor. In all statical investigations the components may be replaced by their resultant, and vice versa. 10. Prop. The resultant of several conspiring forces is a single force equal to their sum. and acting in the same direction. STATICS. This is obvious from the admitted mode of measuring forces. 11. Prop. The resultant of two unequal forces acting in op~ posite directions is a single force equal to their difference, and acting in the direction of the larger. For the smaller will obviously be expended in annulling in the larger a quantity equal to itself, and thus leave an effective balance equal to the excess of the larger over the smaller, and •icting in the direction of the larger. 12. Prop. The resultant of any number of forces acting in the same right line is equal to their algebraic sum. By Art. 10, the resultant of each system of conspiring for- ces is equal to their sum, and by Art. 11, the resultant of these two resultants is equal to their difference with the sign of the greater prefixed. 13. Prop. In a system of points invariably connected, any point in the direction of a force may be taken as the point of application. If the force P be applied at the point B, and to any othei . r> p point A in its direction two j j — >■ other forces be applied, each equal to it, but opposite to each other, since B, by hypoth- esis, is invariably connected with A, one of these forces {Art. 7) will be in equilibrium with P. There will remain, there- fore, a single effective force equal to P and applied at A. In the same manner it may be transferred to any other point in its direction. 14. Prop. 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 symmet- rically situated on the other side of the plane ; and since no reason can be assigned why it 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. Prop. The resultant of two equal concurring forces is in CONCURRING FORCES. 5 the direction of a line bisecting the angle formed by the camp > 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 equilib- rium, they will remain so in whatever ratio their intensities be varied. Let P,, P 2 , P 3 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. Prof. Three equal concurring forces, inclined at angles of 120°, will be in equilibrium. Since each force is inclined in the same Pl 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 oth- ers will prevail. D STATICS. 19. Prop. Two equal concurring forces, inclined at an angle of 120° , have for their resultant a force which will be repre- sented 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 P, and P„, mak- ing the angle BAD=120°. To the point A apply a force P 3 = P, or P 2 , and making, with Pj and P 2 , angles of 120°. By Art. 18, these three forces will be in equilibrium. Now since Pj and P 2 equilibrate P 3 , their resultant must also. Therefore produce FA to C, making AC=AF; and since AC=R equilibrates P 3 , it will be the resultant of Pj and P 2 . Join DC and BC, and ABCD will be a rhombus, of which AC is the diagonal. For since CAD=60°, and AC=AD, the triangle CAD is isosceles ; and since each angle is equal to 60°, it is equilateral. .*. DC=AD=AB. In the 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. Prop. If tltree 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 direc- tion, 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 represented in magnitude and di- rection by the two adjacent sides of a parallelogram, the diago nal 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 two 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, acl CONCURRING FORCES. at A in the directions AD and AB respectively, and be repre- sented in magnitude by these lines, a Q, ^B R ^ C Suppose the force R to act at B (Art. 13), a point in its direction, and to be represented in magnitude by BC. Complete the parallelo- D E F 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 acting at E in the direction EF. Transfer P to B and Q to F, without changing their directions. Then P at B, in BE, and R at B, in BC, will also, by hypothesis, have their resultant in the direction BF, which may be supposed to act at F. We now have all the forces acting at F, and this without disturb- ing their effect upon the point A, supposed to be invariably connected with F. Hence, if the assumption be correct, F is a point in the direction of the resultant of the forces P and Q +R. But the assumption is correct when Q, and R are each equal to P (Art. 15). Therefore, the proposition for the direc- tion 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 wP. Also, putting nP for P, Q,=P, and R = P, it will be true for ?iP and 2P, and so on for mP and nY (m and n being positive in- tegers), or for all commensurable forces. 3°. When the forces are incommensurable. Let AB AC represent the forces. Complete A v B the parallelogram, and draw the diag- onal AD. AD will represent the di- rection of the resultant. If not, let some other line, as AE, be its direc- tion. Divide AB into a number of equal parts less than DE, and on AC C D take as many of these parts as possible. Since AC and AB are incommensurable, there will be a remainder GC less than DE. Draw GF parallel to AB, and join AF. AB and AG nd 8 STATICS. representing two commensurable forces, AF will represent thj 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 re- sultant 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 resultant. Let P, Q, and R be three forces in equilibrium, and AD, AB, AF represent their magnitudes respective- ly. Complete the parallelo- grams BD and DF, and draw the diagonals AC and AE. AC being the direction of the resultant of P and Q,, must, by Art. 20, be in the same straight line with AF, and is therefore par- allel 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 par- allelogram, 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 P 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 are in equilibrium, and a triangle be formed by lines draum in their directions, the sides of the triangle, taken in order, will represent the forces. Con- versely, if the forces can be represented by the sides of a triangle, taken in order, they will be in equilibrium. Let the forces P, Q, and R be in equilibrium, and be repre sented by AB, AD, and AE respectively. Produce EA, draw BC parallel to AD, and complete the parallelogram. AC will represent the resultant of P ai.d Q, {Art. 21) ; and since R equilibrates P D and Q, it must be equal and oppos ;, *.e to their resu.tant. .'. EA=AC. And since BC is equal and CONCURRING FORCES. 9 also parallel to AD, that is, has the same direction, the 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 tri- angle ABC, taken in order, represent the direction and magni- tude of the three forces P, Q, and R, they will be in equilibri- um. 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. 23. Cor. 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 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 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 com- plete the parallelogram. AE will be the other component. "When the directions of the components are arbitrary, then- valuation will be most easily effected by assuming these di- rections 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 in magnitude and direction by the sides of a polygon, taken in 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, P 2 , P 3 , P 4 , and P s . These forces will be in equilibrium. 10 STATICS. Draw the diagonals AC and AD. By Art. 22, AC, the third side of the triangle ABC, repre- sents a force equivalent to the two forces AB and BC. AC is therefore the resultant of Pj and P 2 , and may be substituted for them. In the same manner, AD is equivalent to the forces AC *D and CD, or to P„ P 2 , and P 3 ; and AE to the forces AD and DE, or to the forces P l5 P 2 , P 3 , and P 4 . That is, AE is the re- sultant of the four forces P 1? P 2 , P 3 , and P 4 , and acts in the direction AE. But P 5 is represented in magnitude by EA, and acts in the direction EA. Hence P s is equal and opposite to the resultant of the other four, or the forces are in equilibrium. Cor. The proposition is true of forces which do not lie all in one plane. For the proof is independent of this supposition. 25. Scholium. When three forces are in equilibrium, any three lines, taken parallel to their directions, will form a trian- gle, 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 be- tween the sides and angles of triangles does not obtain in poly gons of more than three sides. 26. Prop. To find, graphically, the resultant of any number of forces, acting at different points in the same plane. Let the forces P x , P 2 , P 3 have their points of application at A, B, and C. Producing the directions of the forces P, and P 2 till they meet in D, construct the parallelogram AB on the lines representing them, and the diagonal will represent their resultant R a . Pro- ducing Rj until it meets the direction of P 3 in F, and constructing the parallelo- gram on the lines representing R l and P 3 , the diagonal will represent their resultant CONCURRING FORCES. 11 R 2 , or the resultant of P 1? P 2 , and P 3 . 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 different 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 rep- resent the three forces. Then AE, B the diagonal of the face ACED, represents the resultant of the forces AD and AC. Compound- ing this with the third force, rep- resented by AB, we have AF, the c~ diagonal of the parallelogram AEFB, for the resultant of AE 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. 28. Prop. If three forces are in equilibrium, they are propor- tional 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. (180° — A), =sin. DAE : sin. EAB : sin. BAD ; =sin. QR : sin. Pit : sin. PQ. 29. Prop. If P and Q be two concurring forces, 6 the angle made by their directions, and R their resultant, then R 2 =P a + Q 2 +2PQ cos. 6. 12 STATICS. By trig, we have, from jig., Art. 22, AC 2 =AB 2 +BC 2 -2AB.BC cos. ABC, and cos. ABC= — cos. BAD= — cos. 6. R 2 =P 2 +Q 2 +2PQcos.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 made by the forces. Let AB, AC represent the two forces P and Q, and AD the diago- nal of the parallelogram constructed on AB and AC, their resultant R. H Take any point, O, for the origin ol moments ; join OA and draw AG per pendicular to OA ; draw Ol, 0>.*., On respectively perpendicular to AB, AC, AD, and BE, CF, DG perpendicular to AG ; also CH parallel to AG. The triangles O/A, OmA, OnA are respectively similar to the triangles AEB, AFC, AGD. Whence AE:AB=OZ G OA, or AE= AF : AC =Om : OA, or AF= AG : AD=0?i : OA, or AG= AB.Ol OA ' AC.Om ~OA~~' AD.Orc OA ' But the triangles ABE and CDH oeing equal, AE = CH = FG. .-. AE+AF=AG, or AB.0l+AC.0m=AD.0n, or designating Ol, Om, and On by p, q, and r respectively, Y.p+Q.q=Tl.r. 2°. When the origin of moments is taken within the angle. CONCURRING FORCES. 13 In this case the moments of the forces tend to produce rota- tion in opposite directions. As- suming one direction as positive, to distinguish them we must regard the other as negative. Let the positive direction be that of the hands of a watch ; then Q.Om will be positive, and P.O/ negative. The proof is the same in this case, except that AG=AF— AE, and AC.Om-AB.OZ=AD.Orc, or Q-q— P.p=R.r. 32. Cor. 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 proposi- tion 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.O/=Q.Om, or ?.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 rota- tion 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 op- posite direction, and there is no tendency to motion either ot translation or rotation. 35. EXAMPLES. 1. When the component forces are P and Q, and the angle made by their directions 6 ; what is the magnitude of the re- sultant R when 0—0 and 6=tt 1 Ans. (P+Q) and (P-Q). 14 STATICS. 2. Show that R is greatest when 6—0, least when 0— n, and intermediate for intermediate values of 0. 3. When P=Q and 0=60°, find R. Ans. R=PV3. 4. When P=Q and 0=135°, find R. Ans. R=Y\J2- V~2. 5. When the three concurring forces 3m, 4m, and 5m are in equilibrium, find the angle 3m, 4m. Ans. 90°. 6. If P=Q and 0=120°, find R. Ans. R=P. 7. If P=G, Q=ll,and 0=30°, find the magnitude of R, am of the angles P,R and Q,R. Ans. R= 16.47, P?R=19°.30', q7R=10°.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 gon taken in order. 9. 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,Q=0, when the pressure on the pin is R- P+Q 2PQ-3(P+ Q 2 ) Ans. Cos. 0= -pQ . 10. A cord, whose length is 21, is tied at the points A a'nd B m 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=- V-4 11. Given the four concurring forces Pj = l, P 2 =2, P 3 =3, P 4 =4, and the angles P^P 3 = 90°, P^P 4 =90°, and V^? 2 = 60°. Find the magnitude of the resultant R, and its direction pTX Ans. R=6.889 and P^k=102°.16'. W & z t \ W " , CHAPTER II. PARALLEL FORCES. Having considered the subject of forces which have a com- mon point of application, or which are reducible to this condi- tion, we next proceed to consider forces which act on differ- ent points, connected together in some invariable manner, as in rigid bodies, and whose directions are parallel. 3G. Prop. The resultant of two parallel forces, acting at the extremities of a rigid right line, is parallel to the components, equal to their algebraic su?n, and divides the right line, or the right line produced, into parts reciprocally proportional to the forces. 1°. 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' l ; these will not disturb the system. P and P' at A will have a resultant P", and Pj and P', at B, a re- sultant P",. The direc- tions of P" and P", will meet in some point D, at which we may suppose them to act. Replacing P" by P and P', and P", by P, and P'j, we now have, acting at D, the four forces P, P', P,, and P'j, of which P' and P', are equal and opposite, while P and Pj act in the direction DC, and have a resultant R = P+ Pj {Art. 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 V we have, >P 16 STATICS. P : P'=DC : CA, and similarly from triangle BCD, P', :Pj=BC:DC. Compounding these ratios, we have, since P'—P',, P : P,=BC : CA, or P.AC=P,.BC. 2°. When the forces act in opposite directions. Let P and P x (Pj>P) be two parallel forces, acting at A and B in opposite directions, and apply to these points, aa before, the equal and opposite forces P' and P'j. The re- sultant P" of P and P' will meet the resultant P", of P, and P'j in some point D, since (Art. 16) P", makes a less angle with P, than P" does with P. We shall then have, as before, the four forces P, P,, P', and P'j, acting at D, of which P' and P'j are equal and opposite, while P and Pj are opposite and unequal, and have a resultant R=P, — P, act- ing at any point in its direction, as C in AB produced. To determine the point C, we have (Art. 22), from the tri- angles ACD and BCD, as in Case 1°, P : P' -DC : CA and P'j :Pj=BC : CD ; compounding P : Pj=BC : CA, or P.AC = Pj.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 P,.CB are the moments of the forces about the point C. 38. Cor. If to the point C we apply a force R,, equal and opposite to R, the forces P, Pj, and Rj 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 transla- tion. And if we draw through C a line perpendicular to the PARALLEL FORCES. 17 direction of the forces, calling the parts intercepted by P and Pj respectively p and p x ; from the similar triangles, thus formed, we should have BC:CA=^, :p. .: PiP^p, :p, or P. j p=P 1 p 1 . Hence the moments of P and P a , which measure the tendency to rotation in opposite directions, being equal, there can be no motion of rotation. 39. Prop. To determine the point of application of the re- sultant 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 1 .(AB-AC), or AC= F ?^-.AB; and P.AC=P 1 .BC=P(AB-BC), or BC= F J^-.AB. 2°. When the forces act in opposite directions, we have P.AC^Pj.BC^P^AC-AB), or AC=p^-p.AB; P and similarly, BC=p p.AB. 40. Cor. 1. When the forces act in opposite directions, the resultant lies without the components and on the side of the larger. 41. Cor. 2. When the forces are equal and opposite, we have R^-P^O, or there will be no motion of translation. Also, AC=p^p.AB=^-.AB=x. B *<\ 18 STATICS. Hence an equilibrium can not be produced except by the ap« plication of an infinitely small force at a point whose distance is infinite ; that is, two equal and parallel forces, acting in op- posite directions, can have no single resultant. 42. Def. Such forces are called a statical couple. Their ef- fect is, tendency to rotatory motion only, and all tendency to rotatory motion can be referred to forces forming such couples. 43. Prop. To find the resultant of any number of parallel forces acting at any points in the same plane. P2 1°. Let the parallel forces P t , P 2 , P 3 . . . P„ all act in the same direction. Draw any two rectangular axes OX, OY, and let x, y , be the co-ordinates of A, the point of application of F^ x 2 y 2 x n y n also, x' y' x" y" B, D, N, C, E, Po p 3 p„ R. x y R, R. Draw Ac and Cb parallel to OX. The triangles ACc and CB6 are similar, and give AC : BC=Cc : B&. By Art. 36, R^P^P,, PARALLEL FORCES. 19 and P, : P 2 =BC : AC=B& : Cc, or P 1 .Cc=P a .B6, P 1 .(CH-AF)=P .(BG-CH), (P, +P 2 )CH=P 1 AF+P 2 .BG ; Compounding R, with P 3 , we shall find the second result- ant, R a =R 1 +P, = P 1 +P a +P„ and R 2 .EL=R 1 .CH+P 3 .DK, R 2 .EL=P 1 AF+P 2 .BG+P 3 DK; ™ R 2 y , =Piy 1 +P 2 y 2 +P 3 y 3 - By continuing the same process, we should find, ultimately, R=P 1 +P a +P 3 + P n (a), Ry=P iyi +P 2 y 2 +P 3 y 3 + .... ?„y n (c). By drawing lines from A, C, B, E, D; &c, parallel to OX we may find, similarly, Ri=P 1 x,+P 2 a; 2 +P 3 a: 3 +, &c +V n x n (b). * 2°. When some of the forces act in opposite directions. To these the negative sign must be prefixed. If Rj, the resultant of y Pj and P 2 , as found in the previous case, be com- pounded with P 3 acting at D, in a direction oppo- site to that of Pj and P 2 , by joining CD and pro- ducing it {Art. 40), in the direction of the greater force, say R,, we have R 2 , the second resultant. R 2 =Rj— P 3 , and E being its point of application, R,.EC -P 3 .ED. Drawing Ce, Ec/, parallel to OX, and meeting DK in d and LE produced in e, the triangles ECe, EDc? are simi- lar ; and EC : DE=Ee : Drf. .-. R 1 .Ee=P 3 .Drf; or R 1 (CH-EL)=P 3 .(DK-EL), w STATICS. or (R 1 -P 3 ).EL=R 1 .CH-P 3 .DK 5 or R 2 .EL=P 1 .AF+P 2 BG-P 3 .DK, and so on for any other forces acting in directions opposite to P, and P 2 . 3°. When the point of application of either force lies in any other angle than YOX, to its co-ordinates must be given their appropriate sign. The ordinate y 2 of B, the point of application of P 2 , will be negative. Draw Aa, Cb parallel to OX. Y Pi TB P 1 .CA=P 2 .CB. By sim. triangles, P!.Ca=P 2 .B&; or P 1 (AF-CH)=P 2 (CH+GB), or (P 1 +P 2 ).CH=P 1 .AF-P 2 .BG, or R^P^-P^. . . 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 2 as the sign of summation, P to represent either force, and x, y its co-ordinates ; thus, R.=s(P) (O, R..r = 2(P.x) (&'), R.y=2(P.y) (C). When, therefore, the magnitudes of the forces and Jie co-or PARALLEL FORCES. 21 dinates of their points of application are given, (a) will give the magnitude of the resultant, and (b) and (c) the co-ordinates x, y of its point of application. 44. Def. The point whose co-ordinates are x, y is called the center of parallel forces. Its position depends on the magni- tude of the forces and the co-ordinates of their points of appli- cation, but is independent of their common direction ; for, by turning the forces around their respective points of applica- tion, at the same time preserving their parallelism, it will be seen that the position of the point E is not thereby affected. 45. Prop. The moment of the resultant of any number of par- allel forces, acting in the same plane, is equal to the algebraic <*um of the moments of the components. Since the origin of the co-ordinate axes and their direction are arbitrary {see figure 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 Pj^, P 2 a: 3 , &c, will represent the moments of P,, P 2 respectively, and Rx the mo- ment 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 P,^, is the moment of the force P,, in reference to the plane passing through OY perpendicular to OX, and P 1 y l is the moment of the force P,, with reference to the plane through OX perpendicular to OY. 47. Prop. To determine the conditions of equilibrium of any number 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 appli- cation, {a'), (&'), (c') will still hold. Then, 1°. We must obviously have R=0 in {a'), or P 1 +P 2 +P 3 + , &c. = 0. 2°. If this value of R be substituted in (6'), we have __ S.Pa _S.Pa; X ~~BT~ * 22 STATICS. Therefore x will be infinite unless 2.Rr=0. But when x is in. finite, we have a couple {Art. 41), and, consequently, there can be no equilibrium. Hence we must also have 2.P#=0, or P 1 x 1 +P 2 .x 2 +P 3 a;3+, &c.=0. And in the same manner it will appear that 2.P1/--O in case of an equilibrium. But 2.P.r 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, 1°. The sum of the forces must be equal to zero. 2°. The sum of their moments, with reference to two planes parallel to their directions, ?nust each be equal to zero. 48. Cor. 1. If R=0, but S.Yx and'S.Py are not equal to zero, there will be no motion of translation, but simply a mo- tion of rotation. For if Rj be the resultant of all the positive forces, and x , the abscissa of its point of application, R 2 the resultant of all the negative forces, and x 2 its abscissa, then Rj-z:,— R 2 i 2 ^0; or, since R,=R 2 , R l (x 1 — £ 2 );>0. But Rj being finite, x 1 — x 2 must also be finite. That is, the points of application of Rj and R 2 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 a;. 49. Cor. 2. If the equilibrium subsist, one force, as P x , must be equal and opposite to the resultant of all the others. 50. Cor. 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 PARALLEL FORCES. 23 and B 6 feet distant. Find the magnitude of their resultant, and its point of application G. Ans. R=18, AC=4i BC=lf. 2. Find the resultant, and its point of application, when the forces in the last question act in opposite directions. Ans. R=8, AC=9f, BC=3f. 3. If two parallel forces, P and Q, act in the same direction at A and B, and make an angle d with AB, find the moment of each about the point of application of their resultant. P.Q „ Ans. p—rrAB sin. d. 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=2.P=l + 2+3+4+5=15 lbs. And, taking the extremity of the line for the origin of co-or- dinates, R£=15.£=2.Pz=lXl+2X2+3X 3+4X4+5X5=55. .-. 5=3| feet. 5. Let the weights 4, —7, 8, and —3 lbs. act perpendicu- larly 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 equi- librium, 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. Def. The arm of a couple is the perpendicular distance between the directions of the forces. 54. Def. The moment of a couple is the product of the arm by one of the forces. 55. Prop. A couple may be turned round in any manner in its own plane without altering its statical effect. Let PjABP 2 be the original couple. Suppose the arm AB turned around any point in it to the posi- tion ah. Apply the equal and opposite forces P 3 and P 4 per- pendicularly to ab at a, and similarly P 5 and P 6 at b, and let each be equal to Pj or P 2 . These forces, being in equilibrium, will not disturb the system. The two equal forces P, at A, and P 4 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 4 . Similarly, the resultant of P 2 and P 6 at D will be an equal force in the direction CD. These forces, being equal and opposite, may be removed; that is, we may remove from the system the forces P,, P 2 , P 4 , and P 6 , and we have remaining the forces P 3 and P 5 at a COUPLES. 25 and b, forming the couple P 3 a&P 5 , 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 chang- ing its statical effect. Let the arm AB of the couple Pj ABP 2 be removed in its own plane to the parallel po- sition ab, and let the forces P 3 , P 4 , P s , P 6 , each equal to the original forces, be applied perpendicularly to «&, at the extremities a and b, in oppo- site pairs. Join Kb and aB. These lines will bisect each other in C. Pj at A and P s at b will have a resultant 2P, at C, parallel to the original direction. Similarly, P 2 at B and P 4 at a have a resultant = 2P 2 at C, opposite to the former; these will consequently balance each other, and may therefore be removed, or the forces P 19 P 2 , P 4 , P s may be removed, &nd we have remaining the couple P 3 «&P G , equivalent to the original couple, removed parallel to itself in its own plane. 57. Prop. A couple may be removed into any other plane, par illel to the original one, without altering its statical effect, the parallelism 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 P a or P 2 . Join Kb and «B. These lines will bisect each other in C. The forces P, and P 6 will have a resultant —2? l at C, and P 2 and P 4 a result- ant = 2P 2 at C. These resultants will be equal and opposite and may therefore be removed without disturbing the statica. effect of the system. We have then remaining the couple 26 STATICS P 3 a&P s , equivalent to the original couple, and transferred to the plane RS. 58. Prop. All statical couples are equivalent to each oilier whose planes are parallel and moments equal. Let PjABP, 0? E Q4 AP2 Q3 y and QjDEQ, 2 be any two couples whose planes are parallel and moments equal. If they I> are not in the same plane, by the foregoing propo- Yqj^ sitions let the latter be * transferred to the plane of the former, and moved in that plane until the extremities D and A of their arms coincide, and DE take the position Pl=P'+Ql AC. Apply at C, perpendicularly to AC, two equal and op- posite forces Q 3 and Q 4 , each equal to Q K Resolve P, into two forces F and Q lf so that P^P'+Q,, or P 1 -Q 1 =P / . Then, by hypothesis, COUPLES. 27 P 1 .AB=Q 1 .DE=Q 1 .AC=Q 1 .AB.+Q 1 JBC=Q r AB+Q 3 .BC and (P 1 -Q 1 )AB=F.AB=Q 3 .BC, or the resultant of P' and Q 3 (Art. 36) passes through B. It is also equal and opposite to P 2 . Hence the forces P', Q, 3 , and P 2 may be removed, and there will remain the couple Q,ACQ 4 , equivalent to P X ABP 2 . 59. Prop. Any statical couple may be changed into another, which shall be equivalent and have an arm of given length. Let PjPj be the moment of any couple, of which Pj is one of the forces and p , its arm, and let p be the given arm. Find a fourth proportional P'j to p, p lf and P, ; or take p : p x = P, : P',, which gives Y'\p=Y 1 p 1 . Hence, by Art. 58, their moments being equal, and in the same plane, the couples are equivalent. 60. Def. The aris of a couple is a line perpendicular to the plane of the couple. If the length of the axis be taken proportional to the mo- ment 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. 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. 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 Pj, P 2 , P 3 , &c, be the forces ; p lt p 2 , p 3 , &c, their arms respectively. The couples may all be removed into one 28 t a r i c s. Pi 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- main the same. Let p be the com- mon arm AB ; F,i F 2 , F 3 , &c, the forces; so that V\p=Y^p l , P' 2 p= | pi P 2 P 2 » ?'zP=?zPv &c - Now the forces P'jjP^jFg at A are equiva- Y Ej lent to a force \(£ F 1 +F 2 +F 3 +, &c.,=- L? - L + 2 +^ J? +, &c. Vx p p P And the forces at B are equal to the same sum. We hav8 then for the moment of the resultant couple, (F , +F a +F 3 +, &c.) AB= (F x +F a +F , +)p, =P I p 1 +P 3 p 2 +P 3 p 3 +,&c. If either of the original couples, as P 3 p 3 ,tend to pro- p 3 duce motion in the opposite A 'fl direction, its sign will be neg- ative. The sum of the forces will be P'j+F.-F.+.&a, and the moment of the result- ant couple (F 1 +F 2 -F 3 +,&cO J p=P^ 1 +P 2 jo 2 -P 3 /'i+,&c. 64. Cor. 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 find the resultant of two couples in different 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, and let the arms be made coincident and of the COUPLES. 2, = v'P.AB 2 +Q 2 .AB 2 +2P.ABxQ.ABcos.0: (since OK=P.AB, &c), = v / OK 2 +OL s +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. 80 STATICS. If L and M represent the axes or moments of the compo- nent couples, and G the axis or moment of their resultant, then G 2 =L 2 +M 2 +2LM cos. 6. 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 parallelo- piped 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, G 2 =L*+M a +N\ 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 an- gles to each other. 68. Prop. To find the magnitude and direction of the result- ant of any number of concurring forces in the same plane. Let Pj, P 2 , P 3 , &c. . . P„ be the n forces, and let the point at which they act be taken for the origin of co-ordinates. Let P, make the angle a, with OX, and /3 X with OY. P 2 " a 2 " a P 3 « a 3 " 03 « &c, &c, &c, P„ " a n " A, If OA represent the force P^and the parallelogram OMAN be completed, OM will be the component of P, in OX, and ON that in OY. And cos. a j, Y 1= P, cos. (3 t , cos. a 2 , Y 2 =P 2 cos. /3 a , cos. a s , Y 3 =P 3 cos. p 3 , &c, &c, &c, ;os. a„, Y„=P„ cos. |3 B . 32 STATICS. OM^X^P, cos. a lt ON=Y 1 =P 1 cos. 0, Pursuing the same course with the forces P 2 , P 3 , &c, and calling the resolved parts of the forces respectively in OX, X 2 , X 3 , X 4 , &c. . . . X„, and those in OY Y 2 , Y 3 , Y 4 , &c Y n , we shall obtain X 1= P-, X 2 = P 2 X 3 = P>) &c, X„=P n c But the components in OX are equivalent to a single force, =X 1 +X 1 +X,+, &c X„=2.X, and those in OY to a single force, =Y 1 +Y 2 +Y 3 H-. &c. . . . Y„=2.Y, the Greek letter 2 being used as the sign of summation. Hence we have 2.X=S.P cos. a, (1) 2.Y=S.P cos. (3. (2) Now if R be the resultant required, and 6 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. .-. R cos. 0=2.X, (3) R sin. 0=2.Y. (4) 2.Y Hence Tan. 6= -p^-. (5) Also, R 2 . cos. 2 0+^sin. 2 0=R 2 =(2.X) 2 +(Z.Y) 2 ; or R=V(2.X) 2 +(2.Y) 2 . (6) Equations (1) and (2) give the values of 2.X and 2.Y, by which we obtain from (5) and R from (6) ; or R may be ob- tained, numerically, more readily from the equation R cos. 0=2.X. For R=-^r=2.Xsec.0. (7) cos. 69. Schol. It is readilv seen that the signs of the compo- ANALYTICAL STATICS. 33 nents along either axis are involved in the trigonometrical ex- pressions of their values, (1) and (2). Assume the directions OX and OY positive, and OX', OY' negative. The components Ox ,, Oy 1 of Pj are positive ; their values, Pj cos. «,, Pj cos. (3 lt will also be positive, since the cosines of angles in the first quadrant are positive. The com- ponents of P 2 , which lies in the angle X'OY, are Ox 2 , Oy 2 , the former negative and the latter positive. The value of 0.r 2 = P 2 cos. a 2 is negative, since a 2 , reckoned from OX around to the left, will be in the second quadrant ; and that of Oj/ 2 =P 2 cos. (3 3 positive, since j3 2 , 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 Ox 3 , Oy 3 of P 3 , in the angle X'OY', are both negative, and their values P 3 cos. a 3 , P 3 cos. |3 3 are neg- ative, because the angles a 3 and (3 3 are both in the third quad- rant. The components Ox 4 , Oy i of P 4 , which lies in the angle Y'OX, are, the former positive and the latter negative. The value of 0.r 4 =P 4 cos. a 4 is positive, because a A is in the fourth quadrant, or, if taken negatively, is less than 90° ; that of Oy 4 = P 4 cos. /3 4 is negative, because /3 4 is in the second quadrant. By the above mode of reckoning the angles, we al- ways find a+/3=90°, or one the complement of the other. 70. Prop. Required the conditions of equilibrium of any number of concurring forces in the same plane. Since the forces are in equilibrium, their resultant must be equal to zero, or R=0. This gives, by (0), (2.X) S + (2.Y) 2 =0. 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 STATICS. Hence 2.X=0, 2.Y=0, (8) are the two necessary and sufficient equations of equilibrium 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. 71. Prop. Required the expressions for the resultant force and resultant couple of any number of forces acting at different points in the same plane. Let P,, P 2 , &c. . . . P n be any number offerees in the same plane. Take the co-ordinate axes OX, OY in the plane of the forces, and let a l , a 2 , a 3 . . . a n be the angles the forces make respectively with OX, and (8,, j3 2 , |3 3 . . . j3 n the angles they make with OY. Let s,,!/, be the co-ordinates OM , , MjAj of A,, the point of application of the force P x , and x 2 . y„, &c. . . . x n , y n be the co-ordinates of the points of application of the others respect- ively. Resolving P, in the directions of OX and OY, we shall Y^P, cos. j3 lf have for the components Xj— P, cos. a,, at the point Aj. Now apply at O, 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 lf we have X l at A„ or N, and —X, at O, which together form a couple, and Xj at 0. That is, for the force X, at A,, we may substitute X, at O, and the couple X 1 .ON,=X 1 y 1 . In like manner, applying at O, in OY, two forces, opposite and each equal to Y x , we should have Y, at A, equivalent to Y, at O, and the couple -Y.OM 1 = -Y.a: 1 . ANALYTICAL STATICS. 35 This last couple will be negative, since it tends to produce ro- 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,+X 2 +X 3 +,&c X„=2.X, (9) and at O, in OY, Y 1 +Y 2 +Y 3 +,&c Y„==2.Y, (10) and the couples X 1 2/ 1 +X 2 y 2 +X 3 3/ 3 X„y a =S.Xy, (11) —Y l x l —Y 2 x 2 —Y 3 x 3 .... -Y n x,=2. — 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. 03) 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=2.Xy-2.Yx=X.(Ky-Yx). (13) Now if R be the resultant force acting at 0, and 6 the angle it makes with OX, we have, as before, R cos. 0=2.X, R sin. 0=2.Y. s.Y Tan. 0= S , (14) and R 2 =(2.X)'+(2.Y) 2 . (15) 72. Schol. 1. The equations (13), (14), and (15) determine the magnitude of G and the magnitude and direction of R. To construct these results, draw through the origin O the line OR, making the angle with OX, to represent R. Then put G= £.(Xy— Yx)=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 in an 36 STATICS. opposite direction to the resultant force ; draw OA perpendic ular to OR and equal to ;•, 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 6 with OX. 73. Schol. 2. To find the equation of this final resultant. It will be of the form y=ax+b. „ 2.Y But a= tan. 6=- and 6=OB= 2.X' OA r G G cos. d cos. d R cos. d 2.X' Therefore, by substitution, we have for the equation 2 -Y G y = s3* + s3 (16) - 74. Prop. To determine the conditions of equilibrium of any number of forces acting at different points in the same plane. In order that there may be no motion of translation, we must have R=0, which gives, as before, 2.X=0, (17) and £.Y=0. (18) And, in order that there may be no motion of rotation, we must also have G=l(Ky-Yx)=0. (19) Equations (17), (18), (19) are the three necessary and suffi- cient conditions of equilibrium. 75. Cor. When there is a fixed point in the system, if this point be taken for the origin, its resistance will destroy the ef- fect of the resultant force R, and the sole condition of equilib- rium will then be G=0, or l.(Xy-Yx)=0; or there must be no tendency to rotation around the. fixed point. ANALYTICAL STATICS. 37 EQUILIBRIUM 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 re- solved 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, be- ing unopposed, would cause the point to move. Hence, 77. Prop. To determine the conditions of equilibrium of a point, retained on a plane curve or line, by forces acting in 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 2.X, 2.Y be the sums of the components of all the other forces resolved parallel to each axis respectively, and R their result- ant. 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+2.X=0, (a) N sin. a+2.Y=0; (&) or, transposing, squaring, adding, and reducing, N= v'(2.X) 2 +(S.Y) 2 =R, or the reaction is equal to the resultant of all the other forces. Now let i be the inclination to the axis of x, of the tangent to the curve through the point, then a=90° + z. b. T and .'. cos. a=cos. (90°+i) = — sin. i, sin. a=cos.i. Substituting these values of cos. a and sin. a in (a) and (b), and dividing, we get 38 STATICS. 2 X Tan. 1=-™™ or 2.X+2.Y tan. i=0. (20'. But the differential expression for the tangent of the angl dxi which the tangent makes with the axis of a: is -£-. Therefore. & dx substituting and reducing, 2.X.dz+2.Y.di/=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 gen- eral, be necessary, and -~ must be deduced from the equation of the given curve. VIRTUAL VELOCITIES. 78. Def. If any forces P x and P 2 act at the point A, and this point be displaced through an in- 1 definitely small space Aa, and the perpendiculars ap x , ap 2 be drawn from a on the directions of the forces, then Aj9j and Ap 2 are called the vir- tual velocities of the forces Pj and P 2 ; Ap lf measured in the direction of the force P 15 is positive, and Ap 2 , measured in the direction of P 2 produced, is negative. 79. The principle of virtual velocities is thus enunciated : If any number of forces be in equilibrium at one or more points of a ~igid body, then if this body receive an indefinitely small displacement, the algebraic sum of the products of each force into its virtual velocity is equal to zero. 80. Prof. To prove the principle of virtual velocities for concurring forces in one plane. Let A be the point at which the forces P,, P 2 , P 3 , &c P„ act ; a x , a 2 , a 3 , &c. . . . a n the angles they make respect- ively with OX, and j3,, j3 2 , (3 3 . . . (3 n the angles they make with OY. Let be the angle which the direction of the dis- placement Aa makes with OX ANALYTICAL STATICS. sy ■X Let »,, v a , v 3 , &c. . . . v n be the virtual velocities of the forces respectively. Then u 1 =Ajo 1 =Aacos.p 1 Aa=:Aa. cos. (a,— 0), =Aa(cos.a 1 cos. 6+s'm.a 1 sin.0), and ¥ l v l =l > l .Aa(cos.a 1 cos.0-f sin.aj sin.0), = A«(cos. 0.P , cos. a , +sin. 0-Pj sin. a ,). In like manner, for P 2 we should have P 2 u 2 = Aa(cos. 0.P 2 cos. a 2 +sin. 0.P 2 sin. a 2 ), and so for the other forces ; and taking their sum, we should get, remembering that sin. «=cos. 0, 2.Pu=P 1 v 1 +P 2 v 2 +P 3 i> 3 +, &c V n v n , = Aa[cos. (P, cos. a, +P 2 cos. a 2 +, &c P„ cos. a„) + sin. (Pj cos. 13, +P 2 cos. /3 2 +, &c P„ cos. ft,)]. But when there is an equilibrium at a point (8), Pj cos.a,+P 2 cos.a 2 +P 3 cos.a 3 +,&c P„cos.a„— 2.X=0. and P, cos. |3, +P 2 cos./3 2 +P 3 cos./3 3 +,&c. . . . P„ cos.ft =2.Y=0. Hence 2.Pv=0, or the principle is true when the forces all act at a point. 81. Prop. To prove the principle of virtual velocities for forces acting at different 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 STATICS. Let A,, A a , A 3 , &c. . . . A„ be the particles to -which the forces Pj, P 2 , P 3 , &c. . . . P n are applied, v x , v a , v 3 . . . v„ the virtual velocities of the forces respectively. Let r a a a be the action of the particle Aj upon the particle A 2 , ra 2 a\ " reaction of " A 2 " " A,, ra]a t " action of " A~ " " A 3 , ra^ " reaction of " A 3 " " A,. &c, &c, &c. Let Va x a 2 , va^, Va^, Va^, &c, &c, be the corresponding virtual velocities. Then Va^^va^, t7tf 1 o 3 =«a s o i , &c., &c, from the nature of action and reaction. Also, ra x a ;t ——ra i a 1? r ai a a =—ra 3 a x , &c, &c. For, let A, R ' A1P1 A2P2 J82 M ' ■ M- . ax a.2 and A 2 be the particles displaced to a x and a 2 . Draw the perpendiculars a x p l , a 2 p 2 ' Then, if the line a x a 2 is par- allel to AjA 2 , it is obvious that A^p^=A 2 p 2 . But A,p, is the virtual velocity of R l5 and A 2 j9 2 of R 2 , and they have op- posite signs. .*. Va y a=—Va 2 a x . If a l a 2 is not parallel to A X A 2 , let them meet when pro- C— OT= ~ Aijh Al [ u - «» a2 duced in some point C. Since the displacements are indefin- itely small, the perpendiculars a J p 1 ,a 2 p 2 coincide with cir- cular arcs whose center is C, and Ca l =Cp 1 , Ca 2 =Cp 2 . But A 1 ^ 1 =Cp 1 -CA 1 =Ca 1 -CA 1> and A„p a =Cp a —CA a =Ca a —CA = (Ca 1 +a 1 a )-(CA 1 + A.A,), =Ga l —CA l . .: A lPl = A 2 p 2 , Or Va l a u = ~" Va^. Let the sum of the products of all the forces P,, P 2 , &c . into their virtual velocities, acting on the particle ANALYTICAL STATICS. 41 A j be 2.(P ai .v a ), those on A 2 be S.(Pa a .ra a ), those on A 3 be 2.(Pa 3 .Va 3 ), &C, &C, those on A„ be I,.(Pa H .va t ). 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, = 2. (Pa^.Va) + 1\a 2 .V ai a 2 + 1'a^.Va^ + , &C, 0=2.(Pa a .Va^+ra ;i a 1 .Va a a l +ra a a 3 .Va 3 a 3 + i &C, = 2.(Pa 3 .Va a )+7'a 3 a l .Va 3 a l + ra a a a .Va a a a + , &C, &c, &c, &c, = 2.(P« > .Va n ) +ra Oj.r^fflj + ra a 2 .Vo n a 2 + , &C. In taking the sum of the products for all the particles, the products of the reactions into their virtual velocities will dis- appear, being in pairs, equal in magnitude with contrary signs ; therefore we have 2.(Ya 1 .Va 1 )+2.{¥a 2 .Va 2 )+2.(Pa 3 .Va 3 )+, &C 2.(P O| ,.V O J = : or, generally, when there is an equilibrium, S.(Pu)=0. 82. PRor. Conversely. If the sum of the products of the forces into their virtual velocities be equal to zero, or 2.(P.u.) =0, then there will be an equilibrium. For if the forces are not in equilibrium, they will be equiva- lent either to a single force 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 equi- librium ; 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, S.(P.»)+R.«=0. But by hypothesis, 2.(P.v) = 0. .'• Ru=0, which, being true foi all small displacements of the bodv, we must have R = 0, or 42 STATICS. the body was in equilibrium from the action of the original forces. In the second case, if the forces were equal to a resultant 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, s.(P.v)+Q?+Qy=a But 2.(P.u)=0. .-. Qq+Q'q'=0, 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 ever ij particle of matter 13 subject to a force which attracts it in a direction perpendicu- lar 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 con- sidered together, we may, without sensible error, regard their directions as parallel. This force is cr.lled gravity. 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 from the center. But for any points in the same sys- tem, or any bodies nearly in the same place, this variation of intensity, as well as difference of direction, may be neglected without error. Hence, 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. 86. Def. The weight of a body is the resultant of all the ef- forts 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 STATICS. intensity at another place where it is assumed as unity, \vc 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, D=y; (23) in which D, M, and V represent the number of units of each kind. 90. Cor. Since W=gM, and M=VD, we have W=gYD. (24) 91. Prop. The masses of two bodies of the same density are in the direct ratio of their volumes. If M, V, and D be the mass, volume, and density of one body, and m, v, and d the same of the other, by (23), M=VD and m=vd; v M : m=VD : vd ; (25) or, since D=d, M : m=Y : v. 92. Prop. The masses being equal, the densities are inversely as the volumes. Since M : m=VT> : vd; if M=m, YD=vd; or D : d=v : V. (26) 93. Prop. The volumes being equal, the masses are directly as the densities. Making, in (25), V=t>, we have M : m=D : d. (27) 94. Def. The center of gravity of a body is the center of the parallel forces of gravity on each of its component particles. 95. Cor. 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. CENTER OF GRAVITY. 45 1°. The resultant of all the vertical efforts of gravity ou 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 body. 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 propor- tional to its mass, and passing vertically through its center of gravity. Hence we have only to consider a system of heavy 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 plane. Thus, let m, m' be two symmetrical par- ticles, so that the line mm' may be bisected in b by the plane aa' . Letting fall the perpen- diculars ma, m'a', the equality of the triangles mab, m'a'b gives ma=m'a'. Hence the par- 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 from it. 97. Prop. The center of gravity of every homogeneous body of uniform density, symmetrical with respect to a plane, is situ- ated in that plane. For any two particles, symmetrically placed, will be at the 46 STATICS. same distance from the plane, and their moments will be eqna 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 symmetrical with respect to an axis when it is symmetrical with respect to two planes pass- ing through that .axis. 99. Prop. The center of gravity of a homogeneous body, sym- metrical with 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 to,, w 2 , w z , &c, act at their respective centers of gravity, vertically downward, and therefore con- stitute a system of parallel for- ces. w x and w 2 have a result- ant =w 1 +iv 2 acting at some point a, such that iv 1 .Aa—iv 2 . Ba. The distance AB being given, the distanceA a is de- termined by taking {Art. 39) w„ w, +u)„ : w =AB : Aa=AB. -. — . Compounding the weight w l +w 2 at a with another weight w 3 acting at C, they will have a resultant w 1 +w 2 -\-w 3 acting at some point b, such that (mj, +w 2 ).ab=w 3 .Cb. W3 CENTER OF GRAVITY. 47 First determining aC from 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 iv, +w„ +U3, : w~=Ca : ab=Ca. ; • . 1233 w l -\-w 2 +w 3 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 of particles in the same plane whose positions are giver by their co-ordinates. Since the weights of the particles constitute a system of par- allel forces, let P,, P 2 , P 3 , &c, represent the weights of the particles which may be supposed collected in their respective centers of gravity, x , , y x ,x 2 ,y 2 , &c, their co-ordinates. We shall have for the co-ordinates x, y of the center of parallel for- ces (Art. 44), or center of gravity of the whole body (Art. 94), _ = P 1 x 1 +P 2 x +P 3 .r 3 +,&c. ^ S.Px ^ X P 1+ P 2 +P 3 +,&c. S.P I -_ P,y,+P 8 y a +P,y 3 +.&c- = s-Py [ { } y ~ P,+P 2 +P 3 +,&c. S.P'J 104. Cor. 1. If the particles all lie in a straight line, this line maybe taken for the axis of x, and i/, =0, y 2 =0, &c. .*. 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) will be proportional to the volumes ; and if «,, v 2 , u 3 , &c, denote the volumes of the particles, and V the whole volume, we have X= ~ ^^~ "^ I _ 13,y i +13 2 y 2 +13 3 l/ 3 +, &C. 2.13?/ r y= i t) Hence the sum of all the particles, or the whole volume, multi~ plied by the distance of its center of gravity from a flane, is 48 STATICS. equal to the sum of each particle into the distance of its cenle* of gravity from the plane. 106. Cor. 3. If the center of gravity of the whole volume be given, and the center of gravity of one of its parts, the cen- ter of gravity of the other is readily obtained. Let V equal the whole volume, and v lf v 2 the volumes of the two parts, x, y, as,, y lt x 2 , y 2 , the co-ordinates of their re- spective centers of gravity. Then _ !),.£, +V 2 X, , and y= »,yi+« 9 y 3 y ' "~" y y But, by hypothesis, V, v l and x,y,x l ,y l are given. Therefore, u„=V — v., and Vx- ■v.x -, and y. Vy-ViVi (30) A v 2 "" i> £ 107. We shall now proceed to apply the foregoing princi- ples to specific cases. Ex. 1. To find the center of gravity of a uniform physical straight line. If AB be the uniform straight line, and C its middle point, C will be its center of gravity ; 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 paral- lel 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 pass- ing through C. The center of gravity, therefore {Art. 97), is in jo this plane, and as it is also in the line AB, it must be at their intersection C. 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 inconsider- able. Bisect AB in D, and AC in E. Join C, D, and B, E intersecting in A G. G will be the center of gravity of the plate. CENTER OF GRAVITY. 49 K 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. For the same reason, it will be in the line BE, and will therefore be 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=iDC. In the same way it may be shown that EG=^EB. Hence the center of gravity of a triangle is one third the distance from the middle of either side to the opposite vertex. Ex. 3. To find the center of gravity of a parallelogram of uniform density and thickness. Bisect the sides AB and DC in H and K ; also the sides AD and BC in E and F. The plane passing through H and K will divide the parallelogram sym- metrically, since it will bisect all lines parallel to AB. The center of gravity will lie in this plane, and 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 diagonal bi- sects all lines drawn parallel to the other, it will be at the inter- section of the diagonals. Ex. 4. To find the center of gravity of a thin, polygonal plate, of uniform density and thickness. Let ABCDEF be the poly- gon. Draw the lines AC, AD, AE, dividing it into triangles. D H 50 STATICS. When the polygon is given, these triangles will be known, and their centers of gravity, g x ,g 2 , g 3 , g t , 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), gi+g 2 -g a =gig a '-gi G '^g,g 2 'Z~tr' 6 1 '62 This determines the point G', the center of gravity of the portion ABCD. In the same manner, we may find the point of application G", of the resultant of g t +g 2 acting at G', and g 3 acting at g 3 , and so on; the last point so determined will be the center of gravity G of the polygon. Ex. 5. To find the center of gravity of a triangular pyramid of 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 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 BD will also divide the pyr- amid symmetrically ; and since the center of gravity of the pyramid will be in both these planes, it must be in their inter- section D/. But the point/ is the center of gravity of the face ABC {Ex 2). Hence the center of gravity of the pyramid CENTER OF GRAVITY. 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 D/l To find fG, join fe, which will be parallel to AD, since it divides the sides ED and EA of the triangle AED proportion- ally. Now the similar triangles fGe and AGD, with/Ee and AED, give fG : GB=fe : AD=Ee : ED=1 : 3. .-. 3/G=GD. Adding fG to both members, 4/G=/G+GD= /D, or /G=i/D. Hence, the center of gravity of a triangular pyramid is one fourth the distance from the center of gravity of one face to the 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 will pass through the cen- ter of gravity of every sec- tion parallel to the base, and the center of gravity of the whole pyramid will be in AG. Join GB, GC, GD, GE, b< and GF, and conceive planes to pass through A and each of these lines, thus dividing the whole pyramid into as many triangular pyramids as the base has 52 STATICS. sides. The centers of gravity of these pyramids will be at one fourth the distances, respectively, from the centers of grav- ity of their triangular bases to the common vertex A. These distances being thus divided proportionally, the points of di- vision 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 inter- section g. But the plane divides all lines drawn from the ver- tex A to the base proportionally ; therefore, the center of grav- ity g of the whole pyramid is one fourth the distance from 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 dis- tance from the center of gravity of the base to the vertex. Ex. 1. To find the center of gravity of a frustum of a cone or pyramid cut off 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 frustum. Then (30) we have vx—v,x, in which x—^a, x x =\a'. Now the part of the cone or pyra- mid 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 : u,=a 3 : a'\ or v—v x : v=a s —a" : a 3 . a" and v a = v—v 1 =v[l -J, CENTER OF GRAVITY. 53 __ *fr- p fe)* , ' _ a q'-q'' , * X "~ ( a /3 \ a —a' 4 a 2 +aa'+a" ' 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 3a' 3 a—i- ,(a+a')(a i +a n )_a cf+aa'+a'* ~4~ ' 4(a*+aa'+a n )' O X Ex. 8. To find the center of gravity of the perimeter of a tri- angle in terms of the co-ordinates y of the angular points. In the triangle ABC, let a, b, c represent the sides respective- ly opposite to the angles A, B, C. Their centers of gravity will be each at the middle point of the side; as g lt g 2 , g 3 . Let x l y l be the co-ordinates of A referred to the origin O, x 2 y 2 " " " B, x 3 y 3 « "• » C. Then the co-ordinates of g z are ^(x^+x^, \(V 1+2/2)' g 2 » \{x 2 +x~ 3 ),\{y 2 +y 3 ), g 3 « ^(x 3 +x,),\{y 3 + yi ), and x, y, being the co-ordinates required, by (28), we have a(x 2 +x 3 )+b (x l +x 3 )+c(x 1 +x 2 ) 2(a+b+c) -__^ a{y 2 +y 2 )+b(y,+y 3 )-\-c{y,+y 2 ) y 2(a+b+c) Ex. 9. To find the co-ordinates of the center of gravity of a triangle. Let x x y^, x 2 y 2 , x 3 y 3 be the co-ordinates of the points A x=- 54 STATICS. O L or and N M B, C respectively. Draw AD bisecting BC in D, and take AG=|AD : G is the center of gravity of the triangle. The co- ordinates of D are ^(x 2 + a; 3 )»i(y a +y 3 ); and if £=ON, y=GN, be the co-ordinates of G, we have ON=OL+f(OM-OL), GN=AL+f (DM-AL) ; y=yi+%{Uy 2 +y 3 )-yA=Wji+y 2 +y 3 )' CONDITIONS OF EQUILIBRIUM OF BODIES FROM THE ACTION OF GRAVITY. 108. Prop. If a body have a fixed point in it, the condition of 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 force acting at g, in the direction Ag, will be resisted by the reaction of the fixed point A. If the center of gravity be at any other point g', then drawing the 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. a couple of which the forces are, the weight of the body at g', and the reaction of the fixed point A lf and the arm Am. Therefore (19), in order to equilibrium, S(Xy— Yx)=w.Am=0. .'. A?n=0, 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 unstable when, being disturbed, it tends to re move further from its original position ; and neutral when, after being disturbed, it still remains in equilibrium. 110. Prop. When the equilibrium of a body containing a fixed point is stable, the center of gravity is in the lowest posi- tion it can take ; when unstable, in the highest. Let A be a fixed point in the bodies M and N, g their ren- ter of gravity. The center of gravity can only move on the surface of a sphere whose center is A and radius Ag. When the body M is dis- turbed, its center of gravity g being re- moved to g', will rise through the versed sine of the arc gg', and when at g', the moment iv.Am will obviously tend to bring it back to its orig- inal position. The equilibrium is therefore stable, and the center of gravity the lowest possible. In the body N, the cen- ter 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, and the equilibrium was therefore unstable. 111. Cor. 1. The pressure on the point by which a body is suspended is clearly, in the case of equilibrium, equal to the weight of the body. 112. -Cor. 2. If a body is suspended from two points, the po- sition of equilibrium is that in which the center of gravity is in 56 STATICS. the vertical plane passing through the two points of suspem sion, since it is then the highest or low est possible. To determine the pres- sures on the fixed points A and B, let GC represent the weight of the body acting vertically at G. Resolve GC 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 po- sition 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 verti- cal 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, there will not be an equilibrium unless the vertical through the center of grav- ity G is in the same plane, or, which is the same thing, meets CENTEROPGRAVITY. 57 the line PQ. The pressures on P and Q may be determined by the theory of parallel forces, and P : Q : W=WQ : 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 tri- angle formed by joining the three points, and W the point where the vertical through the center of gravity meets it. We must re- solve the weight acting at W into three others parallel to it, acting at P, Q, and R. Join PW, QW, and RW, and produce them to the op- posite sides. Then, by the theory of parallel forces, P : W=WK : PK= triangle WRQ : triangle PRQ, Q : W=WM : QM= " WRP : " " PRQ, and R:W=WH:RH= " WPQ: " PRQ. \ P:Q:R : W=WRQ : 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 weight of the body. 117. Cou. 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 will turn over. 58 STATICS. 118. Prop. The stability of a body is measured by the excess of the shortest line that can be drawn from the center of 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 GH, since G must be elevated a distance equal to this dif- ference, in order to turn the body over the edge of the base at P. P H Cor. 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 arc, through which the center of gravity must move from rest to its highest point. For GP-GH=GP-BP=AP-BP=AB — versed sine of arc GA to radius PG. 119. Prop. If a body be placed on an inclined plane, it will descend when there is no resistance from friction. For the weight of the body, represented by GA, may be re- solved into two others, GB and BA,' one perpendicular to the plane, and the other parallel to it, of which GB, the one perpendicular to the plane, can alone be counter- acted by the plane. In this case, if the vertical GA fall within the base of the body, the body will slide ; if it fall without, it will slide also. 120. EXAMPLES. 1. If two right cones have the same base and their vertices 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. \ sum of their altitudes CENTER OF GRAVITY. 59 2. The center of gravity of a paraboloid being in the axis at a distance from the vertex equal to § of the axis ; fina the center of gravity of a frustum of a paraboloid from the base, a and b being the radii of the two ends, and m the parameter to the axis (30). a 6 -3a i b i +2b B AUS ' 3//i(a 4 -i 4 ) • 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. a'+a?b+ab*+b 3 Ans. 5 — t- — . a'+ab+b 2 4. The distance of the center of gravity of a hemisphere from its base being § the radius, find that of a hemispherical bowl whose internal radius is a and thickness c. 3 4a 3 + 6a"c+4ac"+c 3 AnS - ~ 6 ' 3a 2 +3ac+c 2 '" 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. \a. APPLICATION OF THE PRINCIPLES OF THE INTEGRAL CALCULUS TO THE DETERMINATION OF THE CENTER OF GRAVITY. 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 do. In this case formulas (29) will take the form fxdv _ fydv 7&T' y== JdV' in which x and y denote the distances of the center of gravity of dv from the co-ordinate axes. 122. Prop. Required the differential expressions for the co-or- dinates 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 %F=-Z£r> y=-7vAT' ( 31 > 60 STATICS. _ fxds _ fyds *=— > y=— ■ (32) If the arc is symmetrical with respect to the axis of x y the center of gravity will be in that axis {Art. 99), and y=0. _ fxds s s sufficient. 123. Prop. Required the differential expressions for the cu~ ordinates of the center of gravity of a plane area. Since the differential element of a plane area is dxdy, dv= dxdy. By substitution in (31), we have - Jfxdxdy _ = ffydxdy ffdxdy ' y ffdxdy ' Integrating in reference to y, we have ^_fxydx -JMte m fydx ' y fydx If the area is symmetrical with respect to the axis of x, the center of gravity is in that line {Art. 99), and y—0. _ fxydx ' ' x ~ fv dx is sufficient. 124. Prop. Required the differential expressions for the co- ordinates of the center of gravity of a surface of revolution around the axis of x. The center of gravity will obviously be in the axis of x, and therefore ^=0; and since, for a surface of revolution, dv= 2-nyds, the first of equations (31) become, - f x y ds (34) and this equation is sufficient. 125. Prop. Required the differential expressions for the co* ordinates of the center of gravity of a solid of revolution. In this case dv=-nifdx. Hence, from the first of equations (31), we have • CENTER OF GRAVITY. 61 _ fy^xdx ' r ~7pzp (35) which alone is sufficient. By proper substitutions for do in the fundairental equations (31), we may find expressions for the co-ordinates of the cen- ter of gravity for other forms of bodies. 126. Prop. The surface generated by the revolution of a curve around 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 2TT.y.s=2rrfyds. Now 2rry is the circumference of which y is the radius, and 2ny.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 2nfyds, which is the area of the surface generated by the revolution of the curve. Hence, &c, 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, from the second of equations (33), we have 2nyfydx = -nfy^dx. In this equation, fydx is the generating area, 2iry is the cir- cumference described by its center of gravity, and -nfifdx is the volume generated. Hence the truth of the proposition. These last two propositions comprise the theorem otGuldin, 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 dis- tance of the center of gravity from the axis, and the magnitude generated, any two being given, the other may be determined. 62 STATICS. 128. EXAMPLES. 1. Required the center of gravity of a circular arc. Let the axis of a: bisect the arc MAM in A, the origin being at the center of the circle, and let MAM' =25. From the x equation of the circle ?/ 2 =r 2 — a; 2 , we ob- . dy 2 x 2 tain -~= dy 2 dx 2 r 2 —x 2 rdx But ds=dx\/ 1-f dx 2 Vr-f fxds r P xdx ry Vr'-x* +-Vr 2 -x 2 +C=+^+C. s s ry When y—0, x=r, and — =r. .'. C=0 Hence *=CI=^=^ ; s 2s ' or, the distance of the center of gravity of a circular arc from the center of the circle is a fourth proportional to the arc, the radius, and the cord of the arc. If the arc be a semicircle, y=r, and s=^7rr 2r .'. CI=— =0.63662/-. 2. Required the center of gravity of a circular segment. Putting CP=a {Fig., Ex. 1), and taking the center for the origin, we have y= Vr 2 —x 2 . Hence (33), ^_f: xydx _ f:x{r 2 -x 2 )hxJt{r 2 -a 2 Y-JsW? X ~ f' a ydx~ MAP ~ MAP "MAP' __ ? V(2MP) 8 _ T y(chord) 3 """ ^--^MAM'P" segment * If the segment is a semicircle, _ T V (2r) 3 4r a:= V / =— = 0.42441r: T>7TV 3T ici. CENTER OF GRAVITY. G3 3. Required the center of gravity of the surface of a spher- ical segment. Taking the origin at C (Fig., Ex. 1), the center of the gen- dy 2 x 1 erating circle, we have x*+y 2 =r\ j^=—, and yds=rdx. (34) x-- dx 2 fyxds fxdx fyds fdx ' Integrating between x=r and x=a=CP. r—a Hence the center of gravity is at the middle of PA. 4. Required the Center of gravity of a spherical segment. Taking the origin at A, the vertex of the generating circle, we have, for its equation, y i =2ax—x i . .-. (35) x= Hence AG= fxy^dx f (2ax — z*) xdx fy'dx f(2ax—x~)dx ' %ax 3 — \x* 8 ax— 3a: 2 3 x 12a— 4x' If the segment is a hemisphere, x=a, and x— fa. 5. Required the surface of a hemisphere. By the centrobaryc method (Art. 126), we have The generatrix =^Ttr, the ordinate of its center of gravity 2r y= — (Art. 128, Ex. 1), and the circumference described by 77 2r the center of gravity =27t.— = 4r. Hence the surface =\irrAr =2nr\ 129. EXAMPLES ON THE PRECEDING CHAPTERS. Ex. 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 makino 64 STATICS. M C N 32 » ^^^\ B <1 \ f )V2' with each oth or the angle a. Since C is a fixed point, the only condition of equilibrium is, that the sum of the moments about C is zero (Art. 75). Let g z , g 2 be the centers of gravity of the beams, and £jC =a,g 2 C=b. Also, w,= weight of AC, acting at g-^and w 2 = weight of BC, acting at g 2 . Draw through C the line MCN horizontally, meeting the vertical directions in which w x and w 2 act, at M and N. By(19),u) 1 CM-MJ 3 CN=0. Let BCN=0. The determin- ation of 6 will fix the position of the compound beam. Since CM=g 1 C. cos. MCA and CN=g- 2 C. cos. BCN, we have w 1 .Cg l cos. ACM-w 2 .Cg 2 cos. BCN=0, 0. or a. cos. (180 — a+6) — w 2 .b. cos. 6- _ w n b+w.a. cos. a .-. Ian. 6=— - —. . w x a. sin. a Ex. 2. When a given weight W is hung from the end of one of the beams, A (Ex. 1), find 6 in case of equilibrium. iv n .b+(w i +2W)acos. a Ian. 0= — 11 —. ^r— - — ; . (w 1 +2W)asm. a Ex. 3. Two beams, as in Ex. 1, are suspended from one end B ; find the angle which the upper one makes with a horizontal line. (2iv x +w 2 )b— w x a. cos. a Tan. 6=- 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 -CG=Cm-2NB=fl. cos. (a-0)- 2b cos. 0. EXAMPLES ON THE PRECEDING CHAPTERS. 65 .*. w 2 b. cos. 6—w l a. cos. (a -6)+2w l b. cos. 0=0. Ex. 4. Two spheres of unequal radii, but of the same ma- terial, are placed in a hemispherical bowl ; find the position they take when in equilibrium. Since the reactions of the bowl upon the spheres are in the directions of the radii of the spheres through the points of contact, and since these radii produced pass through C ; if C was a fixed point, and connected with A and B by a rigid rod without weight, the bowl might be removed without disturbing the equilibrium. The ques- tion, then, is reduced to finding the position of equilibrium ot 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 is known. Let R be the radius of the bowl, r,, r 2 the radii of the spheres A and B respectively, and ACB=a. Then AB=r 1 +r 2> CA=R-r 1 , CB=R-r 2 , AC 2 +CB 2 -AB 2 M C 1ST r Jo \ 1 A / B ) / > 1Vl Y -JV2 and cos. a- 2AC.BC ' (R_ ri )M-(R.-r,)"- ! , which gives a. 2(R-r 1 )(R-r 2 ) Then (19) iu 1 .CM-u? 3 CN=0 > or rJ.CM-»*CN=0 ; since the weights of the spheres are as the cubes of their radii. Substituting the values of CM=(R-r,) cos. (180— (a+0)) and CN=(R — r 2 ) cos. 6, expanding and reducing, we get Ian. V= jy~ r— : . 7j(lt— r,) sin. a Ex. 5. A heavy beam rests upon a smooth peg with one end against a smooth, vertical wall ; find the position of equi- librium. E 66 STATICS. Let ACB be the beam, resting at A against the wall ADK and upon the peg C. The center of gravity g, when there is an equilibrium, will evidently be at some point beyond C from A. Let Ag=a, DC = &, w= the weight of the beam acting at g, R= the reaction of the wail per- pendicular to itself at A, ana R'= the reaction of the peg perpendicular to the beam at C. The angle 6, 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' " " " =-R'sin.0. .-. (17) 2.X = R-R'sin. 0=0. (a) Also, w estimated vertically = w, R = 0, R' " " =-R'cos. .-. (18) 2.Y=w-R'cos.0 = O. (&) Also, the moment pf w about C=w.CF=w.(DF— DC). But DF=AH = Ag-cos.0=acos.0, and DC =6. . . w.CF = w.(a. cos. 6— b), the moment of R about C=R.CK=R.AK.tan.6»=R.&. tan.0, R' o. (19) 2.(Xy-Y:c) = u>.(a.cos.0-&)— R.&.tan.0=O. (c) Multiplying (a) by cos. and (b) by sin. 0, and subtracting, we have R. cos. 0— w sin. 0=0 .-. R = w tan. 0. EXAMPLES ON THE PRECEDING CHAPTERS. G'< Substituting this value of R in (c), w(a cos. 6—b) — w.b. tan. 2 0=0, or a cos. 0= ft (1 + tan. 2 0) = ft. sec. 2 0= b. — . v ' cos." cos, 3 lb 0= Vj and ft(2r cos. 6— a) cos. 6=0. Substituting the value of R, and reducing, 2r tan. 6. sin. 6— 2r cos. 0+a=O, 2r—4r cos. 2 6+a cos. 6=0. (a) (b) or cos. 6- a±V32r i +a i '' 87 in which the + sign only is admissible. Ex. 8. Solve the last example by following, step by step, the method of Art. 71, taking A for the origin of co-ordinates, and AB for the axis of x. Ex. 9. Find the horizontal strain on the hinges of a given door, and show that the vertical pressures are indeterminate. Let the annexed figure represent the door, of which A and B are the hinges. Let g be its center of gravity at which the weight w acts. The door is kept in equilibrium by the EXAMPLES ON THE PRECEDING CHAPTERS. O'J weight acting at g, and the reac- tions of the hinges represented by the oblique arrows at A and B. Let A be the origin of co-or- dinates, AX the axis of a:, AY the axis of y ; and let x = a, y=b be the co-ordinates of g ; x = 0, y=h those of the hinge B. Let the resolved parts of the reactions at B beQ,, horizontally, and R l vert- ically, and Q 2 , R 2 those at A re- spectively. Ri _S4- =a» A Q 2 -X y, Then 2.X -Q.-Q^CorQ^Q,, 2. Y =uj-R 1 — R a =0, S.(Xy-Ya;)=M).a-Q 1 A=0. («) (6) (P) iv a From (a) and (c) we have Q,=-t- = Q 2 ' which gives the horizontal strain ; and it is the same in mag nitude at each hinge, but opposite in direction. Again, from (b) we have R,+R 2 =i«, but w T e have no oth- er relation by which we may determine the values of Rj and R 2 , which are therefore indeterminate. Ex. 10. Two given smooth spheres rest in contact on two smooth planes, inclin- ed at given angles to the horizon ; find their 2 position of equilibri- um. Let the planes AB, AC make the angles a and respective- ly with the horizontal line through A; O, and 2 be the cen ters of the spheres at which their weights w y ,w 2 respective!) act: Rj, R 9 the reactions of the planes at the points of con 70 S T A T 1 t: s. tact, perpendicular to themselves, and therefore passing through the centers of the spheres to which they are tangents. Let S be equal to the mutual pressure of the spheres at their point of contact, acting in the line passing through their centers, and making the angle 6 with the horizontal line AD.0 is required. Each sphere is in equilibrium from its own weight, the re action of the plane against which it rests, and the pressure of the other sphere. By resolving the forces in the direction of each plane for the equilibrium of each sphere, we shall avoid equations involving the unknown reactions R, and R„, and have, in the direction of AB, w 2 sin. a— S cos. (a—6)=Q, (a) in the direction of AC, w 1 sin. 0-S cos. (0+0) = O. (b) Eliminating S, we have w 2 sin. a. cos. (j3+0) = i« 1 sin. (3. cos. (a— 6). Expanding and reducing, w sin. a. cos. (3. — w. sin. /3. cos. a w cot. jQ— w l cot. a tan. 6 = — 7 ; r— : : 5 =— : . (w 1 J rW 2 ) sin. a. sin. p iv 1 +w 2 Ex. 11. A sphere is sustained upon an inclined plane by the pressure of a beam movable about the lowest point of the in- clined plane ; given the inclination of the beam to the plane, required that of the plane to the horizon. Let AgB be the beam movable about A, w= weight of beam acting at g, B -Pj the point of contact with the sphere whose center is C, m>'== weight of the sphere. The sphere is in equilibrium, from the reaction R of the plane at the point of contact F, from the pressure P of the beam at B, and from its own weight w'. These three forces all act through the center C. Ag=a, AB = b, BAD — a are given, or, EXAMPLES ON THE PRECET/NG CHAPTERS. 71 instead of either of the two latter, the radius of the sphere may be given. The angle DAE = 0, the inclination of the plane when in equilibrium, is required. » For the condition of equilibrium of the beam, take the mo- ments about A. "P.AB=w.Ag. cos. (a+6). Hence ^~~h cos " ( a ~^)* For the condition of equilibrium of the sphere, resolving the forces in the direction of AD, we have w' sin. 6— P sin. a=0, or whence u • w' sin. 6—Wj sin. a cos. {a-\-0) ; tan. 6- wa cos. a. sin. a wa sin. 2 a+u;'6' which gives 6 the elevation of the plane, as required. Many statical problems require, for the determination of all the unknown quantities, equations to be formed by geometrical relations. Take, for illustration, the following : Ex. 12. A heavy beam turns about a hinge, and is kept in equilibrium by a cord attached to the lower end ; the cord passes over a pulley in the same horizontal line with the hinge, and sustains a given weight ; find the position of equilibrium of the beam. Let A be the hinge, C the pulley, AC=c, AB the beam =1 in length, g its center of gravity at which its weight w acts, Ag=a and P= the weight hung from the cord and measuring its tension t. Let 6>=ABC and £=BAC, ® both unknown. Taking moments about A, we have, t.AB. sin. 6=w.Ag cos. (t>, 72 STATICS. or w a tin. 0=^.-. cos. 0. PT From the geometrical data, we have AC_ sin. AB~sin.ACB' c (a) or sin. 0=7- sin. (0+0). (*; Equations (a) and (b) suffice to determine and 0. i?x. 13. A uniform beam rests with its lower end in a smooth, hemispherical bowl, and its upper end against a smooth, vertical plane. Find the position of equilibrium. The beam AB rests against the ^ vertical plane AD at A and upon the bowl at B, and is sustained by its weight w acting at g, the reac- } tion R of the bowl in the radius BC, and the reaction R' of the plane perpendicular to itself. Let r= radius of the bowl ; AB =2a, Ag=a, since the beam is uniform, w= weight, and CD=d, ad supposed known. Resolving vertically, we have R sin. O—w—Q, or R: or sin. 0' Taking the moments about A, we have R.AB. sin. (0— 0) — w.Ag - . cos. 0=0, R.2a. sin. (0—. tan. a. Ex. 16. A heavy beam, AB = Z, of which the weight is w, lies with the end A against a smooth vertical wall AD, and the end B on a smooth horizontal plane DB, making with it the angle 6. The distance of its center of gravity g from B is a, and it is kept in equilibrium by a cord attached to it at B and fastened at D. Required the tension of the cord, the re- action R of the wall, and the pressure P on the plane. Ans. R = t = w.j. cot. 6. V=w. Ex. 17. A body (weight =w) is suspended by a cord (length = /) from the point A in a horizontal plane, and is thrust out of its vertical position by a rod without weight, acting from another point B in the horizontal plane, such that AB = d, and making the angle 6 with the plane. Find the tension t of the cord. I t=w- cot. 6. a Ex. 18. A triangular plate of uniform thickness and density is supported horizontally by a prop at each angle. Find the pressure on each prop. Ex. 19. A uniform beam rests on two planes inclined at an- gles a and fi to the horizon. Find the inclination of the beam to the horizon. sin. (0—a) tan.0= . . 2 sin. a sin. p £ %cl >» z^6c / 7 -3 /— // . '. £~=- J? e w e ( / ) . 5L2 EXAMPLES ON THE PRECEDING CHAPTERS. Ex. 20. A uniform beam AB hangs by a string BC from a fixed point C, with its lower extremity A resting on a smooth, horizontal plane. Show that, when there is an equilibrium, CB must be vertical. Ex. 21: A uniform beam AB is placed with one end A in- side a smooth hemispherical bowl, with a point P resting on the edge of the bowl. If AB= 3 times the radius R, find AP. AP=1.838R. Ex. 22. A beam, whose weight is w and length 6 feet, rests on a vertical prop CD (= 3 feet) ; the lower end A is on a hor- izontal plane, and is prevented from sliding by a string DA (= 4 feet). Find the tension of the string. lension =— — w. 125 Ex. 23. A uniform beam, CB=2a, has one end C resting against a smooth, vertical wall AC, and the other sustained by a cord, whose length is c, fastened to the point A. Find AC when the beam is in equilibrium from its weight w, the tension t of the cord, and the reaction R of the wall. (Vide Ex. 14.) Ex. 24. Given the inclination i of the right line AC to the horizon and the weight of a heavy body W, to determine what force or weight Pj acting in the given direction WM will be sufficient to sustain W on the line. The body W is kept upon the line by the action of the force Pj in the direction WM, and its weight acting in the direction WE. The reaction of the line normal to its direction is a third force (Art. 76). Resolving P and W horizontally and vert- ically, we have E| D S.X=P, cos. (e+i)+Wcos. 270°=P, cos. (e+i), S.Y=P 1 sin. (e + i) +W cos. 180° = P 1 sin. (e+i)- W 76 STATICS. Hence (20) becomes P, cos. (e+z*) + (P 1 sin. (e+i)-W) tan. i=0; or, expanding and reducing, P, cos. e=W sin. i, Ws'm.i or P,= . (a) cos. e If the reaction of the line be required, resolve all the forces horizontally (8). This gives N. cos. (90° + i)+P 1 cos. (e+i) = 0. ^ cos^+0 1 sin. i v ' or, by (a), N=W .^hSS+l>. (e) J v ' cos. e v ' Ex. 25. A given weight W is kept at rest on a circular arc by a weight P attached to a cord which pass- es over a point M in the vertical line MX iw through C, the center of the circle. Required the position of W, supposing no friction at M. Resolving the forces P and W vertically and horizontally, we have, calling the angle PMW, e, 2.X=W-Pcos.e, 2.Y= -P sin. e. Let the co-ordinates of the point W be x, y, and call the dis- tance MC, a, and MW, I ; then the equation referred to M as the origin, is (a- xy+y*=r\ or y—^/r i — {a—x) n - Differentiating, we obtain dy a—x a—x dx Vr 2 -(a-.r) 2 V x Substituting in (21), and recollecting that cos. «=7 and sin. e=j, we get w _P^_pL^ =w ._ P :L_p^=o l Idx I I Hence EXAMPLES ON THE PRECEDING CHAPTERS. 77 aP b W an equation which determines the distance of W from M. M Ex. 26. Instead of a circle, as in Ex. 25, let the curve be a hyperbola with its transverse diameter m vertical, the point M being at its center. Ex. 27. Required a curve such that a given weight P, by a cord passing over a fixed point m without friction, will balance another given weight W at every point of it. W Ex. 28. Required the co-ordinates of the center of gravity of a semiparabola whose equation is y°=px, height =a, and base =b. Ans. x=%a, y—lb. Ex, 29. Required the center of gravity of the surface of a right cone. Ex. 30. Find the center of gravity of a paraboloid of revo- lution whose altitude is a. Ans. x=fa. Ex. 31. Find the center of gravity of a segment of a hyper* boloid whose altitude is a. Jr j> £ W CHAPTER VI. ON THE MECHANICAL POWERS. 130. The general object of machinery is to transmit and to economize the action of certain forces at our disposal. The specific end is, sometimes to augment the action of which the power employed is capable when applied without the inter- vention of machinery ; sometimes, merely to change the direc- tion of the action ; and sometimes to regulate the velocity of the point to which the action is transmitted. The most simple machines are denominated Mechanical pow ers, and are reducible to three classes, viz., the Lever, Cord. and Inclined Plane. The first class comprehends every machine consisting of a solid body capable of revolving on an axis, as the Wheel and Axle. The second class comprehends every machine in which force is transmitted by means of flexible threads, ropes, &c, and hence includes the Pulley. The third class comprehends every machine in which u hard surface inclined to the horizon is introduced, as the Wedge and the Screw. The force which is used to sustain or overcome any oppo- sition is called the Power ; the opposition to be overcome is called the Weight. This distinction in the names of the forces employed implies none in their nature. § I. THE LEVER. 131. Def. A Lever is an inflexible rod capable of motion about a fixed point, called a fulcrum. The rod may be straight or any how bent. It is generally regarded, at first, as without weight, but its THE LEVER. 79 weight may obviously be considered as another force applied in a vertical direction at its center of gravity. Def. The arms of a lever are the portions of it intercepted between the power and fulcrum, and between the weight and fulcrum. 132. Levers are divided into three kinds, according to the relative positions of the power, weight, and fulcrum. In a lever of the first kind, the fulcrum lies between the points at which the power and weight act. In a lever of the third kind, the point of action of the power is between that B ] 2 of the weight and the ful- -A- crum. In a lever of the sec- ond kind, the weight acts at a point between the fulcrum and the point of action of the power. 133. Prop. Required the condition of equilibrium and pressure on the ful- crum when two parallel forces act on a straight lever. A^ HO STATICS. Since the fulcrum C is a fixed point, by Art. 75, the sum ol the moments of the forces about C must be zero. Aw Let a be the angle made by / the. direction of the forces with A. / V7 the lever. From (19), we B C have W.BC. sin. a— P.AC. sin. a=o, the moment of P being nega- r P tive, since it tends to produce motion in a direction opposite to that of W. W AC Hence -5-=^, or, in case of equilibrium, the weight and power are reciprocally proportional to the distances at which they act from the fulcrum. Hence ( Art. 36) the resultant of P and W must pass through C, and the pressure on the fulcrum is equal to the algebraic sum of P and W, and acts in the direction of the greater. Cor. If the power equal the weight, the distances of their points of action from the fulcrum will be equal. For P=W gives AC=BC. 134. Prop. Required the condition of equilibrium and the pressure on the fulcrum when any two forces in the same plane act on a straight lever. jy Let the forces P ^ "' / \\ and W make the an- /' \ \ gles a and /3 respect- / '• \ ively with the lever, / \ \ and let their direc- /\ q/\ -X tions, when produced a / ': C \ i \-q if necessary, meet in ^T \ D. Since C is a fixed 1 w point in the lever, the * R ' sum of the moments wf P and W about C must be zero. Hence W.CB. sin. /3-P.AC. sin. a=0, THE LEVER. W AC. sin. a or (2.) \a P CB. sin. ft' But AC. sin. a= per- pendicular from C on AD, and CB. sin. 0= perpendicular from C onBD. Hence the con- dition of equilibrium requires that the pow- j± ,-''' er and weight should be inversely as the perpen- diculars from the ful- crum on their respective directions. Since the lever is in equilibrium by the actions of P and W and the reaction of the fulcrum, the resultant of P and W must be equal and opposite to that reaction. It will therefore pass through C, and be equal to the pressure on the fulcrum. To find R, we have {Art. 29) R 2 =P 2 +W 2 +2PW. cos. ADB in Fig. (1), and R 2 =P 2 +W 2 +2PW cos. ADW in Pig. (2). But ADB=180-(a+j3), and ADW=180-ADB=180- (P-a). ... R 2 =P+W 2 -2PWcos.(a+0), or R 2 =P 2 +W 2 -2PW cos. (0-a). To find the angle 6 made by R with the lever, resolve par- allel and perpendicular to the lever, the reaction of the fulcrum being equal and opposite to R. Hence we have, in Fig. (1), P. cos. a— W cos. j3+R cos. 0=0, P sin. a+W sin. /3-R sin. 0=0. Tan. 6= P. sin. a+W sin. (3 in Fig. (2), W cos. j3— P cos. a ' P cos. a— W cos. /3— R cos. 6=0, P sin. a- W sin. /3+R sin. 6=0. W sin. (3— P sin. a Tan. 6= P cos. a—W cos. (3' F 82 S T A 1 ICS. 135. Cor. Whenever the lever is bent or curved, the conch tion of equilibrium is the same. For, since the moments of the forces about the fulcrum in oppo- site directions must be equal, in case of equilibrium, we have or P.CM=W.CN. WCM T~CN' 136. Prop. Required the condition of equilibrium and the pressure on the fulcrum, when any number of forces act in any direction in one plane on a lever of any form. Let Pj, P 2 , P 3 , &c, be the forces in one plane, Pi,p 2 ,p 3 , " the distances of their points of applica tion from the fulcrum, a,, a„, a 3 , " the angles made by the directions of P and p respectively. Then p x sin. a 1 ,p 2 sin. a 2 , p 3 sin. a 3 are the perpendiculars from the fulcrum on the directions of the forces, and Pj p^ sin. a,, &c, the moments of the forces. When there is an equilibrium, the sum of the moments of the forces about the fulcrum will be zero, or V l p l sin. «i+P 2 p 2 sin. a 2 +F 3 p 3 sin. a 3 + ,&c., =0. The signs of the moments will depend on the direction in which they tend to produce rotation. To find the pressure on the fulcrum, we must determine the resultant of all the forces supposed to concur at the fulcrum. For each moment F 1 p 1 sin.«, is equal to a couple of which the forces are P and the arm p sin. a, and a single force P act ing at the fulcrum. The magnitude of R will then be determ- ined by the equation (0), R=v / (2.X) 2 + (2.Y) 2 . THE LEVER. 83. Cor. If the lever be straight and the forces parallel, a 1 =a 2 =a 3 = , &c. Hence P 1 je> 1 +P 2 ^ 2 +P 3 ^ 3 +, &c.,=0. The same result will also be obtained by following, step b.v step, the method of Art. 71. EXAMPLES. Ex. 1. On a straight lever AB of the first kind, without weight, 36 inches in length, a weight W=15 lbs., acting at B, is balanced by a power P=3 lbs. acting at A. Required the distance of the fulcrum C from A. LetAC=x; then BC = 3G-x. By Art. 133, I\AC=W.BC, or 3.z=15(36-x). Hence .t=30 inches =AC. Ex. 2. On one arm (=p 1 ) of a straight lever of the first kind, without weight, a body counterpoises a weight (=a lbs.), on the other (=p 2 ) a weight (=b lbs.). Required the weight of the body. A straight lever of the first kind, with unequal arms, and hav rag the fulcrum at its center of gravity, is called a false balance. Let x— the unknown weight. By Art. 133, x.p^—a.p. and x.p 2 =b.p By multiplying the equations member by member, x*=ab, x = Vab. Ex. 3. On a straight lever, without weight, are suspended five bodies, Pj-4 lbs., P 2 = 10 lbs., P 3 = 2 lbs., P 4 =3 lbs., P 5 =7 lbs., at the points A, B, C, D, and E, such that AB=4 feet, BC=2 feet, CD=6 feet, and DE=8 feet. Required the posi- tion of the fulcrum F, about which they will balance. A B C D E '2» 84 STATICS. Let AF— x. Then, by Art. 136, Cor., we have P,= 4and^,=AF=.r. .\V l p l = 4.x. P 2 = 10 « jo„=BF=AF-AB=x-4. P 2j p a =10(z- 4). p 3 = 2 " / 3 =CF=AF-AC=:c-6. P 3 />3= 2(x- G) P 4 = 3 " p 4 =DF=AF-AD=.x-12. P 4j p 4 = 3(x-I2) P.= 7 " j9 5 =EF=AF-AE=x-20. P 5 p 5 = 7(z-20). And 2.Pp=4x + 10(x-4)+2(a;-6) + 3(.r-12)+7(x-20)=0 =26:c— 228. Hence z=AF=8if f ee t. Otherwise : Since the weights are parallel forces, their re- sultant R is equal to their sum. The whole system being in equilibrium, the resultant must pass through the fulcrum, and the moment of the resultant must be equal {Art. 45) to the sum of the moments of the components. Taking A for the origin of moments, we have R.x=P 1 .0+P 2 .AB+P 3 .AC+P 4 .AD+P 5 .AE, or 26.^=4X0+ 1X4 + 2X6 +3X12+7X20=228. .-. z=AF=8ff feet. Ex. 4. A uniform lever AB of the first kind, 12 feet long, whose weight w—Q lbs., has a weight W=100 lbs. suspended from the shorter arm CB=2 feet. Required the power P which must be applied vertically at A, to equilibrate W. The weight of the lever has the effect of a weight equal to itself, applied at its center of gravity, which, since the lever is uniform in size and density, is at its middle point. The dis- Ap "Dp tance of this point from C is , and its moment about r nu AC ~ BC G will be w. . Hence P.AC+w- ~— =W.BC, W.BC->C-BC) 100X2 _ 3X8 or P= m = = 17.6 lbs. Ex. 5. The arms of a bent lever are a=3 feet and b=5 feet, and inclined to each other at an angle 0=150°. To the arm THE LEVER. 85 a a power P=7 lbs. is- applied, and to the arm b a weight W = 6 lbs. Required the inclination of each arm to the horizor when there is an equilibrium. Let a be the inclination of the arm a, and (3 the inclina- tion of the arm b to the ho- rizon. Then P.MC = W.NC, or P.a. cos. a=W.b. cos. p=W.b. cos. (180°- {a+6)), = -W.b.cos.(a+6) ' = — W.b. cos. a cos. 0+W.&. sin. a. sin. 0. P.a+W.&.cos.0 21+30. cos. 150° wQ Tan. a- W.b. sin. 6 21-30 cos. 30° 30. sin. 150° 21-30xiv / 3 4,98 ~15 30. sin. 30 30Xf Log. Tan. «=9.5211381 n , a=-18°.22', and 0=180°- (a+6») = 180 o -r-18 o .22'-150 o =48°.22 . Hence the arm AC is inclined at an angle of 18°.22' above the horizon, and BC at an angle of 48°.22' below the horizon. Ex. 6. The whole length of the beam of a false balance (Ex. 2) is 3 feet 9 inches. A body placed in one scale coun- terpoises a weight of 9 lbs., and in the other a weight of 4 lbs. Required the true weight W of the body, and the lengths a and b of the shorter and longer arms. Ans. W=G lbs., a=\ ft. G in., 6=2 ft. 3 in. Ex. 7. A false balance has one of its arms exceeding the 1 • other by — th part of the shorter arm. Supposing a shop- keeper, in using it, puts the weight as often in one scale as the other, does he gain or lose, and how much per cent. 1 Ans. Loses — r- — per cent. m +m V-w;_ ( \A/~ Vv_ XU^JL ^-&L JlS S6 STATICS. Ex. 8. The anus of a bent lever are 3 feet and 5 feet, and inclined to each other at an angle of 150°, and to the shorter arm is suspended a weight of 7 lbs. Find what the othei weight must be in order, 1st, that the shorter, and 2d, that the longer arm may rest in a horizontal position. Ans. 1st, 4.85 lbs. 2d, 3.64 lbs. § II. WHEEL AND AXLE. 137. The Wheel and Axle consists of a cylinder or axle, perpendicular to which is firmly fixed a circle or wheel, whose center is in the axis of the cylinder. The whole is supposed to be perfectly rigid, and movable only round the axis of the cylinder. In the ordinary applications of this ma- chine, the power is applied tangentially to the surface of the wheel, and the weight, in the same manner, to the surface of the axle. 138. Prop. Required the condition of equilibrium of the wheel and axle when two forces are applied tangentially to the circum- ference of the axle and of the wheel. Let ADE be a section of the wheel perpendicular to the axis, and BCA a horizontal line through the center of the axis, terminated at A by the circum- ference of the wheel, and at B by the circumference of the axle. The power P acts in this vertical plane at A, and the weight W acts in a plane par- allel to it. Since the axle and wheel are firmly connected, the action of the weight may be transferred to the plane ADE WHEEL ArfD AX I.E. 81 {Art. 57), and supposed to act at B. Then, since the sum of the moments must be zero for equilibrium, P.AC-W.BC=0. W_AC__radius of wheel P BC radius of axle ' or, the Power : Weight— radius of axle : radius of wheel. 139. Cor. 1. The same relation will exist in all positions ot the wheel, so that this machine may be called a perpetual lever. 140. Cor. 2. When the power and weight act vertically on opposite sides of the axis, the pressure on the rests or Ys equals the sum of the two ; when vertically on the same side, it equals their difference ; when in any other directions, it equals the diagonal of a parallelogram, whose sides represent the power and weight in magnitude and direction. If ropes are used to transmit the action of the power and weight, we must suppose the forces applied to the axes of the ropes. Hence, if r and R represent the radii of the axle and wheel respectively, and t and T represent half the thickness of the ropes. W.(r+0=P-(R+T), or P : W=r+t : R + T. 141. Prop. Required the condition of equilibrium of any num- ber of forces, acting in any direction in planes perpendicular to the axis. Let the actions of all the forces be transferred (Art. 57) to the same plane ; then, since the intersection of the axis with this plane is a fixed point, the condition of equilibrium is given by Art. 74 ; or, the algebraic sum of the moments of all the forces about the axis ?nust be zero. 142. Def. Toothed or cogged wheels are those on the circum- ference of which are projections, called teeth or cogs. When two wheels have their cogs of such form and distance that those of one will work between those of the other, the motion of one wheel will be communicated to the other by the p?-essure of the cogs. S8 STATICS. When the teeth are on the sides of the wheel instead of the circumference, they are called crown wheels. In the preceding instance, the axes of the wheels and pin- ions are parallel or perpendicular to each other. When the axes of two wheels make an acute angle, the wheels take the form of frusta of cones, and are called beveled wheels. Axles on which teeth are formed are called pinions, and the teeth leaves. 143. Prop. Required the condition of equilibrium when the action of the power is transmitted to the weight by means of cogged wheels. Let S be the mutual pressure of one cog upon the other. This pressure takes place in the direction of the line S;/z, m'S normal to the cogs at the point of contact, Cm,C'm' being per- pendiculars from the centers v-> ^ 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'w', and about C, W.CB=S.Cm. Dividing the latter by the former, we have W C'A Cm P ~~ CB 'Cm Now, if the radii of the axles are equal, or C'A = CB, we shall have WCm Y~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 TVm'~C'o~ 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 va- riable. But when the cogs are of such form that the normal S??i m'S at their point of contact shall always be tangent to both circles, the lines Cm and Cm' will become radii, and their ratio con- stant, and the point o a fixed point, in which case W_Co_ Cm _ R T~Co~Cm 7 ~R" R and R' being the radii of the circles C and C respectively. 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 cir- cumference of C to the circumference of "C', or as the radius of C to the radius of C. W_number of cogs on the wheel of W P number of cogs on the wheel of P ' Scholium. 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 the extremity H of the flexible line FaH 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- 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. 90 STATICS. Let R, R,, R 2 , &c, be the radii ol the successive wheels; r,r l ,r 2 ,&,c.,the radii of the corresponding pinions ; P, Pj, P 2 , &c, the powers applied to the. circumferences of the successive wheels. Taking the moments about the center of each wheel, we have P.R=P 1 r, P.R^P.r,, P a R 3 =P s r a , &c. ; since the power applied to the circumference of the second wheel is equal to the reaction on the first pinion. Multiplying these equations member by member, and re- ducing, P„_ R.R,.R 2! &c. P r.r 1 .r 2 ,&c. or, the power is to the weight as the product of the radii of the 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 3f 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 balance 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. W=1234f. Ex. 4. There are two wheels on the same axle ; the diam- eter of one is 5 feet, that of the other 4 feet, and the diameter THE CORD. 91 of the axle is 20 inches. What weight on the axle would 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 communi- cation of force. It is regarded as perfectly flexible and with- out 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 advan- tageously. 147. Prop. Required the condition of equilibrium of a cord acted upon by three forces. Let the forces Pj, P 2 , P 3 be applied at the ex- tremities A and B of the cord ACB, and to a knot at C. Draw any line CD in the direction of the force P 2 , and DN, DM parallel" to CA and CB respectively. In case ot equilibrium {Art. 28), CM : CD : CN=P 1 : P 2 : P 3 = sin. DCN : sin.MCN : sin. MCD, = sin. a' : sin. (a+a') : sin. a ; or, the forces are each as the sine of the angle contained between the directions 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 Pj and P 3 , and are equal to the tensions of the two parts of the cord re spectively. 149. Cor. 2. If the force P 2 be applied to a running knot or ring, the points A and B being fixed, the condition of equilib- rium requires that the direction of P„ 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 P 2 could not be in equilib- rium except when normal to the curve, in which case it bisects the angle ACB. Hence, sin. a= sin. a', and P a =P 3 ; and since Pj : P,= sin. a' : sin. («+a')= sin. a : sin. 2a=l : 2 cos. a. .'. P 2 =2P, cos. a. Otherwise : since the tension of the cord is the same through- out, when the cord passes over an obstacle without friction. P,=P 3 ; a=a' and P 2 =2P! cos. a {Art. 19). 150. Pitor. Required the conditions of equilibrium when any 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 l5 P 2 , P 3 , &c, at the points A, B, C, &c, in the directions AP,, BP 2 ,CP 3 , &c. The force P, at A may be considered as acting at B in its di- rection BA ; and since B, when in equilibrium, must be so from the ►action of P,, P 2 , ana the tension of BC, the resultant of P, and P 2 must be in the direction of CB, and may be considered 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 Wo have thus transferred the forces P, and i v , to act at C parallel to their original directions. In the same manner, the resultant of P a , P 2 , and P 3 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 P 1? P 2 , and P 3 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 re- solved 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. Prop. Required the relations of the forces which, acting on a cord in one plane, keep it in equilibrium. Produce P 2 B to N, P 3 C to M, &c, and let ; the Z-ABN=«, Z.NBC=a', z.BCM=i3, z_MCD=/3', &c, &c. Let t x , t 2 , t 3 , &c, be the tensions of the sev- eral successive portions of the cord. Then {Art. 147) t 1 : P 2 : t 2 = sin. a' : sin. {a+a') : sin. a; {a) also, t 2 : P~ 3 : t 3 = sin. (3' : sin. (j3+j3') : sin. (3, (a') &c, &c. „ , . , . ~ sin. d . _ sin. a From {a) we obtain <.=P - — -. — ; — x, and t n =r„- — -. — ; — -^ v ' l - sin. {a+a') - -sin. {a+a') sin. 13' _ sin. /3 {a') ^=P 3 - sin>(i3+J g,y and ^ = ^3 siR<(/ 3 +i g,y &c, &c. Equating the values of t 2 , t 3 , &c, we have sin, a _ sin.^ 2 Vm. {a+a') 3 "sin. (P+0')' W 94 STATICS. P,.-. sin. /3 -=P sin. y **sin; (y+y')' (b>) 'sin. ((3+(3') &c, for the relations of the forces. 152. Cor. 1. If the cord be fixed at the points A and E, and the forces P 2 , P 3 , P 4 , &c, be parallel, we have sin. a'= sin. (3, sin ,(3'= sin. y, &c. Multiplying these equations by (b), (b') member by member in their order, we obtain sin. a sin. a' sin. j3 sin. (3' ? sin.ysm.y' But ' sin. (a+a') 3 sin. ((3+(3') * 4 sin. (y+y') 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! sin. a sin. a' cot. a+cot. a' cot. a+ cot. a' cot./3+cot./3' cot. y+cot. y' A . N :, &C. 153. Cor. 2. If the cord be fixed at A and E, and the forces P 2 , P 3 , P 4 , &c, be weights, the horizontal tension of each por- tion of the cord is the same. For, by resolving the tension of each part horizontally, wo have the horizontal tension of AB=^. sin. ABN=£, sin. a, BC=£ 2 sin. BCM=f 3 sin. (3, &c. Substituting for t Jf t 2 , &c, their values found above, sin. a sin. a! P„ horizontal tension of AB = P n BC=P 3 sin. (a+a') cot. a+ cot. a 1 ' sin. (3 sin. j3'_ P 3 sin. ((3+13') ~cot.j3+ cot. (J 1 &c, &c , which, by Co?: 1, are all equal. THE CORD. 95 154. Cor. 3. Since the reactions of the points A and E equi- librate the resultant of all the weights, the lines AB, ED pro- duced, must meet in some 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 press- ure on each. Since the tension of the cord PjABCDEP £ throughout, each point is acted upon by three forces, viz., two equal ten- sions on each side of it and the reac- 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., Art. 19, the pressure on A=2P T cos. ^a, B=2P 1 cos.l&, &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 cord 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 bv 96 STATICS. ,B three equal forces, and, by Art. 18, must make angles of 120° with each other. Draw the horizontal line AD; and, since ACD— 120°, 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 cir- cumference 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 trans- ferrins; the friction from the cord and circumference of the wheel to the axle and its supports, which may be highly pol- ished or lubricated. The effects of rigidity are diminished by turning the cord in a circular arc instead of a sharp angle. The pulley is called fixed or movable, according as the block is fixed or movable. , " IT 7 " ' -• B| 157. The fixed pulley serves merely to change the direction of the forces trans- mitted by the cord, since, neglecting the friction of the pulley, the tension of the cord is the same in every part of it. Hence the power equals the weight, and the press- ure on the axis of the pulley equals their sum. THE PULLEY. 9-? 158. Prop. Required the relation of the power to the weight m the single movable pulley. The tension t of the cord, being the same p 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 opposite to the weight W, must be vert- ical. Let a be the angle made by the cords with this vertical, and, resolving the tensions vertically, we have 2t cos. a=2F cos. a—W. W .'. P=: 2 cos. a' the same as obtained in Art. 149. Cor. 1. If the weight w of the pulley be taken into account, P= . 2 cos. a Cor. 2. If the cords are parallel, a=0 and W+m; • Jl — z — . Cor. 3. If a=90°, 2a=180°, or the cord becomes straight and horizontal. In this case n 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 sys- tems of pulleys. 160. Prop. Required the relation of the power to the weight in the first system of pulleys. G 98 STATICS. The annexed figure represents this system. Neglecting friction, the tension t of the cord is the same throughout, and equal to P. The 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=nP=W+w p = W+to n If the weight of the lower block be neg- lected, n Of this system of pulleys there are various modifications. The annexed form is the one in most common use. 161. Prop. Required the relation of the power to the weight in the second system of pulleys. The annexed figure represents this system with three mov- able pulleys, each pulley having its own rope. Designate by a,, a 2 , a 3 , &c, the pulleys re- spectively in their order from the weight W, by w x , w 2 , iv 3 , &c, their weights, and by t lt t 2 , t 2 , &c, the tensions of their respective cords. Then, for the equilibrium of a,, we have W+10. 2t 1 =W+w 1 , or *,= -— K For the equilibrium of a 2 , 2t 2 =t 1 + v: _W+w l +2w, or to = W+u>,+2io s For the equilibrium of c 3 ,2t 3 =t 2 + to. W+w^+2w 2 +2'u> i or THE PULLEY. W+w 1 +2w 2 +2*w, 99 For the equilibrium of a n . W+io 1 +2M> 3 +2 9 to 3 +2'M> 4 + . . . 2"-'mv. P=t n 2" Cor. 1. If the weight of each pulley is the same, and equal w ^ W+w 1 (l+2+2 2 +2 3 +....2- 1 ) _W 10,(2"-!) ~ 2 n+ 2" W 1 Cor. 2. If the weights of the pulleys be neglected, W P=_, or W=2".P. 162. Prop. Required the relation of the power to the weig.u 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 x , a 2 , a 3 , &,c, their weights respectively by io l , w 2 , u) 3 ,&c, and the tensions of the successive cords by £j, t 2 , t 3 , &c, we have t 2 =2t 1 +w 1 =2V+w 1 , t 3 =2t a +w a =2 , l?+2w 1 +u^, t i =2t 3 +w 3 =2'V+2''w 1 +2w 2 +w 3 . And if there be n cords, t n =2"- 1 V+2 n -ho l +2"-'w 2 + &c 2w n ^+u\^. But W=t,+t 2 +t 3 +,&tc U =P(1 +2+2 2 +2 3 +, &c. . . . 2"- 1 ) +iv , (1 +2+2 2 +, &c . . . 2'- 2 )+u; 2 (l+2+2 2 +.&c 2"- 3 )-h&c. 100 STATICS. = P(2"-l) + u> 1 (2 n - 1 -l)+w 2 (2"- : -l)+,&c w„_ 1 (2-l). If the (ft— 1) pulleys are of the same weight u>,, W=P(2 n -l)+w 1 (2 n - 1 +2- 2 +2'- 3 +,&c 2 "+2 + l-n) ='P(2 n -l)+w 1 (2 n -l)-iu 1 n = (Y+w l )(2 n -l)-iv 1 n. If the weights of the pulleys be neglected, W=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. 1 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 Ex. 5. Find the ratio of the power to the weight in the annexed modification of the second system of pulleys. § V. THE INCLINED PLANE. 163. The Inclined plane, 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 reac- tion, 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. Pkop. 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 P height, and AC the base. Let i be the in- clination of the plane to the horizon, e the angle made by the direction of the power P with the plane AB, W = the weight of the body a y and R= the reaction of "W~V 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 Pcos. e— W sin. ?'=0, R+P sin. e— W cos. i=0. From (a) we obtain W COS. £ P sin. 7 From (b), R=W cos. i— P sin. £=W cos. z- (a) (P) W sin. i sin. e COS. £ W. cos. (i+£) W COS. £ COS. £ or (d) R cos. (i+e)' the same relations as obtained in Ex. 24, Art. 129. 165. Cor. 1. If the force P act parallel to the plane, £-0 and (c) becomes W_ 1 AB P _ sin.2~BC' or, the power is to the iveight as the height of the plane to iti length. 102 STATICS. o W 1 AB From(^veget- r --^, or, the reaction of the plane is to the weight as the base to the length. 166. Cor. 2. If the power act parallel to the base of the plane, f— — if and (c) becomes W cos. i AC T == sin~£ == BC' or, the power is to the weight as the height to the base. W cos./ AC From (d), -^ =— = 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. Let W and W, be the weights of the bodies, and i,i, the inclinations of the planes. If t be the tension of the cord, we have for equilib- rium on the plane AB c {Art. 165), t=W. sin. i, on the plane BC <=W, sin. £ I . .*. Wsin. i=Wj sin. f f , AB Vl BC WAB W^BC' or, the weights are proportional to the lengths of the plana on which they rest respectively. or or THE WEDGE. 103 EXAMPLES. Ex. 1. What force acting parallel to the base of the plane is necessary to support a weight of 50 lbs. on a plane inclined at an angle of 15° 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. TEIE 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 perpendicu- larly 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 the wedge. Let ABC be a section of the wedge perpendicular to the angle or edge A. Draw AD bisecting 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, or in the direction AD, we have 104 STATICS. or or P . Y7= sin. a K 2R.sin.BAD-2P=0, P=R. sin. a, BD BD.Z \ the back of the wedge B A BA.Z face of the wedge where I— the length of the edge or breadth of the face. 170. In the foregoing investigation of the theory of equilib- rium 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 prob- lem 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. Let DE be the beam and BAC the wedge, Z.BAC = a, z.ADE = /3, /= the length of the beam, g the center of D | 1V A 33 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=0, or w.a. cos. j3— R.Z. cos. (a— 13)=0. a cos. (3 / cos. {a—li) But the principles of the wedge give P=R. sin. a. THE SCREW. 105 _ a sin. a cos. /3 .. P = tt?.y. — . / cos. (a—fj) By an examination of this value of P, it will be seen 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 di- agonals of the rectangles thus m formed, and if we suppose the whole rectangle to be wrapped round the surface of a cylinder, the perimeter of whose base is equal to AB and altitude to BC, 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 external 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 the screw. From the construction of the screw, it appears that the thread of it is an inclined plane, of which the base is the circum- ference of the cylinder and the' height the distance between the threads. The force is generally applied perpendicularly to the end of a lever inserted into the 1 06 STATICS. cylinder, and in the plane perpendicular to ti\e axis of the cylinder. The power P thus applied, in turning the screw 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, «-*^"-*£- » By Art. 166, V =W ^X£> Q, W distance of two threads or 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 _ w distance of two threads ' circumference of the cylinder' t-. , . ^ -^ 27ra _ circumference described by P By (a), Q=P.-— =P. : ? e — Fj — • J v 2nr circumierence ot 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 revolu- tion 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 powei BALANCES, ETC. 10*3 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 ex- tremity a force of 1 lb. will support a weight of 1000 lbs. on a screw, whose threads are f inch apart? § VIII. BALANCES AND COMBINATIONS OF THE MECHANICAL POWERS. 174. The common balance, in its best form, is a bent lever, in which the weight of the lever must be taken into consider- ation. In the annexed figure the points A and B, from which the scale-pans and weights are suspended, are call- ed the points of suspen- sion ; C is the fulcrum, being the lower edge of a prismatic rod of 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 surfa- ces, so that the beam turns freely about C as a fulcrum. 175. The requisites of a good balance are, 1°. That the beam rest in a horizontal position when loaded with equal weights. , 2°. That the balance possess great sensibility. 3°. That it possess great stability. 176. Prop. To determine the conditions that the beam rest in a horizontal position 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 STATICS. But taking into consideration the weight of the beam, its cen 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 sym- metrical 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 -y in the preceding figure, and join Cg, the centers of mo- ■N tion and of gravity. Cg is perpendicular to AB and bi- sects it, if the beam is con- structed in accordance with the first requisite. Let M C E N be a horizontal line through C, meeting AB in E making with it an angle equal to 6. The sensibility is measured by the amount of deflection 8 of the line AB from a horizontal position by a given small dif- ference 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 =iv. Now MN is bi- sected in d, and Md=a cos. 0, Cd=d. sin. 6, and Ca=h. sin. 6. If the system is in equilibrium, the moments about C give P.CM-Q.CN-u>.Ca=0, or P.{M.d-Cd)-Q,(Nd+Cd)-w.Ca=0, or (P-Q)a. cos. 0.-(P+Q)tfsin. 6-w.h. sin. 0=0, or (P-Q)a-{(P+Q).rf+u>./i}tan.0=O. . tan P- (P ~ Q) - g Hence the angle 6, and, therefore, the sensibility, is increased for given values of P and Q, by mc?*easing the lengths of the BALANCES, ETC. 109 arms (a), by diminishing the weight of the beam (w), or by di- minishing the distances of the fulcrum from the center of grav- ity of the beam (Ji) and from the line joining the points of sus- pension (d). 178. Prop. To determine the conditions that the balance may possess great stability. If the balance be loaded with equal weights and disturbed from its position of equilibrium, the rapidity with which it re- turns 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+w.Ca, or ¥.2.Cd+w.Ca, or (2P.d+w.h) sin. 0. Hence, for given values of P and d 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 increas- ing 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 sacri- ficed to stability ; but for philosophical purposes great sensi- bility is required, and stability is of little comparative import- ance. THE STEELYARD BALANCE. 180. The steelyard balance, or Roman steelyard, is a level of the first kind, with un- equal arms. The body W to be weighed is hung \^j|Y \ f=§ § j at the shorter arm A, and j*? p a given constant weight P is moved along the other arm till it balances W; then the weight of W is known from the place of the counterpoise P. W 110 STATICS. 181. Prop. To determine the law according to which th 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=jo, OC=p l , OG=g, and taking moments about the fulcrum O, in case of equilibrium we .have Wp — Wg— Pj9 j = 0. wg Wp Taking Ox, a fourth proportional to P, w, and g, we have 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 pro- gression. THE BENT LEVER BALANCE. 182. This balance is represented b\ 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 determine 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 weight (W) of a body placed in it will act vertically through A. Let the horizontal line through B meet the vertical lines through g and A respectively in D and K. Then the moments about B, in equilibrium, give w.DB-(E+W)BK=0. (a) BALANCES, ETC. Ill As greater weights are put into the scale E, the point A ap- proaches more nearly the vertical through B from 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 DBg=6. When a weight is put into the scale the point A will descend through some angle 0, and the arm Bg will rise through the same angle. In this new position the angle DBg- will become 0—0. Let Bg=a and BA=b ; then DB — a cos. (0—0), and BK=& cos. 0. These substituted in (a), give w.a. cos. (6— 0) — E.&. cos. 0— W.b. cos. 0=0, or w.a. cos. cos. (p-\-w.a sin. 6 sin. 0— E.5 cos. 0— W.b. cos. 0=0, or w.a. cos. 6+w.a sin. 6 tan. — E.&— W.b—0. tan. (p- W.b ;+■ E.6 — w.acos.d w.a. sin. u w.a. sin. Hence, tan. varies as W, and the limb GF must be di- vided 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 paradox- ical 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, with equal cross-beams turning about pivots at a and b. These cross-beams are connected by pivots at c, d, e, and/, with two oth- er equal pieces in the form Li M A 112 STAT' C S. weights are suspended from the horizontal arms ofaT. The of the latter pieces 185. Prop. In RobervaVs balance, equal weights balance at all i distances from the upright stem. Let the letters in the annexed figure" indi- cate the same parts as in the former. Let equal and oppo- site forces P, and P 2 each equal to P act in ec, and, similarly, let P 3 and P 4 act in df. Q2<^ These forces P a , P. P„ and P 4 do not dis- Lurb the equilibrium. Now P at B, as in Art. 71, is equivalent to P, at e and the couple P, B/j, P 2 , and, similarly, P at C is equivalent to P 4 at / and the couple P, Ck, P 3 ; and since P, at e balances P 4 at f, we have only the two couples to dis- pose of. Now for the couple P, Bh, P 2 , Arts. 58 and 59, we may sub- stitute a couple Q,,,ec, Q 2 in its own plane and of equal mo- ment, in which the forces Q x and Q 2 , 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, P 3 may be replaced by the equivalent couple K 1 ,fd, R 2 , in which Rj and R 2 are destroyed by the resistance of & and a. These new couples, therefore, do not disturb the equilibri- um, and the original forces P at B and P at C must be in equi- librium. 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 BALANCES, ETC. n& B C" B" W B' "W" A"' c _S7 >T A" W" 186. Prop. To determine the ratio of the power to the weight in a combination of levers. Let the power P act at A, and the weight W at B'". The first three levers are of the second kind, and the last one of the first kind, the ful- crums being at C, C', C", and C". Let BA', B'A", B"A'" be rigid rods connecting the levers, and let the action of the first lever on the rod BA' be W, which becomes the power acting- on the second, W" and W" the weights to the second and third respectively, and powers to the third and fourth. P CB W C'B' W" C"B" , W" C"B' a" B' ThenTT77=; C"B" , W — , and W CA' W" C'A" W" C"A"' "~" W C'"A' Taking the continued product of these equations member Dy member, we have P _ CBxC'B'xC"B"xC'"B'" W - CAxC'A'xC"A"xC"'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 han- dle of the winch, W the pressure of the screw on the teeth of the wheel, and W the weight suspended from the axle of the wheel. Then H and 114 STATICS. P distance between the threads of the screw W circumference described by the power ' W'_ radius of axle VV radius of wheel' P distance between two threads of screw W circumference described by the power radius of axle V ; • radius of wheel ' or, the ratio of the power to the weight in the endless screw is equal to the product of the ratios of the power to the weight in the screw and in the wheel and axle. 188. Prop. To determine the ratio of the power to the weight in any combination of the mechanical powers. Let P= the power for the whole combination, w »_ (( we ight " " " W = " " to the 1st in the series and power to the 2d, W"= " " " 2d " " " 3d, &c, &c, &c. Let a 1 = the ratio of the power to the weight in the 1st, a 2 = " " " " " 2d, a 3 = " " " " « 3d,&c. JL_ w ' _ W" _ W"-'_ lhen yy, — a i> ^y//— a 2' W"/" - a 3 ~yy^~~ a " » and, taking the product, P ypi=a x .a 2 .a 3 a H . Hence the ratio of the power to the weight in any combina- tion of elementary machines is equal to the product of the ratios 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 AB BALANCES, ETC. 115 2*-P turns about a firmly fixed pivot at A, and is connected by anoth- er pivot at B to the rod BC, whose extremity C produces the pressure on the obstacle at E. Let the power P act horizon- tally at some point F in the le- ver AB, ANC be a vertical line meeting the direction of P in N, and DE a horizontal plane, on which, at E, is the substance subject to pressure. Let R= the reaction of the rod BC in the direction of its length, AM, DL perpendiculars upon its direc- tion from A and D, and W the vertical resistance of the sub- stance at E. Taking the moments about A and D in equilibrium, we have P.AN=R.AM, and W.DE-R.DL, P AMxDE ° r W~ANxDL* When BC becomes nearly vertical, DL becomes nearly equal to DE, and AN to AF, while AM becomes very small. T , . . . P AM . , _ . In this situation, Tx7=~rFr nearly, so that P is a very small fraction of W. CHAPTER VII. APPLICATION OF THE PRINCIPLE OF VIRTUAL VELOCITIES TO THJ 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 2.P.u=0. Also, the virtual velocities of the support- ing 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* displace- ments must be taken indefinitely small. 191. Prop. The principle of virtual velocities obtains in the wheel and axle in equilibrium. The forces which act on the wheel and axle are the power P, the weight W, and the reac- tion R of the steps which sup- port each end, C, of the pivot about which it turns, and, in con- sequence of the rigidity of the PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 117 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' after- ward. Then W ascends through the space WW- arc BB', and P descends through PP'=arc A A'. PP' is the virtual ve- locity (Art. 78) of P, and positive ; WW is the virtual veloc- ity of W, and negative. Hence (Art. 81), P.PP'-W.WW=0, or P.arcAA'-W.arcBB'=0, or P.AC.z.A'CA-W.BC.A.BCB'=0. P.AC-W.BC: W AC °' ° r P = BC' the condition of equilibrium found in Art. 138. 192. Prop. The principle of virtual velocities obtains in a pair of toothed wheels. Let the circles in the annexed figure represent the pitch- lines of the wheels (Art. 144), andD 1? D 2 the points which were in contact in the line CC before displace- ment. Since the pitch- lines roll on each oth- er without slipping, arc DDj=arc DD 2 , md P's displacement =PP=AC'.^D 1 C , D=AC W's displacement =WW =CB.z.D 2 CD=CB By the principle of virtual velocities, P.PP'-W.WW =0, arcD.D CD ' arc D 3 D ~CD~ ilS STATICS. or or RAC ,i^-W.CB.^^=0, CD CD „AC Tir CB p w ■ — =0 DC CD W AC CD ■ W CD ; and when AC = BC, " P BC'C'D as found in Art. 143. P ~'C'D' 193. Pkop. TTie 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'=iPF, 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 nega- tive. .-. P.PP-W.WW'=0, or P.PP'-W.|PP'=0. W Hence -p-= 2, as found in Art. 158. 194. Prop. The principle of virtual velocities obtains in the first system of pulleys. In the figure of Art. 160 we see that, if W be raised through a space WW', 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'=0 becomes P.7i.WW'-W.WW'=C or Y= 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 de- scends through the space PP', pp, the pulley a n would rise through a space =_ it~> PP' •i > 2 &c, &c, pp/ (( n « « » " 3 2" -2 ' pp PP' " a j or the weight W " -r^-, And the equation of virtual velocities P.PP'— W.WW'=0 becomes PP' P.PP'-W.— r =0, w or -p-=2", 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 a n _ x will descend a distance =WW. The next pulley a„_ 2 will descend a distance =2. WW' by the descent of a n _ lf and a distance WW' by the elevation of W, oi will descend on the whole (2+1) WW'. Similarly, the pulley a„_ 3 will descend through {2(2 + l) + l SW.W'=(2'+2 + l)WW'. Proceeding in the same way, we find that pulley a„ Hn _^ )t or a 3 , will descend through the space 120 T A T I C S. (2"- 4 +2"-*+, &c 2+1) WW, and a through the space (2"- 3 +2 n - 4 +,&cc 2+1) WW, and a", " « " (2"- 2 +2"- 3 +,&c 2 + l)WW, and P will descend through twice the last found space by the descent of the pulley a l , and through the space WW' by the elevation of the weight ; or PF=WW'.{2(2"-+2"- 3 +, &c 2 + l) + l} = WW.(2"- 1 +2"- 2 +, &c 2 2 +2 + l) =WW(2"-1). The equation of virtual velocities is P.PP'-W.WW'=0, which becomes P.WW(2"-1)-W.WW=0, W or as in Art. 162. =2"-l 197. Prop. The principle of virtual velocities obtains in the inclined plane. Let the force P make any an- gle e with the plane, a= Z-BAC, a the first position of the body whose weight is W, a' the posi- tion of it after displacement. Drawing the perpendiculars av, a'u, we have —a'v=—aa' cos. £= the virtual velocity of Yw' V w~ P, and au=aa'. sin. a— the vir- tual velocity of W. By the equation of virtual velocities, ¥.a'v-W.au=0, or P.aa'. cos. e—W.aa'. sin. c=0. W sin. e Hence as found in Art. 164. sin. a 198. Prop. The principle of virtual velocities obtains in the wedge. PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 121 Let 2P be the whole power, R and R the pressures perpendicu- lar to the faces of the wedge ABC, which produce equilibrium. Let the wedge be displaced to the posi- # tion A'B'C. The displacement of tne point of application of Pis aa'= AA' ; that of b, the point of applica- tion of R, is bb'=Am, a perpendicu- lar from A on A'B', and Am=AA' . BAC sin. — - — . The equation of virtual velocities is Y.aa'-R.bb'=0, BAC or P.AA'-R.AA'. sin. — — =0. R= BAC sin. as found in Art. 169. 199. Prop. The principle of virtual velocities obtains in a lev cr of any form. Let ACB be the lever be- fore displacement, A'C'B' its position afterward. From A' draw A'u perpen- dicular to AP, and from B', B'w perpendicular to BQ, produced. Au is the vir- tual velocity of P, and Bu that of Q. Now when the displacements are indefinitely small, the circular arcs AA', BB' become straight lines, and kv=KA.i cos. A'Au-AC.Z.ACA'. cos. (PAC-90 ) = AC. sin. PAC.Z.ACA'; Bw=BC. sin. QBC.Z.BCB', and aACA' = ^BCB'. The equation of virtual velocities is 122 STATICS. ~P.Av-Q,.Bu=0, or P.AC, sin PAC-Q.BC. sin. QBC=0. _P_ BC. sin. QBC *'• Q~AC.sin.PAC as found in Art. 134. 200. Prop. The principle of virtual velocities obtains in th& single movable pulley with cord's inclined. 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 P', when the weight is raised to W, or the point f of meeting of the cord is raised to A'. Draw the 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'=Aw, BAC PP'=Am+A7i=2AA' cos.-—-, WW=AA*. The equation of virtual velocities is P.PP'-W.WW'-O, which becomes BAC P.2AA'. cos. W or =2 cos. -W.AA' BAC 0, P 2 as found in Art. 158. Scholium. In the preceding propositions, the expression P.PP'=W.WW, PRINCIPLE OF VIRTUAL VELOCITIES, ETC. 123 WW' P or PF ~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 that P is diminished. CHAPTER VIII. FRICTION. 201. The surfaces on which bodies pressed have hitherto been regarded as perfectly smooth, so that they offered no re- sistance to motion parallel to themselves, their only reaction being perpendicular. When rough surfaces are in contact, the motion, or tenden- cy to motion, parallel to the surfaces, is affected by the rough- ness, and the effect is called Friction. Friction may be divided into two kinds : sliding friction, when one rough surface slides on another, and rolling friction, when one rolls on the other. The former only will be consid- 3red here, under the term Statical Friction. 202. The laws of friction are determined by experiment. . If the body A rest upon the perfectly smooth >-F plane BC, the smallest (27 © 35 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=W. 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 ma- terial, when the weight is the same, is independent of the extent FRICTION. 125 of the surfaces in contact, except in extreme cases, whete 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 ex- cepted 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. Cor. The friction is therefore some function of this pressure, and we may represent it by juR. ju. is called the coefficient of F friction, and is equal to ^5-, or the ratio of the friction to the re- action of the plane. 204. Prop. The coefficient of friction between two given sub- stances is equal to the tangent of the inclination of the plane formed of one of the substances, when the body formed of the oth- er 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 from the normal reaction of the plane R, the friction juR acting up the plane, and its weight W acting vertically downward. Resolving parallel and perpendicular to the plane, we have [iR — W. sin. a=0, R— W cos. a=0. Eliminating R, we have or ft cos. a — sin. a=0, ju=tan. a. 205. Prop. To determine the limits of the ratio of P to W on an inclined plane, when friction acts up or down the plane. 126 STATICS. Let the power P make an angle e with the plane whose inclination is a, and W the. weight of the body. 1°. To determine the great- est value P can have without causing the body to move up the plane. In this case, the friction juR, opposing the motion up the plane, will act down it, and, resolving parallel and per pendicular to the plane, we have P. cos. e—fiR—W sin. a—0, (a) P. sin. e+ R-Wcos. a=0. (b) Multiplying (b) by /z and adding to (a), we have P(cos. e+n sin.s) — W. (sin. a-\-\i cos. a) — 0, _, W. (sin. a -\-fxcos. a) or P= * : : . COS. £+jU sin. £ 2°. 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. e+j"R— W sin. for the increment of velocity, or the velocity generated by the force in one second,

the more nearly will Au 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 veloc- ity. 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 num- ber 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=MV. [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 of mass. When so measured, they are called accelerating forces. If the mass moved differs from that which we have called the unit of mass, and it is taken into consideration in estimating DEFINITIONS. 135 the effects of the forces, they are then called moving forces. Let 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 the unit of time, or $=0M. [VI.] $ Hence 0=tT|» or the accelerating force, is equal to the moving force divided by the mass. 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, esti- mated 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 matter, and 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 motion in a body is proportional to the force impressed and in the direction of that force. 227. When one body impinges on another at rest or in mo- tion, 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 mo- tion. 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 follows, and is called the third law of motion : Action and reaction are equal and in oppositt directions. CHAPTER f. 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 uniform motion. X . Let OB be the path of the body (Art. 218), v its veloci- ; ty (Art. 213), and t the time of its motion in seconds. : Let O be the origin, or point from which we estimate the ^ : successive positions of the body, s the distance of the body from O at the end of the time t, and OA=Sj 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, B or s=s,+vt, (36) which is the general equation of uniform motion. If the body, instead of receding from the origin O, approach it, then v will be negative ; or if the body begin to move from A', then s z will be negative. Cor. 1. From (36) we obtain s— s, AB v= -r = — ■ 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 MOTION. 137 point where the body is when t=0, or suppose the space and time to commence together, then s 1 =0, and s=vt. (37) 230. Prop. 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 : 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 u— v', s : s'= t : V. 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', ifs=s', vt=v't'. .'. v : v'=t' : t. 233. Prop. An impulsive force is measured by the momentum /it can produce in any mass. If v be the velocity produced by the force F, in a body con- taining M units of mass, by [V.] the momentum will be Mr. 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, ad- mitting the principle that effects are proportional to their causes, if A be the constant ratio of the force to the momentum, then F 1 =A.ilftj. 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 F, to A, we have F F=-^ = Mu. ( 38 ) Cor. If M=l, F,=A.u, in which X is constant. Hence the force is proportional to the velocity produced in the unit of mass. 138 DYNAMICS. Schol. Since we know nothing of the nature of forces, we can not determine their effects a priori. The foregoing propo- sition 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 transla- tion, 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, what- ever be the direction of these vibrations in reference to an east and west line. 234. If any number of forces act upon a body in the same 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, »>', v" . . . the velocities imparted by each respective- ly, (j) the resultant force, and u the velocity due to ", &c. Since the forces are conspiring, their resultant will be equal to their sum. .-. F+F'+F"+ &c, =cp=X(v+v'+v"+ &c). But (p — Xu. Hence u=v-\-v'\-v"-±- &c. 235. Prop. If the adjacent sides of a parallelogram represent in magnitude and direction the velocities which two forces, by their separate action, would respectively produce in a body, the. diagonal of the parallelogram will represent the actual velocity produced by their joint action. Let AB and AC represent the velocities which the forces P and Q respectively im- part 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 will represent the velocity produced by their result- ant, 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. .-. s=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). 236. Prop. The velocity of a body being given, to find the com- ponent 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, AX and AY the rectangular directions in which the components are required, and 6 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. Then AB=AD. cos. 0, AC=AD. sin. 0, or a=v. cos. 9, b=v.s'm.O. (39) Again, if a and b represent the given component velocities in AX and AY, to find the resultant velocity v we have v = Va'+b% (40) b (41) and as required. tan.0=-, a l40 DYNAMICS. Cor. Hence velocities may be compounded and resolved like forces in statics. 237. Prop. Two bodies, A and B, describe the same path with the velocities v and v', and at the commencement of their motion are at a distance a from each other : to find 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=0 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'=±6. Subtracting (b) from (a), and putting t=t', we have s— s'=vt— a— v't=±b. a±b , x .-. t= -. (42) v — v a±b Also, from (a) s=vt=v. ;, (43) . . ,_. a±b va±v'b , . and irom (b) s'—a+v't—a+v'. -= — . (44) » ' v — v V — V Cor. 1. If the bodies move in opposite directions, v' will be negative, and a±b a±& va^iv'b v+v v-\-v v+v Cor. 2. If the time when the bodies are together be required, then 6=0, and a va v'a t= .-, s= ,=s', s' — a= -. v — v v — v v — v 238. Prop. When two bodies A and B move in the circum ference of a circle with uniform velocities, to determine the cir- cumstances 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 6=0 in (42), we have UNIFORM MOTION. 141 t. = = the time of their first meeting. v — v a+c t a = 7 = " " second » a+2c ,. , * Q = r = " " third " '3 v— V a+(n— l)c t n = ^ — -!-= " " nth v~v' also, s=v — ; — = space described by A, v— v . v'a+v(n—l)c _ and s-a= —, — —= " " B. v—v' Cor. The interval between two successive conjunctions is 239. EXAMPLES. 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. 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 1G min- utes 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 veloci- ty 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 extin- guished, in what time would the intelligence reach us, the velocity of light being 191,000 miles? Ex. 4. Suppose 964 tons of ice to be floating directly to the east at sunrise on the 21st of March, with a velocitv of 12 feet 1 42 DYNAMICS. 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, 21,840 miles, in whai 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 4°.09239 ; that of Venus 1°.60216; that of the earth 0°.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 velocity of the drops ? Ans. 68 feet. Ex. 10. When the path of the earth in its orbit is per pen 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 20",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. IMPACT OF BODIES. 240. Def. When two bodies in motion impinge, if their cen- ters of inertia move in the same straight line perpendicular to a plane tangent to the bodies at their point of contact, the im- pact is said to be direct and central. If the straight line described by the center of inertia of one of the bodies is not perpendicular to the tangent plane, the im- pact 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 ap- proach each other, and during the remaining portion will re- cede 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 comj)ression ; the opposing force causing them to separate again is called \\\q 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 perfectly 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 call- ed 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 DYNAMICS. fectly inelastic, but these states may be considered as limits to the various degrees of elasticity presented in nature. 243. 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. 1°. Let m l and m 2 be the masses or bodies moving in the ©„ same direction with the velocities mz >■ Q >- v l and v 2 {v^v^, 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^mjVj, and F'=m 2 v 2 , and their resultant, F+F'=m l v l +m 2 v 2 . After impact the bodies move on together as one mass, and its momentum (m l -\-m 2 )v must be a measure of the force F+F'. .-. {m l +m 2 )v=m x v 1 -\-m 2 v 2 , m l v 1 +m 2 v 2 or v= ; . (46) m l +m 2 2°. If the bodies move in opposite directions, the resultant of the forces will be F— F'. .'. F— F'=m 1 v 1 — m 2 v 2 = (m 1 +m 2 )v, m l v l —m 2 v 2 or u= , — —. (47) The same result will be obtained by changing the sign of v 2 in (46). Hence the velocity of two inelastic bodies after impact is equal to the algebraic sum of their momenta before impact, di- vided by the sum of their masses. Cor. 1. If two bodies move in opposite directions, with veloci- ties reciprocally proportional to their masses, they will rest after impact. For m l : m 2 —v a : v, gives m v 1 =m 2 v 2 , which, substituted in (47), gives v=0. IMPACT OF BODIES. 145 Con. 2. If m 2 is at rest before irr pact, v 2 =0, and m , v , ?n 1 +m 2 ' and if, at the same time, the masses are equal, v =r Cor. 3. If the masses are equal, and move in the same di reotion, f m opposite directions, 2 244. Prop. In the impact of inelastic bodies there is a loss of living force, and this loss is equal to the sum of the living forces due to the velocities lost and gained by the bodies respectively. For the living force before impact ^m^l+m^l (Art. 224), and " " " after " = (m l +m 2 )v\ .'. the loss of living force by impact is m 1 v 2 +m. 2 v 2 2 —(m 1 +m 2 )v*=m 1 v 2 -{-?n ;i v 2 2 — 2(m 1 +m 2 )v 1 + (m^+m^v* (46), =m l v 2 +m 2 v 2 2 — 2(m l v l +m 2 v 2 )v-\-(m 1 +m 2 )v' 1 =m l (v i — vy+m 2 (v 2 —v)\ which is necessarily positive, and in which v x —v is the veloci- ty lost by m,, and v—v a the velocity gained by m 2 . 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 machinery is impaired. 245. Prop. To find the velocities of two imperfectly elastic bodies after direct impact. Let ?n 1 and m 2 be the masses, v x and v 2 their velocities be- fore impact, and v' lt v' 2 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 l at this instant wil be v l — v. K 14G DYNAMICS. Let e be the modulus of elasticity, or the ratio of the torce 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 l by the force of resti- tution will be e(v 1 —v). Hence the whole velocity lost by m, will be v l — v+e(v 1 — v) = (l+e) (v^—v). This, subtracted from the velocity of m l before impact, will give its velocity after impact, or v\ = v 1 — (1+e) (p 1 —v)=v—e(v 1 —v). (a) In like manner, the velocity gained by m 2 during compres- sion will be v— v 2 , and the velocity gained by the force of restitution e(v—v 2 ). Hence the whole velocity gained by m Q will be (1+e) \V— v 2 ). This, added to the velocity before impact, will give the velocity after impact, or v' 2 = v 2 + (l+e) (v—v 2 )—v+e(v—v 2 ). (b) Substituting in (a) and (b) the value of v (46), and reducing, m 1 +m 2 7n 1 +?n 2 (48) 2 m l +?n 2 ?n 1 +m 2 As in Art. 243, if the bodies move in opposite directions, we must change the sign of v 2 , or, if one of them be at rest before impact, make u^O. Also, if we put e=l, the results will be those for perfectly elastic bodies, or make e=0, the result will be that for inelastic bodies. Cor. 1. If the bodies are perfectly elastic, their relative velocities before and after impact are the same. For, making £=1 in (a) and (b), and subtracting the latter from the former, we have v' l — v' 2 =v 2 —v l . (50) Cor. 2. In the impact of bodies no motion is lost. For, multiplying (48) by m l , and (49) by m 2 , adding and reducing, we have m l v' l +m 2 v' 2 —m l v l -\-m 2 v 2 , (51) IMPACT OP BODIES. 1 47 in which the first member is the sum of their momenta iftei impact, and the second member the sum of their momenta be- fore impact. Cor. 3. If the bodies are perfectly elastic and equal, they will interchange velocities by impact. For, making ?n 1 =?n 2 and £=1 in (48) and (49), we have v 2 = i( v i+ v 2) + 2( v i— v 2 ) = v i- Cor. 4. The velocity which one body communicates to an ithev at rest, when perfectly elastic, is equal to twice the ve- locity of the former divided by one plus the ratio of the masses. Making in (49) m 2 =rm 1 , v 2 —Q, and e=l, we obtain 2 r+1" 246. Prop. When, 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, r^m .... r n ~ 1 m be the bodies, and v the velocity of m. By Cor. 4, Art. 245, 2v the velocity of the second will be third " fourth " nth r+1' 2v 1 2*v 1+r'l+r (1+r) 2 ' 2 2 t> 1 2 s v '(l+r) s, l+r (1+r) 3 ' 2 , - 1 u (1+r)'- 1 * Cor. 1. The momentum of the nth body is T' l v ( 2r Cor. 2. If the bodies are equal, r=l, 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 in contact after impact, and the last will move off with the ve- locity of the first. 1 48 DYNAMICS. 247. Prop. The motion of the common center of gravity vf two bodies after direct impact is the same as before impact. Let v be the velocity of the common center of gravity of the bodies m 1 and m 2 , moving in the same direction with veloci- ties v l and v 2 respectively before impact, and v' the velocity of their common center of gravity after impact. Let x x , x t > x be the distances respectively of the centers of gravity of m, and m 2 , and of their common center of gravity from any fixed point in their line of motion at any instant ; x' l , x' 2 , xl the same quantities after an interval t, so that (36) x' 2 =x 2 +v 2 t, x' =x -\-vt. By (29), Art. 105, (m 1 +)n 2 )x =m 1 x l +m 2 x 2 , (a) (m l +m 2 )x'=m 1 x' 1 -\-m 2 x' 2 . (b) Substituting in (b) the values of x', x' 2 , x\ above, and subtract ing (a) from the result, we have m 1 +m 2 v 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 to- gether. If the bodies are elastic, and v\ , v' 2 are their velocities after impact, then x' 1 =x 1 +v' l t, X 2 X 2 ~T~V 2 t, x' =x -\-v't; from which we deduce, as before, m 1 +m 2 v ' 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 equal ; 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 mo- tion 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 inelastic body after oblique impact upon a smooth, hard, and fixed plane. Let the body m impinge on the plane AC at B, with the velocity v, making the angle of incidence mBN—6. The component of the velocity v parallel to the plane (39) is v. sin. 6 ; and this velocity will not be changed by impact, since the body and plane are smooth. The com- ponent of v perpendicular to the plane, viz., v. cos. 0, will be destroyed by the plane, and, since the body and plane are in- elastic, there will be no vertical velocity after impact. Hence the body will slide along the plane with the velocity v. sin. 0. 251. Prop. To determine the motion of an elastic body after oblique impact upon a smooth, hard, and fixed plane. Let the body m impinge on the plane RS at B, with the velocity v, making the angle of incidence PBN=0. Let PB rep- resent the velocity v before impact. Draw PR and PN perpendicular and parallel to the plane RS. PN=RB=u. sin.0 is the component of the velocity parallel to the plane, and is not af- fected by the impact. PR = NB=u. cos. 0, the component ot the velocity perpendicular to the plane, will be destroyed by 150 DYNAMIC the plane. But the body being elastic, the force of restitution will give it a velocity e.v. cos. 6 in the direction BN. Take BM=e.t\ cos. 6, BS=u. sin. 0, and, completing the parallelo- gram, draw BQ. BQ/rc' is the direction, and BQ, the velocity of the body after impact. Now BQ 2 =v' 2 =:BM 2 +MQ 2 =e.V cos. a 0+w a sin. 9 also, BQ=v'=v Ve* cos. s 0+sin. a ; MQ_ PN _ tan.0 tan - _ MB _ lNB - e ' Cor. If £=1, 6=6', 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'=oD, and 6' =90°. 252. Prop. To determine the direction in which a body of given elasticity must be projected, in order that after reflection from a given plane it may pass through a given point. Let MB be the given plane, A the position of the body before projection, D the point through whi'ch it is re- quired to pass after reflection, and e the modulus of elasticity of the body, supposed known. Draw AMC perpendicular to the plane, and take AM : MC = 1 : e 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 e.AM=MC, MB MB eTAM-MC tan. MAB or or £ tan.0 =tan. MCB ; =tan.0\ as in Art. 251. IMPACT OF BODIES. 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 (52) u, — v 2 254. Schol. 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 lowest points. If, therefore, two spher- 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 veloci- ties v x ,v 2 ,v\, v' 3 will be known, and these, substituted in (52), will give e the modulus of elasticity. The following table exhibits the results of experiments, per- fect elasticity being unity : 2 1 1 2 Substances. Moduli. Substances. Moduli. Glass. . . . Hard-baked clay Ivory. . . . Limestone . . Steel, hardened Cast iron Steel, soft . . 0.94 .89 .81 .79 .79 .73 .67 Bell-metal Cork Bras9 Lead Clay, just malleable by the hand 0.67 .05 .41 .20 .17 255. EXAMPLES. Ex. 1. Two inelastic bodies, weighing 12 lbs. and 7 lbs. re- spectively, 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 tVat gained by the other. Ex. 2. A mass w,, with a velocity 11, impinges on m 2 moving in the opposite direction with a velocity of 5; by impact m l 1 52 DYNAMICS. loses one third of its momentum. What are the relative mag- nitudes of m l and m 2 1 Ex, 3. m,, weighing 8 lbs., impinges on m 2 , weighing 5 lbs., and moving in m^s direction with a velocity of 9; by impact m 2 s velocity is trebled. What was m,'s velocity before im- pact? Ex. 4. Two bodies m x and m 2 are moving in the same straight line with velocities u, and v 2 . Find the velocity of each after impact when 6??i 1 =5m 2 , v l =l, 4o 1 +5t> 2 =0, and C 3* Ex. 5. Two bodies m, and m 2 are perfectly elastic and move in opposite directions; in 1 is treble of m a , but m 2 's velocity is double that of m 1 . 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. m 1 (=3m 2 ) impinges on m 2 at rest; yn/s velocity after impact is f of its velocity before impact. Required the value of e, the modulus of elasticity. Ex. 8. Two bodies m i and m„, whose elasticity is f, mo-ing in opposite directions with velocities 25 and 16 respectively, impinge directly upon each other. Find the distance between them 4t\ 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 e, 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. Re- quired the angle 8 made by the direction of projection with tr>e radius at the given point. 3 e 2 Ans. Tan. 6= .-— -3 . 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 ve- locity 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 succeed- ing 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. Putting/ for the force, and v for the velocity acquired in the time t, we have, as before, v=ft. (53) 257. PRor. To find the space in terms of the force and time when a body moves from rest by the action of a constant force. 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 intervals reckoning from the commencement of the time, will be t 2t 3t 4d (n—\)t nt ~i > j t CZtC.f , . n n n n n n By (53), the velocities at the end of these intervals will be> 1 n ,J n' J n' J n ' ' * ' J ' n ,J 'n' 154 DYNAMICS. If now the body moved uniformly during each interval - 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 - with the velocity it had at the beginning of the interval, we should have „ f r 2f „3f „ „(n-l)* 8 ' s=0+f.~+f—+f~+, &c, . . . f> ~, J n J n J n J n f =/.-{l+2 + 3+, &c, . . . (w-1)}, / n t~ n(?i—l) n*' 2 ' f f = f-2-f'¥n ^ But if the body moved uniformly during each interval - with the velocity it had at the end of the interval, we should have , * a , , M 3t\ (n-l)f nt J n - n J n J n J n f =/.^(l+2+3+,&c, n), f n(n+l) ~~J'n v 2 ' f f Since the velocity is uniform during no sensible interval, the 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=±f? and sx t\ (54) Hence the space described from rest by a oody from the action of a constant force is equal to half the product of the force by the square of the number of seconds, and the spaces vary as the squares of the times. 258. Prop. To determine the space in terms of the force and velocity ; also in terms of the time and velocity. 1°. Eliminating t from (53) and (54), we have v 2 s=-7, orsx^. (55) 3 2°. Eliminating/ from (53) and (54), we obtain s=^vt, or socvt. (56) Cor. 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 ac- quired velocity. For the space s„ described in the time t, with a uniform ve- locity v, is, by (37), s 1 =vt, which, compared with (56), gives 259. 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 x =\ft\ (a) The space described in the time (t—n) is Subtracting (b) from (a), we have, for the space s described in the last n seconds, g=s x -s a =%f(2nt-n 3 ). (57) Cor. If the space described in the m seconds next preceding the last n seconds is required, we have, for the space s 3 , de- scribed in the time t— (n+m) seconds, s a =^f(t—n—my, which, subtracted from (b), gives for the space s required, s=s 2 -s. J =$f(2mt-2mn-m u ). (58) 156 DYNAMICS. 2G0. Prop. 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 im- pulsive and constant forces will be equal to the sum or differ- ence of the velocities due to each, according as they act in the same or opposite directions. If, therefore, v J be the velocity of projection, the whole velocity v, at the end of the time t, will be »=P t ±/t (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 1 t±^ff. (60) 261. Prop. 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 x by the action of the constant force. Then (55) . v\ = 2fh, and for the space (h±s) v*=2f(h±s), =2fh±2fs, =v\±2fs, (61) the signs to be taken as in the last proposition. 262. Prop. When a body is projected in a direction opposilt to that in which a constant force acts, the velocity acquired in returning to the point of departure is equal to the velocity of projection. If t), be the velocity of projection in the direction AB, from (61) we have *=~w (a) ACTION OF A CONSTANT FORCE. 157 Now the actual velocity v of the body is continual- ly diminished by the action of the force/, and when the body has reached its greatest distance from A, as B, v=0. Hence (a) gives s=^jr, or v 1 =V2fs. But the velocity v acquired in moving from rest through BA=s is, by (55), v= Vlfs. , Cor. 1. The velocity of the body at any given dis- tance from A is the same in going and returning. For in the expression (a), since u, and 2f are constant, v is always the same for the same value of s=AC. If u>Uj, s is negative, or on the opposite side of A from B. 2v Cor. 2. The whole time of flight T=-^. Making v=0 in v (59), we have t=-j-, when the body is at B. But, in returning, it acquires the velocity v l =ft. Hi /' V Hence the time of return is t—-^r, and the whole time 2£=T= 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 uni- form density, or one whose density is the same at equal distan- ces from the center, attracts a body exterior to it as if the mat- ter of the sphere were collected at its center, and with a force varying inversely as the square of the distance of the body from 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 DYNAMICS. 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 li)77 ' i:!!)1 , or about the 2000th part of that at the surface ; a variation too small to affect sens- ibly the circumstances of the motion of a falling body com puted on the hvpothesis 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 in- sensible, 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. Schol. 2. Representing the intensity of gravity at the surface of the earth, as before, by g, we have, from (54), s=\gf, and if t=l, s=hg, or g=2s from which it appears that the acceleration is equal to twice the space described ir. 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 substi- tuting g for /, we have the following relations, supposing no resistance from the air : ^=W=~=U V , (54') • 2s v = gt=~r = V2gs, (53') v 2s / 2s , fflrt t= - =— ==V— , (56') S v V g 0*5') g V V g v _2s _o 5 1 = F = 2? ACTION OF A CONSTANT FORCE. 159 Also, when a body is projected with a velocity v 1 vertically upward or downward, (260) and (261), s=v 1 t±lgt\ (60') tf=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=i£i'=16 T VXl21 = 1946 T V feet. By (53'), v= gt=32} X 11= 353| feet. Ex. 2. Find the time in which a falling body would acquire a velocity of 500 feet, and the height from which it must fall. ^ v 500 . By (56'), t= - =—- = 15.544 seconds. „. , , v 2 250000 By (54'), s=— =-^-=3886 feet nearly, or *=4*u=£x 15.544X500=3886 feet, as before. Ex. 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=3811136lbs. 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 fall 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- 1 60 DYNAMICS locity of 100 feel. What is its velocity at the end of 5 sec« onds? ■ = - .' o. Ex. 9. A body is projected vertically upward with a veloci- ty of 100 feet. Find its velocity at the end of 5 seconds, and its position at the end of 8 seconds. Ans. v=— 60f feet. s=— 229^ 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 veloci- ty »,. Find the time at which it is at a given height h in its ascent. If t be the time required, (60) gives h=v 1 t— \gf\ , 2v. 2h whence t—t — H =0, g g v,±Vv*—2gh or t =— — . g 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 in- terval between the times of its passing a point whose height is h in its ascent and descent is 21. Find the velocity v of pro- jection, and the whole time T of its motion. Ex. 13. A body whose elasticity is £ is projected vertically upward to the height h 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 force 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. Prop. The path of a body, moving under the joint in- fluence 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- jection from the point A, in the direction ANY, and t the time in which the body will describe AN=y, with the uniform veloci- ty v, if gravity do not act. Let AM=a; be the space through which the body will fall in the time t by the action of gravity. Completing the parallelogram MN, the actual place of the body at the end of the time t is ^' P. By (37), AN=y=vt; and by (54),AM=NP=x=%* 2 . Elim- X inating t from these two equa- tions, we get y — — x. g («) If h be the space due to the velocity v, or the space through 1G2 DYNAMICS. which the body must fall to acquire the velocity of projection (55), v i —2gh, which, substituted in (a), gives y'=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 parame- ter to the diameter at that point, which is the distance from that point to the directrix or to the focus. 267. Prop. To find the equation to the path of a projectile when referred to horizontal and vertical co-ordinate axes. Let v be the velocity of pro- jection in the direction AN, which makes with AX the angle of elevation NAX=a, APB the path of the body, AM=z, PM=?/, the co-or- dinates of the point P, t the time in which the body de- scribes the arc AP, and pro- duce MP to meet the direc- tion of projection in N. Then AN =vt, NP=%* 9 , and MN=vt sin. a. Also, AM.=x—vt. cos. a, (a) and ~PM.=y=vt sin. a— \gf. (&) Eliminating t from (a) and (b), we obtain v=x tan. a— x" 1 —* p-, (63) y 2v cos. a and, substituting the value of v*=2gh, PROJECTILES. 163 ^ Xian - a -ih^ra (64) 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 Jlight is the time occupied in describing APB. The height through which the body must fall to acquire the velocity of projection is call- ed the impetus. 269. Prop. To find the time of Jlight of a projectile on a hori- zontal plane. At the points A and B the ordinate y=0. This value of y, substituted in (b), Art. 267, gives vt sin. a— hg?=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 time 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 u, in (60), and we make s=0, we get, as before, 0=vt sin. a— ^gt 7 , . 2v sin. a t =0, and t= . g 270. Prop. To find the range of a projectile on a horizontal plane. Put y=0 in (64), and we have x* 0—x tan. a— —. : — ; 4/i cos. a from which we obtain for the point A, x=0, for the point B, x=4h sin. a cos. a, or AB=R=2A sin. 2a, (66) the horizontal range required. Cor. 1. The horizontal range is greatest when a=45°. For in this case 2a = 00°, and sin. 2a=l. .•. x=2h, 1G4 DYNAMICS. or the greatest horizontal range is equal to twice the height due to the velocity of projection, or twice the distance from the point of projection to the focus of the trajectory. Cor. 2. The range is the same for any two angles of eleva- tion, the difference between which and 45° is the same, or for (45±0). For sin. (90° +26) = sin. (90° -26), or sin. 2(45°+ 6)= sin. 2(45°- 6). If, therefore, we put either 45°+0 or 45° — 6 for a in (66), the value of R remains the / 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 elevation 45°— 6, for which B ' B the path is APB', and to 45° + 6, for which the path is AP"B'. Cor. 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=2/*sin. 2a, and R'=2/i sin. 2a'. _ sin. 2a' , k ••• R '=inr^- R - < 67 > Cor. 4. Since the horizontal velocity of a projectile is urn- form, the range is equal to the horizontal component of the ve- locity into the time of flight, or _ 2v sin. a 2v* sin. a. cos. a. K=v. cos. aX = =2h sin. 2a, as before. g g 271. Prop. To find the greatest height which a projectile at- tains. 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 sin. a . ,,. flight. Putting t= in (b), Art. 267, we have PROJECTILES. 165 H= v sin. a g v sin. a tv sin. a g h sin. 2 a. (68) 272. Prop. To find the co-ordinates of the point where a pro- jectile will strike an inclined plane passing through the point of projection, the range jn the inclined plane, and the time of flight. Let y=x tan. (3 be the equation of the line AC, which is the in- tersection of the inclined plane, with the vertical plane of the path of the body. Substituting this value of y in (64), we obtain, after reduction, for the abscissa of the point C, M 33 _4h cos. a. sin. (a— (3) cos. (3 (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. (3. sin. (a— (3) y ~ cos. 2 /3 To find AC=R', multiply (a) by sec. (3= cos. j3' (*) and we have kri „. _ 4/j. cos. a. sin. (a— 0) 4 . AC=R'=a;sec.|3= — -i ^. (G9) cos. (3 v 7 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 ab- scissa AM with the horizontal component of the velocity. Hence (37), dividing (a) by v cos. a., we get 4/tsin. (a— (3) v. cos. (3 ' 2vs'm.(a—(3) g. cos. (3 (70) 166 DYNAMICS. 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, \/^ feet. (71) V w u=1600 w 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. , x , u 3 3600 From (54'), h=— =-—^-=55.96. 2g 64g- From (66), R=2A sm.2a=2h. sin. 30°=2./L|=/*=55.96 From (68), H = A sin. 9 a. Log. sin. a= 9.4129962 " " = 9.4129962 h = 1.7478777 H = 3.7486 " H = .0.5738701 „ _ 2v sin. a From (65), T= . Log. 2v= 2.0791812 Log. sin. a= 9.4129962 a.c. " g= 8.4925940 .-. T = 0.96554 " T= 9.9847714 Ex. 2. A body is projected at an angle of elevation of 45° 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 and the time of flight is 15 seconds. Required the angle of eleva- tion, velocity of projection, and greatest altitude. Ans. a =74 o .33'.09". v =250.29 feet. H=904.69 " PROJECTILES. 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 that 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 the altitude? Ex. 7. A body is projected at an angle of elevation of 60°, with a velocity of 150 feet. Find the co-ordinates of its posi- tion, 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 45°, fell upon the plane at the distance of 250 feet from the point of pro- jection, which is also in the inclined plane. Required the ve- locity 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 de- scent 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 of 168 DYNAMICS. elevation of the plane being 28°.45', and that of the mortar be- ing 48°.30' ? Ex. 12. Find the velocity and angle of elevation that a pro jectile may pass through two points whose co-ordinates are x=300 feet, ^=60 feet, z'=400 feet, and y'=40 feet. Also find the horizontal range, greatest altitude, and time of flight. CHAPTER V. CONSTRAINED MOTION. § I. MOTION ON INCLINED PLANES. 275. Prop. 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 the inclined plane CA, whose in- clination is a, M be the position of the body at any time t from rest at C, CM=s, and v— the velocity at M. The force of gravity g, acting in the direction Mg, may be resolved into two forces, one f=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. If, therefore, this value of/ be substituted in (53), we have v —gt sin. a, (72) in (54), we have s =%gf sin. a, (73) in (55), we have v"—2gs sin. a, (74) from which all the circumstances of the motion may be de- termined. 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 /i=BC, the height, by (74), v' ! =2g.s. sin. a =2g.CB. 170 DYNAMICS .*. v = V2gh t 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=k' 5 sin - «> BC or AC=%«' •• <= AC V; 2 ^:bc x AC, when BC is constant. 278. Prop. To find the relations of the space, time, and veloc- ity 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 t dtgt sin. a, (75) in (GO), we have s =v l t±^gt' 1 sin. a, (76) in (61), we have v*=v\±2gs sin. a, (77) which give ail the circumstances of the motion of the body. 279. Prop. The times of descent down all the chords of a cir- cle in a vertical plane, drawn from either extremity of a vertical diameter, are the same, and equal to that down the vertical di- ameter. Let AB be the vertical diameter, BD and AD any chords drawn from its ex- tremities, and DC perpendicular to AB. To find the time of descent down BD we have, from (73), BD: „BC ■ \gf sin. B DC ,_2BD 2 ~>.BC 2.BA 8 CONSTRAINED MOTION. 17] or t- which, by (5C), equals the time down the diameter In the same manner, DA=fcf Cj DA' from which we obtain, as before, 2.BA g 280. Prop. To find the straight line of quickest descent from a given point within a vertical circle to its circumference. Let P be the given point. Draw the 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 AC. Since BC=AC, BO=PO. With O as a center, and radius PO, describe the circle PBD, which will be tangent to the circle C at B. Now the time down PB will equal the time down the verti 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 circum- ference of O, 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 Q 172 DYNAMICS. = ECA, the triangle EAC is isosceles. With E as a centei, and EA or EC as a radius, describe the circle ACD, to which BCG will be a tangent. Now the time down i\.C 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 suspended from the extremities oj a cord passing over a fixed 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 — & j g g respectively. Neglecting the rigidity of the cord, the inertia and friction of the pulley, if P=Q, they 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 their weights. But this being a moving force {Art. 225), is equal to the accelerating force into the mass moved, =/( ) , where / represents the accelerating force. Hence p_Q=/?±5 g , 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, \g is an additional force to be overcome by the difference of the weights, and in this case P-Q ® © f=g-v P+Q+Ig-" CONSTRAINED MOTION. 173 This is the formula for Atwood's machine, an instrument fo* illustrating the laws of falling bodies. The friction of the pul- ley 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 ren- der the time, space, and velocity easily determinable. § II. MOTION IN CIRCULAR ARCS. 284. Prop. When 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 from the point D {Fig., Art. 279) down the arc DA. Since the reaction of the curve is always per- pendicular 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 C A of the arc. Hence, by (55), u a =2g€A AD S = 2 *AB_ ocAD. When the arcs are small, the velocities are nearly proportional to the arcs, the principle referred to in Art. 254. 285. Prop. When a body is constrained to describe the sides of a polygon successively, to find the velocity lost in passing from one side to the succeeding one. Let AB and BD be two adjacent sides of the polygon, w the angle made by their directions, and v the velocity of the body at the point B. In passing from AB to BD some of the velocity will necessarily be lost. Resolving the velocity in the direction BD, we have, for the velocity in BD, 174 DYNAMICS. V COS. W. This, subtracted from the primitive velocity v, will give for the velocity lost, Uj=U — v cos. w =v(l — cos. w) = v versin. w v sin. 2 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, iv, and consequently its sine, becomes indefinitely small. In this case, versin. w, small in comparison with 2, may be re- iected, and v ■ a v , =- sin. w. But if sin. w is infinitely small, sin. 2 u> is infinitely smaller, and the velocity lost 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. Let ABC ... be the polygon, and let D the material point describe AB with the velocity v in the unit of time. If, when \z 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 direc- tion BN, such as to cause it to describe BN in the same time that it would describe BD, the point will describe the diagonal BC, the succeeding side of the polygon. CONSTRAINED MOTION. 175 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 trian- gle 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 cir- cumscribing 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, 2r RC 2 and /=BN= . But BC = AB is the space described in the unit of time, and is therefore represented by v. Hence or, the intensity of the impulse is equal to the square of the ve- locity in the polygon divided by the radius of the circumscribing circle. 288. Cor. The force which must continually urge a material point toward the center of a circle, in order that it may de- scribe the circumference with a uniform velocity, is equal to the square of 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 frequency of the impulse in- creased 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 be- comes 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 176 DYNAMICS. V If the mass of the body be taken into consideration, the moving force {Art. 225) will be F=mf=^. (80) It is obviously immaterial whether the body be retained in ts path by the reaction of a smooth curve, or by an inextens- ible 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 centripetal force. The tendency which the body has to recede from the center, in con- sequence of its inertia, or the resistance which it offers to a de- flection from a rectilinear path, the resistance being estimated in the direction of the radius, is called a centrifugal force. 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. 1°. When the masses are equal, by (79), j r Tf r be constant, / ac v- ; or, when a body moves in the circumference of a circle by means of a cord fixed at the center, the tension of the cord, ot the centrifugal force, will vary as the square of the velocity. 2°. If v be constant, /4 r or, if equal bodies describe different circles with the same ve- locity, the centrifugal forces will be inversely as the radii of the circles. or CONSTRAINED MOTION. 177 3°. Let T :e the periodic time, or time of one revolution. Then (37) vT==2nr t , 4ttV 2 (79), =fr. 4?rV ••■ f=- T r. (81) r or, the centrifugal force varies directly as the radius of the circle, and inversely as the square of the periodic time. 4°. If T ac r\ by substitution in (81), , r 1 or, when the squares of the periodic times are as the cubes oi the distances from the center, the centrifugal force will be in- versely as the square of the distance. 5°. If o) be the angle subtended by the arc described by the body in the unit of time, w is called the angular velocity. As- suming for the angular unit the angle whose arc is equal in length to the radius, v=ru. Substituting this value of v in (79), f=ror- (82) or, the centrifugal force 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 addi- tion to the other causes of variation. The foregoing principles admit of a very simple and satis- factory illustration by means of an instrument called the whirl- ing table. 291. Prop. To find the relation of the centrifugal force 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 i —2gh. This value of d s , substi tuted in (79), gives M 178 DYNAMICS. f J*gh J- r > f 2h or -= — . g r Hence the centrifugal force is to the force of grivily as twice the height due to the velocity in the circle is to the radius of the circle. Cor. Uf=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. Prop. To find the centrifugal force at the equator. By (8i), /=^?- The equatorial radius R of the earth is 3962.G 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 5 =86164 9 . These values, substituted in the above, give /=0.1 11255 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+0.1 1 1255=32.200795, f 0.111255 1 and G = 32.200795 = 289 nearly; (83) or the centrifugal force at the equator is — — the force of 289 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 COXS T R A i N E U MOTIO N. 179 R. JT f'-f-'F' T"' But in this case R=R', and/' is to become equal to G. .-./:G= 1 1 "fT12 • rp/2» or and .J rp2 rp2 TH G ~289 ' T' =— f nearly. 17 J 1 Hence the earth must revolve in y-th of its present period. 295. Prop. The centrifugal force diminishes gravity at dif- ferent places on the earth's surface in the ratio 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 the axis, AC = R the equatorial radius, B any point whose latitude is the arc AB, measured by the angle ACB=0, and BG=R' the radius of the parallel of lati- tude passing through B. Let the centrif- ugal force at B, which acts in the direc- tion GB, be represented by BD. 47T 2 R' By (81), BD^ 4 ^. Resolving this in the direction of the vertical CZ, opposite to that in which gravity acts, and calling this component/', we have /' = B6=-7pj-cos.0. But R'==R cos. (p, ■'. f- 4tt ! R COS.* 0, T , ™ r, (84) in which the coefficient of cos. 2 is constant for all latitudes. Hence f ac cos. 2 0, 180 DYNAMICS. f being the diminution of gravity by the actio.Q of the centrif- ugal force. The latitude of Middletown being 41°.33'.10", the centrif- ugal force/', 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' =£'+/'= 32.224385. 296. Cor. Resolving the centrifugal force at B in a direc tion perpendicular to the radius CB, we have 4ttR f'=Bd=-~r cos. <(> sin. , 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 horizontal parallax, is found by astronomical observations to be at a mean 57'. 1". This angle, when subtended by the radius of the earth in the , , . R i latitude of 35°.15'.52", is 7r=56'.57".33. Now -5-= sin. *=-, from which we find ?i=60.3612. P=2360585\ These values, substituted in (a), give J=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 '~}V : 'rfR* ; g whence g' =-5. The intensity of gravity in the latitude of 35°.15'.52", uti 182 DYNAMICS. diminished by the centrifugal force of the earth in its diurna. motion, is £•=32.24538. Substituting for n and g their values, we find g g' = 0.00885. The difference between/ and g' is less than one ten thousandth of a foot, a difference which may be attributed to errors of ob- servation. § III. 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. Pkop. To find the force by which the pendulum is urged in the direction of its path. Let M be the material point suspend- ed from C by the line CM. When re- moved from its vertical direction CM to the inclined position CM', gravity will cause the point to descend and de- scribe the arc M'M. Resolving the force of gravity g in the directions M'B and M'T, parallel and perpen- dicular to the radius CM', the com- ponent g cos. 6, 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 sin. 6. Let CM.=l, the length of the pendulum and the arc M'M=* the semi-arc of the vibration; then s=W. .-. f=g. sin. -. CONSTRAINED MOTION. 183 Now if x be any arc of a circle, it is shown in trigonometry that x» Hence ^(j_^+_±__), &c. But when the arc 5 is small compared with I, the cube and higher powers of j may be neglected, and 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 fixed point 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 in- tensity of the force and the time of its action, or will be meas- ured by the product of the force by some function of the time. Hence s—f.(ji{i). Since s varies as/, let s=nf, n being the constant ratio of the space to- the force ; then, by substitution, n= As the circumstances of the material point at the end of the time T are the same as at the commencement, it will then per- form another vibration in the same time, and so continuing, all its vibrations will be isochronal. 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 of different pendulums. 1°. Since in (86) it is constant, TocXl; 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. 2°. If g be constant, which is the case in the same latitude, and at the same elevation above the surface of the earth, T x Vl; or, the time will vary as the square root of the length. 3°. If / be constant, or the length of the pendulum remain the same, then T.-L vg or, the time of a vibration will vary inversely as the square rooi of the intensity of gravity. CONSTRAINED MOTION. 185 4°. If T be constant, since lecg; or, the lengths of pendulums vibrating in the same time vary as the accelerating force. 304. Cor. 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 intensi- ties of gravity at those places. Since these intensities of grav- ity at different places are dependent upon the figure of the earth, they will serve to determine it. The pendulum, therefore, be- comes 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 vibra- tion by each will be T=|andT'=|, But T : T'= Vl : V7=^ . H Hence 1: Z'=N' 2 : N 2 . Also, Z'N' 2 W' (87) 306. Sciiol. 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 / of whose pendulum is required, be regulated to vibrate seconds. Sus- pend in front of the clock a pendulum of known length /'(/' N N'' 1 1 ° r (r+h) .: N : N'=r+h : r, and N' N-N'= r+h AN r+h N =— ;- (89) 1+ h N— N', the loss in a day, is the rate of the clock. Hence, divide 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 denomina- tion. 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 . r(N-N') , x h= 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 /= the length of the cord AP, fixed at A and attached to the body at P, VC = r the radius of the circle which the body i88 DYNAMICS. •describes with the uniform veloc- ity v, m— the mass of the body, and the angle PAC = 0. Since the body is in the same circum- stances 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 Ail) 2 PA, the centrifugal force /=—^-, in the direction CP, and the weight of the body nig downward. Resolving horizontally and vertically, we have mv" t sin. 6 — 0. t cos.O—mg =0. From (b), we have the tension in the cord. mg cos. d' Eliminating t from (a) and (b), a gr sin. d (a) =gl cos. sin. 2 cos. d' The time t of performing one revolution is _2nr cos. 6 g AC g ' 313. Dr. Bowditch, in the second volume of his translation of the Mecanique Celeste, gives the following formula for com- puting the length / of the seconds pendulum in any latitude /=A+wsin. 2 0, in which A=39.01307 in. and w=0.20G44 in. CONSTRAINED MOTION 189 314. EXAMPLES. Ex. 1. The length of an inclined plane is 400 feet, its height 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 25°. 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 in- clination 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 fths 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 W 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 from the bottom of the plane with a velocity equal tc 190 DYNAMICS. that acquired in falling down a similar inclined plane n times its length. Where will they meet ? Ex. 10. A ball having descended lo 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 the 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 from 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 mount- ain, lost 48.6 seconds in a day. What was the height of the mountain? Ex. 18. The length of a pendulum vibrating sidereal sec- onds, 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 RIGID BODIES. 315. Prop. When a rigid body, containing a fixed axis, is acted upon by a given force in a plane perpendicular to that axis, to determine its motion. Let the annexed figure be a section of the body by a plane perpendicular to the axis at C, and the given force F act at B, in the direc- tion AB, at a given distance from C. If v be the velocity which the force F can impart to a mass M when free, F=Mu. When the force acts on the body, each parti- cle is constrained to move in a circle whose center is in the axis through C. If w 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 rw, and its momentum f=mro). The mo- ment of this force, in reference to the axis through C, is rf—mr^di. The moment of any other particle will have the same form, and hence rf+r'f'+, &c, =»ir 2 (,)+m'r' 2 w+, &c, or 2.r/*=S.mr 2 w=w27?ir 2 , since w is the same for each particle. But the moment of F is rF=Mu.CB, and, as the former is a measure of the effect of the latter, w2.mr=Mu.CB. Mu.CB .% "=^ — r « (°i) z.mr Thus, if the body consisted of two particles m and m' at the distances r and r from the axis, and the force F would impart to m, when free, the velocity v, then the angular velocity 192 D Y N A M I C 3. mvr mr +m'r' and the linear velocity of m will be mvi fi mr % -\-m'r ri ' If the two particles were equal, and at equal distances from the axis, then mr^-rn'r' 2 , and Hm=m', and r'—2r, v v r«u=— . 5 Cor. If the body were m motion before the force acted on it, then the change of angular velocity will be given by (91). 316. Def. 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 from the axis. Thus, if m be the mass of a particle of a rigid body, r its distance from a fixed straight line, 2.??zr 2 is the moment of in- ertia pf the body about that line. The definition given above is to be regarded as a verbal enunciation of this analytical ex- pression, which has required nomenclature by the frequency of its' occurrence in dynamical investigations. 317. Prop. 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. Let m be a particle of the body, and let the plane ?nCG, passing through m, cut the two axes at C and G. Through the axis at G pass a plane perpendicular to CG, intersect- ing the plane of the figure in AB. Let mC=r, mG=r„ and CG=/*. Draw mF per- pendicular and niD parallel to AB. Then r=r;+/( 2 ±2/i.GD, according as the angle mGC shall be obtuse or acute. Hence ROTATION OF RIGID BODIES. 193 mr^=mr\+mli i ±2mh.GT>, and 2.mr*=2.mr 2 1 +h 2 2m±2h2.m.GT). Now, since the plane through the axis at G passes through the center of gravity, by (29), 2.m.GD=2.m.Fm=0. .-. 2.mr 2 =2.mr?+A 2 2m =Zmr\+Mh\ M being the whole mass of the body. 318. Schol. Assuming k, such that Mk i =2.mr a 1 , we have 2.mr 2 =M(F+A 2 ). MA- 2 is the moment of inertia of a body about an axis through t he center of gravity, and M(F+/j 2 ) is the moment of inertia about a parallel axis at a distance h from the former. The moment of inertia about any axis will, therefore, be easily de- termined when the moment of inertia about an axis through i r ■ • i «■ 2.mr 2 . the center of gravity is known. Since k — M , the determ- ination of k will, in general, require the aid of the integral cal- culus. « The length VA 2 +/j 2 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 k is the least value of V/f+h*, it is sometimes called the prin- cipal 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 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 if free in each instant of time, the moment of N 194 DYNAMICS. gravity will be Mu.CA. By Art. 315, Cor., tne change of an- gular velocity produced in each instant of time is _Mi?.CA 2.mr a ' 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.CO 3 , and the change of angular velocity in the same instant of time would be w '~"M.co a ' But the position of the point O being arbitrary, CO, and there- fore CB, will vary at pleasure. We may, therefore, put Z.mr' : M.C0 3 =CA : CB=CG : CO, (a) which gives CA CB S.mr* M.CO" or w=w' ; that is, the change of angular velocity is the same in the two cases, and, therefore, the motion from rest will be the same. 320. Cor. The point O may be found from (a), which gives S.mr* M(& 2 +/i 2 ) CO: M.CG - M.h _k 2 +h 2 h =h+j. (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 de- fined : 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. ROTATION OF RIGID BODIES. 195 322. Cor. Her.ce, when a body makes small oscillations about a fixed horizontal axis, it is only necessary to calculate the value of the expression /j+-t-=/, and the time of a small oscillation will be *\J 1 -- (86). 323. Def. A body of any form suspended from a fixed axis, about which it oscillates by the force of gravity, is called a compound pendulum. 324. Prop. The centers of oscillation and suspension are con- vertible. 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), GO-h=h'=j, and h=- ; (93) so that if h' be the distance from 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 of gravity, perpendicular to the axis of motion, points be taken at distances h and h' from the center of gravity, and on opposite sides of it, then the length of the equivalent simple pendulum is h+h' ; 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. Cor. 2. From (93), we have hh'=1?\ or, the principal radius of gyration is a mean proportional /»; 196 DYNAMICS. tween the distances of the centers of oscillation and suspension from the center of gravity. 327. Cor. 3. Since h : Vk*+h*= Vk*+h* :h+-r=CO, we in* h 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 consequent- ly 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. Weights 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 ob- tained 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 ac- curacy 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 force 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 per- pendicular to its length : required the length of the isochronal simple pendulum. ROTATION OF RIGID BODIES. 197 Let 2a be the length of the line, and h the distance of the point of suspension from its center of gravity, which is its mid- dle point. The radius of gyration of a straight line about an axis through its center of gravity, perpendicular to its length, a is ~7r. 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), vJL* h 3k' 2 and {Art. 325), l=h+h'=h+-r. Oil 1°. Let the point of suspension be at the extremity of the line, in which case h=a. Then h'=\a, and /=a + i« = |.2«, that is, the center of oscillation is two thirds of its length below the axis of motion. 2° Let h=\a; then h'=\a and Z=|.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 oscilla- tion and of suspension. 3°. If h=%a, h'=$a, and Z=}fa=f.2a+i.2a; or, when the center of suspension is three eighths of its lengtn from one end, the center of oscillation is one sixth of its length below the other end. 4°. lfh=0, h'=ao, and 1= ao ; or, when the center of suspension is at the center of gravity, the length of the equivalent simple pendulum is infinite, and 198 DYNAMICS. 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 from the action of gravity. Ex. 2. A sphere being made to oscillate about a given axis : required the length of the equivalent simple pendulum. Let r be the radius of the sphere, and h the distance of the axis of motion from its center. The principal radius of gyra- tion of a sphere is rV%. 2r 2 Hence h'——r . bh 1°. Let h=r, or the spnere vibrate about an axis tangent to its surface. Then &'=§r, and /= r +| r; or, the center of oscillation is two fifths of the radius distant from the center of the sphere. r 2°. If/*=10r, 4 '=35' and ;=10 ''+i 3°. If h=\r, h'=2r, and r o 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 2.mr*, m being the portion of the mass which is at the distance r from the axis of rotation ; when the number of points becomes in- definite, the expression will evidently become fr'dM, dM be- ing an element of the mass at the distance r. Hence (Art. 318), Mk*=fr*dM. (a) ROTATION OF RIGID BODIES. 199 We shall now illustrate the method of determining M/t a 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 thick- ness and density of the line are supposed to be uniform, eacl may Vv j taken =1. .-. M=a+b and dM=dr. r 3 Hence (a) (a+&)& 2 ==/VWr=— +c, o and the integral being taken between the limits —a and +b. . v= aa+b ° 3(ci+b)' If a=b, or the axis be at the center of gravity, we have, for the principal radius of gyration, a h= V3' __„ Ma 2 2 , and MF=— =-a 3 , o o 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=27rr, and since all the points are at the same distance from the axis, r is constant. ... Mk^fr i dM=rydM=2nr 3 , and k=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 na?, and at a distance r from the axis the area will be 7rr 2 . Hence, in this case, dM=2nrdr and MA a =7raT=2rr/r*rfr=$7rr 4 =£7ra 4 , when r=a, and 200 DYNAMICS. Ex. 4. To find the moment of inertia of the circumference of a circle about a diameter. Taking the proposed diameter for the axis of a:, the distance of any point (x, y) from the axis of rotation is y. From the equation of the circle, y i =a 2 —x', we find Vdx i +dy i =dM=-dx. y .: r-dM=y*dM=aydx, and Mk' i =2nak' i =afydx=aX area of circle=Tra 9 . a ' " ~V2' Ex. 5. To 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 generated by a circle of variable radius, y, whose center moves along the axis. By Ex. 3 the moment of this generating circle is ^y\ Hence the whole moment will be the sum of the moments of the generating circle in all its posi tions. .-. Mk' i =\7ify i dx. Buty^tf-x*. .'. Mk i =\-nj\a i -x i ydx, and this integral, taken between the limits +a and —a, is But M=A*a'. .-. A=aV§. CHAPTER VII. 331. The methods employed in the preceding chapters to 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. We now proceed to show, to a limited extent, how the circum- stances of the motion of a body or point, in such cases, may be determined. In order to this the fundamental formulas will need some modification and extension. 332. By Art. 214, variable velocity at any instant is meas- ured by the limit of the ratio of the space to the time, that is, from the nature of the differential coefficient, it is the differen- tial coefficient of the space regarded as a function of the time Hence In like manner, it appears from Art. 223, that the force, when variable, is measured at any instant by the differential coeffi- cient of the velocity regarded as a function of the time. Hence *=-. [VIII.] From [VII.], by differentiation, we obtain dv =[ix n . (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.], where represents the velocity generated by the force in the unit of time. If v l 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 MOTION OF A FREE POINT. 203 Substituting this value of v in [VII.], we have ds=v l dt±(ptdt, and, integrating, in which c is an arbitrary constant depending on the position of the point when 2=0. If the space be reckoned from the position of the point when t=0, then c=0, and s=v 1 t±l(pf. (60). If the point move from rest, v l =0, and s=%>t\ (54). - All the other relations are readily deduced from these. 335. Prop. To determine generally the velocity of a particle ?noving in a right line to or from a fixed point by the action of a variable force. • By [X.], vdv=(f>ds ; or, measuring the line on the axis of x, vdv== — \ix; the sign being negative, because the force which is directed to the origin tends to diminish the distance x. Substituting this value of in (95), and integrating, we get v i =c— [IX 1 . If a be the distance of the particle from the origin when the motion commences, when x=a, v=0. Hence and v'=fi(a a -tt 9 ), 97) from wjiich it appears that x can not pass the limits ±a. 338. Cor. Since the force by which a material point, at iiberty 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, \i will be found from the proportion r: l=g:p. Hence u 2 =^(r 5 -.<). r As this velocity must be spent before the particle stops, if we make v=0, 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 vfie distance. dx From [VII.], we have dl= — dx V[J> Va?—x*' To integrate this, let x=a sin. 6, whence a 1 — x t =a' i cos.'B. RECTILINEAR MOTION OF A FREE POINT. 205 1 a.d sin. 6 \ dt=- i/fi a cos. u =-L de, and t=—=.6+c. Hence 6= V[t(t—c), and x=a.s'm.V[t(t—c). Suppose the time to commence when the pai'icle is at tne origin, or x=0; then 0=— a.sin. Vp.c, or c—0. .-. t=^.d. {a)' lfx=a, sin. 0=1, and d=±n or §tt, &c, 1 1 o and t=±n—= or #7r-^=, &c, V[j> " Vf* from the commencement of the motion. If x= — a, sin. 0= — 1, and 0=§7r or |7r, &c, 1 1 . and i=^7r — = or ^tt^=, &c. Hence the particle moves from +a to — a, while t changes from |7r— r= to f tf-7=f or the time of one vibration is Vp " vjti T=tt-4=, (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, in g which case w==-. is r 200 DYNAMICS. m it /r T '= 2 \/v Q =2i'.5".a Cor. 3. Let S be the center of force, AS=a the distance of the particle at the commence- ment of motion, P the position of the particle at any instant, and t the time of describing PS=a;. On AS describe a quadrant, and draw PQ per- pendicular to AS. Then £=a.sin.QSB. .-. QSB=0 and QB=«.0. V>. V~rf^a?= V> .PQ a PQ, 1 „ 1 QB_ QB _ nn a By (97), and by (a), t=- V[i Vfi' a vel - atS When the particle is at A, arc BQ becomes BA and i : r rv = 2v> 340. Prop. 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), and [X], Integrating, If v=0 when x=a, ~l^r- \{ax— x r — -versin. \-c\. \2fi/ iV ' 2 a s _2x When t=0, x=a, and versin. J — =tt. a (a \ ' la 2x\ ) -) {(ax-xy+-(ir-verBm.-*—)\. (100) Cor. 1. If x=0, we have 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 draw PQ, per- pendicular to AS. Then PQ= Vax--x*, arc AQ,S=47m, arc SQ=SC.z.SCQ=| versin.-^. 208 DYNAMICS. Hence the time through AP is, by (100), x t= (J-\ 2 (PQ+ arc AQS- arcQS) i |^(PQ+arcAQ). If SQ, be produced to meet in R a tangent at A* SA.PQ aVax—x 2 AR= PS '2p /a—x\ \ ax ) . by (99), v= f-Pj 2 . AR ac AR. When P coincides with S, AR is infinite. 343. Prop. To find the velocity and time when the force varts* inversely as the cube of the distance. In this case, Such are the general equations by which the motion c f a point in a plane may be determined. 346. If the velocity is given, we must, by resolution, 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- O 210 DYNAMICS. locity in each axis, and thence, by composition, the velocity of the point in its path. If the path is required, each of equations (103) must under- go two integrations, at each of which operations a constant must be introduced. The constants introduced at the first in- tegration 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 off, an equation between the co-ordinates will be obtained, which is the equation of the path of the point. 347. Prop. To determine the velocity of a point in its path. The velocity may be found as above indicated, but the fol- lowing is the preferable method. Multiply the first of equations (103) by 2dx, the second by 2dy, and, adding the results, we have The first member of this equation is the differential of dx iJ rdy" ds 2 a ~~de = de =v ' Therefore, integrating, we find v*=2f(Xdx+Ydy)+c. (105) 348. Cor. If this expression ~K.dx-\-Ydy is integrable, the ve- locity 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 « 9 =2/(*,y)+c, and if u, is the velocity at the point whose co-ordinates are a and b, so that v\=2f{a,b)+c; then v* - v\ = 2f(x, y) - 2f(a, b) . Hence it appears that the velocity acquired by the particle CURVILINEAR MOTION OF A FREE POINT. 211 in passing from one point (a, b) to any other (x, y) is the same, whatever be the curve described between these points, since the change of velocity is independent of the co-ordinates of any intermediate point. 349. Prop. The expression Xdx+Ydy is always integrable whenever the forces are directed to fixed centers, and their intens- ity is a function of the distance from those centers. Let F, Fj . . . be two forces directed to fixed centers, " a, b, a l , &j be the co-ordinates of the centers, " x, y .... be the co-ordinates of any position of the particle, " r, r x . . be the distances of the particle from the cen- ters, " a,(3,a 1 ,(5 1 be the angles which r and r, make with the axes. Resolving F parallel to each axis, we have for the com- ponents F.cos. a, F cos. (3. x — a v — b But cos. a= and cos. (3=- — . r r Hence the components are r ' r The other forces may be resolved similarly. Therefore the expression Xdx+Ydy=F i^dx+ y -^) \ r, r, J But r*=(x-ay + (y-b) 2 , x — a v — b and, by differentiation, dr— — — dx-\ — dy. c. ., , , x—a. . y—b. Similarly, di\= dx-\ ■ — dy. .-. Xdx+Ydy=Fdr+F l dr l . ' ind since, by hypothesis, each fc. T ce F is a function of the dis- 212 DYNAMICS. tance r, each of the terms of the second member of the equa- tion is an exact differential, and therefore the first is also. 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 cen- ter, called the radius vector. Multiply the first of equations (103) by y, and the second by x, and subtract the former from the latter. This gives xd"y—yd"x ,_ __ ,,«„* y J =xY-yH. (106) Resolving F in the direction of the axis, we have X=F-, Y=F^. r r Multiplying the first by y, and the second by x, and taking the difference, we get xY-yX=0. Therefore the first member of (106) is zero, and since it is the xdv — i/dx differential of — , by integration we obtain / 'xd~y—yd*x xdy—ydx__ W dt ~ C ' or xdy—ydx=cdt. (107) Let o) be the angle made by r with the axis of x ; then x=r cos. o), (a) y=r sin. co ; (b) also, dx— cos. udr—r. sin. u.do), (c) dy=s'm. o)dr+r cos. w.cfcd. (d) Multiplying (c) by (6), and (d) by (a), and subtracting, we get xdy— ydx = r'dco — cdt. r 2 do) ••• -^r-c. (io8) CURVILINEAR MOTION OF A FREE POINT. 213 But |rWw is the differential of the area of a plane curve re- ferred to polar co-ordinates. Hence the ratio of the element of the area to the element of the time is constant, and the area described by the radius vector in any time is proportional to the time. By integrating \r^da>=\cdt, we have for the area described in any time k=\ct, (109) and if 1=1, we have c=2A, or the constant c is twice the area described in the unit ot 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 the times, we have (107) xdy—ydx dt Differentiating, we get xd*y—yd 2 x -c. ?>J> =0. dt" . . (106) zY-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-an- gled triangle PMS, in which PM=y and SM=x. As an application of the general equations of motion of a free point (103), we may take the following cases. 352. Prop. 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) d*z „ . d*y dt 2 :X = 0, df :Y-t0. 214 DYNAMICS. dx du Integrating, ^=„ lf ^=« a ; (a) Uj and r 2 being constants, added to complete the integral. Compounding these velocities (Art. 236), we have for the resultant velocity /dx*+dy 9 ds —— — - v =V—de-=dt= Vv ° +v " The velocity is therefore uniform. Integrating equations (a), x—v 1 t+s 1 , y—v 2 t+s 2 , (b) the constants s l and s 2 being the co-ordinates of the point when the time commences. Eliminating t from equations (b), we have ij=-f-x+v 2 (s 2 -s,), u i the 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 determine 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 vei- t.ical, (103) give d 2 x _ d 7 y , r Multiplying by dt, and integrating, dx dy -7-=!),, —=v„—gt. dt x dt 2 * Multiplying by dt, and integrating again, we have x=v l t+s l , y=v 2 t-lg?+s 2 . (a) But if the point of projection be at the origin of co-ordinates, and t be reckoned from the commencement of motion, s, ana s a will each be zero. Putting v 2 =av x , s 1 =0, s 2 = 0, and elimi- nating t from the equations (a), we get . g y=ax—x—- 2 -. Compare this with (63). CONSTRAINED MOTION OF A POINT. 215 § III. CONSTRAINED MOTION OF A POINT. 354. Prop. To determine the velocity of a point moving on a given curve. If a point is constrained to move 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 3 the angles which the normal makes with the axes, and X and Y the sums of the components of all the other forces in the direction of the axes. Resolving N in the same directions, we shall have from (103) d'x di -=X+N cos. a §=Y+Ncos.0 But cos. a— sin. rtm=+-r-, supposing the motion in the direction tV, and cos. 8= sin. a- N M _ dx cos. rtm= — — . ds lf- x+N Ts de ds' Multiplying (a) by 2dx, and (b) by 2dy, and adding 2dxd 7 x+2dyd*y de = 2(Xdx+Ydij). The first member is the differential of dx*+dy *_ds*_ a df ~~df~ V ' Whence, integrating, v 9 =2f(Xdx+Ydy)+c. (110) Cor. 1. Since this result is exactly the same as that obtained for unrestrained motion (105), we may conclude. 216 DYNAMICS. 1°. That, if no accelerating force act on the point, or X=0 and Y=0, its velocity will remain constant, and be not at all retarded by the action of the curve. See Art. 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. Cor. 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. (Ill) Cor. 3. If in this last case the force, or resultant of the forces, is constant, and equal to/, we have u 2 =±2fx+c, 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 from .the origin, x—a, when v=0. c=2fa, and v-=2f(a-x); (112) and since only the ordinates on the axis of x are involved, the velocity of the point on the curve depends not on the curve de- scribed, but on the difference of the ordinates on the axis of a:. 355. Prop. To find the time of motion of a point on a given 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) and substituted in that of dt, the time may be found by integration Cor. If the forces act in parallel lines, and their resultant is constant, by (112) ds dt= =. (113) V2f(a-x) 356. Prop. To find the reaction of the curve. dxi dx Multiplying (a), Art. 354, by -j-, and (b) by -=-, anc subtract iog the latter from the former, we get CONSTRAINED MOTION OF A POINT. 217 N— Y— - X— 4- ' lyd * X ~ (/ "' d '' ' ds ds dsdt" t,,. . . , , , . ds Eliminating dt by the equation. v= -r-, dx dy a dyd 2 x — dxd~y ds ds ds 3 But if p be the radius of the osculating circle, ds 5 P dyd^x—dxd^y __ ^ T dx v dy v" ds ds p v\ = Y cos. (3— X cos. a-\ — 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 v~ the action of these forces, while the other term — is the reac- P 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), Art. 354. Hence N Jdx'+dy'\ =N ,Jd*xy + (d*yy dt' ds or, eliminating dt by the equation dt= — , N=B • 5? • But the coefficient of u 2 is the reciprocal of the radius of curvature when s is the independent variable. .-. N=- (114) Compare this with (79). As an application of the formulae of constrained motion, we may take the following rases : 218 DYNAMICS. 357. Prop. To determine the motion of a body descending by the force of gravity down the arc of a vertical circle. 1°. To find the velocity. Formula (112) applies to this case, and if A=SA be the height from which the body at P descends, and x any distance from the lowest point on SA, the axis of x, we have v= V2g(h—x). When x=0, we have for the velocity at the lowest point, u= V2gh, which is the same as that due to the vertical height h. 2°. To find the time, we have from (113) ds dt=- V2g(h—x) The equation of the circle is y*=2ax— x 2 . Hence _(*-*) W But ds°-=dx 2 +dy 2 =dx\l + V2ax—x' 1 This, taken negative, because the arc is a decreasing fur.c tion of the time, and substituted in the above, gives a dx dt= — V 2g V (h — x) (2ax — x 2 ) a dx V2g V(hx-x*) (2a-x) a (2a—x)~2dx V2g Vhx—x' 1 («) — *J± " v 2a J g Vhx- CONSTRAINED MOTION OF A POINT. 219 Expanding (1 — — - J by the binomial theorem, we have 1 fa dx , \(x\ \Z(x\* lZ5(x\ % &c.}. — x"dx Thus the terms to be integrated are of the form Vhx— x 1, the exponents n being natural numbers beginning with 0. Per- forming the operations and taking the integral between the limits x=h and x—Q, we find Hsv/f * l + G) '• (s) + GD 3 GD 2+ Gil) ' (£) '+* &c.|. If /i be small, the first term will give an approximate value for the time, viz.. Otherwise. If in (a) # be rejected as small in comparison with 2a, the equation reduces to i fa _J^__ dt=-W~. .-j- , . v g- Vnx—x . /a . 2x .'. *=-£y-.versin. l ^-+c. But 2=0 when x=h; and ^=^-\/— (tt— versin. _1 -r- ), and when a;=0, £=£tt\/-. (See (8G) ). © 358. Prop. To determine the motion of a body descending by gravity down the inverted arc of a cycloid whose base is hori- zontal and axis vertical. The cycloid being inverted, let the origin be at the lowest 220 ]) Y K A M 1 C point, the axis of x being vertical, and that of y horizontal Let the axis AB = 2a, AM=.r, PM=?/, and AP=s. The velocity, as in the preceding propo* ~J sition, is t;= V2g(h—x), h being the abscissa of the point K from which the boay begins to descend. The differential equation of the curve, referred to the vertex as the origin, is (2ax — x)dx dy= — , 9 V2ax-x' 2 /(2a— xY But ds* = dtf + dx* = ( ^ - s + 1 ) dx\ .'. ds=dx\/ — . V x Substituting this value of ds with its proper sign in [VII.], we find /a ~ dx V g' Vhx—x 2 ' Integrating, t= y — . (tt— versin. -1 -r- ) , and therefore for the time of descent to the lowest point, where x=0, /a tt /4a , , . t=n\/-= b a/— • ( 115 ) V g 2 V g Cor. 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 circular arcs. On account of this remarkable property the cycloid is called tautochronal. 359. Schol. The property of taatochronism which attaches to the cycloid was supposed to give peculiar advantages to a pendulum made to oscillate in this curve. The method of ac- complishing this is naturally suggested by the property of the evolutes ; and as an evolute of a cycloid is another equal cy- CENTRAL FOKCE. 22 1 cloid, wo have only to take two equal semicycloids and place the extremities of their bases contiguous in s the same horizontal line as in the figure. If we then suspend a body at the point S, common to the two semicycloids, by a flexi- ble string equal in length to either semicy- cloid, or, which is the same thing, to twice the axis, and make it oscillate between the two, it will generate 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 importance to justify a distinct discussion, especially as the formulae are susceptible of considerable simplification, when the force is given as a function of the distance from the center. The fun damental equations A -~dt" X ~df" 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 DYNAMICS. 361. Prop. To transform the expressions foi the force from, rectangular to polar co-ordinates. Suppose the origin to be at S, the center of force, and let 7'=SP, the radius vector or dis- y^^\ tance of the particle, and w=PSX, the angle made by r with axis of x. Then X= — F cos. w, Y=— F sin. u, 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. w, and the second by cos. w, and subtracting, we have X sin. w— Y cos. u=0. (a) Similarly, multiplying the first by cos. w, and the second b\ sin. o), and adding, we get X cos. w+Y sin. «— — F. (b) But x and y being the co-ordinates of the position of the particle x=r cos. o), y=r sin. w. Differentiating each twice, we obtain d*x=d*r.cos. (o—2dr.do).s\n. cj—rdo^.cos.oi—rd'^.s'in. w, d^y^d^r. sin. u+2dr.du.cos. a— rd^.s'm. (o+rcPw.cos. which are two polar equations to determine the motion. 362. Cor. I. The radius vector of a point revolving about a center of force describes equal areas in equal times. For, multiplying (116) by rdt, we obtain , d<*> „d 2 (o ,rWw t CENTRAL FORCES. 223 Integrating, ~dT =Ci ( 118 ) and fr'd(o=ct-\-c i . But \fr~du is the area described by the radius vector in the time t, and it varies as t. 363. Cor. 2. The angular velocity of the radius vector varies inversely as the square of the distance of the particle. „ d<*> c From (118) we obtain -77=-. (119) dt r i<3 the* nnmilnr vpI dt But -j7 is the angular velocity of r, and it varies inversely as r . 364. Prop. To determine the velocity of the -particle in its orbit. Multiply (116) by rdu, and (117) by dr, and add the results. This gives ■ d'r 7 dtf 57 d 2 o> ^ 7 , fdr 2 rfw s \ _ , d \-d7 + >^) +2Fdr =°- Therefore, integrating, we have dr 2 +r'du df +2fFdr=c'. (120) . ds 2 rfr a +rW a 3* But de = — dt* — * ( 367 ' 4 °^ .. v 2 =c'-2fFdr. (121) Cor. Since the value of v depends only on the distance 01 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. Prop. To find the time of describing any portion of the orbit. 224 DYNAMICS. By (119), which, substituted in (120), gives dr* c 2 - +p+ 2/F. (124) When the law of force is given, we can obtain from this ex- pression, by integration, the relation between u and w, or the polar equation of the orbit ; or, if the relation between u and &> is given, the law of force may be determined. 367. Schol. A familiarity with the geometrical representa- tives of the terms and factors in the foregoing formula? will conduce to facility in their application. In the annexed figure let the center of force be at S, the origin of co-ordinates, SP=r the radius vector of the particle at P, W=ds x an element of its path coinciding A s " with the tangent PT, w=PST the angle made by the radius vector with the axis of x, P / SP=c/oj the angle described by the radius vector in the indefinitely CENTRAL FORCES. 225 small time dt, and mV =dr the increment or decrement of the radius vector in the same time. Let the arc Vm be described with S as a center, and radius SP, and the arc nn' with the radius Sn=l. Then, 1°. Since Sn : SP=wn' : Vm or 1 : r=doj : Vm, Vm=rdw. 2°. The area of the sector SPm is A= — - — =h-*do). 2 2 3°. If the point P' be indefinitely near to P, PP' and Vm may be considered straight lines, and the triangle PmP' a plane rectilinear triangle, right-angled at m. Hence P'P^Pm'+P'm 2 , ds*=r*du?-\-dr i . 4°. If Sp—p be a perpendicular on the tangent, when the triangle PwzP' is in its nascent or evanescent state, the angle PP'm=^PS and the triangle PP'm is similar to joPS. Hence PP : Pm=PS : Sp, or ds : rda=r : p. .: P=^, and [VII.], r*doi ?-"d(o . />= — =-. .*. vp——rr = , by (118), twice the area described in the unit of time. ? J d(i)' r^dui 2 A1S °' P *-'ds> ~dr*+rW 5°. Since S/j : pV : PS=Pm : m?' : V'V=rdo) : dr : ds, Sp _Vm rd(ji_ . •*' SP^PP^Is"" 8111, F » PS~PP~ rfs ~ C0S - F ' »S raP rc?w and — s =— T57=-j-=tan. P. pr mr ar 6°. If the force F be resolved in the direction of the curve, «re shall have d*s ^ „ „di- = _p C os. P^-F^. dC ds £26 DYNAMICS. Multiplying both sides by 2ds, and integrating, we have ds* —=v*=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 direc- c , , rdbf r'do)- __ rdo) . . e tion oi v ; the second term —rz-= — tt~- But —7- is the arc 01 at rdt dt a circle of radius r, divided by the time. It is, therefore, the linear velocity d, of a body in a circle. Hence 7 „ =—=f. J 1 J dt r - the centrifugal force in a circle. .*. (117) becomes -7-7-=/— F, F being the force by which the point describes any orbit,' and / the centrifugal force in a circle whose radius is r, the dis- tance of the point at the instant, -jj, the difference of these forces at any time, is that by which the radius vector is in- creased or diminished, and is called the paracentric force. In d*r dr like manner, the integral of -7-7 or -=- is the velocity of ap proach or recession from the center, and is called the para centric velocity. The paracentric force is the difference be tween the centrifugal and centripetal forces. rdui 2 1 r 4 e?« s c a Cor. 1. Since df r 3 * dt 2 r 3 .'. /=£ (125) Cor. 2. The quantity by which c~ is multiplied in (123) is, by (4°), equal to — j. Hence -=+2fFdr=&. P * Differentiating, p being variable, 2cVd „ , J-+2Fdr=0. CENTRAL FORCES. 227 Hence F=^r> (120) p dr v an expression most useful in finding the law of force by whict a given orbit may be described. /-. o- ■ 2 ds 2 dr+r\lu? r'du 2 e 2 Lor. 3. Since v =-^=- 7^ — „ ,,■> =—; dt dt p'dt" p~ c 2 Fpdr „ pdr „ _ pdr and (126), -=^-=2F.^, ,. »'=2F^-. (127, pdr But 77-r -=i chord of curvature through the pole, and (if F 2dp * o l \ be constant) is (55) the space through which a body must fall 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 formulae to the motions of the planets. 368. Observation has established three facts respecting the motions of the planets, which, from their discoverer, are called Kepler's Laws. 1°. The areas described by the radius vector of a planet art proportional to the times. 2°. The orbit of a planet is an ellipse of which the center of the sun is one of the foci. 3°. 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. 3G9. Prop. The accelerating force by which a planet describes its orbit is directed to the center of the sun. 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 DYNAMICS. 370. Prop. The force by which a planet describes its orbit 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. ail-e 2 ) 1+e cos. (to— 6) v ' in which a is the semi-major axis, called the mean distance, and e the eccentricity. The angle 6 is an angle made by a fixed line, through the focus with the major axis, reckoned from the lower apsis, or least distance, and called the longi- tude of the perihelion, the lower apsis being called the perihel- ion, and the higher the aphelion. The angle w— 6, which ex- presses the angular distance from the perihelion, is called the true anomaly. Putting r=- in the above equation, we get 1+e cos. (o— 6) ■ . . U== a(l-e>) ' <& Differentiating in reference to w, du_ e.sin. (to— 6) dto a(l—e 2 ) Differentiating, again, d'u e cos. (to— 6) dto' a(l-e') Adding (b) to (a), d'u 1 (P) dto 2 a(\—e) y ' Differentiating (124), d'u F dto' cu Hence F =^-r=^- } } ? . (129) The coefficient of — being constant, the force varies inverse r Iv as the square of the distance. CENTRAL FORCES. 229 Ifr=l, F =~7i i\- a(l—e) Hence the coefficient of — 2 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='na" Vl— e\ .-. cT = 2rraWl-e\ c _4ttV and a(l-e a ) = ~T T ~- Similarly, if a', e', c', T' express the same quantities in any other orbit, we shall have c' 2 _4ttV 3 a'(l-e' 2 )~ T' 2 * By Kepler's third law, T : T /2 = « 3 : a'\ c 2 c' 2 ' a(l-e 2 ) a'(\-e'"-)' Hence the force does not vary from one planet to another except in consequence of change of distance. 372. Prop. To find file velocity of a planet at any point in its vrbit. By (121), v 2 =c'-2fFdr. Substituting in this the value of F (129), we have »- 2c 2 rdr V ~ C '~~a(l-e 2 )V V' 2c 2 1 Integrating, v = c + a n_ e *y ? To determine c', we must know the velocity at some given 230 DYNAMICS. distance, or at what distance and with what velocity the planet was originally projected into space. ' Let Uj be this primitive velocity, and ;-, the corresponding distance. 2c 2 1 Then v\=c'+- or c'=v\ 2c" 1 1 ^l-e 2 )';-,' , . 2c 2 (\ 1 \ , x •*• v =v i+ n »x -( — ;■ ( 13 °) a{\ — e) \r r,' v Otherwise. From Art. 367, Cor. 3, we have c" and (Analytic Geometry) j5"-— 2<2— r 2a— r c 2 2a —r (131) ,, and is independent of the angle of projection ip. Hence, in what- ever direction the body is projected, the major axis of the orbit will be the same. The orbit will be an ellipse, hyperbola, or parabola, accord- ing as the value of e is less than, greater than, or equal to unity, or (d) according as >s negative, positive, or equal to zero. Or, since (99), v*= ^(1--), if a=cc, v 2 =— . Therefore x x a/ x when a body is attracted from an infinite distance to a center of force which varies inversely as the square of the distance the square of the velocity at any distance p is CENTRAL FORCES. 23li n 2W . _ 2/4 v; = — and v. =0. P P Hence the orbit will be an ellipse, parabola, or hyperbola, ac- cording as the velocity of projection is less than, equal to. 01 greater than, that acquired from an infinite distance. 374. Prop. To find the time of describing any portion of th? orbit. Substituting the value of F=— in (122), we get dr 2 c a 2[i_ df ' V~~~ C ' The values of the constants found in terms of the elements of the orbit in the preceding proposition are fi c'= — and c'=ua(l— e~). a Substituting these values, and multiplying by — , wc obtain ar*dr* v , „. £-5r+.(«-r)-a«=0. a\2 rdr Hence dt= . #) vVe 2 -(a-r) 2 In order to integrate this expression, let {a— r).=aez, then i /a.\ 2 (1— ez)dz dt=—a' V-J Vl-z 2 a 3 \ 2 ( — dz zdz j P' ' I Vl-z 2 Vl-r \ ' Integrating, t=[—J {cos. l z— e(l— z a ) 3 ]+k. When z=l, or r=a—ae, and the body is at the nearer apsis, then t l =k= the time at the perihelion. Hence the time through any portion of the arc from the perihelion is t " t ^ = (7) 3{C0S " 1 *- e < 1 " Z ^' • (133) 234 I) Y N A M I C S. If z= — l,r=a+ae» and the body is at the further apsis, in which case t—ti is the time of half a revolution, and fa* l-t =71 I - 1 \/* Also, if T be the periodic lime, T a - *a\ I" or the square of the periodic time varies as the cube of the mean distance. 375. Schol. The quantity \i 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 in the same of any planet, and if we assume for the unit of force the attraction of a unit of mass at a unit of dis- tance, 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=/i ; and for any other distance r M+m_fi 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 m x its mass, and since the forces are directly as the masses, and inversely as the squares of the distances, we have m, M_ Mr; ~r\ : ~7~ S ' m^P g ' the fourth term being the attractive force of the sun at any dis- TIXV tnnce r. Similarly, the attractive force of the planet is l -g: J l m^ . n . M+jfi r\ fi r\ hence their joint influence is — = — . — g=~i- — g, in whicli J r ?n° r m° CENTRAL FORCES. 235 — L .iT is that which was assumed above for the unit of force, 01 the attraction of the unit of mass at the unit of distance. 376. EXAMPLES. Ex. 1. Required the velocity and time when a materia] particle is attracted to a fixed point by a force varying in- versely as the square root of the distance. II -LI Ans. v=2fi s (a 3 —x 2 ) 2 2 t= ^J^+ 2a h i^-x^f. Ex. 2 Find the velocity and periodic time of a body re- volving in a circle at a distance of n radii from the earth's center. Ans. v= (£. \ n 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. A. Given the velocity, distance, and direction of pro- jection, when the force is attractive and varies as the distance, to find the orbit. - 3 Ex. 5. A body is. projected in a direction which makes an c angle of 60° with the distance, with a velocity which is to the n velocity from infinity as 1 : VS : the force varying inversely as the square of the distance. Find the major-axis, the posi- - tion 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 per- pendicular to the distance, when the force is repulsive and varies inversely as the cube of the distance. Find the path of the body. HYDROSTATICS. CHAPTER I. 377. Matter 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 perfect fluid is one in which the cohesion of the con- stituent particles is so feeble that the least force is capable of effecting 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, incompressible or liquid, and compressible or aeriform. 379. Incompressible fluids 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 PRESSURE. 381. Prop. A force or pressure applied to a given portion of the surface 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 direc- tion of its action, and is sustained by an equal force impressed in an opposite direction. Applied to a fluid the force is effect- ive in every direction, and can be sustained only by forces ap- plied to every point in the surface of the fluid. Let the annexed figure represent a section of a vessel filled with a fluid without weight. In this case, supposing the press- ure of the atmosphere 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. Let an aperture of a giv- 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 friction. Now, since the particles are without friction, a force P, of one pound, applied to this piston, will act upon the stratum of particles in contact with its base, and this stra- tum 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 P' l5 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 commu- nicate motion to them, unless resisted. A motion will therefore be communicated to any equal piston T'[, unless sustained by a 238 HYDROSTATICS. pressure of one pound. Hence both pistons P', and. P',' will sustain the same pressure by the action of P,. Further, if the pistons P', 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 P 4 be four times that of Pj, then a force of four pounds would be necessary to keep P 4 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 Y x and P„ on any two portions A, and A„ of the surface will be proportional to their areas, or £=£=«. (134) Also, the normal pressure^ on a unit of surface will be P P 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 ap- plied 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 P'. If h and h' be the spaces through which the pistons move, a and a' the areas of their bases, then ha=h'a'. But P and P' being the equilibrating pressures on these areas, «_P a'~P' ; hence hP=h'¥', or AP-/i'P'=0, or the principle of virtual velocities applies to fluids in equilib- rium. SURFACES OF EQUILIBRIUM. 239 SURFACES OF EQUILIBRIUM. 382. Prop. The free surface of a fluid subjected to the action 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 acceler- ating force, a containing vessel will be necessary in order to preserve a coherent mass. Let ABCD be the section of a ves- sel containing a fluid subject to the ac- tion of gravity, the base DC being hori- zontal. If any portion of the free sur- face, as abc, have a direction not per- pendicular to the direction of gravity, 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 in- clined plane, they will yield to this component. But whenever every portion of the surface is perpendicular to the direction of gravity this force will 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. Prop. If a vessel containing a fluid be moved horizontal- ly 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 uniformly accelerating force P. Then any ele- ment, k, of the surface, whose weight is w, and mass - (22). O will be urged by its weight iv in the direction kg, and by it? 240 HYDROSTATICS. inertia P— in the direction kk, and g therefore by their resultant in the direction ko. As the same is true of every element of the surface, the whole surface (Art. 382) will be per- pendicular to ko. Let the ansrle of inclination to the horizon HIM=I. Then since HIM= okg, tan. o/co-=tan. 1= P » w 384. Prop. If a vessel containing a fluid be made to revolve uniformly about a vertical axis, the surface of the fluid ivill take the form of a paraboloid of revolution. Let the vessel ABCD revolve uni formly about the vertical axis XX j. Any element k of the surface will be urged horizontally by a centrifugal force directed from the axis, and vertically downward by gravity. Let w be the angular velocity of the fluid, w the weight Xi of the element k, and y its distance AM from the axis. Then (82) we have, for the centrifugal force P of this element, g Being urged vertically downward by its weight w, the surface of the element will be perpendicular to the resultant AR=R of these two forces. If this resultant be produced to meet the axis in N, we have and or LRkg=Lk'NM, Rg AM kg 0) v— g . w 'MN' y "MN" PRESSURES ON IMMERSED SURFACES. 241 .'. the subnormal MN=— = a constant, a property of the parabola. Hence the surface is a paraboloid of revolution. NORMAL PRESSURES ON IMMERSED SURFACES. 385. Prop. The pressure on the horizontal base of a vessel containing an incompressible fluid is proportional to its depth below the surface of the fluid, and independent of the form of the vessel. Let HO be the free surface of the fluid in the vessel ABCD, whose base DC is horizon- tal, and suppose the fluid divided into hori- zontal strata of small but equal thickness. Let «,, a 2 ,a 3 , &c, denote the successive strata, A , A,, A 2 , the units of surface in each, P , P,, P 2 , the whole pressure on each, p , p x , p 2 , 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 indef- initely small, the upper and lower surfaces of each may be re- garded as equal. Now the weight of a l is w x —gpaA Q (24), " " " " a 2 is w 2 =gpaA lf " " " " a n is w n =gpaA n _ t . The pressure on a l is P , P on a unit of a l is -r^—p (135), on a 2 is P,=P +^ 1 =P +5 , paA , . . .P. P gpaA. on a unit oi a n is -r- =-r--\ — r or p 1 =p +gpa, A, A, Aj on a 3 is P a =P 1 +w 2 =P 1 +gpaA lf . P 2 P, , gpaA, * on a unit of a 3 is t-=t — 1 — t — or Pa ~P i +gp-a=p 9 +gp.2a. on a„ +1 is P ( ,=P,_ I +u; Il =P fl _ 1 -f-g , paA l _„ Q 242 HYDROSTATICS. . r . P„ P,,-! gpaA n _! on a unit 01 rt„ +1 is -7- =-7 — ' a or p,=p +gp.?ia. But na—h is the depth of a, l+l below the surface of the fluid, and if the upper surface of a„ +l 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=P +gph> and the whole pressure on the base V=Ap=Ap +gphA=V Q +g P hAorh. (136) Cor. 1. Since h A is equal to a prism whose base is A and height h, and gp/iA is its weight, if we disregard P , 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 ivhose base is that 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. = 62.5 lbs., we have, for the pressure on the base of any vessel con- taining water, P=62.5 AA 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. Cor. 3. The pressure on every portion of a horizontal stra turn 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, I, 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 vertically PKESSUKES ON IMMERSED SURFACES. 243 A W C D upward as that to which the stratum fg of the fluid is subject. Moreover, the stratum fg being in equilibrium, must be press- ed 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. Prop. The normal pressure on any plane surface in- clined to the horizon is proportional to the depth of its center 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. By the pre- ceding proposition, p—gpxk; 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= I,gpxk=gplxk. Let x be the depth of the center of gravity of the immersed surface below the surface of the fluid, and A its area. Then (29), Ax=lxk; ... Y=gpAx=gpZ.h'k=gpV. In the same manner, the vertical upward pressure p x on the, corresponding element k x of the lower surface is equal to the weight of a filament ak l of fluid whose base is k x and height ak^—J^, or p i —gph 1 k. But the sum of all these filaments is the volume V, of the solid and fluid above the upper surface, and the whole upward pressure P, is the weight of this volume of fluid, or Now h 1 —h'=h is the length of the filament k'k^ of the solid p } -ip'=p its weight, and V l —V'=Y is the volume of the solid. Therefore, the difference P of the upward and downward pressure is P=P 1 -P / =2.p 1 -2.p'=:£.;?, and 'P=gp2.h 1 k—gp2.h'k=gp2hk=gpV; * ... Y=gpV=2p, (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. Let the distance of the point of application of p=hk from the ver- tical plane HK be x, and the distance of the point of applica- tion of the resultant P be x. Then, since the moment of the resultant equals the sum of the moments of the components, l*x=2.px, Z.px S.px and * = -£- = ~^f < 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 vertical- ly upward, and its point of application is the center of gravity of the displaced fluid. 248 HYDROSTATICS. 390. Prop. To find the conditions of equilibrium rf an xm mersed solid. The conditions of equilibrium involve a consideration of the weight of the body. Let V be its volume, and a its density. Its weight will be goV. The body will be urged downward by a force equal to goV 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, 1°. These forces must be equal; and, 2°. Their lines of direction coincident. The first condition gives g a V-gpV> or o=p ; that is, their densities must be the same. Cor. 1. If o be not equal to p, the body will ascend oi; de- scend by a force equal to g(p—o)V, according as p—o is posi- tive or negative. If p>a, the body will rise to the surface, and be but partially submerged. Let Vj be the displaced fluid, or the part of the solid immersed when the equilibrium is restored Then gpV,=g°V, (142) and V : V , =p : a ; 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 pla?ie of flotation. The line joining the centers of gravity of the solid and of the displaced fluid is called the axis q 'flotation. DEPTH OF FLOTATION. 392. Prop. To find the depth of flotation when the volum* and density of the body are known. DEPTH OF FLOTATION. 249 Let V be the volume of the body, o its density, Vj the volume of the displaced fluid, and p the density of the fluid. Then (142) g oY=gpV„ and V.=-V. p Now whenever V, 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 Y l =irr'x, and Y—nr'a; o .'. 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 T Then the radius of the plane of flotation is -x. Hence a r 2 V =l7T— X 3 V 1 3 Q* ' and Y=^Trr > a; 3 /a .'. x=a\/ -. V p If the vertex of the cone be upward, 3/ a y x—a\ I— . cn v p 393. Prop. To find the positions of equilibrium of a right 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, and 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 im- mersed 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 HYDROSTATICS. the ratio 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 im- mersed, and Fig. 2 when two angles are immersed. LetAB=c,AC=6,BC=a, AQ,=;r,AP=?/,andlet<7be the ratio of the density of the solid to that of the fluid. Now from (142) we have, in Fig. 1, AQP \xy sin. A_xy ABC \bc sin. A xy=cbc. be' (a) In Fig. 2, BQPC ABC-AQP bc-xy ABC be Q>) ABC .'. xy=bc(\— a). Since (b) is derived directly from (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 AM and Km. Take MM'=plA and mm'=\mK. M' 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'm' should be vertical, and since Mm is parallel to M'm', Mm is perpendicular to QP and MQ=MP. Let AM=A, MAQ=0, and MAP=0. Then MQ?=x"-+h 2 -2hx cos. 6, and MF=y s -fA*-2Aycos.^. Hence x* — y 1 — 2hxcos.d+2hy cos. 0=0. (c) Substituting in (c) the value of y from (a), and reducing, x i -2h cos. d.x z +2abch cos. 0.ar— ff'&V=0, (d) DEPTH OF FLOTATION. 251 and changing a into 1 — c, we have for Fig. 2, x i -2h cos. 6.x 2 +2(1 -o)bch cos. <£.a;-(l-a)W=0. (e) The values of a; deduced from (rf) 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 absol- ute 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 will be positive and one negative. Hence, 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 a: must also be less than b, and when subtituted in (a) and (b) give for y values less than r. Let us take the case of an equilateral triangle. Then 6=1, then x>a, which is inconsistent with tl# supposition that only one angle is immersed. The greatest value, therefore, which V9— IGo can have is unity, and in this case the second angle will lie in the surface of the fluid. Putting V9— 16a=l, 1 we get „ and <7 there will be nine positions with ? tyyf ff f . f CENTER OF PRESSURE. 253 one angle immersed and three with two immersed ; if ct<- 7 7 and o>—, nine of the latter and three of the former; if a<— lo lb 9 or 0>— , there will be only three of each. lo D A CENTER OF PRESSURE. 394. Def. 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 immersed plane surface. Let the immersed surface be the inclined side AE of the vessel ABE supposed to be filled with a fluid. Let x and y be the co-ordinates of any element k referred to the rectangular axes AY and AX, and hk=h the vertical distance of the element be- low the surface of the fluid. Then the normal pressure jo 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=2./>. The 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 perpendicular to AY, will be py. Hence, if x and y are the iy f> 7/l/X;.- ii^_ h V B N^C /E n X 254 HYDROSTATICS. co-ordinates of the point of application C of the resultant R, we have (Art. 43) R = Z.p, Rx='Zpx, Ry=2.py. If 6 be the inclination of the side of the vessel or immersed surface to the surface of the fluid, we have h=x sin. 6. Hence p=gpxk sin. 6, and R=2. gpxk sin. Q=gp s'm. O.l.xk, Rx=2.gpx"k sin. 0=gp sin. 6.2.x*k, Ry=2.gpyxk sin. d=gp sin. d.l.xyk. _ 2.x"- k ••• « S =I3' < 143 > and v= Y^k- ( 144 > When the upper boundary is below tne surface of the fluid and at a distance a from it, then, since the pressures p are limited to the immersed surface, h=(a+x) sin. 6, and we shah have _ S(ax+x')k , , x= _\ , ' , (145) _ E(a+x)yk y-k+^k- < I46 > Further, if the axis of x is so taken that it will bisect ever} 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 (143) is the mo ment of inertia of the surface, and that the denominator .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 ic, this depth, _ Z.x*k *Tf? < 14? ) (148) » Ax, ' 396. Prop. To find the center of pressu re of a rectanguitu CENTER OF PRESSURE. 255 surface vertically immersed, and having one side in the surface of the fluid. Let YY, be the line of intersection of the immersed surface ABCD with Y,- the surface of the fluid, and conceive ABCD divided into n rectangles by horizontal lines, n being a very large number. Let AB=& and AD=/i, and draw OX bisecting the rectangle. The center of pressure is obviously in OX and y—Q. To find x, we have, since the height of each h small rectangle is -, for the area of each, , bh n The distances of these rectangles from the surface are h 2/i 3/t . -, — , — , &c. ; n n n If 2Vi' 3V*' B — ri -T7-> &c - their squares, n n n „ bh 3 bh 3 .2* M 3 .3 2 B Hence 2.xk=—r-\ =— H = — K &c n 3 n 3 n 3 =^(l 2 +2 2 + 3 2 + n). But when n is very large, the sum of the mth powers of thft natural numbers 1,2, 3, &c, to n is m+l Also, .-. X.x*k=—.--=±bh 3 . n 3 3 3 Ax l =bh.\h=\bh\ ibh' or the center of pressure is two thirds the height of the rect- angle below the surface of the fluid. 397. Prop. To find the center of pressure when the immersed 250 HYDROSTATICS. surface is a triangle, having one side horizontal and the opposite vertex in the surface. A p b Let BDC be the triangle, h its height, and b its base, and suppose the triangle di- vided as before into n horizontal divisions. The heights of the successive divisions h the lengths, the depths, are - ; n b 2b 3b c -, — , — , &c. ; 71 n n h _2h _3h n n - n Their areas, considering the horizontal sides of the success- '•ve trapezoids as differing insensibly from each other, are _bh _2bh _3bh 1 n* a n 3 n Multiplying these by the squares of their depths, and taking their sum, we have „ bh* 2 3 bh* 3 s bh* c n'bh 5 S.a;-A=— j-+— j — H— 1 — b &c — — , n n n n bh 5 =— (l 3 +2 3 +3 3 +, &c n 3 ), n bh'n* „, =—.-=\bh\ n 4 Also, K=\bh, and Sj =§ A. Hence *=#&=& If now BE be drawn bisecting the base, the center of press- ure will be in BE, and at a distance from the surface BO=fBE. 398. Prop. To find the center of pressure when the base of the triangle lies in the surface. In this case the heights of the successive divisions of the tri- angle ADB (Fig., Art. 397) are - ; lL . . . b , 2b . 2b „ the lengths, b — , b , , &c. : & nun CENTER OF PRESSURE. 257 the depths, x l =-, x 2 = — , x. 3h n n n , &c. bh bh bh 2b h bh 3b h the areas, k= r, h,= r> «3 = i~s & c, > n n which, multiplied by the squares of the depths, and their sura taken, give bh' bh 3 2*bh 3 2 3 bh 3 3W 3"6/** 2.z*k=— H i ~ ^ i —+» &c -> n n n n n n =^(l 2 +2 2 +3 2 + . . . ?i 2 )-^ r (l 3 +2 3 +3 3 + ...»•) Wn» M 3 n 4 „, 773 71 s 3 % 4 3 4 and H.x'k- ±bh.\h=\bh\ ^blf Mi- =u. B Draw DF bisecting. AB, and take FO'=iFD. 0' is the cen- ter of pressure of DAB. 399. Prop. To find the center of pressure when a rectangle is immersed vertically, having a side parallel to the surface of the fluid and at a given distance below it. Let a be the distance of EF, the upper side of EFCD from AB the surface of the fluid, and conceive the rectangle divided as before into n divisions whose heights h . bh TT . . . . are — , and areas — . Using (145), and n n reckoning x from EF, the origin of the surfaces, the depths are h 2h 3h „ a-\ — , a-\ — , a-\ — , &c. n n n v , abh" bh 3 2abh* 2*bh* 3abh* 3*bh 3 Hence 2.{a+x)xk=-^~\ -7^-+— 3 — ~Zi — ~ir~H — n — h&c- E D M n abh bh 3 — r (l+2+3+....n)+— (l 2 +2 2 +3 2 +...n s ) ablf ?i 2 bh n* R 258 HYDROSTATICS. A.nd Z(a+x)k=Kx l = bh(a-\-\h) = abh-\-\bh' i __ \abh*+\bh' i ] 2h+3a _ h h+2a = d 3 -a 3 abh+±bh* , 2A+3a j^ft. , . _ — l-a=f. MO. /j+2a 3 'cf-a' J ' Hence LO=a+x: in which d=AD=a+A. 400. Prop. To determine the conditions of equilibrium oj fluids of different density. When fluids of different densities which do not mix or unite chemically are contain- ed in the same vessel they will arrange them- selves in the order of their densities, the most dense taking the lowest position. This is ob- vious from the ready displacement of their particles and Art. 390, Cor. 1. The limiting surfaces of each will likewise be horizontal ; for, in order to equilibrium, any horizontal stratum must sus- tain in every part of it the same pressure. But if the common surface, as EF, of two of them be inclined, then a horizontal stratum, as HR, will sustain at different points, as O and O', superincumbent columns of fluid of different weights, and the equilibrium will not subsist. 401. Prop. In communicating tubes, the heights at which fluids of different densities will stand above their common base when in equilibrium, are inversely as their densities. In the bent tube AED, let one fluid oc- cupy the portion AR, and the other the por- tion RED, the fluids having the common base HR. Let h be the height of the fluid in AR above HR, and p its density, h' the height of the surface CD above HR, p' its density, and let the common base HR=A. The pressure of the fluid in AR on A is gphA, and that of the fluid in RED on the same base gp'h'A. In equilibrium these pressures will be equal. H B D EXAMPLES. 259 Therefore gphA=gp'h'A, * P' and 77=-. (149) « P 402 EXAMPLES. jG'.r. 1. A cubical vessel, each side of which is ten feet square. is filled with water, and a tube thirty-two feet long is fitted to an aperture in it, whose area is one square inch. If the tube be vertical, of the same diameter as the aperture, and filled with water, what is the pressure on the interior surface of the vessel, neglecting the weight of the water it contains ? Since the weight of a cubic foot of water is 1000 oz., the • i r , . . , . 1000 , . . . weight ot one cubic inch is , „ , =.5787 oz., and the weight 1728 of 32X12=384 cubic inches is 384X. 5787=222.2208 oz. = 13.8888 lbs. This is the pressure on the aperture, or one square inch of the surface of the water in the vessel* The number of square inches in the interior surface of the vessel is 6X 10 2 X 144=86,400. Then, Art. 381, the pressures being in the ratio of the areas pressed, the whole pressure on the interior surface will be P=13.888SX86399=1,200,000 lbs. nearly. Ex. 2 What is the pressure on the bottom of the vessel in Ex. 1 when the weight of the water in the vessel is taken into account ; 1°, without the vertical tube, and 2°, with it. 1°. The base of the vessel is 100 square feet, and its depth 10 feet. Hence, Art. 385, P, = 10X 100X62.5=62,500 lbs. = the weight of the water. 2°. The pressure transmitted to each square inch of the base by the water in the tube is 13.8888 lbs. Hence the whole transmitted pressure P 2 = 13.8888 X 10 2 X 144 = 200,000 lbs. nearly, and the entire pressure on the base is P=262,500 lbs. Ex. 3. What is the pressure on each vertical side of the ves- sel without the tube ? The depth of the center of gravity of each side of the vessel is 5 feet. Hence, Art. 386, P 3 = 5X10 2 X62.5 = 31,250 lbs., equal one half the pressure on the base. 260 HYDROSTATICS. Thus we have the whole pressure on the interior surface of the vessel, including that produced by the water in the tube, P+P,+4P 3 = 1,200,000 + 62,500 + 125,000 = 1,387,500 lbs. Ex. 4. Required the pressure on the base of a conical ves- sel filled with water, the radius of the base being r=5 feet, and the altitude a=10 feet. By (137), P=62.5AA lbs., in which h=a, and A=nr\ .-. P=62.5:rr«=62.5X3.1416X25X 10=49,087.5 lbs. Ex. 5. Required the normal pressure on the concave surface of the cone of Ex. 4. The slant height of the cone is Vcf+r*, and the concave surface is 2m . The vertical depth ot the center of gravity of the concave surface is — . Hence (137), P x = 62.5X^* vV+7 o =62.5XfXl0X3.1416X5Vl25=73,175.11bs. Ex. 6. Required the vertical pressure on the concave sur- face of the same cone. Since the normal to the side of the cone makes the same an- gle with the axis of the cone that the side of the cone makes with the base, the cosine of the inclination of the normal to the axis is vV+r a Hence the vertical pressure {Art. 387) P a =P. -=^==62.5Xfa:rrvV+r.-=^==62.5Xf7rra, vV+r a vV+r 3 =|P {Ex. 4) = 32,725 lbs. Ex. 7. Required the resultant of all the pressures on the in- terior surface of the same cone. By Art. 388, the resultant of the horizontal pressures is zero EXAMPLES. 261 and the resultant of the vertical pressures is equal to the ex- cess of the downward pressure over the upward pressure. .-. R = P (Ex. 4)-P a (Ex. 6) = 16,362.5 lbs., and this is the weight of the water in the cone. Ex. 8. A rectangular parallelogram, whose sides a and b are 26 feet and 14 feet respectively, is immersed in water with the side b in the surface, and is inclined to the surface at an angle =56°.35'. Required the pressures Pj and P 2 on the parts into which the parallelogram is divided by its diagonal. Ans. P,= 82286.5 lbs. P 2 = 164573.0 lbs. Ex. 9. When a=30 feet, 6 = 20 feet, and 0=59°.38', re- quired the pressures on the equal parts into which the paral- lelogram is divided by a line parallel to the horizon. Ans. P, = 121332| lbs. P 2 =363998 lbs. Ex. 10. When the parallelogram of Ex. 9 is vertical, how far from the surface must the dividing line be drawn that the pressures on the two parts may be equal ? Ans. z=21.213 feel. Ex. \\. A sphere 10 feet in diameter is filled with water. Required the entire pressure on the interior surface, and the weight of the water. Ex. 12. A vessel in the form of a paraboloid, with vertex downward, is filled with water, and revolves uniformly on its axis. Required the time of one revolution when the angular velocity is just sufficient to empty it. Ex. 13. The concave surface of a cylinder filled with fluid is divided by horizontal sections into n annuli in such a man- ner that the pressure on each annulus is equal to the pressure on the base. Given the radius of the cylinder. Required its height, and the breadth of the mth annulus. Ex. 14. A regular tetrahedron, with one face horizontal, is filled with water. Compare the pressures on the base, and on the other sides with the weight of the water. 262 H V I) R n s T A T I C S. ¥r? Ex. 15. An irjn vessel 40 feet long, every transverse sec- tion of which is an isosceles triangle whose base is 16 feet and altitude 20 feet, floats with its vertex downward. If a cubic foot of iron weighs 487.5 lbs., required the depth to which the vessel will sink when the sides and ends are one quarter of an inch thick and there is no deck. Ex. 16. An embankment of brick- work, of which ABC is a section, weighs 120 lbs. to the cubic foot. Its height AB is 14 feet, and its base BC is 6 feet. Find whether or not the embankment will be overthrown by the pressure of water on the surface AB. ND B Draw AN bisecting CB, and take NG=|NA. G will be the center of gravity of the section and the weight W will act at G. Take AM=|AB, and M will be the center of pressure (Art. 396). The resultant P of the pressure of the water will act at M and will pass through G. The moment of P to turn the embankment over C will be P.CO, while the moment of W to resist the overturn will be W.CD. If P.CO> 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=1X14X6X|X120=5040 lbs., and P=lXl4X7X62.5=6125lbs., OC = BM = ixl4 and CD = f X6=4. .•-. W.CD=20160, and P.CO=28583i. The latter being greater than the former, the embankment will be overturned. Ex. 17. A wall of masonry, a section of which is a rectan- gle, is 10 feet high, 3 feet thick, and each cubic foot weighs 100 lbs. Find the greatest height of water it will sustain without being overturned. Ex. 18. If the height of the wall be 8 feet, its thickness 6 feet, and each cubic foot weigh 180 lbs., find whether it will stand or fall when the water is on a level with the top. CHAPTER II. SPECIFIC GRAVITY. 403. Def. The specific gravity of a body is the ratio of the weight of the body to the weight of an 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 u, p, o, and w be the volume, density, specific gravity, and weight respectively of one body, and v p p,, o 1 , w^ the same of another body; then, since v=i\, o w g-p-v p —=—=-*-*— =r~, (150) °, w, g.p x 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 1 its weight or the force with which it will descend when immersed in water. Then the loss of weight in water is w — w Jf 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 o= =- — ^— . (151) w — w x loss ot weight v The absolute weight as well as the loss of weight is ascer- tained by the hydrostatic balance, which differs from the or- dinary 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 by Z(f4 HYDRO&TATICS a weight w, and then suspended to the hook, and when irn mersed in the water its counterpoise w 1 again determined. 405. Prop. To find the specific gravity of a body less dense than water. Since the body A is less dense than water, it will not de- scend in the water by its own gravity. Let a more dense body B be attached to it, and call the compound body C. Let iv = absolute weight of A, w' = " " " B, and w\= its weight in water w"= " " " C,and w'[= " " " " Then w' — w\ =loss of weight of B, w"—w" l = " " " " C, and (w" — w'() — (w'~ w\) — \oss 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 w"—w'+w. and, by substitution, (w"— w") — {w'— w' 1 )=w'+w— w"— w'+w\=w+w\ —to','. .-. «t=-— jj. (152) w-\-w l —w{ v 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. Prop. To find the specific gravity of a liquid. Let a body whose weight is iv be weighed both in water and in the liquid, the weight in the former being w , , and in the latter w 2 . Then w— to, and io—w 2 are the weights of equal volumes of the two liquids. Hence w—w„ o= 2 -\ (153) w — iv l or, if an empty bottle whose weight is w, weighs when filled with water w l , and when filled with the liquid w 2 , then iv n — 10 c=-^ . (154) SPECIFIC GRAVITY. 265 407. Prop. To find the weights of the constituents in a me- chanical composition when the specific gravities of the compound., and constituents are known. Let w, to,, w 2 be the weights of the compound and constitu- ents respectively. a, a,, a 2 their respective specific gravities, v, u,, v 2 their volumes. In all merely mechanical combinations, v = v t +v 2 (a), and w=w l +w 2 (b). t> w w. . w„ but v= — , v= , and u = . gP gPi gP 2 , . W 10, w •'. (a) -=—+^-; P Pi P 2 or, since their densities are as their specific gravities (150), w vk w n -= — +—. ° G l °2 Substituting in succession the values of w l and w 2 , obtained from (b). we have 10, (\ 1\ /l n (o n -a)a. IV, . =W (I_IW±_1) = £^%„. (I56) \o o 2 J \a 2 o l J (cr,— o 2 )a x ' 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 hol- low globe, one of whose diameters is prolonged in a flat oi 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 200 HYDROSTATICS. 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 o the specific gravity of the second fluid. Then p and p x being the densities of this fluid and water, the weight of the water displaced by the in- strument will be gP-tV, 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-vx). V o V v (a) a=- V-vx V If c be previously known, and x be observed, we can de- ,'mine the value of V — =x.- . v \—a V Substituting in (a) this value of—, and putting for x, 1,2, 3 &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, r x-^-(l Giving to a a small increment do, the corresponding increment of a; is v o that is, for any small increment of the specific gravity of the fluid the corresponding change in the depth of immersion of V 1 the instrument varies as — .— , which may be considered a meas- ure of the susceptibility of the instrument. STECIFIC G E A V 1 T Y. 26" 2 J . Areometer of a constant volume. The principal obstacle to the use of the areometer of a con- stant 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 construct- ing the stem of that perfectly uniform thickness which is nec- essary 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 Y and w the weights respectively necessary to sink it to the same point a of the stem in water, and in the fluid whose ^ specific gravity a is required, V the constant volume of the portion immersed, p 1 and p the densities of the writer and the fluid. Then gpY=Yi+w and gp 1 Y=W+w 1 . _ p _ W+w ~~ p~ W+ttJ,' Cor. Differentiating, we get dw={W +w 1 )do=gp t Ydo ; 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 of the specific gravity of solids as well as t yy liquids. For this purpose, a metallic basket is attached to the lower extremity, in which the body may be placed, and its weight in water ascertained. The basket admits of reversal, so that the body may be retained under water when specifically lighter. Let w be the weight in C necessary to sink the instrument to/. Replacing w by the body, let w, be the weight which must 26*3 HYDROSTATICS. be added to sink the instrument to the same depth/. Removing the body to the basket beneath, let w 2 be the weight in C req- uisite to sink the instrument a third time tof. Calling W the weight of the instrument, W x 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=W l , (a) gpV+W+w^W,, (b) and gpV+W+w 2 =W i +gp i y. (c) Subtracting (a) from (b), and (b) from (c), gpV=w— w,, gp l y=w 2 -w 1 . p w—w. a=- p, w 2 —w 1 410. EXAMPLES. Ex. 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, llu=22 or u=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, m/,'=8, and w\=20. _ w _ 12 1 °~ w+w\ —w" x ~ 12+20— 8~2' Ex. 2. A diamond ring weighs 69i grains, and when weigh- ed in water G4| grains. The specific gravity of gold being 161, 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 /I the weight of a unit of volume of water. Then u/l is the weight of a volume of water equal in bulk to the gold, and 3 3 3 tvt the weight of the gold. In the same manner, |u'A is the weight of the diamond. Hence SPECIFIC GRAVITY. 269 69±=yi>A.+|ij'A, (a) and 64i= : y j ?;A+^ , A-(y+y')A. (&) Subtracting (b) from (a), we have 5=Ay+Au'. .-. from (a), 69i=y(5-A»')+|A« / f 139=165-26Au', Au'=l, fAy'=3i, 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 A', A" be the number of grains in the volumes of water equal to the volumes of A and B respectively, and a', a" their specific gravities. Then, by the conditions of the question, ( or the elastic force is directly as the density. 415. PRor. To estimate the effect of heat on the volume and tension of atmospheric 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 in ArL 411, and by the same method the increase in the ten- sion 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 tension remains the same, the volume will increase 36.7 pei cent. Let v be the increment of volume, the tension remain- ing the same, for one degree of Fahrenheit's thermometer ; then S 274 HYDROSTATICS. v=0.002039, and for an increase of * degrees of temperature v the increase of volume will be r*=0.002039*. If now V be the original volume of a given mass of air, and it be heated *, degrees, the tension remaining the same, the new volume V, will be V,=(1+,*,)V ; •md when heated *„ degrees the corresponding volume will be Y a =(l+vt 2 )Y . V, 1+v*, 1+0.002309*, " Y 2 ~l+vt 2 1 +0.002309*/ But the densities p, and p 2 are inversely as the volumes. P, V 2 l+v* 2 P a V, l+vt,' If, also, a change take place in the tensions at the same time, letp be the tension at 32°,/?, the tension at 32°+*,,andp 2 that at 32°+* 2 . Then, since the tension is inversely as the volume, V, = (1+,*,)^V and V 2 = (1+.* 2 )^V . "Y 2 (l+vt 2 ) Pl ' and pT~(i+^-?:~(T+^)v (160) in which &, and b 2 represent the measures of the tensions/?, and p 2 , or the corresponding heights of the barometer. 416. Prop. To find the density of the air at different tempera- 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=0.081 12 lbs. avoirdupois. Hence, for the tem- perature 32°+*°. p 0.08112 „ .,„,. ^ = T+^Tl +0.002039* IbS ' (161) If, also, the barometer, instead of &=30 inches, should stand at some other height, as b l inches, the density will be express- ed by p &,_ 0.08 112 b ± _ 0.002704&, 9 ~ 1 +vt b ~ 1 +0.00204*'30~ 1 +0.00204* S * ( ) COMPRESSIBLE OH AElUfOKM FLUIDS. 275 Whenever the elasticity ol the air is expressed by the press- ure p= 14.75 lbs. on the square inch, instead of the barometric height b, the density for any other tension p 1 will be jt?,_ 0.0 8112 p, 0.0055;?, 1 +0.00204*'14.75~ 1 +0.00204* lbs. (163) ri 1+vp p 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, 0.00344/? , 1+0.00204* lbs. 417. Prop. To determine the height corresponding 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 o 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 into portions of equal weights w, and heights x , x , , x 2 , &c, beginning at AB, so small that the den- eity of each may be regarded as uniform. Then x +x l +x 2 , &c. . . . x n —^.x=h is the height of the nth stratum. Let p be the tension of the lowest stratum, and p n that of the n\\\, p and p n their densities respectively. Then the weight of the nth stratum is A B x n p n p n P (a) Po Pn But the tension of the (n— l)th stratum is p„-\~ip, and the height, therefore, Po w 2*n-; = P b 'Pn+w' 276 HYDROSTATICS. in like manner, ar„_ 9 =='- 2 -. — rr— , p p„+2w ^ Po w = p, w °" X "~ n P 'Pn+nw p > ' since p n +nw=p Q . Now, in the exponential series, when y is very small, we have e^l+y, or y=yle=l(l+y)=2.3025SL.(l+y), (b) in which 7 denotes the Naperian logarithm, and L the common logarithm of a number. w If in (b), for y, we put — , we get — =7.(i +— )=Z. M- )=/.(/vfu>)-Z..p.. .*. from (a), we have P Now, substituting j9 n +w successively for J9„, we obtain x n =—\l(p n +w)-lp n \, Po x n _ l =^{l(p n +2w)-l(p n +w)\, r r n a: =x n _ n =^{/.(7>„.+ww)-Z.(p n +(n-l)u ? )}. r By taking the sum of these equations, the terms in the brackets will all cancel, except h(p„+nw) = lp t in the last and lp„ in the first, and we shall have Po r Po P» COMPRESSIBLE OR AERIPORM FLUIDS. 27? To find p n when h is given, we have P» Po or ^=e*°. P* _p 1 h Hence p,=p .e Po , (165) where e=2.71828, the base of the Naperian system of loga- rithms. 418. Schol. Formula (164) may be adapted to the determ- ination of heights by the barometer. To this end let b and b„ be the heights of the barometrical columns at the lower and v b upper stations respectively. Then, since — °— ir* /*=2.30258^.l| 5 , (a) The value of — may be determined from the consideration Po J thatp expresses the weight or pressure of a column of the at- P mosphere on a unit of surface as one square foot, and — - must Po express the height of this column on the supposition that its density is uniform. Now a cubic foot of air at 32° 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 ' —769. Hence the height of U.Uo 1 I & a homogeneous atmosphere at a temperature of 32° is 34x769 =26146 feet. .-. 2.30258— =2.30258X26146=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 varia- tion of gravity at different elevations. From numerous obser- vations made at different elevations above the sea, and at known differences of height, this coefficient is found to be 60345 fee/ 278 il Y D a O S T A T US. at a temperature of 32°. But the actual temperature of the an at both the lower and upper stations will, in general, offer 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 t and t n i)e the indications of Fahrenheit's thermometer; then the mean temperature will be l(t +t,), and the deviation t from the standard will be t=l(t +t„) -32. 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 /*=60345(1 +. 002220 L.t^. (166) It is further obvious that a change in the length of the mer- curial column will be produced by a change of temperature of the mercury. Let r and t„ be the temperatures of the mer- cury, 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' lt (l+. 0001) {r -r n ), where b' n is the observed height of the barometer at the upper station. Using this value of b n , the difference of elevation be- tween two stations or the height of a mountain may be determ- ined with considerable accuracy. In the determination of the constant coefficient, the variation of gravity at different elevations is allowed for in the assump- tion 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 45°. When the latitude differs from this, it will be necessary to multiply the result by (1 +.002837 cos. 2t/>), i/> being the latitude at which the observations are made. COMPRESSIBLE OR AERIFORM FLUIDS. 279 419. EXAMPLES. Ex. 1. A cylindrical tube 40 inches long is half rilled with mercury, and then inverted in a vessel of mercury. How high will the mercury stand in the tube, the pressure of the ex- ternal air being equivalent to 30 inches ? Let / be the length of the tube, a the length of the portion occupied by the air before it was inverted, h the height of mer- cury due to the pressure of the external air, x the height of the mercury after the tube is inverted, and h' the column of mer- cury equivalent to the tension of the air in the tube. Then (157), ha=h'(l-x). But h=h'+x or h'=h—x. .*. ha=(h—x) (l—x), whence we get x=\{l-\-h)±\ V(l— xy+4ah. Using the data of the question, /=40, a— 20, and h=S0, we get a,-=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 reached 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 : Pyrometer. Det. Ther. At Ther. At the base, 6 = 30.09in. t =83.0 r = 84.5 On the summit, b'„ =29.65 in. t n =85.0 t„ =83.5 280 H Y D 11 O S T A T I C S. Hence Also, t—%(t +L)—32°=52° and 1 +.00222/ = 1.1 1544. t -t„=1° and 6 n =6:,(l+.OOOlXl) = 29.653 in. , „ T 30.09 .'. // = 60345 X 1.11 544. XL 29.653 30.09 log. 1.4784222 29.653 a.c. " 8.5279314 0.0063536 log. 7.8030199 1.11544 " 0.0474462 60345. " 4.7800413 427.67 feet " 2.6311074. Ex. 7. The following observations were made by Humboldt at the Mountain of Quindiu, New Grenada, in lat. 5° : Barometer. At the level of the Pacific, 30.036 in. IT On the summit, 20.0713 6^ Required the height of the mountain. Ans. 11500 feel neaiiv. Det. Ther. n " 1r0 .54 At. Ther. 77°.54 75 08°.00 £ HYDRODYNAMICS. 420. Prop. The velocity of a fluid in a tube of variable di* ameter, kept constantly full, is in different transverse sections in- versely as the areas of the sections. Since the tube is supposed constantly full, and the fluid in- compressible, 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 quan- tity 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 bottom 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 .£F represent a very small orifice in ^ lm b the bottom of the vessel ABCD, filled with a fluid to the level of AB, GF a stratum whose thickness FH=/i' is indefinitely small, and FM=A the whole height of the column ver- tically above the orifice. If now the stratum GF fall by its own weight through HF=/j', the velocity will be T) G- E F 282 HYDRODYNAMICS. v= V2gh'. 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 v will be v '= V2g'h'. But the forces g and g' are as the weights of the columns GF and LF, or as their heights h' and h. Hence g' h gh — =— or £•' = —. g h' 8 h 1 Substituting this value of g 1 in that of v', we have v'=V2gh, (168) which is the velocity of a body falling 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, horizon- tally, or vertically upward, the velocity of efflux will be the same, since the pressure of fluids is the same in every direc- tion. In the first case, its path will be a parabola whose equation is (G4). In the second case, the angle of elevation a=0, which re- duces the equation to x > =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 orifice 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 O, the equation of its path is x* =4 hy, in which /j — 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 DISCHARGE 1 OM SMALL ORIFICES. 283 x=CP=2 y/hy=2 Vha=2.0M, or the horizontal distance is equal to twice the ordinate at the orifice, in a semicircle whose diameter is the vertical distance 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- tances above and below the mid- dle the range will be the same. 423. Prop. To find the quantity of fluid discharged in a given time from a small orifice in the bottom of a vessel when the fluid 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=ktV2gh. (169) If the orifice be circular, and r its radius, then k=nr*, and Q=n V~2g.t?'W7i, =25.195.tr Q Vh. The time being in seconds, and r and h in feet, Q, will be in cubic feet. If the weight W be required, W=62.5Q. Cor.' 1. The time required for the efflux of a given quantity is nr* V2gh Cor, 2. Since t and g are constant in (169), Hence the quantity discharged in the same time from orifices differing in size and distance from the surface, varies as thf size of the orifice and square root of its depth. If the orifices are the same, QxVh. In order, therefore, (hat the discharge from one orifice maybe twice that from an- other, its depth must be four times as great. 284 HYDRODYNAMICS. 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= V2gh, 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), h fa _ v'=y, v — FT V r 2^ r /t oc Vh. K k 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 up- ward. 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 time 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 in the vessel (169), .=-3= kV2gh Therefore, in the former, *=-?S=. (170) kV2gh If r' be the radius of a tranverse section of the vessel, r thai of the orifice, Q=7rr' 2 A and i=nr*. .\ t= — =Vh. rW2g 425. Schol. The preceding deductions are founded on the hypothesis that the fluid particles descend in straight lines to the orifice, and ah issut' with a velocity due to the height of the fluid surface. Experiment shows that this is true only of DISCHARGE FROM SMALL ORIFICES. 285 those vertically above the center of the orifice, that those sit- uated 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 to- tal 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 itself, 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 hori- zontal, 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 slight- 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 ve- locity 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 l =0.62^=0.62 V2gh, and for the quantity in a unit of time (169), Q=kv 1 =0.G27iV~2gh. When the orifice or pipe through which the discharge is made has the length and form of the vena contracta, the velocity will be v , =0.97u=0.97 V2gn, and the quantity Q=kv l =0.97kV2gh, h being the smallest section of the contracted veir. 280 II Y D R O D i' N A M I C S. By the use of cylindrical or conical adjutages, the quantitv 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 fric- tion. The discharge is greater when the adjutage is conical and the larger end is the discharging orifice. 426. Prop. To determine the quantity of water which will flow from a rectangular aperture. 1°. When one side of the aperture coincides with the sur- face of the water. Let h be its height and b its breadth, and conceive the aperture to be divided horizon- tally into a very large number n of equal di- visions so narrow that the velocity of the fluid in every part of each may be regarded as the same. The larger n is, the more near- ly will the hypothesis be satisfied. The depth of these successive divisions be- low the surface will be h 2h 3h n n n The velocities in each will be , &c. /hi 2h / 3h and since the areas of these divisions are each b-, the quanti* n ties discharged by each in a unit of time will be (169) bh f~h bh I 2h bh J 3h „ tv^ tv 2 =v tv ^t &c - and their sum, or the whole quantity Q, discharged, will be But ^ bhV2gh i i i Q= -^=(l 2 +2 2 + 3 2 +, &c. . nvn l 2 +2 2 +3^+, &c. . I /r 4+i 1 + 1 In 2 . DISCHARGE FROM RECTILINEAR APERTURES. 28T bh ... Q=|6/ t V2gh=°~b V2gh\ Or, if 15=7^ be the mean velocity, 2°. When the upper side of the rectangular aperture is not coincident with the surface of the fluid. Let its depth EL=/j i5 and the distance EG=A, 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 q=lbV2gh 3 -^V2ghl = ^bV2g(h 2 -h 2 1 ); and if Q=b(h— A,)u, v=%V2e. h—h 1 427. Prop. To determine the quantity of water which will flowf?'om a triangular aperture. 1°. When the vertex of the triangle is in the surface. Let the base GH=&, and the height HF=/j, v ^ 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 II lengths, beginning at the vertex of the trian i b 2b 3b ffle, are -, — , — , otc. 1 heir areas, regard ° ' n n n ing them as parallelograms, which we may do, since n is in definitely large, are bh 2b h 3b h c -.-, — .-, — .-, &c. nn n n n n Hence the discharges through each are bh /~~h 2bh /~~2k Sbh /~~3h . and the whole discharge will be 288 HYDRODYNAMICS. Q = bhy/2gh a § g = —-(l 2 +2 2 +3 2 +, &c. n* Vn 3, . -. bhV2srh/n 2 n 2 ) Also, hV2gh/7i- \ nWn Vf+1' fMV2gh=%bV2gh 3 . 2°. When the base of the triangle is in the surface. In this case the quantity which will flow through EFG wil, equal the discharge through EFGH minus that through FGH, or Q / =fbhV2gh-^bhV2gh= T %bhV2gli, and v=^jV2gh. Cor. If the aperture be a trapezoid, as EFGH, whose upper base EF=&j lies in the surface, lower base HG=fe 2 , and alti- tude KH=/i, we may find the discharge by regarding the aperture as made up of the rectangle KHGM, and the two triangles EKH and MGF. Hence Q^=^bJiV'2^+ T %(b 1 -b 2 )V'2gTi== T 2 j{2b 1 +3b 2 )hV'2gh. Also, through 'the triangle HOG, whose base HG=& 2 , and depth of vertex hO—h l , the quantity of the discharge will be Q=tW'i V2^7- T 2 J (2& 1 +3& 2 )Av / 2i : A. In a similar manner, rectilinear orifices of other forms may be divided into triangles, trapeziums, &c, and the discharge determined. 428. Prop. To determine the velocity with which an elastic fluid will issue from a small orifice into an unlimited void when urged by its own weight. By reference to Art. 421, it will be seen that the result there obtained is independent of the density of the fluid, and will 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 : io'=gh : gh'=h : h'. MOTION OP FLU iDS 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 vacuum is that which a heavy body would acquire in falling from the height of a homogeneous atmosphere. The height of a homo- geneous atmosphere {Art. 418), when the temperature is 32° and the barometer stands at 30 in., is /i' = 26146 feet, which gives u=1297 feet. But for a temperature of 60° the height A, will be (159) h 1 =h'(l+vt), in which *=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= V2gh l = V2gh'(l +0.00222X28) = V2X32|X2614GX(1+0.00222X28) = 1336 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 in- volving 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 al- lowed to flow through pipes whose diameters were li 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 kOO HYDRODYNAMICS. 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 from a pipe of given dimensions, it was found that a pipe 30 feet long and li 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 fath- oms in length and 12 inches in diameter, with a head of 12 feet of water, discharged T Vth, and a pipe of the same diameter and 2340 fathoms in length, with a head of 20 feet, discharged T Vh 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 sec- 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 determine generally the velocity with which a fluid will issue from an orifice of any size in the bottom of a 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' Let the distance of the lamina pp'qq' from the surface A be x, and K the area of its surface. Then, if dx be its thickness and p its density, pKdx will be its mass, and gpKdx its weight, or the force with which it wili descend if free. Now the resistance with which it meets in its descent is the difference of pressures on its lower and upper surfaces. Let/> represent the pressure on a unit of tht f c D B E GENERAL METHODS. 291 upper surface by the water above it ; then p+dp will be the up- ward 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 lamina. Hence the moving force will be gpK.dx-K.dp, and the acceleration, [VI.] and [IX.], _gpKdx— Kdp gpdx — dp d 2 x 0= ~pKdx = pdx = ~dJ' (a) Let v be the velocity of discharge at B, k the area of a sec- tion of the issuing fluid, and dt the element of time in which the surface K descends a distance equal to dx. Then (167) we have „ , 7 7 dx kv Kdx=kvdt or -jr—r?-- (o) dt K v ' By differentiation, regarding dt as constant, we get d 2 x __kdv ~dT = Kdt kdv _gpdx — ap Kdt pdx _ , pkdv dx g P dx-dp=- w . Tt . k* Hence (b) gpdx—dp— p^.vdo. Integrating, gpx—p=p^.—+C. Let now a;=A=AB and k=K ; then gp h-p=p-+C ; and subtracting the preceding from this, we have g{ h- x )=- { l--), -v 7 ^. ( „„ 1_ K a 292 HYDRODYNAMICS. If k be very small, v= V2g(h—x) nearly; and if the vessel be kept constantly full, x=0 and v= V2gh (168). If k=K, v= ac, or the velocity must be infinite. From which 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, the portion of the fluid at the lower extremi- ty, being urged by the pressure of all above it, will neces- sarily have a greater velocity than those portions which are higher, and therefore (167) a section of the issuing fluid is nec- essarily less than a section of the tube. 431. Prop. To determine the time in which a cylindrical ves sel will exhaust itself by a small orifice in the base. Since k is very small, we have v— V2g(h—x) for the veloc- ity at the end of the time t when the surface of the fluid has descended through the space x. Let this velocity be sup- posed constant during the indefinitely small time dt. Then the quantity discharged in this time will be tfQ, = kdt V2g(h—x) = Kdx, dx being the descent of the surface in the time dt. K _i Hence dt— — =.(h—x) -dx. kV2g 2K i Integrating, t= —.(A— x) 2 +C. kV2g 2K - If x =0, *=0, and C= — =:Vh. kV2g 2K , - .-. t= — =(Vh-Vh-x), (172) kV2g and when x=h, or the whole is discharged, kV2g kV2gh kV2gh The time is therefore twice as great as that which is required to discharge the same quantity when the vessel is kept con- stantly full. EXAMPLES. 293 432. Pitop. To determine the quantity which will issue from an aperture of any size and form in the side of a prismatic vessel. Let the Fig., Art. 430, represent one face of the vessel, and DCE the aperture. The element of the area of the orifice will now be ydx. .-. dQ=tydxV2g(h 1 +x), in which h l =AC, x=Cm, and y=mn. This, integrated between the limits x~0, and x=h—h l —CB will give the quantity discharged in the time t. If the aperture be rectangular, y—a. constant=&. Then dQ=bt V2g(h 1 +z) dx, and Cl=^btV : 2g(h 1 +xf+C, and, between the limits above named, Q=f&*V%r(A f -4). (1^4) If the orifice extend to the top of the vessel, then h , =0, and . 3 Q=f6f V2g.h 2 =%bht V2gh, v= m =%V2gh, 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 constant- ly full, or the surface of the fluid to be very large compared with the aperture. 433. EXAMPLES. 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 discharged through a small hole at the vertex. The equation of the semi-cubical parabola is ay 1 — x\ Hence (167) kvdt= — Kdx, Trx^dx or k V 2gxdt = — ny^dx = . 294 HYDRODYNAMICS. .*. at= ^.x~dx. kaV2g Taking the integral between the limits x=0 and v=a we have ~ i 7:V2a i t== z=z.^a- = =r. kaV2g Ik Vg Ex. 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. 3 2-rrph 2 Ans. t= =. 3kV2g 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. 7rr 2 F(2"-l) Ans. t— j . 20kg 2 Ex. 4. Find the times in which a fluid contained in a vessel, formed by the revolution of the curve whose equation is y i =a s x about the axis of x, will descend through equal distances h , supposing a small orifice at the vertex, and the axis vertical. 3 _ , -na'h Ans. .hacri t= — ^. kV2g Ex. 5. The horizontal section of a cylindrical vessel is 100 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=n m 3G'.5. HYDROSTATIC AND HYDRAULIC INSTRUMENTS h A- B- C4 M D MARIOTTE'S FLASK. 433. This piece of apparatus was devised to illustrate several successive positions of equilibrium of air and water, and variations in the velocity of dis- 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 the reverse. The apertures at first are all closed, the vessel is filled with water, and the tube DF is inserted at M and filled with water to the height D. 1°. Let the aperture B be opened. If h be the height of a column of water equiva- lent to one atmosphere, and x represent, in every 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 press- ure of the column of water DF,—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 v— 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. 2°. 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 press are of the air at A, and to the weight 296 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. of a column of water equal in height to AB=HE=:r, or the upward pressure on E is h+x. The resultant upward press- ure 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 h 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 y=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. 3°. Let. A and B be closed and C opened. The pressure at C inward is h, outward is h+x, represent- ing by x, as before, the vertical distance of the surface of the water in the tube from the open orifice C. Therefore the re- sultant 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 sur- face 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 en- trance of the air, and the descent of ho to F, the value of £ has remained unchanged, and therefore the velocity of discharge has remained constant. After this the value of a; will diminish, and the velocity of discharge also, till the water in the flask de- scends to the level of C. BRAMAH'S HYDROSTATIC TRESS. 434. In the annexed section of this instrument L and H are vertical cylindrical cavities, in a solid mass of metal or other BRAMAHS HYDROSTATIC PRESS. 297 strong material. The diameter of H is con- siderably less than that of L, and they commu- nicate through an aper- ture N, in which is a valve opening into L. A is a strong piston or solid cylinder of iron fitting closely to the surface of the hollow cylinder, and movable m. it. IH is a piston similarly applied to the other cavity H, and movable by means of a lever KP, whose fulcrum is at P. At O 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 press- ed 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 re- peated. The pressure on the base of A (134) is to the force impress- ed 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 / the longer and shorter arms of the lever KP, p the power ap* plied at K, and P the resulting pressure on the base of A Then the force p', applied to the piston H, will lie But (134), PttR 2 pi -ftp"* 298 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. R 2 L R a *■ r J I r This press seems to present the simplest and most effective of all contrivances for increasing human power. T R If ^=100 lbs., -r=10, and -=10, then P= 100,000 lbs. HYDROSTATIC BELLOWS. 435. This instrument presents an illustration of what is term- ed the Hydrostatic Paradox : that fluids, however unequal in quantity, may be made to equilibrate. It consists of two circular boarcte, A and B, firmly connected by a cylindrical coating of leather or other flexi- ble 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, / \ and a weight W, very large compared with ■H- ~7 that of the water in CM, may be supported / \ on B. The fluid in CM which counter- ) ( poises 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 cyl- inder, or that of B, and p the pressure of the fluid NP above P on k. Then (134) ^=|. If o be the specific gravity of the weight W, V its volume, and NP=/i, then gp,hk _k gpV ~K* . PV V •*■ ^pTk^'k- If V be the whole volume of the fluid in the instrument, and II the height of the cylinder AB, then DIVING BELL. 299 V'=H(K+A)+M=H(K+A)+ffV~ V'K-aVk •"' H_ K(K+A) * Let now the quantity of fluid be increased by v, and the cor- responding increment aH in the height of W will be K(V'+v)-aVk KV'-aYk v aH- K(K+k) K(K+A) K+A* DIVING BELL. 436. The diving bell is commonly a hollow cylinder or par- allelepiped, 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 con- tinually exercises a greater pressure on the contained air, con- denses 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 thst of the air in the bell, and p x that of water. Then the pressure p, on a unit of surface of air in the bell, will be p=g{hp' + (R+x) Pl }. But the unit of pressure arising from the elasticity of the in? rn. al air is gph. .-. g{hp' + Ql+x)p l }=gph. The quantity of air in the bell being constant, and a horj na- tal section of it the same throughout (155) and (156), px=p'a, or p=p'.-. x , TT h .'. hp'+Hp i +xp 1 =p'a-, or — k+Hx+x^ — ah, Pi P, and g being the specific gravity of air, ohx-\-Hx+x' i =oah. a:=-i(H+(j;t)±v r i(H+a/*)M-ffiiA. 300 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. B Pii xMi SEA GAGE. 437. The vessel AB is perforated with holes, and has with- in it, fixed in a vertical, position, a glass tube, hav- ing 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, 'ftie instrument is allowed to sink in the water whose depth is to be determined, and by a mechanical contrivance the weight is de- tached when it strikes the bottom, so that the gage will ascend by the buoyancy of the ball. The height MP to which the mercury has risen in the j— L-i 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, I the length of the tube above the surface of the mercury in the cup, x the depth of the water, and MP=Aj. If h 2 be the height of a column of mercury which would be sustained by the elastici ty of the air in NP, then (155) h 2 (l—h 1 )=hl or h 2 = -, — =—. Now the elasticity of the air in NP+ the weight of the col umn MP= the pressure of the atmosphere on the surface of th& fluid + the weight of a column of water extending to the bottom; or, p being the density of mercury, and p x that of the water, ph a +ph 1 =ph+p 1 x. ■ \ x— — (h 2 +h l —h) Pi -oh 2 +o(h 1 —h) hi ah 1 in which o is the ratio of the specific gravity of mercury to -hat of the water of the ocean. SIPHON. 30] SIPHON. 439. The siphon is an instrument for transferring fluids from one vessel V to another V in which the surface of the fluid is lower. It consists 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 1 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 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 exert- ed 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 gp 1 ak. This excess of pressure will urge the fluid from V to V. Hence the resultant pressure is equal to gak(p l —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 equiva- lent 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 with the atmosphere, and AB=.r, then the upward pressure on k in the shorter branch, at the level of the fluid, is that due to h, while the downward pressure is that due to x. Therefore the fluid is urged in the direction DAO, at A, by a force represent- ed by h—x. As long as x4ah, there will be two points. If a = 10,1 = 36, and h=32, z=20 and 16. AIR PUMP. 305 V 13 v p' -a\£ R 1H AIR TUMP. 442. B and B' are cylinders or barrels, commonly of the same size, in which pis- tons P and P', with valves V and V opening upward, are movable by means of rack-work, the one ascend- ing as the other descends. At the lower extremity of the barrels there are aper- tures with valves opening upward, and communica- ting by the pipe ac with the 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, through which the contents of B escape into the external air. The action of both pistons is mani- festly 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 1 the capacity of either barrel, and p„ the density of the air in the machine after the wth stroke. Now the quantity of air in the machine after the nth. stroke is p„V, and by the (?i + l)th stroke the volume V l5 or quantity p n Y lf is expelled. There will re- main then in the machine after the (?i+ l)th stroke the quan- tity p„V— p„Vj, and this being diffused through the space V, we have, by (155) and (156), PH-iV=P»V- Prt V 1 =p n (V-V 1 ), V or P-+i=P»(l' -)• If p be the initial density of the air, by making n equal to 0, 1, 2, &c, successively, we have U 306 HYDROSTATIC AND HYDRAULIC INSTRUMENTS. Pi=PoO— y-J v \ v\ ! P 2 =PAl—Y)=Po(l-y 1 ) Pn=P (l—f)- 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, v', were made to open downward, it would represent a condenser. By each descent of the piston a volume of air equal to that of the 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 p„ and p B+1 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, V, that of either barrel, and p the density of the external air. Then P» +1 V=P»V+ Po V 1 . •'. P„+1 = P„+P -y- If n be made successively equal to 0, 1, 2, 3, &c, we hav« p 1 -Po(i+ : y-) V 2V \ P 2 =P i +P o~Y=P o (1 +_ V/ CLEPSYDRA. 307 CLEPSYDRA. 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 con- taining 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 beginning of the motion. Equation (172) gives, using x instead of h — x, kV2ga k 2 g . x=a — tA — —t Let &x and M be corresponding increments of re and t; then k V 2sa , . k*g , x+Ax=a %*-{t+U)+2gi(t+My. Subtracting the preceding equation from this, and we have F°- ( K / 2a The time t is in seconds, and, to determine the divisions cor- responding to successive minutes of time, put A£=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 vert- ical distance of the successive divisions will all be equal, or the surface will descend equal distances in equal times. THE END. CVl CO