•iiil' mmmi¥< mm r p r --i nr: 3DE 3DE Tl GIFT TO THE LIBRARY Civil Engineering Department University of California BY PROFESSOR FRANK SOULE 1912 □< f,^ Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsanalytOOwoodrich THE ELEMENTS OF ANALYTICAL MECHAXICS SOLIDS AND FLUIDS. DeYOLSOX ^OOD, a.m., C.E., VBOF£SSOB OF MiTHEMATICS -tSD MECHA2fICS IX STEVII^S INSTITrTE OF TEOHKOL OGT j ArTHOB OF "resistance OF MATEEIALS ; " " EOOFS AKB BRIDGES;" REVISED EDITIOS OF " ilAHAX'S CIVIL EXGIXXXRISG," ASD " EI.TIMENTABT ICECHAXICS." FOURTH EDITIOX. BEVTSED, CORRECTED, A>D EXLARGKD. NEW YORK : JOHX WILEY & SOXS, 15 AsTOR Place. 1884. Engineering Library COFYBIGHTED, 1876, BT deVOLsok wood. QASOS' iingineeritig tibjcary PREFACE. The plan of this edition is the same as the former one. It is desiirned especially for students who are beginning the study of Analytical Mechanics, and is preparatory to the higher works npon the same subject, and to Analytical Physics and Astro- nomy. The Calculus is freely used. I have sought to present the subject in such a manner as to familiarize the student with analytical processes. For this reason the solutions of problems have been treated as applications of general fornnilas. The solution by this method is often more lengthy than by special methods; still, it has advantages over the latter, because it establishes a uniformity in the process. My experience has shown the importance of applying the fundamental equations to a great variety of problems. I have, therefore, in Article 24, and Chapters lY. and X., given a large number and a considerable variety of pn^blems to be solved by the general ecpiations under which they respectively fall. In the revision I have been aided not only by my own expe- rience wath the use of the former edition in the class-room, but also by the friendly advice and criticism of several professors of colleges who have used the work. The result has been that several pages have been rewritten, some definitions changed, and the typographical errors corrected. Several new pages in the latter part of the work have been added. I am especially indebted to Professor E. T. Quimby, of Dartmouth College, Hanover, K. II., for his valuable suggestions and for assistance in reading the final proofs. The nature of force remains as much a mystery as it was 788302 IV PREFACE. when its principles were first recognized. Of its essential nature we shall probably remain forever in ignorance. We can only deal with the laws of its action. These laws are deter- mined by observing the effects produced by a force. Force is the cause of an action in the physical world. The results of the action may be numerous and varied. Thus, force may pro- duce pressure, tension, cohesion, adhesion, motion, affinity, po- larity, electricity, etc. Or, to speak more properly, since force may be transmuted from one state to another, we would say that the above terms are names for the different manifestations of force. The question " what is the connect measure of force" has taken different phases at different times. During the last century it was contended b}' some tliat momentum {Mv) was tlie correct measure, while othei-s contended that it should be the work which it can do in a unit of time [jiMv^). But as one has happily expressed it, " theirs was only a war of words ; " for the real measure of force enters only as a factor in the expressions. Thus, if i^ be a constant force, the value of the momentum is Ft, see page 51, and of the work Fs, see page 45. At the pre- sent day some contend that the only measure of force is the motion which it produces, or would produce, in a unit of time. This is called the absolute measure, and the absolute unit of roKCE is the velocity which the force produces, or rooxild jpro- duce^in a unit of mass in a unit of time if it acted during the U7iit with the intensity which it had at the instant con- sidered. If the intensity of the force were constant, the velo- city which it produced at the end of the unit of time would be the required velocity. Hence, tlie absolute measure of any force acting on any mass is the 2y^'<>duct of the mass into the acceleration^ and is the second member of equation (21). This is a correct measure, and is accepted as such by all writers on mechanics. But those who contend that this is the only measure, neces- sarily deny that weight, or more generally pounds, is a mea- sure. I contend i\\Ai pounds is, a measure of the intensity of a PREFACE. "V force both statically and dynamically. Many authors maintain the same position. Indeed, it is probable that the position Avhich I have taken can be deduced from any standard work on mechanics ; but in some it is left to inference. Thus, in Smith's Jlechanics, page 1, we find this terse and correct defi- nition, " The intensity of a force may be measured, statically, by the pressure it will produce ; dynamically, by the quantity of motion it will produce." I say this is correct, but I will add that the intensity of a force which produces a given motion is also measured by a pressure, or by something equivalent to a pressure, or to a pull. To those who will look at it analyti- cally, it is only necessary to say that the first member of equa- tion (21) is measured in pounds. If we know the absolute measure, we may easily find its value in 2)oimds. The pound here referred to is the result of the action of gra- vity upon a certain quantity of matter. The amount of matter having been fixed, either by a legal enactment or by connnon consent, and declared to be one pound at a certain place, its weight, as determined by a standard spring-balance at any other place, becomes a measure of the force of gravity as compared M-ith the fixed place. This standard spring-balance may meas- ure the intensity in pounds of any other force, whether the body upon which the force acts be at rest or in motion. If a perfectly free body were placed in a hollow space at the centre of the earth, at which place it would be devoid of weight, and pulled or pushed by a constant force, whose intensity, measured by a standard spring-balance, equaled the weight of an equal body on the surface of the earth, then would its motion be the same as that of a falling body. See page 24, Problem 7. In the forces of nature producing motion, there being no visible connection between the point of action of the force and the body upon which it acts, we are unable to weigh their intensity except by calculation. If the absolute measure is known, the jyounds of intensity may be computed. Tiie absolute measure of the force of gravity on a mass rti is mg, and the weight of the'lTody being IF, we have TF= mg. The sun acts upon the earth with a force which may be expressed by the absolute vi PREFACE. measure, and also by a certain miinber of pounds of force. More than half of the examples in Article 24 involve an equality between j)oimds ofiniensiti/ and the absolute measure of the force. The fact is, that, in case of motion, these quanti- ties are co-relative. Since, then, it is correct to use the term pound as the measure of the intensity of a force Avhether the body be at rest or in motion, and since it is in common use, and the student is familiar with it, I prefer to consider a force as measured by a certain number of pounds. See Article 9. It is more simple, containinc; as it does oidy one element, than the absolute measure, which contains three elements — mass, velocity, and time. There is anotiier advantage in thus measuring force. Stu- dents frequently, and in some cases writers, use the expressions, " quantity of force," " amount of force," " force of a blow," etc, when they mean (or should mean) momentum, or work, or vis viva. In such cases an attempt to answer the question "how many pounds of force " would show at once that the quantity re- ferred to was wot force. So much ambiguity, or at least indeiiniteness, has arisen in regard to the term force, that I have rejected the terms " Im- pulsive Force" and "Instantaneous Force," and used the term " Impulse " instead of them. We know nothing of an instan- taneous force, that is, one which requires no time for its action. I also reject the expression /6»r6^e of inertia. I do not believe that inertia is ^ force. To the question " The inertia of a body is how many pounds of force " there is no answer. The term moment of inertia has no physical representation. The nearest approach to it is in the expression for the vis viva of a rotating body. In such problems the moment of inertia forms an important factor. The energy of a rotating body hav- ing a constant angular velocity is directly proportional to its moment of inertia in reference to its axis of rotation. See page 202. But motion is not necessary for its existence. See page 165. The expression appears in the discussion of numeroua PREFACK Vll statical problems, siicli as the flexure of a beam, the centre of pressure of a fluid, the centre of gravity of certain solids, etc. It is not the moment of a moment, although it may be so con- strued as to appear to be of timt/'orm. Some other term might be more appropriate. Even the expression moment of the mass would be less objectionable. The subjects of Centrifugal Force and Unhal-anced Force have been discussed of late in Engineering. Some assert that there is no sucli thing as a centrifugal force. Much unprofita- ble discussion may be avoided by strictly defining the terms used. If it is defined to be a force equal and opposite to the deflecting force, it will, at least, have an ideal existence, just as t!ie resultant in statical problems has an ideal existence. But the vital question is, is the centrifugal force active when the deflecting force acts ? Or, in other words, do both act upon the body at the same time? It seems, however, quite evident that if both acted upon the body at tlie same time they would neutralize each otlier, and the body would move in a straight line. Hence, in the movement of the planets, or of any free rotating body, there is no centrifugal force. But in the case of a locomotive running around a curve there may be both cen- tripetal and centrifugal forces ; the former acting against the locomotive to force it away from a tangent to the track ; the latter, against the track, tending to force it outward. "Wher- ever the force is conceived to act, whether just between the rail and wheel or at some other point, it is evident that both do not act upon the same body. Similarly in regard to the unhalanced force. It is a conveni ent term to use, but, in a strict sense, an unbalanced force does not exist ; for action and reaction are equal and opposite. But in reference to a particular body, all other conditions being ignored, the force may be unbalanced. Tlius, when a ball is fired from a cannon, the force of the powder, considered in the direction of the motion of the ball only, is unbalanced ; but tlie powder exerts an equal force in the opposite direction, and in that sense also is unbalanced. But when the entire effect of the Vlll PREFACE. force in all directions is considered, the algebraic resultant is zero. In other words, the centre of gravity of the system, for forces acting between its integrant parts, remains constant. These are some of the fundamental questions which will arise in the mind of the student as he studies the subject. Fortu- nately, it is not necessary for him to settle them beyond the question of a doubt before he proceeds with the subject. On some of these points scholars, who have made the subject a specialty, differ ; and it is only after a careful consideration of the points involved that one can take an intelligent position in regard to them. DeYolson Wood. HOBOXEN, AuffUtt, 1817. GREEK ALPHABET. Letters. A a B 13 Ty Names. Alpha Beta Gamma A 8 Delta z r Epsilon Zeta H 7? Eta ^e Theta I I Iota K K A \ Kappa Lambda M ti Mu Letters. IV V KameiU Nil S 1 Xi micron IT TT Pi X 0-9 Kho Si2;ma Tt Tail •T V - Upsilon ^ g; Phi 'Xx- Chi w ^ Psi fl (0 Omega TABLE OF CONTENTS. CHAPTER I. DEFINITIONS AKD THE LAAVS OP FORCES WniCn ACT ALONG 4 STRAIGHT LINE. PAoa ARTICLES 1-14 — Definitions 15_20— Velocity ; Gravity ; Mass 6 21-23— Force ; Mass ; Density 1*^ 24— Examples of Accelerated Motion 21 25-27 — Work ; Enerjnr ; Momentum ^'* 28-33— Impact ^* 34-36— Statics ; Power ; Inertia 37_38— Newton's Laws of Motion. Eccentric Impact 62 CHAPTER II. COJIPOSITION AND RESOLUTION OF FORCES. 39-42— Conspiring Forces "_ 43-40— Composition of Forces ""^ 47— Resolution on Three Axes "" 4y_49_Constrained Equilibrium of a Particle J^^ 50-CO— Statical Moments '^ Examples CHAPTER III. PARALLEL FORCES. 86 61_63— Resultant °" G4— Moments of Parallel Forces ^ 65-68— Statical Couples ^^ 69-74— Centre of Gravity of Bodies J^^ ■ 75_Centrobaric Method || 77_81— General Properties of the Centre of Gravity 10° 82— Centre of Mass. , XU TABLE OF CONTENTS. CHAPTER IV. XONCONCURRENT FORCES. ARTICLES PAOI 83- 85 — General Equations Ill 8G— Problems 115 CHAPTER V. STRESSES. 87- 90— Stresses Resolved 142 91- 92 — Shearing Stresses — Notation 144 93_ 94— Resultant Stress 14G 95- 96 — Discussion 158 97- 98 — Conjugate Stresses. General Problem 156 CHAPTER VI. VIRTUAL VELOCITIES. 100 — Concurring Forces 159 101 — Noncoucun-ing Forces IGO Examples 161 CHAPTER VII. MOMENT OF INERTIA. 102-104 — Examples. Formula of Reduction 165 105 — Relation between Rectangular Moments 169 106 — Moments of Inertia of Solids 172 107— Radius of Gyration 173 CHAPTER VIII. MOTION OF A FREE PARTICLE. 108 — General Equations 175 109 — Velocity and Living Force 176 110-117— Central Forces 180 CHAPTER IX. CONSTRAINED MOTION OP A PARTICLE. 118-121— General Equations 191 122-124— Centrifugal Force on the Earth 198 TABLE OF CONTEXTS. ^" CHAPTER X, HOT - -- ■ ' ATION OP A BODY WHEN THE FORCES ARE IN A PLANE. PAGE ARTICLES _ ^ 200 125-131— General Equations ^^^ 132-Reduced Mass. ._. . ^ ■-.•■•. -^^;;^ I •^;^ instantaneous " 133-135— Spontaneous Axis ; Centre ot l:'ercussioii , aju ^^^ Rotation ^-^q Examples "'*'"■ ^ oi o 136-138-Compound Pendulum. Captain Kater s Experiments -^^^ 139-m-Empticity of the Earth. Torsion Pendulum '. ". '. ] 224 143— Density of the Earth ~~^ 144-146— Problems CHAPTER XI. GENERAL EQUATIONS OF MOTION. 230 147_D'Alembert's Principle ^gx 148_General Equations ^g_ 149— Conservation of the Centre of Gravity ~'^_ 150— Consei-vation of Areas ^.^^ 151_Conservation of Energy ^^^ 152— Composition of Angular Velocities ~^^ 153— Moments of Rotation of the Centre ~^J 154_Motion of the Centre of a Body ~^^ 155— Motion of Rotation of the Centre ~^'^ 156— Euler's Equations 2^^ 157— Principal Axes 040 158_Xo Strain on Principal Axes ~ 150-Relation betsveeu the Fixed and Principal Axes ~^^ 160— Relations between Angular Velocities ~_^ 161— ^\^is of Spontaneous Relation '_";;; 102-ReIation between Spontaneous and Central Axes -^^ 164— Centre of Percussion "^^^ 105— Conservation of the Centre of the Mass ~ "- 166-Cnnservation of the Moment of the Momentum ^^J 167— The Invariable Plane ^^^ 168— Mutual Action between Particles ~^^ 169— Attraction of a Homogeneous Sphere ~^^ 169«-Mass and Stress 273 169&-Repulsive Forces 273 _ 1:0— Principle of Least Action ■ ~^.^ 170a-Gauss' Theorem* of Least Constraint XiV TABLE OF CONTENTS. CHAPTER XII. MECHANICS OF FLUIDS. ARTICLES PAQj. 171 — Introductory 275 172 — Definitions 273 178 — Elastic and Xon-Elastic Fluids 273 174 — Equality of Pressures 27G 175 — Perfect Fluid at Rest 277 176— The Law of Equal Transmission 277 177 — Pressure on Base of Prismatic Vessel 278 178— Pressure on Base of any Vessel 278 179 — Static Head 27J 180— Free Surface .* 279 181— Pressure on Submerged Surface 330 182 — Resolved Pressures 231 183 — Resultant Pressure 281 181 — Centre of Pressure 283 185— Condition of Flotation 285 186— Depth of Flotation 28;; 187 — Specific Gravity 238 188 — Examples 289 189— Fluid Motion. Definitions 291 190— Bernouilli's Theorem 292 191— Discussion of 294 192 — Graphic Representation 295 193 — Flowing through Varying Sections. Examples 296 19^t — Velocity of Discharge from Small Orifice 000 195— Discharge from Large Orifice 301 196— Vertical Orifices. Examples 303 197— Conical Adjutages 307 198— Reaction of Fluids 310 199— Centrifugal Force T 310 200— Resolved Centrifugal Force 311 202— Pressure Due to Discharge from the Side of a Vessel 314 203— Pressure Due to a Discharge Vertically Upward 315 205— Pressure of a Stream Impinging Normally Against a Plane Sur- face. 316 206— Case of Cup Vane 316 207— Case of Bent Tube 317 208— Case of Inclined Surface 319 209— Remark. Examples 319 210— Case of Impinged Surface in Motion. . .^ 322 211 — Work Done on Vane 303 212— Energy Imparted to a Vane 304 TABLE OF CONTENTS. XV ARTICLES PAGE 213 — Expression for Efficiency 325 21-i — Resultant Pressure on a Vane 325 215 — General Case 320 216 — Discussion 327 217— Hydraulic Motors 3^2 218— Case of a Single Vane 332 220— The Undershot Wheel 333 222— The Poncelet Wheel 3y6 223— The Breast Wheel 337 224 — The Overshot Wheel. Examples 337 225— Effective Head 342 226— The Jet Propeller 343 227— Barker's Mill 343 228— Pressure Due to Centrifugal Force. Examples 348 229— Detinitions of Turbines 350 230— Analysis of Turbines 351 231 — Special Cases 355 232— General Case. Examples 359 233 — Resistance Due to Sadden Enlargements 362 234 — Resistance in Long Pipes 363 235 — General Case of Flow in Long Pipes. Examples 364 236— Gases 369 237 — Boyle's Law 370 238 — Tension of Gas in a Vertical Prism 371 239— Numerical Values in Regard to Air 372 240— Tension of the Atmosphere Gravity Variable 372 241 — Heights by Barometer 374 242— Gay Lussac's Law 375 243— Absolute Zero 376 244— Heat a Form of Energy 377 245— Thermal Unit 378 246 — Specific Heat 378 247 — Dynamic Specific Heat 379 248— The Adiabatic Curve. Examples 379 249— Velocity of a Wave in an Elastic Medium 382 250— Value o f ^ 387 251— Remarks. Examples 387 252— Velocity of Discharge of Gases 388 254— Weight of Gas Discharged 390 APPENDIX I. Solutions of Problems 391-463 "- . APPENDLX II. The Potential 467-470 ANALYTICAL MECHANICS. CHAPTER I. DEFINITIONSj AND PRINCIPLES OF ACTION OF A SINGLE FORCE, AND OF FORCES ACTING ALONG THE SAME LINE. 1. Mechanics treats of tlie laws of forces, and the equi- librium and motion of bodies under the action of forces. It has two grand divisions, Dynamics and Statics. 2. Dynamics treats of the motion of jnaterial bodies, and the laws of the forces which govern their motion. 3, Statics treats of the conditions of the equilibrium of bodies under the action of forces. There are mauy subdivisions of the subject, such as Hydrodynamics, Hy- drostatics, Pneumatics, Thermodynamics, Molecular Mechanics, etc. That part of mechanics which treats of the relative motion of bodies which are so connected that one drives the other, such as wheels, pulleys, links, etc., in machinery, is called Cinematics. The motion in this case is independent of the intensity of the force which produces the motion. Theoretic Mechanics treats of the effect of forces applied to material points or particles regarded as without weight or magnitude. Somatology is the application of theoretic mechanics to bodies of definite form and magnitude, 4, Matter is tliat which receives and transmits force. In a physical sense it possesses extension, divisibility, and impene- trabihty. Matter is not confined to the gross materials which we see and handle, but includes those substances by which sound, heat, light, and electricity are transpiitted. It is unnecessary in this connection to consider those refined speculations by which it is sought to determine the essential nature of matter. According to some of these speculations, matter does not exist, but is only a conception. 1 2 DEFINITIONS. [5-6.] According to this view, bodies are forces, within the limit of which the attrac- ti-ve exceed the repulsive ones, and at the limits of which they are equal to each other. But observation, long continued, teaches practically that matter is inert, that it has no power within itself to change its condition in regard to rest or motion ; that when in motion it cannot change its rate of motion, nor be brought to rest without an external cause, and this cause we call FORCE. One also learns from observation that matter will transmit a force, as for instance a pull applied at one end of a bar or rope is transmitted to the other end ; also a moving body carries the effect of a force from one place to another. 5. A Body is a definite portion of matter. A particle is an infinitesimal portion of a body, and is treated geometrically as a point. A molecule is composed of several particles. An atom is an indivisible particle. 6. Force is that which tends to change the state of a body in regard to rest or motion. It moves or tends to move a body, or change its rate of motion. We know nothing of the essential nature of force. We deal onlv with the laws of its action. These laws are deduced by observations upon the effects of forces, and on the hypothesis that action and reaction are equal and opposite ;. or, in other words, that the effect equals the cause. In this way we find that forces have different intensities, and that a relation may be established between them. It is necessary, therefore, to establish a unit. This may be assumed as the effect of any known force, or a multiple part thereof. The effect of all known forces is to produce a pull, or push, or their equivalents, and may be measured by pounds, or by something equivalent. The force of gravity causes the weight of bodies, and this is measured by pounds. We therefore assume that a standard POUND is the UNIT of force. The standard pound is established by a legal enactment, and has been so fixed that a cubic foot of distilled water at the level of the sea, at latitude 45 degrees, at a temperature of G2 degrees Fahrenheit, with the barometer at 30 inches, will weigh about G2.4 pounds avoirdupois. The English standard pound was originally 5,7G0 troy gi-ains. The grain was the weight of a certain piece of brass which was deposited with the clerk of the House of Commons. This was destroyed at the time of the burning of the House of Commons in 18R4, after which it was decided that the legal [7.] STANDARD UNITS. 3 pound should be the weight of a certain piece of platinum, weighing 7,C0C gi-ains. This is kno\\-n as the avoirdupois pound, and the troy pound ceased to be the legal standard, although both have remained in common use. The legal standard pound in the United States is a copy of the English troy pound, and was deposited in the United States Mint in Philadelphia, in 1827, where it has remained. The avoirdupois pound, or 7,000 grains, is used in nearly all commercial transactions. The troy pound is a standard at 02 de- grees Fahrenheit and 30 inches of the barometer. The weight of a cubic inch of water at its maximum density, as accepted by the Bureau of Weights and Measures of the United States, is stated by :Mr. Easier, in a report to the Secretary of the Treasury, 1842, to be 252.7453 grains. Mr. Easier determined the temperature at which water has a maxi- mum density, at 39.83 degrees Fahrenheit, but Playfair and Joule determined it to be 39.101° F. The (xact determination of the equivalent values of the units is very difln- cult, and has been the subject of much scientific investigation. — (See The Me- tric System, by F. A. P. Barnard, LL.D., New York, 1872.) "SMien a quantity can be measured directly, the unit is generally of the same quality as the thing to be measured : thus, the unit of time is time, as a day or second ; the unit of length is length, as one inch, foot, ynrd, or metre ; the unit of volume is volume, as one cubic foot ; the vnit of money is money ; of weight is weight ; of momentum is momentum ; of work is work, etc. When dissimilar quantities are used to measure each other a proportion must be established between them. It is commonly said that "the arc mea- sures the angle at the centre," but it does not do it directly, since there is no ratio between them. The arc is a linear quantity, as feet or yards, or a num- ber of times the radius, while the angle is the divergence of two lines, and is usually expressed in degrees. But angles are proportional to their subtended arcs ; hence we have an equality of ratios, or angle subt ended arc ^ unit angle ~ arc ichich subtends the unit angle' and since a semi-circumference, or ir, subtends an angle of 180°, it is easy from the above equality of ratios to determine any angle when the arc is known, or tice versd. Similarly, the intensity of heat is not measured directly, but by its effect in expanding liquids or metals. The magnetic force is measured by its effect upon a magnetic needle. The intensities of lights by the relative shadows produced by them. Similarly with forces, we measure them by their effects. Dissimilar quantities, between which no proportion exists, do not measure each other. Thus feet do not measure time, nor money weight. Pounds for commercial purposes represents quantities.of matter ; but when applied to forces it represents their intensities. In a strict sense, pounds does not -measure directly the quantity of matter, but is always a measure of a force. 7. The line of action of a force is the line along whic-li the force moves or tends to move a particle. If the particle is FORCE, SPACE, TIME. [8-14 ] acted tipon by a single force, the line of action is straight. This is also called the action-line of the force. 8. The point of application of a force is the point at which it acts. This may be considered as at any point of its action-line. Thus, if a pull be applied at one end of a cord, the effect at the other end is the same as if applied at any intermediate point. 9. A FORCE is said to be given when the following elements are known : — 1st. Its magnitude {pounds); 2d. The dii'ection of the line along which it acts {action-line) ; 3d. The direction along the action-line (+ or — ); and, 4th. Its point of application. A force may be definitely represented by a straight line ; thus, its magnitude may be represented by the length A B^ Fig. 1 ; its position by the A ^ B position of the line A B\ its dii'ection along ^.^^ .,_ the line by the arrow-head at B^ which indi- cates that the force acts from A towards B\ and its point of application by the end A. 10. Space is indefinite extension, finite portions of which may be measured. 11. Time is duration, and may be measured. Probably no definition will give a better idea of the abstract quantities of lime and space than that which is formed from experience. 12. A BODY is in motion when it occupies successive portions of space in successive instants of time. In all other cases it is at ahsolute rest. Motion in reference to another moving body is reUitive. But a body may be at rest in reference to surrounding ob- jects and yet be in motion. Thus, many objects on the surface of the earth, such as rocks, trees, etc., may be at ]-est in refer- ence to objects around them, while they move with the earth through space. Observation teaches that there is probably no body at absolute rest in the universe. 13. Motion is uniform when the body passes over equal portions of space in equal successive portions of time. 14. Yariable motion is that in which the body passes c er unequal portions of space in equal times. fl5.] VELOCITY. * 15. Velocity is the rate of motion. When the motion is uniform it is measured by the linear distance over which a body would pass in a unit of time ; and when it is variahle it is the distance over which it would pass if it moved with the rate which it had at the instant considered. The path of a moving particle is the line which it generates. For uniform velocity, we have ^=^. (1) y t ' iu which s = the space passed over ; t = the time occupied in moving over the space s ; and' V = the velocity. J^or varialle velocity, we have Examples. 1. If a particle moves uniformly thirty feet in three seconds, what is its velocity? ^ ' — 2. If -s = at, what is the velocity? "^" ^ 3. If s = af + ht, what is the velocity at the time t^ or ^t the end of the space s% • ir -t^ ^f'-t^' j -Vtv ^f Here -S^ 7^ dt which is the answer to the first part. Find t from the given equation, and substitute in the expression for v, and it gives the answer to the second part ; or v= VU' + 4 as. yj___,^,^,, 4. If s = 4^, required the velocity at the end of five seconds. '^- 5 If 3,s-^ = 5^2 required the velocity at the end of ten sec- 6 If s = haf, what is the velocity in terms of the time and * ,w epace? ^^ -.. l^^,f^ . T^ — ■■ ^ .•', ■— - '/ " 1. ' .- 7. If at = e^'—l, required the velocity in terms of the^faime} _ and space. - -2 ANGULAE VELOCITY. [16, 17.J 16. Angulak velocity is the rate of angular movement. II a particle moves around a point having either a constant or a A'ariable velocity along its path, tlie angular velocity is meas- i> 11 red hy the arc at a unifs dis- tance which suhtends the angle sioept over in a unit of time hy that radius vector which passes through the 'particle. If 5 = AB = the length of the path described ; V = the velocity along the path AB /iv/'t». ;: a body describes the arc ^6' with such a varying velocity that its pro- jection on BI), a tangent at B, is constant, required the velocity and the acceleration parallel to BE. From the equation of the curve we have dx y ' From the conditions of the problem we have dy , -rr = constant = v ; dt ' , ^ ^^ dy dn , y , /2.» dt d'l v ' /> dx dy di: dt dt dij which is the velocity parallel to x ; dt dt dy jp ' J} 12 EXAMPLES. riS.j dPx v' di/ v'^ " df ~ jp dt ~ j^ ' hence the acceleration parallel to the axis of x will be constant. Let ds be an element of the arc, then will the velocity along the arc be ds Ida? dir\\ , /, 2,r\* 3. Determine the accelerations parallel to the co-ordinate axes X and y, so that a particle may describe the arc of a parabola with a constant velocity. Let the equation of the parabola be ' ' dx y' The conditions of the problem give ds -y- = constant = v. dt ds _ dij ds _ dy V da^ + dif' __ dy ./ dx'' ' dt ~ dt dy ~ di dy ~~ dt dy'' -dt^ ^ ^f-""' dy j)v " dt |/y4.2/2' and differentiating, gives d?y pvy dy d^ ~ ~ (^pi J^ V/2^| dt 29v~y / o , 0,2' which being negative shows that the acceleration perpendicular to the axis of the parabola constantly diminishes. Similarly we find d'^x jp^i? [lai EXAMPLES. 13 4. A M'lieel rolls along a straight line with a uniform velocity; compare the velocity of any point in the circumference with that of the centre. Let V = the velocity of any point in tlie circumference, v'= the uniform velocity of the centre, r = the radius of the circle, X = the abscissa which coincides with the line on which it rolls, and y = the ordinate to any point of the cycloid. Take the origin at A. The centre of the circle moves at the same rate as the successive points of contact jB. The centre is vertically over J^. The abscissa of the point of contact cor- responding to any ordinate ?/ of the cycloid, is r ve)\sin-'^ -; r , d I . 1 y\ ^ dy dt \ rl |/ ^,.y _ ^2 dt ^ dp dt , V 2/"^ — y^ The equation of the cycloid it • -iV 1 a? = r versm -- — {^ry — ify ; and from the theory of curves ds" = d^ + dif .-. d_2_ dy y 1 + ^ ^ or. and. ds dy '^~2r-y'^ ds dy ds 1 2v 14 GRAVITY. fl9.] If y = 0, ^ = 0; y ■= r^ V = \^2v' ; y = 2r, V = 2v' ; y = ^r, V = v'. Hence, at the instant that any point of the wheel is in con- tact with the straight line, it has no velocity, and the velocity at the highest point is twice that of the centre. The velocity at any point of the cycloid is the same as if the wheel revolved abont the point of contact, and with the same angular velocity as that of the generating circle. For, the length of the chord which corresponds to the ordi- nate y is ^2/'?/, and hence, if v:2v' : : V2ry : 2r ; /2i/ we have v =-\/ -^ v', && before found. 19- Gravitation is that natural force wliich mutually draws two bodies towards each other. It is supposed to extend to every particle throughout the universe according to fixed laws. The force of gravity above the surface of the earth diminishes as the square of the distance from the centre increases, but within the surface it varies directly as the distance from the centre. If a body -vvere elevated one mile above the surface of the earth it would lose nearly -^q-^ of its weight, which is so small a quantity that we may consider the force of gravity for small elevations above the surface of the earth as practically constant. But it is variable for different points on the surface, being least at the equator, and gradually increasing as the latitude increases, according to a law which is approxi- mately expressed by the formula , (/ = 32.1726 - 0.08238 cos 2Z, in which L = the latitude of the place, ff = the acceleration due to gravity at the latitude Z, or simply the force of gravity, and 32.1726 ft. = the force of gravity at latitude 45 degrees. [19.] EXASIPLES. 15 From this we find that at the equator g = g^ = 32.09022 feet, and at the poles g — g,^ — 32.25498 feet. The varying force of gravity is determined by means of a pendnkim, as will be shown hereafter. It is impossible to de- termine the exact law of relation between the force of gravity at different points on the surface of the earth, for it is not homogeneous nor an exact ellipsoid of revolution. The delicate observations made with the pendulum show that any assumed formula is subject to a small error. (See Mecanigue Celeste, and Piiissanfs Geodesie.) Substituting g for/ in equations (11), (12), and (13), we have the following equations, which are applicable to bodies falling freely in vacuo : — , — 2-5 v = gt= V^lgs = J ; s=igf='^^ = ivt; > (16) _ V _ /'2s _ 2.S EXAMPL E S. 1. A bod J' falls throug-h a height of 200 feet; required the time of desceat and the acquired velocity. Let (/ = 32j feet. Ans. i = 3.53 seconds. « V = 113.31 feet. 2. A body is projected upward with a velocity of 1000 feet per second ; required the height of ascent when it is brought to rest by the force of gravity. Ans. s = 15,544 feet, nearly. 3. A body is dropped into a well and four seconds afterwards it is heard to strike the bottom. Required the depth, the velocity of sound being 1130 feet per second. Ans. 231 feet. 4. 'A body is projected upward with a velocity of 100 feet per second, and at the same instant another body is let fall from a height 400 feet above the other body ; at what point wiU they meet? 16 ' DYNAMIC MEASURE OF A FORCE. [30, 21. i 5. With what velocity must a body be projected downward that it may in t Beconds overtake another body which has already fallen through a feet ? a Ans.v=zj+V2a'g. 6. Required the space passed over by a falling body during the w"" second. 20. Mass is quantity oy matter. If we conceive that a quantity of matter, say a cnbic foot of water, eartli, stone, or otlier substance, is transported from place to place, without expansion or contraction, the quantity will remain the same, while its weight may constantly vary. If f)laced at the centre of the earth it will weigh nothing ; if on the moon it will weigh less than on the earth, if on the sun it will weigh more ; and if at any place in the universe its weight will he directly as the attractive force of gravity, and since the acceleration i: also directly as the force of gravity, we have TF — = constant, 9 for the same mass at all places. This ratio for any contem- poraneous values of TFand (/ may be taken as the measure of the mass, as will be shown in the two following articles. The weight in these cases must be determined by a spring balance or its equivalent. 21. Dynamic measure of a force. Conceive that a body is perfectly free to move in the direction of the applied force, and that a constant 'uniform force, which acts either as a pull or jjush, is applied to the body. It will at the end of one • second produce a certain velocity, which call v^i^. If now forces of different intensities be applied to the same body they will produce velocities in the same time which are proportional to the forces ; or foe V(i), in which f is the applied force. Again, if the same forces ai-e applied to bodies having differ- ent masses, producing the same velocities in one second, then will the forces vary directly as the masses, or, f ocM. [21.] DTXA3IIC MEASURE OF A FORCE. 17 Hence, general!}^, if xiniform^ constant forces are applied to different masses producing velocities «J(i, in one second, then f o\\\w^,i\\Q rate of varia- tion of the velocity will be one of the elements of the measure of the force. Hence if F := 9. variable force ; 3f = the mass moved ; — =f= the rate of variation of the velocity; or dt dv velocity-increment ; and, ^ be substituted for v^t, in equation (17), reducing by equation (0), we liave From this we have dt F de (18) hence the value of cM is expressed in terms of the constant ratio of the force F to that of the acceler- ation j^ To determine this ratio experimentally I suspended a weight, IF", 1)y a very long line wire. The wire should be long, so that the body will move practically in a straight line for any arc through which it will be made to move, and it should be very small, so that it will contain but little mass. By _ means of suitable mechanism I caused a fig. s. constant force, F. to be applied horizontally to the bodv, thus causing it to move sidewise, and determined " 2 W F 18 UNIT OF MASS. [23.J the space over wliich it passed duriiii:; the first second. This equalled one-half the acceleration (see the first of equations (12) when t = l). I found when F= -^j^W, that /= 1.6 feet, nearly ; and for F= j\ ^, f= S.2 feet, nearly ; and similarly for other forces ; hence cM = -^j TF, nearly. But the rati-o of F to f is determined most accurately and conveniently by means of falling bodies; for y= ^ = the acceleration due to the force of gravity, and W the weight of the body (wliich is a measure of the statical effect of the force of gravity upoii the body), hence TF oM= — ; (19) g in which the values of W and g must be determined at the same place ; but that place may be anywhere in the universe. The value of g is assumed, or the relation between c and M fixed arbiti-arily. If c = 1 , we ha^e M=—; (20) 9 and this is the expi-ession for the mass, which is nearly if not quite universally adopted. This in (18) gives and hence the dynamic measure of the pkessukb which moves A BODY is the product of the mass into the acceleration. This is sometimes called an accelerating force. If there are retarding forces, snch as friction, resistance of the air or water, or forces pulling in the opposite direction ; then the first member F, is the measure of the unbalanced forces in pounds, and the second member is its dynamic equivalent. 22. Unit of Mass. If it is assumed that c = 1, as in the preceding article, the unit of mass is virtually fixed. In (20) if ir= 1 and ^^ = 1, then i/"= 1: that is, a %init of mass is the quantity of matter which will weigli one pound at tliat [22.] TJXIT OF IVIASS. 19 place in tlie imivei-se M'here the acceleration due to gravity is one.. If a quantitj- of matter weighs 32|- lbs. at a place where g = 32|- feet, we have hence on the surface of the earth a body which weighs 32^ pounds (nearly) is a iLiiit of 'mass. It would be an exact unit if the acceleration were exactly ^^ feet. In order to illusti-ate this subject further, suppose that we make the unit of mass that of a standard jpouiid. Then equa- tion (19) becomes in which g^ is the value of g at the latitude of 45 degrees- This value resubstitated in the same equation gives and these values in equation (18) give the final value of which is the same as (21). Again, if the unit of mass were the weight of one cubic foot of distilled water at the place where g^ = 32.1801 feet, at which i^lace we would have W= 62.3791, and (19) would give 62.3791 ^ • ■'■ " 32.1801' and this in the same equation gives 32.1801 ^ W ^ 62.3791 ' g' and these values in (18) give Jf = r^, , as before, g dt~' 20 DENSITY. 133.] 23. Density is the mass of a unit of volume. li M = the mass of a body ; • V = the voUime ; and D = the density ; then if the density is nniforui, we have If the density is variable, let 8 = the density of any element, then . clM M = fhdY (23) from which the mass may be determined when S is a known function of V, Examples. 1. In a prismatic bar, if the density increases nniformly from one end to the other, being zero at one end and 5 at the other, required the total mass. Let I = the length of the bar ; A = the area of the transverse section ; and X = the distance from the zero end ; then will 5 y = the density at a unit's distance from the zero end *, 5 —X = the density at a distance x ; and dV—- Adx ; 2. In a circular disc of uniform thickness, if the density at a unit's distance from the centre is 2, and increases directly as the distance from the centre, required the mass when the radius is 10. 124.] APPLICATIONS OF EQUATION (21). 21 3. Ill the preceding problem suppose that the density in- creases as the square of the radius, required the mass. 4. In the preceding problem if the density is two pounds per cubic foot, required the weight of the disc, 5. If in a cone, the density diminishes as the cube of the distance from the apex, and is oiie at a distance one fix)m the apex, required the mass of the cone. Having established a nnit of density, we might properly say that mass is a certain number of densities, 24. Applications of Eqcation^ (21). j^Obs. — if^ for any cause, it is considered desirable to omit any of the matter of the following article, the author urges the student to at least establish the equations for the acceleration for each of the 31 examples here given. This part belongs purely to mechanics. The reduction of the equations belongs to mathematics. It would be a good exercise to establish the fundamental equa- tions for all these examples, before making any reductions. Such a course serves to impress the student with the distinction between mechanical and mathematical principles.] 1st. If a hody whose weight is 50 jpounds is moved horizon- tally hy a constant force of 10 pounds, required the velocity ac- sol^ s. ^ ^ quired at the end of 10 seconds ^Bj ^ , and the sjpace passed over during ^^^ g that time, there leing no friction nor other external resistance, and the hody starting from re^t. Here J/"= — = TTTT lbs., and hence (21) gives g 321. F- 10 lbs. : df~~M'~ 30' Multiply by dt and integrate, and dt~ 30 The second integral is ds 193 ,^ f^ V = -^ ^ + (6\ = 0). 22 ACCELERATING FORCES. [24.] ^=60-^ + *^^^^ = ^)' and hence for t == 10 seconds, we have V = 6433 + feet. 8 = 321.66 + feet. 2d. Sujypose the data to he tl\e same as in the preceding example^ and also that the friction hetween the hody and the fJane is 5 pounds. Hcquired the space passed over in 10 seconds. Here i^= (10 — 5) pounds. dh _F _ 193 '''df~M~ 60' 2>d. Suppose thcit a hody whose weight is ^0 pounds is Tnoved horizontally hy a weight of 10 Ibs.^ which is attached to an inex- tensihle, hat perfectly Jlexihle string which passes over a loheel and is attached at the other end to the hody. Required the distance passed over in 10 seconds., if the string is without weighty and no resistance is offered hy the v^heel, plane, or string. soils. 10 Us. Fir. 10. In this case gTavitv exerts a force of 10 pounds to move the mass, or F^^ 10 lbs., and the mass moved is that of both bodies, or JI/= (50 + 10) -T- 321-. dh F _ 193 The integration is performed as before. Alls, s = 268.05 feet. ^th. Find the tension of the string in the preceding example. [24.] ■ MOVING MASSES. 23 The tension will equal that force which, if applied directly to the body, as in Ex. 1, will produce the same acceleration as in the preceding example. Let jP = 10 pounds ; W = 50 pounds ; T = tension ; — = the mass in the former example ; and g — = the mass moved by the tension. Bence, from Equation (21), f=F- and w g-^ Eliminate y, and we find WP Mr + P' .-. r= 8.33 lbs. What must be the value of P so that the tension will be a maximum or a minimum, P -i- W being constant ? 5M. In example 3, what must he the loeight of P so that the tension shall le Qif" ;paH of P f Ans. P = {n-l) W. QtL If a body lohose weight is W falls freely in a vacimm ly the force of gravity, determine the formulas for the motion. Here Mg = TFand the moving force F= IF; Wd^S _ jr^ d^s The integrals of this equation will give Equations (IG), when the initial space and velocity are zero. Let the student deduce them. 24 PROBLEMS OF * [24.] ^th. Suppose that the moving pressure {pull or push) equals the weight of the body, required the velocity and space. Here Mg = W and F= W, hence the circumstances of motion will be the same as in the preceding example. The forces of nature produce motion without aptparent pressure, but this example shows that their effect is the same as that produced by a push or pull whose intensity equals the weight of the body, and hence both are measured by pounds, or their equivalent. %th. If the force F is constant, show that Ft = 3Iv ; also Fs = ^Mv^, and ^Ft^ = Ms. If F is varialle we- have Mv =fFdt. dth. Suppose that a piston, devoid of friction, is driven hy a constant steam- pressure through a portion of the length of a cylinder, at what ptoint in the stroke must the pressure he instantly reversed, 80 that the full stroTce shall equal the length of the cylinder, the cylinder heing horizontal ? At the middle of the stroke. Whatever velocity is gen erated through one-half the stroke will be destroyed by the counter pressure during the other half. 10th. If the pressure -upon the piston is 500 pounds, weight of the piston 60 po'U7ids, and the friction of the piston in the cylinder \00 pounds, required the point in the stroke at which the pressure must he reversed that the stroke may he 12 inches. The uniform effective pressure for driving the piston is 500 — 100 = 400 lbs., and the uniform effective force for stopping the motion i* .500 + 100 = GOO pounds. The velocity generated equals the velocity destroyed, and the velocity destroyed equals that which would be generated in the same space by a foi-ce equal to the resisting force ; hence if F= the effective moving force ; s = the space through which it acts ; V = the resultant velocity ; F' = the resisting force ; and s' = the space through which it acts ; Fig. 11. [24.] ACCELERATING FORCES. 25 then, from the expression in Example S, we have and FW = ^JIv', .-. Fs = FW, or, F: r::s' : s. In the example, F= 400 lbs., and F'= 600 lbs. Let x = the distance from the starting point to the point where the pressure must be revevsed. Then 600 : 400 : : « : 12 — a, .-. « = 7^ inches. 11th. If in the jpreceding exainjple the j)iston moves vertically up and down, required the jpoint at which the jpressure must he instantly reversed so that the full stroke shall he 12 inches. The effective driving pressure upward will be 500 — 100 — 50 = 350 pounds, and the retarding force will be 500 + 100 + 50 = 650 pounds, and during the down-stroke the driving force is 500 + 50 — 100 = 450 pounds, and the retarding force is 500 — 50 + 100 = 550 pounds. Vsth. A string jpasses over a luheel and has a weight P attached at one end, and W at the other. Jf there are no resistances from the string or iDheel, and the string is devoid of weight, req;uired the resulting motion. buppo; then ie TF > P: ir=TF-P, and TF+ P M = 9 . dh_F__ W-P •'• dV'~ M~ W + P^' By integrating, we find and. v — s = \ W-P W+ P W-P W+ P 9^. 26 PROBLEMS OF [24.J 13th. Required the tension of the string in the jyreceding examjple. The tension equals the weight P, plus the force which will produce the acceleration W- P W+ P^ when applied to raise P vertically. The mass multiplied by the acceleration is this movino; force, or P_ 9 W-P hence the tension is If +P TF^-^P „ _ 2TFP W+P^ - TF+P' Similarly, it equals TT minus the accelerating force, or TF_ p 2TFP . W-~ 1 TF= -^ — TF+P TF+P A complete solution of this class of problems involves the mass of the wheel and frictions, and will be considered here- after. 14fA. A string jpasses over a wheel and has a iveight P attached to one end and on the other side of the wheel is a weight IF, which slides along the string. Pequired the friction between the weight W and the string, so that the weight P will remain at rest. Also re- quired the acceleration of the weight W. Via. 13. The friction = P ; Mg= TF; and,i^= TF-P; d?s_F__ W-P " df hence, and, M W W-P , v= -^^9t, rW-P . s = i— ^- 9^' [341 ACCELERATING FORCES. 27 15fA. In the preceding exam/ple, if Wwe?'e an animal whoso loeirjht is less than P, required the acceleration with ivhich it must ascend, so that P loill remain at rest. 16M. If the weight W descend along a rough rope with a given acceleration, reguired the acceleration with which the hody P must ascend or descend on the opposite rope, so that the rope may remain at rest, no allowance heing rhade for friction on the tcheel. (The ascent must be due to climbing up on the cord, or be produced by an equivalent result.) I'^th. A particle moves in a straight line xinder the action of a uniform acceleration, and describes spaces s and s' in t^^ and \!^^ seconds respectively, determine the accelerating force and the velocity of projection. Let -yo = the velocity of projection, and f = the acceleration ; then /=' and ro t' - t' - ^'^-^ - ^) — -<-^ - ^) "" 2{t — t') g' Of _ I If - = ^ ; , then Vo = 0. s 2t — 1 ISth. If a jperfectly fexihle and perfectly smooth rope is placed upon a pin, find in what time it will run itself off . If it is perfectly balanced on the pin it will not move, unless it receive an initial velocity. If it be unbalanced, the weight of the unbalanced j)art will set it in motion. Suppose that it is balanced and let t'g = the initial velocity, 21 = the length of the rope, w = the weight of a unit of length, and t = the time. Take the origin of coordinates at the end of the rope at the instant that motion begins. "When one end has descended t feet, the other has ascended tlic same amount, and hence the 28 PROBLEMS OF [24.] unbalanced weight will be 2^05 2wl -h g ; hence we have dh _F_ _ 2^ws_ ~ M The mass moved will be df '^wl^-f' Multiply by ds and integrate, and we have ^*' _ „2 _ £...2 , in—o> 2i • ds \/-Vo Integrating again, gives 19th. If a particle moves towards a coitre of force whion ATTRACTS dircctli/ as the distance from the force., determine the motion. Let /x = the absolute force ; that is, the acceleration at a unit's distance from the centre due to the force ; and s = the distance then d^'s df=-^'- A force is considered positive in whatever direction it acts, and the plus sign indicates that its direction of action is the same as that of the positive ordinate from the origin of coor- dinates, and the negative sign action in the opposite direction. If a be the initial value of «, we have (see Ex. 5, Art. IT) : -y = V//, (c Ka- — 6-^) ; 1 5 t = iJi 'i (sin - — i'Tr)i r24.] ACCELERATING FORCES. 29 and the velocity at the centre of the force is found by making s = 0, for which we have, V = a Vn ; and 1 1 1 X 3 1 t = — -M~*/7, -fi~-7T, ;;,«"*?r, etc. hence, the time is independent of the initial distance. It may be proved that within a liomogeneons sphere the attractive force varies directly as the distance from the centre. Hence, if the earth were such a sphere, and a body were per- mitted to pass freely through it, it would move with an accele- rated velocity from the surface to the centre, at which point the velocity would be a maximum, and it would move on with a retarded velocity and be brought to rest at the surface on the opposite side. It would then return to its original position, and thus move to and fro, like the oscilhitions of a pendulum. The acceleration due to gravity at the surface of the earth being g, and ^' being the radius, the absolute force is and the time of passing from surface to surface on tlie equator would be /~ n-,i-,n /20,923,161 .o ifi ^ = V^ = ^•"''' y 82.09025 = *-'"• ^•^'''" The exact dimensions of the earth are unknown. The semi-polar axis of the earth is, as determined by Bessel 20,853,662 ft. Airy 20,853,810ft. Clarke 20,853,429 ft. The equatorial radius is not constant, on account of the elevations and depressions of the surface. There are some indications that the general form of the equator is an ellipse. Among the more recent determinations are those by Mr. Clarke, of England (1873), and his result given below is considered by him as the most probable mean. The equatorial radius, is according to Bessel 20,923,596 ft. Airy 20,923,713 ft. Clarke 20,933,161ft. 30 PROBLEMS OF [24.] The determination of the force of gravity at any place is subject to small errors, and when it is computed for different places the result may difiEer from the actual value by a perceptible amount. The force of gravity at any particular place is assumed to be constant, but all we can assert is that if it is variable the most delicate observations have faUed to detect it. But it is well known that the surface of the earth is constantly undergoing changes, being elevated in some places and depressed in others, and hence, assuming the law of gravitation to be exact and universal, we cannot escape the conclusion that the force of gravity at every place on its surface changes, and although the change is exceedingly slight, and the total change may extend over long periods of time, it may yet be possible, with apparatus vastly more delicate than that now used, to measure this change. It seems no more improbable than the solution of many problems already attained — such for instance, as determining the relative velocities of the earth and stars by means of the spectroscope. 'iOth. Suj)j)ose that a coiled spring whose natural length ie A B, is compressed to B C. Jf one end rests against an im- movable hody B, and the other against a hody at C, v)hich is perfectly free to move horizontally ^wJiat will Fig. 14. be the time of movement from C to A, and tohat will be the velocity at A ? It is found by expeiiiiieiit that tliB resistance of a spring to compression varies directly as tlie amount of compression, hence Ihe action of the spring in pnsliing the body, will, in reference to the point A, be the same as an attractive force which varies directly as the distance, and hence it is similar to the preceding example. But if the spring is not attached to the particle the motion will not be periodic, but when the particle has reached the point A it will leave the spring and proceed with a uniform velocity. If the spring were destitute of mass, it would extend to A, and become instantly at rest, but because of the mass in it, the end will pass A and afterwards recoil and have a periodic motion. If the body be attached to the spring, it will have a periodic motion, and the solution will be similar to the one iji the Author's Resistance of ALaterlals, Article 19. Take the origin at A, s being counted to the left ; suppose that 5 pounds will compress tbe sprhig one inch, and let the total compression be a = 4 inches. Let W= the weight of the body = 10 pounds. [U.] ACCELERATING FORCES. 31 The force at the distance of one foot from the orighi being 60 pounds, the force at s feet will be 60s pounds. from which we find that Vl93 and -y = 4.6 + feet. 21st. Suppose that in the jpreceding jprollem, a hody whose weight is M' is at B, and another M" at C, hath heing perfectly free to move horizontally, required the time of movement that the distance letween them shall he equal to AB ; and the resultant velocities of each. Take the origin at any convenient point, say in the line of the bodies and at a distance x to the left of Jf ', and let x" be the abscissa of M" ; h the length of the spring after being compressed an amount «, and fx the force in pounds which vx'ill ^compress it the first unit ; then the tension of the spring when the length is s will be fA{a + I — s) ; hence we have s = x" — x\ jr'^=M{^t + h-s), M''^^^=-J.i{a + h-s). From the first, d-s _ d'-x" cPo^ . W'~df~ "dt- ' 6ubstitutin^•, integl'Stino; df ^ M'M" ^ 82 • REPULSIVE FORCES. [34.] (iir + 3f"){2a + h)h . But V = iov s = h; .'. C\ = — /.i and integrating again gives M'3I" and, making s = a + h, wq have which, as in the preceding example, is independent of the amount of compression of the spring. To find the relation between the absolute velocities, Let s' = the space passed over by M', and s" = the space passed over by M"; then since the moving force is the same for both, we have df dt^ Integrating, gives M'v' = M"v". 22^, Suppose that the force varies directly as the distajwe from the centre of force and is kepdlsive. Then dh s = L«^_^-«v7y 2V^ in which Vq is the initial velocity. 23(^. Sujpjpose that the force varies inversely as the square of the distance from the centre of the force and is attractive. [TMs is the law of universal gravitation, and is known as the law of the inverse squares. While it is rigidly true, so far as we know, for every fa4.] ATTRACTIVE FORCES. 33 particle of matter acting' upon any other particle, it is not rigidly true for finite bodies acting upon other bodies at a finite distance, except for ?tomoge- ?ieom sp/iert s, or spheres composed of homogeneous shells. The earth bein>j neither homogeneous nor a sphere, it will not be exactly true that it attracts external bodies with a force which varies inversely as the square of the distance from the centre, but the deviations from the law for bodies at great distances from the earth will not be perceptible. We assume that the law applies to all bodies above the surface of the earth, the centre of the force being at the centre of the earth.] Let the problem be applied to the attraction of the earth, and r = the radius of the earth ; g = the force of o;ravity at the surface ; jM = the absolute force ; and s = the distance from the centre ; then and fi = gr- ; df~ ~?' Multiply by ds and integrate ; observing that for 5 = c?, ■y = 0, and we have ds^ df ^4-^ («) „ as — s^ \a I {as — sy using the negative sign, because t and 5 are inverse functions of each other. The second member may be put in a convenient form for integration by adding and subtracting ^ a to the numerator and arranging the terms. This gives ■^a — s — ^a {as — 6-^)^ Is a — 2s , _ ads 2{as-6^)i 2{as-s')i ' 34 ATTRACTIVE FORCES, r84.1 the integral of which is . (as —'6^)i — ^a versin"^ — + O. ^ ^ a But M'heii s = a, t = .'. C= ^a^r ; 1 / (5) From the circle we have tx — versin"' _ = tt — cos~'/1 — _ \ = From trigonometry we have 2 cos' y — 1 = cos 2 y. (2s \ Let 2 y= cos ~M II, then cos 2^^ = 1L — 1 ; . •. cos''' y — —, and 2/ = cos—' j/ ^ ; and a a ■ y a /s~ 2s -] 2y = 2 cos"'|/ — ; or TT — versin^'^ I. From (a) it appears that for s = 0,v = cc ; hence the velocity at the centre will be infinite when the body falls from a finite distance. lis = a = oo,v = 0. If a body falls freely from an infinite distance to the eartli, we have in equation («) a= oo ', and s = r = the radius of the earth ; I* _ for the velocity at the surface. But ^ = g; If ^ = 32J feet and r = 3962 miles, we have V 6280 / Hence the maximum velocity with which a body can reach the earth is less than seve-n miles per second. [24.1 ATTRACTIVE FORCES. 3o 24M. Svjypose that the force is ati'eactive and varies in versely as the n'^ j^ower of the distaTwe. Then d?s _ It df and integrating, gives A-ccoi-ding to the tests of hvtegrahility this may be integrated when 5 3 1 T' 5' 3'-l.T-'0r- ....etc., or 71 = . . . . - , ^ , -, , 2 , or - etc. 25 if A. Z^if the force vary inversely as the square root of the distance and le attractive. (This is one of the special casea of the preceding example.) We have ^.s Z'- df 6i' :. d.r _ dt^ 4/i(«* — .*); or. 2 u*( 7.- -'^-^ The negative sign is taken because t and s are inverse functions of each other. Add and subtract — -= — - and we have ^ ._ . _ f — \/r ^ 2v/c7 8v/;r "| 36 ATTRACTIVE FORCES. [34.] _ F— 3 V's" + 2 V^ 2/y/'a~ j .-. « --^ -^^ ^ /«♦ + 2«4 ] (ai - sA 3V 26?fA. Suppose that the force is attractive and varies in- versely as the distance. Hence d^s_ n W~ ~ s ' d&^ c ^ a df ^ ^S ' m wliich s = a for ■?; = 0. Hence the time from « = « to s = 0, is 1 ro ds In \\ ^2//. 'a ('"Sj) Let ( log — I = 2/ ; then for s = a, y = and f or « = 0, j' — oo . Squaring and passing to exponentials, we have log — = zr . *. — = c^ , or s = « (S "^ ; s s .-. ds = — ae ^''. 2y dy ; /GO This is called a gamma-function^ and a method of integrating it is as follows : — Since functions of the same form, integrated between the same limits are independent of the variables and have the same value, therefore [24.1 ATTRACTIVE FORCES. 37 r j e^ dy = rV e~* dt'. e'^ dy I i~^\lt = / e-y dy Also the left hand member will be of the same value if the sign of integra- tion be placed over the whole of it, since the actual integration wUl be performed in the same order ; hence r— p:o —I 3 /•» ,-»xi i dy dt = / / te-'^^^+'^'Kudu; Jo Jo in which y = tu\ .'. dy = t dn. Integrating in reference to t, we have -t^ (1 + m2) 2 (1 + m2) du. du which f or « = 00 becomes zero, and f or f = becomes ■^ , and the in ■ tegral of this is i tan-hi, which is zero for m = 0, and i tt f or ?i = oo ; ••■/ ^'"'"^-W' (See also Mec. Celeste, p. 151 [1534 0].) Or we may proceed as follows : — -hdx Let e-f'" = x.: dt=-{- log x) ^-^ f e-^ dt= -i{-logx)-idx. ^i Let'a; ^. «/ and consider a less than unity ; then log a will be negative, and log x — y{— log a) ; .'. dx = all^^y dy (- log n) ; 38 ATTRACTIVE FORCES. [24] which substituted above gives '0 , , Z*^ 2 J^ e-t^dt ?=y _ |( _ log x)-^dx = ( - log a)^ j aV dy . Dividing by ( — log a)- and multiplying both sides by — ^ (Za, we have / -J(-logar^(Za / -i (- log ^r-fZu; = / / -laV^dyda. Jl Ji •/() J\ Integrating the second member first in regard to fl, gives -\ dy\ 2/2 + 1 dtf whicli between the limits of and 1 gives \ — ^ — ; the integral of wluoh ia 1 + ?/2 \ tan-^?/ which between the limits of ao and gives Jtt. .-. / -iC-logrt)"- da j -i (- log a;)~*fZ« = J -\{-\ogx)-^dx =i,r; •'• / e-^ dt = •!■ v'TT. (See Mec. Celeste, Vol. iv. p. 487, Nos. [8319] to [8331]. Chauvenet's Spherical Asiroiiomy, Vol. i. p. 152. Todhunter's/ziif^rrtZ Calculus. Price's Infinitesimal Calculus.) Sometimes the integration of an exponential quantity becomes apparent by first differentiating a similar one. Thus, to integrate t e~^ dt., first differ- entiate e-^"- We have d e-^ = e-*'d ( — «^) = c"^ ( - 2 tdt) — - 2te-t dt ; I de-^ — -2 I te-* 2 dt. But the first member is the integral of the differential, and hence is the quantity itself, or e ^ , and hence the required integral is — ^ e~^ . 27th. Suppose that ttoo bodies have their ce?itres at A and A' respectively., and _^_ C I A ^^ ATTRACT a ^particle at Fig. 15. p ivith forces xohich vary as the distances from A and A'. [34] ATTRACTIVE FORCES. 39 Let C be midway between A and xV ; 6> = c ; AC= OA' =a', Cb :^ s = any variable distance ; and let fi = fi' = the absolnte forces of the bodies A and A' respectively. Then and integrating again gives s — c cos t V-t^' 2Sth. Supj>ose that ajMi'ticle is projected with a velocity u into a medium which resists as the square of the velocity / determine the circumstatices of motion. Take the origin at the point of projection, and the axis s to coincide with the path of the body. Let jM = the absolute resistance — or the resistance of the medium when tlie velocity is unity ; then n (-V-) = the resistance for ixny velocity ; •** df ~ ^ w) ' —- = -t.d8. 4|) ds dt ds And integrating between the initial limits, 5 = for — = u, and the general limits, we have log ■£- logu = -fis; 40 KESISTING FORCES. [24.1 1 dt or, log — = - M* ; ds di -us lie ^ ; or J "ttdt = e*' ds. Integi'ating again, observing that t — ^ for s = C, we have The velocity becomes zero only when s =r oo . 29^A. A heavy hodtj falls in the air hy the force of c^ravity, the resistance of the air varying as the square of the velocity ; determine the motion. Take the origin at the starting point, and Let K = the resistance of tlie Ixxly for a unit of velocity ; s = the distance from the initial point, positive down- wards ; t = the time of falling through distance s ; then —^ = for t and s = ; at -— = -y for t = t and s = s: and dt, ' (ds V -J- I = the resistance of the air at any point, and acta upwards ; and g = the accelerating force downward ; hence, the resultant acceleration is the difference of the two, oi d^s I ds V d's (dsV . - vJ _^_ w r24.] FALLING BODIES. 41 Kdt Separating this into two partial fractions, and integrating, gives Passing to exponentials gives , K\i S + " dt K« = i(r) log 7^- dt~\K/ ■^U,2«M^_1 ,2<(/c^)*^l' (b) which gives the velocity in terras of the time. To find it in terms of the space, multiply equation (a) by ds and put it under the form , ^2 '0 2k ds. K \dt! Droper Proceeding as before, observing the proper limits, we find (ds\ 2ks = — log - ds ' - —V = dt i/l{^ -'''-')■ ■ <'> If s = 00 , tJ = 4 / - , and hence the velocity tends towards a constant. f K From equatioP {b), multiplying the terms of the frr.ction hy e-Hi^ff)- , and observing that the numerator becomes the differential of the denominator, integrating, and passing to exponentials, we have, 2e'' = e^^'''^^'' + e-*^''^^\ (rf) • which gives the space in terms of the time. A neat solution of equation (a) may be found by Lagrange's method of Varuttion of Parameters. 42 FALLING BODIES. [24.1 ZOth. Supjyose that the hody is jyr ejected ujpxoard in the air, having the same coeflcient of resistance as in the joreceding examjpile. Take the origin at the point of propulsion, u being the initial velocity ; then ^8 d ds [dsV dt^ 'dt dt hence, Kdt = — ds\ \dt) d('-^i K'^Xdt) (Is Solving this equation for — ; we have, _ ^ _ / ff \^ ^< V « — VfJ tan t V Kg ^ dt~\K/ \/g + u "//T tan t V^g' (/) Substitute sin ty/ug-i-cost \/ ng tor tan tVKg and the numerator be- comes the differential of the denominator, and observing that t — for s = 0, we have 1 M Vk sin t^Kg + V g cos tV Kg s = -log ■ j= ; K Vg which gives the space in terms of the time. Multiply equation (e) by els and it may be put under the form /ds\^ g /(UV^ 7 "^ \cu) Integrating, observing that u is the initial velocity, and g + 2 KS = — log "(iF g + K u- [24.] IN" A EESISTING MEDIUil. 43 ds „ — 2«s g f ^ —2''S \ i- At the highest point v = 0. which in {/) and (g) gives I. t ^-{ower, which equals 33,000 pounds raised one foot per minute. Every moving body on the surface of the eaxth does work, for it OTcrcomes a resistance, whether it be friction or resistance of the air, or some other resistance. The same is true of every body in the vmiverse, unless it moves 46 WORK. [25. i in a non-resisting medium. * Animals work not only as beasts of burden, but in their sports and efforts to maintain life ; water as it courses the stream wears its banks or the bed, or turns machinery ; wind fills the sail and drives the vessel, or turns the windmill, or in the fury of the tornado levels the forest, and often destroys the works of man. The raising of water into the air by means of evaporation ; the wearing down of hills and mountains by the operations of nature ; the destruction which follows the lightning-stroke, etc. , are examples of work. "Work may be useful or prejudicial. That work is useful which is directly instrumental in producing useful effects, and prejudicial when it wears the machinery which produces it. Thus in drawing a train of cars, the useful work is performed in moving the train, hut the prejudicial work is overcoming the friction of the axles, the friction on the track, the resistance of the air, the resistance of gravity on up grades, etc. It is not always possible to draw a practical line between the useful and prejudicial works, but the sum of the two always equals tlie total work done, and hence for economy the latter should be reduced as much as possible. - In order to determine practically the work done, the inten- sity of the force and the space over which it acts must be measured simultaneously. Some form of spring balance is commonly used to measure the force, and when thus employed is called a Dynamometer. It is placed between the moving force and the resistance, and the reading maj^ be observed, or autographically registered by means of suitable mechanism. The corresponding space may also be measured directly, or secured automatically. There are many devices for securing these ends, and not a few make both records automatically and simultaneously. If the force is not a continuous function of the sjjace, equa- tion (26) must be used. The result may be shown graphically by laying off on the abscissa, AB, the distances ac, ce, etc., proportional to the spaces, and erecting o)"dinates ah, cd, ef, etc., proportional to the corresponding forces, and joining their upper ends by a broken line, or, what is better, by a line which 1 * All space is fiUed with something, since light is transmitted from all directions. But is it not possible that there may be a sometldng through which bodies may move without resistance ? [35-] WORK. 4T is slightly curved, the ainouiit and direction of curvature being indicated by the broken line previously constructed ; and the area thus inclosed will represent the Avork. The area will be giv- A en by the formula „ ,^ -L- Fig. 16. XF.^s. Simpson's rule for determining the area is : — Divide the abscissa AB into an even number of equal jparts^ erect ordinates at the points of division, and number them in the order of the natiiral numbers. Add together four times the even ordinates, tivice the odd ordinates and the extreme ordinates, and imdtiply the sum by one third of the distance between any two consecutive ordinates. If 3/1, ?/a, 2/3, etc., are the successive ordinates, and I the dis- tance between any two consecutive ones, the rule is expressed algebraically as follows : — Area = il (y, + 4//, + ^y, + 4y, - - - + 7/„)- (21>) If the applied pressure, ^ is exerted against a body which is perfectly free to move, generating a velocity v, then the work which has been expended is, equation (24), ^3fv^. This is called stored ivorlc, and the amount of work which will be done by the moving body in being brought to rest will be the same amount. If the body is not perfectly free the quantity \Mv^ is the quantity of work which has been expended by so much of the applied force as exceeds that which is necessary in overcoming the frictional resistance. Thus a locomotive starts a train from rest, and when the velocity is small the power exerted by the locomotive may exceed considerably the resist- ances of friction, air, etc., and produce an increasing velocity, until the resistances equal constantly the tractive force of the locomotive, after which the velocity will be uniform. The worlr done by the locomotive in produciug the velocity v in excess of that done in overcoming the resistances will be \Mv^^ in which Jf is the mass of the train, including the locomotive. 48 EXAMPLES. [25.] We see that double tlie velocity produces four times the work. This is because twice the force produces twice tlic velocity, and hence the body will pass over twice the space in the same time, so that in producing double the velocity we have 27^2^ = -^Fs, and similarly for other velocities. [We have no single word to express the unit of living force. If a unit of mass moving with a velocity of one foot per second be the unit of living force, and be called a Dynnm, then would the living force for any velocity and mass be a certain number of Dynains.l Since work is not force, but the effect of a force exerted through a certain space, independently of the time, we call it, for the sake of brevity, space-effeot. Vis viva, or living force, is not force, but it equals the work stored in a moving mass. It equals the sjpace-effect. [The expression Jfi^ was called the vis viva in the first edi- tion of this work, and is still so defined by many writers ; but there appeai-s to be a growing tendency towards the genera] adoption of the definition given in the text. It is immaterial which is used, provided it is always used in the same sense.] Examples. 1. A body whose weight is 10 pounds is movhig with a velocity of 25 feet per second ; required the amount of work which will be done in bringing it to i-est. Ans. 97.2 foot-pounds. 2. A body falls by the force of gravity through a height of h feet ; required the work stored in it. Let W — the weight of the body, M -= the mass of the body, g = acceleration due to gravity, and ■y = the final velocity, then 'i^ = 2gh, and 3Tg = W; .-. Wv'=~'2gh = Wh, [36.] ENERGY. 4:9 3. xV body whose weight is 100 pounds is moving on a hori- zontal plane with a velocity of 15 feet per second; how far will it go before it is brought to rest, if the friction is con- stantly 10 lbs ? A71S. = 34.9 -f- ft. 4. A hammer whose weight is 2000 pounds has a velocity of 20 feet per second ; how far will it drive a pile if the constant resistance is 10,000 pounds, supposing that the whole vis viva is expended in driving the pile ? 5. If a train of cars whose weight is 100,000 pounds is moving with a velocity of 40 miles per hour, how far will it move before it is brought to rest by the force of friction, the friction being S pounds per ton, or -g-^y^ of the total weight ? 6. If a train of cars weighs 300 tons, and the frictional resistance to its movement is S pounds per ton ; required the horse-power which is necessary to overcome this resistance at the rate of 40 miles per hour. Ans. 250. 7. If the area of a steam piston is 75 square inches, and the steam pressure is 60 pounds per square inch, and the velocity of the piston is 200 feet per minute, required the horse-power (.leveloped by the steam, 8. If a stream of water passes over a dam and falls through a vertical height of 16 feet, and the transverse section of the stream at the foot of the fall is one square foot, required the horse-power that is constantly developed. Let g =32i feet, and the weight of a cubic foot of water, 62^ lbs. Ans. 58.2 +. 9. A steam hammer falls vertically through a height of 3 feet under the action of its own weight and a steam pressure of 1000 pounds. If the weight of the hammer is 500 pounds, required the amount of work which it can do at tiie end of the fall. 26-. Energy is the capacity of an agent for doing work. The energvof a moving body is called actual or Kinetic energy^ and is expressed by \Mi?. But bodies not in motion may'have 50 MOMENTUM. [27.] a capacity for work wlien the restraining forces are removed. Thus a spring under strain, water stored in a mill-dam, steam in a boiler, bodies supported at an elevation, etc., are examples of stored work which is latent. This is called Potential energy. A m,oving body may possess potential energy entirely distinct from the actual. Thus, a locomotive boiler containing steam, may be moved on a track, and the kinetic energy would be expressed by ^Mv^, in which M is the mass of the boiler, but the jpotential energy would be the amount of work which the steam is capable of doing when used to run machinery, or is otherwise em^yloyed. These principles have been general- ized into a law called the Conservation of energy.^ which implies that the total energy, including both Kinetic and Poten- tial, in the universe remains constant. It is made the funda- mental theorem of modern physical science. The energy stored in a moving body is not changed by changing the direction of its path, provided the velocity is not changed ; for its energy will be constantly expressed by ^Mv^. Such a change may be secured by a force acting continually normal to the path of the moving body ; and hence ' we saj'' that a force which acts continually jperj^endicidar to the path of a moving hody does no work upon the hody. Thus, if a body is secured to a point by a cord so that it is compelled to move in the circumference of a circle ; the tension of the string does no work, and the vis viva is not affected by tho body being constantly deflected from a rectilinear path. MOMENTUM. 27- Resuming again equation (21), multiplying by dt, and integrating gives, X''''-^'fw = ''T^-^^- (30) The expression 3fv is called onomentum, and by comparing it with the first member of the equation we see that it is the effect of the force F acting during the time t, and is indepen- dent of the space. For the sake of brevity we may call the momentum a time-effect. [27.J MOMENTUM. 51 / If the body has an initial velocity we have >t Mt = 3J:{v-Vo); (31) 'to which is the momentum gained or lost in passing from a velocity Vq to v. Momentum is sometimes called quantity of motion^ on account of its analogy to some other quantities. Thus the intensity of heat depends upon temperature, and is measured in degrees; but the quantity of heat depends upon the volume of the body containing the heat and its intensity. The inten- sity of light may be uniform over a given surface, and will !)e measured by the light on a unit of surface; but the quantity is the product of tlie area multiplied by the intensity. The intensity of gravity is measured by the acceleration which is produced in a falling body, and is independent of the mass of the body; but the quantity of gravity (or total force) is the product of the mass by the intensity (or Mg). Similarly with momentum. The velocity represents the intensity of the motion, and is independent of the mass of the body ; but the quantity of motion is the product of the mass multiplied by the velocity. Differentiating (30) and reducing, gives ' dv . . which is the same as (18), and in which ^ is a velocity-incre- ment ; hence the momentum impressed each instant is a measure of the moving force. If the force F is constant we have from (30), Ft = ILv ; and for another force F' acting during the same time F't = M'v' ; .-. F'.F' \\ Ifv.M'v'-, hence, the forces are directly as the momenta produced by them respectively. 52 IMPACT. [28.1 If the forces are variable, let t t' f Fdt= Q = 2fv, and Jf'cU = Q' = M'v' ; then Q\Q' \\Mv\ IPv' ; lience the time-effects are directly as the momenta impressed. We thus have several distinct quantities growing out of equation (21) of which the English units are as follows : — The unit of force, F^\% 1 R). The unit of work or space effect is 1 lb x 1 ft. The unit of vis viva is 1 tb of mass x 1- ft. x 1 .sec. The unit of momentum 1 lb of mass x 1 ft. x 1 sec. IMPULSE. 28. An ilnpulse is the effect of a hlow. When one body strikes another, an impact is said to take place, and certain effects arc produced upon the bodies. These effects are pro- duced in an exceedingly short time, and for this reason they are sometimes called instantaneous forces / which, being strictly defined, means a force which produces its effect in- stantly, Tequiring no time for its action ; but no such force exists in nature. The law of action during impact is not gen- erally known, but it must be some function of the time. Hesuming equation (31), we have / Fdt - M{v - Vo) ; which is true, whatever be the relation between the force i^'and the time t. If the initial velocity of the body be zero, we have Vo = 0, ' r.t and J Fdt = Mr= Q. to "We see from the above equation that as t diminishes F must increase to produce the same effect. We see that in this [^28.] niPULSE. 53 case the first member is tlie time-effect of an impulse, and the second member measures its effect in producing a change of velocity. Calling this value Q, we have Q = M{v-' V,) = MV. (31a) Hence, the measure of an impulse in jyroducing a change of velocity of a, hody is the increased {or decreased) momentum, jproduced in the hody. This is the same as when the foi'ce and time are finite. If the force were strictly instantaneous the velocity would be changed from Vq to v without moving the body, since it would have no time in which to move it. Similarly from equation (23) we have in which for an impulse J^ will be indefinitely large; and hence the work done by an imj>ulse is m^easured in the same ruay as fo r finite forces. All the effects therefore of an impulse are measured in the same way as the total effects produced by a finite force. In regard to forces, we investigate their laws of action ; or having those laws and the initial condition of the body we may determine the velocity, energy, or position of the body at any instant of time or at any point in space, and hence we may determine final results ; but in regard to impulses we deter- mine only certain final results without assuming to know any- thin »• of the laws of action of the forces, or of the time or space occupied in producing the effect. The terms ^'Imjyulsive force, " and ^'Instantaneous forced are frequently used to denote the effect of an Impact ; but since the effect is not a foi-ce, they are ambiguous, and the term Impulse appears to be more appropriate. ^.n incessant force may be considered as the action of an infinite number of infinitesimal impulses in a finite tiine. The question is sometimes asked, " What is the force of a 54: EXAMPLES. [28.] blow of a hammer ? " If by the force is meant the pressure in pounds between the face of the hammer and the object struclv, it cannot be determined unless the law of resistance to compression between the bodies is known during the con- tact of tlie bodies. But this law is generally unknown. The pressure begins with nothing at the instant of contact and increases very rapidly up to the instant of greatest compression, after which the pressure diminishes. The pressure involves the elasticity of both bodies ; the rapidity with which the force is transmitted from one particle to another ; the amount of the distortion ; the pliability of the bodies ; the duration of the impact ; and some of these depend upon the degree of fixed- ness of the body struck ; and several other minor conditions; and hence we consider it impossible to tell exactly what the force is. Examples. 1. Two bodies whose weights are W and IF^ are placed very near each other, and an explosive is discharged between them ; required the relative velocities after the discharge. 2. A man stands upon a rough board which is on a perfectly smooth plane, and jumps off from the board ; required tlie relative velocities of the man and board. [Oes. The common centre of gravity of the man and board will remain the same after they separate that it was before. After separating they would move on forever if they did not meet with any obstacle to prevent their motion.] 3. A man whose weight is 150 pounds walks from one end of a rough board to the other, whicli is twelve feet long, and free to slide on a perfectly smooth plane ; if the board weighs 50 pounds, required the distance travelled by the n;an in space. 4. In example 3 of article 24, suppose that the weight IG pounds is permitted to fall freely through a height A, when it produces an impulse on the body (50 pounds) through the intermediate inextensible string ; required the initial velocity of the body. [28.1 EXAMPLES. 55 Let I'o =: \/^2gh — the velocity of the weight just before the impulse ; and ■y = the velocity iuiinediately afterward, which will be the comuiou velocity of the body and weight ; tli«^^ ^ 50 . 10, The subsequent motion may be found by equation (21), observing that the initial velocity is v. The tension on the string will be infinite if it is inextensible, but practically it will be finite, for it will be more or less elastic. [Some writers have used the expression impuUice tension of the string instead of 7nomentum.'\ 5. If a shell is moving in a straight line, in vacuo, with a velocity v, and bursts, dividing into two parts, one part moving directly in advance with double the velocity of the body ; what must be the ratio of the weights of the two parts so that the other part will be at i-est after the body bursts? 6. Explain how a person sitting in a chair may move across a room by a series of jerks without touching the floor. (Can he advance if the floor is perfectly smooth ?) 7. A person is placed on a perfectly smooth plane, show how he can get off if he cannot reach the edge of the plane. The same impulse applied to a small body will impart a greater amount of energy than if applied to a large one. Thus, in the discharge of a gun, the impulse imparted to the gun equals that imparted to the ball, but the work, or destruc- tive effect, of the gun is small compared with that of the ball. The time of the action of the explosive is the same upon both bodies, but the space moved over by the gun will be small compared with that of the ball during that time. The product 3it\ being the same for both, as M decreases v increases, but the work varies as the square of the velocity. 56 IMPACT. [29.J DIRECT CENTRAL IMPACT. 29. If two bodies impinge upon one another, so that the line of motion before impact passes throno;h the centre of the bodies, it is said to be central ; and if at the same time the common tan2;ent at the point of contact is perpendicular to the line of motion, it is said to be direct and central. If their common tangent is perpendicular to the line of motion, but if the latter does not pass through the centre of the body impinged upon, it is called eccentric impact. In this place, we consider only the simplest case ; that in which the impact is direct and central. When two bodies impinge directly against one another, whether moving in the same or opposite directions, they mutually dis- place the particles in the vicinity of the point of contact, producing compression whicli goes on increas- ing until it becomes a maximum, at which instant they have a com- mon velocity. A complete analysis of the motion during contact in- volves a knowledge of the motion of all tlie particles of tlie mass, and would make an exceedingly complicated problem, but the motion at the instant of maximum compression may be easily found if we assume that the compression is instantly dis- tributed throughout the mass. Let Ml and M^ be the respective masses of the bodies ; t'l and V.2 the respective velocities before impact, both positive and Vi>V2 '■> v( and v-i the respective velocities at the instant of )naximum compression, v\ > Vg, and Qi and ^o the momenta gained respectively by the bodies during compression. Then from (31) ^, = J/o(r/-ro), wliicli will be tlie momentum gained by Jfo on account of tlic action of M^. Similarly Q, = M, (r/ - t\), which, being negati-ve, will be the momeutum lost by Mx on account of the .action .of M>. [30.] I3IPACT. 57 But at the instant of greatest compression ti — t2 5 and, because they are in rantual contact during the same time, their time-effects are equal, but in opposite directions, .-. Q,= - Q,. Combining tliese four equations, we find by elimiuatiou M{i\ + MoT M, + M. (33) which velocity remains constant for perfectly non-elastic bodies after impact, since such bodies have no power of restitution and will move on with a common velocity. DIRECT CENTRAL IMPACT OF ELASTIC BODIES. 30. Elastic bodies are such as regain a part or all of their distortion when the distorting force is removed. If they reo-ain their original form they are called j^er/ec^J^y elastic, but if only a pai-t, they are called imperfectly elastic. After the impact has produced a maximum compression, the elastic force of the bodies causes them to separate, but all the effect which the force of restitution can produce upon the movement of the bodies, evidently takes place while they are in contact. If they are perfectly elastic and do not fully regain their form at the instant of separation, they will continue to regain their form after separation, but the latter effect we do not consider in this place. The ratio between the forces of compression and those of restitution has often been called the modulus of elasticity ^ but as some ambiguity results from this definition, we will call it the modulus of restitution. At every point of the restitution there is assumed to be a constant ratio between the force due to compression and that to restitution. But it is unnecessary for present purposes to trace these effects, for by equation (31) we may determine the result when the bodies finally separate from each other. 58 IMPACT. [31.] Let 6, = the ratio of the force of compression to that of restitution of one body, which is called the modulus of restitution. ^2 = the corresponding valne for the other ; Fi = the velocity of J/i at the instant wlien they separ- ate from each other ; and F2 = the corresponding velocity for M-^. Tlien from equation (31) e, Q, = M,{Y,-v:y, (34) e^Q^=M^{Y,-vIl Fig, m. .-^R _J_^P Fig. 21 Also, if ^ = 90°, and a be the angle between i? and P ; and ^ between B and F'; then (47) P z= F cos a • P' = i? COS |3 = 7? sin a. Squaring and adding, we have as before. The forces P and i^are called coniponent forces. 44. Triangle of Forces. If three concur rent forces are in equilihrlum, they may he represented in magnitude and direc- tion hy the sides of a triangle taken in their order ; and if the direction of action of one he reversed, it will he the res^dtant of the other two. Thus, in Fig. 19, if AB and BC lepresent two forces in mao-nitude and direction, J. 6' will represent the resultant. parallelogram of pressures ; but we have seen that a pressure when acting upon a free body will produce a certain amount of motion, and that this motion is a measure of the pressure, and hence its use in the proof appears to be admissible. But the strongest proof of the correctness of the proposition is the fact that in all the problems to which it has been applied, the resulta agree with those of experience and observation. 68 POLYGON OF FORCES. [45-46.] Since the sines of the angles of a triangle are proportional to the sides opposite, we have F P B A A sin F,P (48) POLYGON OF FORCES- 45. Jf several ion- current forces are re- presented in magni' tilde and direction hy the sides of a closed polygon ta7ce7i in their order^ they will he in equilihriiim. This may be proved by finding the resnltant of two forces by means of the triangle of forces ; then the resultant of that resnltant and another force, and so on. Fio 21o. PAKALLELOPIPED OF FOKCES. 46. If three concurrent forces not in the saine ^lane are represented in tnagnitude and direction iy the adjacent edges of a jparallelojpijpe- don, the resultant ^oill he represented in magnitude and direction hy the diago- nal / and conversely if the diagonal of a parallelopipedon represents a force, it may he considered as the resultant of three forces represented hy the adjacent edges of the parallelopipedon. In Fig. 22, if AD represents the force F^ in magnitude and direction, and similarly DB represents i^, and BC, F^\ then according to the triangle of forces AB will represent the resultant of F^ and F^ ; and A C the resultant of AB and F^ and hence it represents the resultant of i^, F%, and F^. Fio. 22. f46.] EXAMPLES. 09 If /I , i's , and Fl are at right angles with each other, we have and if « is the augle B^F, , § of ^,i^ , and y of B,F^ ; tneu i^ = i? cos a ; J i^ = i?cos^; V (4:9) i^ = ^ cos y. ) Squaring these and adding, we have B^ = F^" -h J^ + Fs\ a& before. Examples. 1. When i^= F, and ^ = 60°, find B ; (See Eq. (46) ). Ans. R = F\/l. 2. If i^= i^ and B = 120=, find U. 5. If i^= /I and e = 135°, find i?. _^ 4. If F= 2i^ = 3^, find 0. ^^. 6. If ii^= IFi = i2, find the angle F,Fi. Alls. 90". A 6. If i^= 7, i^i = 9, and 6 ^ 25°, find B and angle F,JR. 7. A cord is tied around a pin at a fixed point, and its two ends are drawn in different directions by forces i^and P. Find when the pressure upon the pin is ^ == i (P + Fy ^ 2PF- S{F' + F^) A71S, cos u = »rpF '~' 8. When the concurring forces are in equilibrium, prove that A A A^ P:F: Rii&in F,B : sin P,P : sin P,F 9. If two equal raftei-s support a weight W at their upper ends, required the compression on each. Let the length of 70 CONCURRENT FORCES. [47.] each rafter be a and the horizontal distance between their lower ends be h. a Ans. V4a'- I? 7F. 10. If a block whose weight is 200 pounds is so situated that it receives a pressure from the wind of 25 pounds in a due easterly direction, and a pressure from water of 100 pounds in a due southerly direction ; required the resultant pressure and the angle which the resultant makes with the vertical. KESOLrXION OF CONCUKEENT FORCES. 47. Let there be many concurrent forces acting- upon a single particle, and the whole system be referred to rectangular co-ordinates. Let l\,Fi, Fi, etc., be the forces acting upon a particle at A ; 1/ X, y, z the co-ordinates of A ; «i, «2?etc., the angles Avhich the Fio. 23. direction-lines of the respec- tive forces make with the axis of x ; |9i , l^a 5 etc., the angles which they make with y ; 7i , j'3 , etc., the angles which they make with z ; and X, y, and Z, the algebraic sum of the components of the forces when resolved parallel to the axes a', y, and z, respectively. Then, according to equations (45) and (49), we have for equilibrium ; X = i^ cos «i + i^ cos «2 + 7^ cos «3 -I- etc. = 2* /'^cos « = Y- F^ co3^i+i^ cos fJo4-i^3 COS /^g + etc^^-i^cos ^ = Z*=i^cos'/i-fi^cos72+i^cos 73 +etc.=-ri^cos7 = If they are not in equilibrium, let R be the resultant, and by introducing a force equal and opposite to the resultant, the system will be in equilibrium. (50) [48.] CONSTRAINED EQUILIBRIUM. 71 Let «, h and g be the angles which the resultant aiakes with the axes x, y and z respectiveh' ; then X — li cos a ; r = ^cos5; (51) Z — R cos c . Squanng and adding, we liave X2 + Z^ + Z' = E" (52) If i? = equations (51) reduce to (50). "When the forces are in equilibrium any one of the F -forces may be considered as a resultant (reversed) of all the others. Equations (50) are therefore general for concm-ring forces. The values of the angles «, ^, y, etc., may be determined by drawing a XwxQfroin the origin parallel to and in the direction of the action of the force, and measuring the angles from the axes to the line as in Analytical Geometry. The forces may always be considered as positive, and hence the signs of the terms in (50) will be the same as those of the trigonometrical func- tions. In Fig. 23 the line Oa is parallel to i^, and the corre- sponding angles which it makes with the axes are indicated. If all the forces are in the plane x y then y^ , y.^ , etc. — 90°, and (50) becomes X= -iri^cos « = 0; [ ,Ko\ r=:SFcos§ = o. ] ^ ^'^ CONSTEAINED EQUILIBKICM. 48. A body is constrained when it is prevented from moving freely under the action of applied forces. If a particle is constrained to remain at rest on a surface under the action of any number of concurring forces, the resultant of all the applied forces must be in the direction of the normal to the surface at that point. Foij if the resultant were inclined to the normal, it could be resolved into two components, one of which would be ^lo- 24. 72 CONSTRAINED EQUILIBRIUM. [48.] taiio:eiitial, and would produce motion ; and the other normal, which would be resisted by the surface. Let ir= the normal reaction of the surface, which will be equal and opposite to the resultant of all the impressed forces ; 6;c = the angle (iV,«) ; e^ = the angle (iV,?/) ; e^ = the angle (iV",s) ; L = ({){x, y, z) = 0, be the functional equation of the surface ; and Fi, F2, Fi, etc., be the impressed forces. Then from (51) and (52), we have X= iV^co,s6'^; 1 I-=^cos^,; I Z=iYcos^,; 1^ i^V From Calculus we have 1 cos 6^ = and similarly for cos 6^ and cos 6^ . These values in (54) readily give _X_ Y _ _^ /dL\ "■ / ir, it is an ellipse. If P < TF, it is a hyperbola. The orifjin is at the focus. T. A particle is placed on the concave surface of a smooth sphere and acted upon by gravity, and also by a repulsive 78 EXAMPLES. [49.] force, which varies inversely as the square of the distance from the lowest point of the sphere; find the position of equilibrium of the particle. Take the lowest point of the sphere for the origin of coordi- nates, y positive upwards, and the equation of the surface will be Z = g? ^y^ ■\-^- 'iRy = 0. Let r be the distance of the particle from the lowest point ; then Let /z be the measure of the repulsive force at a unit's dis- tance ; then the forces will be if: = ^ , and mg ~w = the weight of the particle. ?'^ 2Iiy ••^~2Ry'~f' 2m/ r '"'^~2By'r' which in (5G) give, after reduction, it which in (b) gives, 7^ = A. To see if tliese values satisfy equation (57), substitute in it the values of X, Y, Z, and the final values of y and r, and we find, osdx + ydy — Rdy + zds = ; which is the differential of equation {b), and hence is true. [This is the theory of the MectrosGope.'] 8. A particle on the surface of an ellipsoid is attracted by forces which vary directly as its distance from the priiicipa planes of section ; determine the position of equilibrium. [50, 51.] Let MOMENTS OP FORCES. Z=cf.{x,y,s,)=^ + -^+-^-l=0, 79 be the equation of the surface ; //. IL dz 23 7J' 1x (([D dijj and let the x, y, and z-coinjyonents of the forces be respec- tivel}', X= — fi^x, Y — — ii^i^j, Z= — ix^; and (56) will give, which simply establishes a relation between the constants ; and hence when this relation exists the particle may be at rest at any point on the surface. The result may be put in the form, ill IH _ /^i + Ma + /^3 MOMENTS OF FOKCIES. 50. Def. The moment of a force in reference to a point is the product arising from multiplying the force hy the perpendicular distance of the action-line of the force from the point. Thus, in Fig. 28, if is the point from wliich the perpendicular is drawn, F the force, and Oa the perpendicular, then the moment of i^is KOa^Ff', in which y is the perpendicular Oa. FiQ. 2R. 51. ISTature or A MOMENT. The moment of a force measures the turning or twisting effect of a foi-ce. Thus, in Fig. 28, if the particle upon which the for(;c F acts is at A, 80 MEASURE OF A MOMENT. [52-54.) and if we conceive that the point is rigidly connected to A^ the force will tend to move the particle abont O, and it is evident that this effect varies directly as F. If the action- line of F passed through it would have no tendency to move the particle about that point, and the greater its dis- tance from that point the greater will be its effect, and it will vary directly as that distance ; hence, the measure of the effect of a moment varies as thej)roduct of the force and jperpendi- cular ; or as cFf; where d is a constant. But as c may be chosen arbitrarily, we make it equal to nnity, and have simply Ff as given above. 52. Def. The point O from which the perpendiculars are drawn is chosen arbitrarily, and is called the origin of mo- ments. When the system is referred to rectangular coordinates, the origin of moments may, or may not, coincide with the origin of coordinates. The solution of many problems is simplified by taking the origin of moments at a particular point. 53. The lever aem, or, simply, the arm, of a force is the perpendicular from the origin of moments to the action-line of the force. Thus, in Fig. 29, Oa is the arm of the force F^ ; Og that of the force F^ , etc. Generally, the arm is the per- pendicular distance of the action-line from the axis about which the system is supposed to turn. 54. The sign of a moment is considered positive if it tends to turn the system in a direction opposite to that of the hands of a watch; and negative, ii in the opposite direc- tion. This is arbitrary, and the opposite directions may be chosen with eqnal propriety ; but this agrees with the direction in which the angle is computed in plane trigonometry. Gen- erally we shall consider those moments as positive which tend to turn the system in the direction indicated by the nat- ural order of the letters, that is, positive from + a; to + y ; from + 2/ to -f 3 ; then from -f s to + a; ; and negative in the reverse direction. The value of a moment may be represented by a straight line drawn from the origin and along the line about which [55-59.] MOMENT OF A FORCE. 81 rotation tends to take place, in one direction for a positive value, and in the opposite direction for a negative one. 55. The coMPOsrnoN- and resolution of moments may be effected in substantially tlie same manner as for forces. They may be added, or subtracted, or compounded, so that a resultant moment shall produce the same effect as any num- ber of single moments. The general pi'oof of this pro230si- tion is given in the next Chapter. 56. A MOMENT AXIS is a line ]iassing through the ori.-rin of moments and perpendicuhir to the plane of the force and arm. 57. The Mo:srENT of a force referred to a moment axis is the product of the force into the perpendicular distance of the force from the axis. If, in Fig. 29, a line is drawn through O perpendicular to the plane of the force and arm, it will be a moment axis, and the turning effect of F^ npon that axis will be the same wher- ever applied, providing that its arm Oa remains constant. If the force is not perj^endicular to the arbitrarily chosen axis, it may be resolved into two forces, one of which will be perpendicular (but need not intersect it) and the other parallel to the axis. The moment of tlie former component will be the same as that given above, but the latter will have no mo- ment in reference to that axis although it may have a moment in reference to another axis perpendicular to the former. 58. The moment of a force in reference to a plane to which it is parallel is the 2^foduct of the force into the distance of its action-line from the plane. 59. If any numher of concurring forces are in eguilihrium the algebraic smn of their moments will he zero. Let i^ , i^ , I'l , etc.. Fig. 29, be the forces acting upon a particle at A / and the assumed origin of moments. Join O and A, and let fall the pei'- pendiculars Oa, Oh, Oc, etc., npon the action-lilies of the respective forces, and let Oa=f; Oh=f', Oc=f- etc. S2 ■RESULTANT MOMENT. r60.] Resolve the forces perpendicularly to the line OA ; and since they are in equilibrium, the algebraic sum of these com- ponents will be zero ; hence, i^ sin OAF^ +E, sin OAF^ + F^ sin OAF^ + etc. = j 0. ^ Oa ^ Oh ^ Oc or,i^^ + i^-^-^+^3^ + etc. Multiply by OA^ and we have F^Oa + F^Ob + F,Oc + etc. = ; or, F,f, + F^f^ + F,f, + etc. = ^Ff= 0. (60) It is evident that any one of these moments may be taken as the resultant of all the others. MOMENTS OF CONCUKRING FORCES WHEN THE SYSTEM IS EEFEKRED TO RECTANGULAR AXES. Y y 60. Let A, Fig. 80, be the point of application of the forces Fi, Fo, Fs, etc., and O the ori- gin of coordinates, and also the origin of moments. Let sc, y, and 3 be the coordinates of the point A. Kesolving the forces parallel to the coordinate axes, we have, from equation (50), Fig. 30, X = :^Fcos a; Y=2Feosl3; Z= :s Fcosy. The X- forces prolonged will meet the plane of ys in B; and will tend to turn the system about the axis of y, in reference to which it has the arm BC=^z\ and also about s, in refer- once to which it has the arm BD = y. Hence, employing the notation already established, we have for the moment of the sum of the components parallel to a?, — Xy, and + Xz. [60.] MOMENTS OF CONCTJRRING FORCES. 83 Similarly for the y-coinponents we find the moments, -{- JTISj'and — Y2 ,' and for the z-coniponents^ — Zx, and + Zy. The moment Xy tends to turn the system one way about the axis of z, and Yx tends to turn it about the same axis, but iu the opposite direction ; and hence, the combined effect of the two will be their algebraic sum ; or Yx — Xy^ But since tliere is equilibrium the sum will be zero. Com- bining the others in the same manner, we have, for the moments of conoxirring forces, in equilibrium: In reference to the axis of a? ... . Zy — Yz = ; " " " " y . . . . X3 - Z« r= ; I (61> " " " " " " s . . . . Z« - Xy^ 0. The third equation may be found by eliminating z froin the other two; hence, when X, Y, and Z are known, they are the equations of a straight line ; and are the equations of the resultant. If the origin of moments be at some other point, whose coordinates are «', y\ and z' ; and the coordinates of the point A in reference to the origin of moments be .cci, yi,and s^; then will the lever arms be x^^x-x'\ y^ = y-y'; When the system is referred to i-ectangular coordinates the arm of the force, referred to the ;?-axis, is X cos ^ — y cos a, in which y and x are the coordinates oi any point of the action-line of the force ; and a is the angle which the action-line makes with the axis of x, and ^ the angle which it makes with y. Fig, 31. 84 EXAMPLES. [60.] Ill Fig. 31, let BF be the action-line of the force F, O the origin of coordinates, A any point in the line AF, of wliich the coordinates x = Oh, aiid y = Ah. Draw Od and he perpendicular to AF, and hd from h parallel to BF. Tlie origin of moments being at O, Oa vrill be the arm of the force. We have Agh = a = chA, cAb = ^ = hOd, cb =^y cos a = ad, Od = X cos yS ; .'. Oa = Od — ad =-- xao^^ — y cos oc . (61«) If there are three coordinate axes, this will be the arm in reference to the axis of z ; and if thei-e be many forces, the sum of their moments in reference to that axis, will be XF^x cos )8 — ?/ cos a). Examples. 1. A weight W is attached to a string, whicli is secured at A, Fig. 32, and is pushed from a vertical by a strut CB / required the pressure i^on j^C' when the angle CAB is 6. Tlie forces which concur at B are the weight W, the pressure F, and the tension of the string AB. Take the origin of moments at A, and we have - W.B C -VF.AC-V tension x = ; :.F= ir4^'= TF tan 6^. AC 2. A brace, AB, rests against a vertical wall and upon a horizontal plane, and supports a weight TF at its npper end ; required the compression upon the brace and the thrust at A when the angle CAB is Q ; the end B being held by a string BC. FlO. 88. [60.1 EXAMPLES. 85 The concurring forces at A, are TT, acting vertically down- ward, the reaction of the wall jV acting liorizontalljj and the reaction of the brace K Take the orio^in of moments at ^, we have .'.jsr=w tune. 0; Taking the orio:in of moments at D, we have W. AD-F. DB sin ^ + iVT". o = ; ,-.F= TFsec^. 3. A rod whose length \% BC~l is secured at a point B, in a horizontal plane, and the end C is held up by a cord AC so that the angle BAC'v^ 0, and the distance AB = a ; required the tension on AO and compression on BO, due to a weight Tr applied at O. A71S. Compression = — TFcot 6. 4. A cord whose length ABC= I is secured at two points in a horizontal line, and a weight W is suspended from it at B ; required the tension on each part of the cord. CHAPTER in. PARALLEL TOKCES. Fig. 34. 61. Bodies are extended masses, and forces may be applied at any or all of their points, and act in all conceivable direc- tions, as in Fig. 34. 62. Suppose that the action- lines OF all the eoeces are paral- lel TO each other. This is a spe- cial case of concurrent forces, in which the point of meeting of the action-lines is at an infinite distance. In Fig. 35, let the points «, h, c, etc., which are on the action- lines of the forces and within the body, be the points of applica- tion of the forces, and the point where they would meet if pro- longed. If the point recedes from the body, while the points of application a, J, c, etc. remain ^ iixed, the action-lines of the forces will approach parallelism with each other, and at the limit will be parallel. Fig. 35. 63. Eesultant of parallel forces. The forces being par- allel, 'the angles which they make with the respective axes, including those of the resultant, will be equal to each other Hence, [64.] MOilENTS OF PARALLEL FORCES. 87 a^^ ai = a-i = 0-3 , etc. = a ; 6 = /3i = /3., -/33,etc. = /3; c = 7i = 72 = 73 , etc. = 7 ; and these, in equations (50) and (51), give X= B cosa = {K + F.+ F,-{- etc.) cos a. r=Iicos/3 = {F,^F, + F, + etc.) cos /3. (62) Z= I2cosy = {F,+ Fi+Fi+ etc.) cos 7. From either of these, \ve have Ii = F, + Fi + Fs + etc. = ^F (63) Hence, ^A^ resultant of jyarallel forces equals tJie- algehraic sum of the forces. From (62), we have which is the same as (52). MOMENTS OF PAKALLEL FORCES. 64. Let i^, i^, Fz, etc., be the forces, and x^, t/i, z^ ; a-^, 2/?, 02, etc., be the coordinates of the points of application of the forces respectively (which, as before stated, may be at any point on their action-lines). Then the moments of F^ will be, according to Article (00), in reference to the axis of «, F^ cos "i .y\ — F^ cos /3 . Si ; « " " " " " y, Fi cos a . 2i — i^ cos 7 . o-i ; " " " " " " 2, F^ cos /3 . «! — i^i cos a . yi ; and similarly for all the other forces. Hence, the siim of the vioments in reference to the respective axes for equilibrium is, {F,y^ + Fiy-i + F^y^ + etc.) cos 7 ; -(i^i^i + i^?2 + ^% + etc.) cos ^ {Fi2i + F^i + 7^3% + etc.) cos « ) _ ^ __ — (i^a?i + i^a32 + i^ga^g + etc.) cos 7 j (i^a^ + F<^ + i^.r2 + etc.) cos /3 ) _ ^ - (7Iy_i 4- F,y^ + 7'ny3 + etc.) cos a 88 THREE PARALLEL FORCES. t^G.] These equations will be trae for all values of a, /3, and 7, if the coefficients of cos a, cos yS, cos 7, arc respectively equal to zero ; for which case we have i'>i + J^i-^z + ^'l^'s + etc. = ^Fx = 0;" F,y, + F^^ + F^j., + etc. = 5"/^// = ; |. (64) F^z^ + F^z.^ -f F,z^ + etc. = 5'i's = ; ^ frcm which the coordinates of the point of application of any one of the forces, as i^, for instance, may be found so as to satisfy these equations, when all the other quantities are given. Let the given forces have a resultant. Let a force, as ^, equation (64), equal and opposite to the resultant, be introduced into the system, then will there be equilibrium. Let 'ZFx^ XFy, XFz, include the sum of the respective products for all the forces exce^n that of the resultant ; R be the resultant, and ~x, y, 2, the coordinates of the point of application of the result- ant ; so chosen as to satisfy equations (64), then we have Bx-XFx=0- Iiy-^Fy = 0; Rz-XFz^^. (65) Substitute the value of i? = XF, in these equations, and we find by which the point of application of the resultant becomes known, and, being independent of a, /3, and 7, is a 2^oint through which the resultant constantly passes, as the forces are turned about their jpoi7its of application, the forces constantly retaining their parallelism. This point is called the centre of parallel forces. 65. If the system consists of tukee fokges only, and are in the plane xy, we have E=.F, + F,; 1 _ F,x, + F,x, ^.-_F,y, + Fly, \ (67) [66.J STATICAL COUPLES. 89 J. 1st. Consider F^ and I\ as;po8itive. The resultant will equal the arith- metical sum of the forces. Take the origin at a Fig. 36, where the resultant cuts the axis of «; then » = 0, and the second of (67) gives FvCx = - F^T-^ ; and hence, if F^ > F^, x.^, will exceed Xi ; that is, the resultant is nearer the ^^ greater force. 2d. Consider F^ as negative. In this case the resultant equals tin difference of the forces. Take the origin at a, Fig. 37, and we have FiXy = F-iXi ; and hence both forces are either at the ^S'^ riii'ht or left of the resultant. ^^<*- ^"- r. X Fig. Sd. aF. R 3d. Let Fi = Fi = F, and one of the forces be negative, then F (jgi ± x^) • _ B^F- F=^0\x = that is, the resultant is zero, while the forces may have a finite moment equal to F{xi ± x^. Such systems are called COUPLES. 66- -A. ooicple consists of two equal parallel- forces acting in ojpposite directions at a finite distance from each other. A statical couple cannot be equilibrated by a single force. It does not produce translation, but simply rotation. A couple, can he equilibrated only hy an equivalent couple. Equivalent couples are such as have equal moments. The resultant of several couples is a single couple which will produce the same effect as the component couples. 90 AXIS OF A COUPLK [67-68.] Y 67. The arm of a couple is the jper^peiidicular distance hetween the action-lines of the forces. Thus, ill Fig. 38, let O be the origin F of coordinates, and the axis of x per- pendicular to the action-line of F', - X then will the moment of one force be FoC]^, and of the other Fx^, and hence the resultant moment will he X, P Fio. 38. F{x,-x^) = F.ab] (69) hence, ab is the arm. If the origin of coordinates were between the forces the moments would be F{xi + x^ = Fab as before. If the origin be at a we have ^0 -f- Fah = Fah as before. 68- The axis of a staticai. couple is any line perpen- dicular to the plane of the couple. The length of the axis may be made proportional to the moment of the couple, and placed on one side of the plane wdicii the moment is positive, and on the opposite side when it is negative ; and thus com- pletely represent the couple in magnitude and direction. If couples are in ])arallel planes^ their axes may be so taken that they will conspire, and hence the resultant couple equals the algebraic sum of all the couples. If the planes of the couples intersect, their axes may intersect. Let = F.ah = the moment of one couple ; Oi= Fx.a^i = the moment of another couple; 6 = the angle between their axes ; and Op, = the resultant of the two couples ; then On = \/0^ + 0;'+200,tosd ; and this resultant may be combined with another and so on until the final resultant is obtained. Examples. 1. Three forces represented in magnitude, direction and 2)ositio7i,hy the sides of a triangle, taken in their order, produce a couple. [68.] EXAMPLES. 91 Fig. 39. 2. If three forces are represented in magnitude and position by the sides of a triangle, but whose directions do not follow the order of the sides ; show that tliey will have a single resultant. 3. On a straight rod are suspended several weights ; F^ — 5 lbs., i^ = 15 lbs., i^ = 7 lbs., F^ = ^ lbs., i^ = 9 lbs., at dis tances AB^Z ft., BD = G ft., DE^ 5 ft., and EF = \ ft.; required the distance AC sit which a fulcrum must be placed so that the weights will balance on it; also recpiired the pressure upon it. 4. The whole length of the beam of a false balance is 2 feet 6 inches. A body placed in one scale balances 6 lbs. in the other, but when placed in the other scale it balances 8 lbs. ; required the true weight of the body, and the lengths of the arms of the balance. 5. A triangle in the horizontal plane x, y has weights at the S3veral angles which are proportional respectively to the opposite sides of the triangle ; required the coordinates of the centre of the forces. Let a?! , iji be the coordinates of A, x^.y^oi B\ x^.y^oi C; X, y of the point of application of the ro'^ltant ; then we have {a + h + c)x =axx + lx2 + cx^ ; and {a + 1) -v c)y = ay^ + hj^ + cy^. 6. If weights in the proportion of 1, 2, 3, 4, 5, G, 7 and 8 are suspended from the respective I\ angles of a parallelopiped ; re- quired the point of application of the resultant. 7. Several couples in a plane, whose forces are parallel, are applied^ to a rigid right line, as in Fig. 40 ; required the re^ Bultant couple. ^^^^ \q^ F, F, F, 92 CENTRE OF GRAVITY [69.1 8, Several couples in a plane, whose respective arms are not parallel, as in Fig. 41, act npon a rigid right line ; required the resultant couple. CENTRE OF GKAVriT OF BODIES. 69. The action-lines of the force of gravity are normal to the surface of the earth, but, for those bodies which we shall here consider, their convergence will be so small, that we may consider them as parallel. 'We may also consider the force as the same at all points of the body. The centre of gravity of a lody is the point of application of the resultant of the force of gravity as it acts upon eveiy particle of the body. It is the centre of parallel forces. If this point be supported the body will be supported, and if the body be turned about this point it will remain constantly in the centre of the parallel forces. Let M — the mass of a body ; * m = the mass of an infinitesimal element; "r= the volume of the body ; D — the density at the point whose coordinates are X, y, and z ; Jl — W — the resultant of gravity, which is the weight ; and ^, ^, and z be the coordinates of the centre of gravity. We have, according to equations (63) and (20), R = W^tgm='Mxg\ and (65) becomes X Xgin = Xgmx ; or Mx = ^mx ; ^ y "Xgni = Xgmy ; or 31 y = Xmy ; > (70) z Sgm = Sgm2^ ; or Mz = Xmz. ) If the density is a continuous function of the coordinates of the body we may integrate the preceding expressions. The [69.] OF BODIES. 93 complete solution will sometimes require two or three integra- tions, depending upon the character of tlie problem ; but, using only one integral sign, (22) and (TO) become ^/DdV^fDxdV; y/DdV^^J'DydV; zfDdY^fDzdV. (71) If the origin of coordinates be at the centre of gravity, then ^ = 0; 2/ = 0;sr=0 and hence, Xmx=J'DxdY=^\ and similarly for the other values. If D be constant, this becomes f''xdV= 0; (71^) (71J) the limits of integration including the whole body. If the mass is homogeneous, tlie density is uniform, and D being cancelled in the preceding equations, we have _ fxdV y = V (72) __ fzdV Z — y— , Many sohitions may be simplified by observing the following principles : 1. If the hody has an axis of symmetry the centre of gravity will he on that axis. 94 CENTRE OF GRAVITY ["0.] 2. If the hody has ajplane of symmetry the centre of gramty will he in that ^lane. 3. If the hody has two or more axes of symmetry the centre of gravity will he at their intersection. Hence, the centre of gravity of a physical straight line ol uniform density will be at the middle of its lengtli ; that of the circumference of a circle at the centre of the circle ; that of the circumference of an ellipse at the centre of the ellipse ; of the area of a circle, of the area of an ellipse, of a regular polygon, at the geometrical centre of the figures. Similarly the centre of gi-avity of a triangle will be in the line joining the vertex with the centre of gravity of the base ; of a pyramid or cone in the line joining the apex with tlie centre of gravity of the base. There is a certain inconsistency in speaking of the centre of gravity of geometrical lines, surfaces, and volumes ; and wlien they are used, it should be understood that a line is ?i physical or material line whose section may be infinitesimal ; a surface is a tnaterial section^ or thin plate, or thin shell ; and a volume is a mass, however attenuated it may be. When a body has an axis of symmetry, the axis of x may be made to coincide with it, and only the first of the preceding equations will be necessary. If it has a plane of symmetry, the plane x y may be made to coincide with it, and only the first and second will be necessary. 70. Centre of gramty of material lines. Let Tc = the transverse section of the line, and ds = an element of the length, then dV — Ms ; and ^71) becomes xfmds - fDhxds\ yfDMs = fDhjds ; \ (73) 'zfDMs = fDlczds. 170.1 OF LINES, 95 If the transverse section and the density are uniform, we iiave fxds fzds • 8 (74) The centre of gravity will sometimes be outside of the line or body, and hence, if it is to be supported at that point, we nmst conceive it to be rigidly connected with the body by lines which are without weight. Examples. 1. Find the centre of gravity of a straight fine wire of uni- form section in which the density varies directly as the distance from one end. Let the axis of x coincide with the line, and the origin be taken at the end where the density is zero. Let S be the density at the point where « = 1 ; then for any other point it will be D = hx; and substituting in the first of (73), also making ds = dx, we have X I hxdx — / h:(?dx ; .-. x=%a. Jo Jo This corresponds with the distance of the centre of gravity *" • ti'iano;le from the vertex. Fig. 42. ^IQ. 43. S. Find the centre of gravity of a cone or pyramid, whether 9G EXAMPLES. [70.1 right or oblique, and whether the base be regular or irre- gular. Draw a line from the apex to the centre of gravity of the base, and conceive that all sections parallel to the base are re- duced to this central line. The problem is then reduced to finding the centre of gravity of a physical line in which the density increases as the square of the distance from one end. Ans. X = ^a. 3. To find the centre of gravity of a circular arc. Let the axis of x pass through the centre of the arc, B, and the centre of the circle 0; then Y = 0. Take the orio-in at £ ; Fia. 44. and let X = BD, 2s = the arc A BC, and r = OC = the radius of the circle ; then, y'^ = 2rx — a?. Differentiate, and we have 7'—x y hence, the first of (74) gives ydy = rdx — xdx ; dy dx. ^ . , ds — ^- = — which = — ; Jo V %\ xdx X — af X = [ = — I — V~2rx — x^+ s Jo ry ry (ry\ hence, the distance of the centre of gravity of an arc from the [71.] CENTRE OF GRAVITY OF SURFACES. 97 centre of the circle is a fourth proportional to the are, the radins, and the chord of the arc. 71. Centre of gravity of plane surfaces. Let the coordinate plane xij coincide with the surface ; then dV=dxdy; .-. V = ffdxdy=fydx ov/xdy; and (71) be- comes xffndxdy=ffDxdxdy', ^ yffDdxdy=ffDydxdy. j ^'^^ The integrals are definite, includino^ the whole area. The order of integration is immaterial, but after the first integra- tion the limits must be determined from the coiiditions of the problem. If D is constant and the integral is made in respect to ?/, we ha^•e _ fyxdx y- fydx' \fifdx (70) and if x be an axis of symmetry, the first of these equations will be sufiicient. If the snrface is referred to polar coordinates, then ^ F = pdpdQ^ and x — pco^d^y = p sin 6, and (71) becomes -^ ffDp\o^Mpde Jfjjpdpdd 5 -^ ffPphmOdp dd ^ ffDpdpdd ' j (77) E X A M P L PJ S. 1. Find the centre of gravity of a semi-parabola whose equation is j'^ = ^px. Equations (7G) become i^2j) x^ dx X sxy Jo ^ J f) '^V^^^ ^xy ^ 7 98 CENTRE OF GRAVITY. [72] 2. Yind tlie centre of gravity of a quadrant of a circle in wliicli the density increases directly as the distance from the centre. Let B = the density at a unit's distance from the centre ; then J) := Sp at a distance p ; and (77) becomes X = frs p^cosOdpdO ttt / r p^dpdO , r _ JO JO 72, Centre of gravity of curved surfaces. "We have for an element of the area » dY=- dx, dy x sec. of the angle letween the tangent plane and the coordinate jflane xy • or, d V=sec. 6 X dx dy; (■" i+Zw5 the mass mid- tijplied hy the square of the distance hetween the axes. Let AB be an axis through the centre, CD a jDarallel axis, D = the distance between AB and CDy dm == an elementary n)ass, y^ = the ordinate from AB to m, y = the ordinate from CD to m, and Jf = the mass of the section. Then f = (y, + Dy = y,' + 2y,D + m Multiply by dm and integrate, and we have ffdm —fyi^dm + 2Dfyidm + Dfdm. But since AB passes through the centre, the integral of y-idm, when the whole section is included, is zero (see Eq. 71J), and,/6?m = M\ hence, ffdm =fy^^dm + MD". (S3) Similarly, if dA be an elementary area, and A the total area, we have fifdA ^fyldA + AL^. Sl.'^j^i any ])lane area, the sum of the products of each ele- mentary area multiplied hy the square of its distance from an axis, is least when the axis passes through the centre. 110 CENTRE OF MASS. [82.] This follows directly from the preceding equation, in which the first member is a minimum for D = 0. CENTRE OF THE MASS. 82. The centre of the mass is such a point that, if the lohole mass he multijplied hy its distance from an axis, it will equal the sum of the products found hy multiplying each elementary mass hy its distance frotn the same axis. Let ^2- = an elementary mass; M — the total mass ; a?x, 2/i, and Zi be the respective coordinates, of the centre of the mass, and x,y, and 3 the general coordinates, then, accoi-ding to the definition, we have Ifxi = ^mx ; ] 3hj,^i:my',V (84) Msi = ^mz ; ) which being the same as (70) shows that when we consider the force of gravity as constant for all the particles of a body, the centre of the mass coincides with the centre of gravity. This is practically true for finite bodies on the surface of the earth, although the centre of gravity is actually nearer the earth than the centre of the mass is. If the origin of coordinates be at the centre of the mass, we have 2mx = ; Zmy = ; liinz = ; (84(z) which are the same as {71a). CHAPTER lY. NON-CONCURRENT FORCES. 83. EqUTLIBRIUM of a solid BODY ACTED UPON BY ANY NUMBER OF FORCES APPLIED AT DIFFERENT POINTS AND ACTINQ IN DIFFERENT DIRECTIONS. Fig. 54. Let A be any point of a body, at M-hicli a force F is applied, and O the'origin of coordinates, which, being chosen arbitrarily, may be within or without tlie body. On the coordinate axes construct a parallelopipedon having one of its angles at O, and the diagonally opposite one at A. Let the tyjoical force Fhe in the first angle and acting away from the origin, so that all of its direction-cosines will be posi- tive ; tlien will the sign of the axial component of any force be the same as that of the trigonometrical cosine of the angle wliicb the direction of the force makes with the axis. 112 GENERAL EQUATIONS [83.) Let a = the angle between i^and the axis of x, Q —- u a a >i u u u a y a a a a a a u a ^ then will the X, J", and Z-cojnponents of the force ^be JT = Fcos a, r=i^cos/3, i^ = 7^ COS 7. The point of application of the X-comjponent^ being at any point in its line of action, may be considered as at D, where its action-line meets the plane yz. At E introduce two equal and opposite forces, each equal and parallel to X. and since they will equilibrate each other, the mechanical effect of the system will be the same as before they were introduced. Combining the force + JTat Z> with — JTat /i', we have a couple whose arm is DE — y = the y-ordinate of the point A. This couple, according to Article 54, will be negative, Jience, its moment is — Xy. Hence, a force ■\- X 2X A produces the same effect upon a body as the couple — Xy, and a force + X at E. At the origin introduce two equal and opposite forces, each equal to X, acting along the axis of x. This will not change the mechanical effect of the system. Combining — X at with + X at E^ we have the couple -f Xz, and the force 4- X remaining at 0. Hence, a single force +X at A is equivalent to an equal parallel force at the origin of coordin- ates, and the two couples, — Xy and + Xz. Treating the Y-component in a similar manner, we have the force -^ IT at the origin, and the moments, + Yx and — Yz ; and similarly for the Z-component, the force -\- Z at the origin, and the moments, — Zx and + Zy. But the couples + Zy and — Yz, have the common axis x, (84.1 OF STATICS. 113 and hence are equivalent to a single couple Avliich is equal to the algebraic sum of the two ; and siniilai^ly for the others ; hence, the six couples may be reduced to the three following : Zy — Ys, having x for an axis ; Xz - Zx, " 2/ " " " ; Tx — Xy, " 3 " " " ; hence, for tlie single force F acting at A tliere may he svhsiv- tuted the three axial corajyonents of the force acting at the origin of coordinates, and three pairs of cowples having for their axes th^ resjpective coordinate axes. If there be a system of forces, in which /^, i^, i^, etc., are the forces, *ij yi5 ^1? the coordinates of the point of application of F[, ^) y-i-) ^2i ■'- 2? etc., etc., etc., a,, oo, Og, etc., the angles made by Fi, Fy, etc., respectively witft the axis of x, ^u ^ij ^i: etc., the angles made by the forces with y, and 7i' 72? 73j titc, the corresponding angles with 3 ; then resolving each of the forces in the same manner as above, we have the axial components X = i'l cos a^ + Fi cos Oo + i^ cos o^ + etc. = 2'i^cos a ; "j T = i^ cos ySi + i^ cos A + i^ COS /33 + etc. = .S'i^cos /3 ; I (85) Z = i^ cos 7i + i^ cos 72 + Fi cos 73 + etc. = XFcos 7 ; j and the component moments Zy - T3 = X{Fy cos 7 - i^3 cos/3) = Z ; I X3- Zx = X{F3 cos a - Fx cos 7) = J/; ^ {SQ) Yx~Xy= ^{Fx cos /? — Zy cos a) = X; ) in whicli Z, JZ, and iVare used for brevity. KESCLTANT FOECE AND EESULTxVNT COUPLE. 84. Let B = the resultant of a system of forces concurring at the origin of coordinates, and having the same magnitudes and directions as those of the given forces ; 114 DISCUSSION OF EQUATIONS. [85J a, h, and c = the angles which it makes with the axes a:, y, and z respectively ; O = tlie moment of the resultant couple ; d, e, and f = the angles which the axis of the resultant couple makes with the axes x, y, and z re spectivelj' ; then 2^ =: R cos a ; (87) Z =^ H cos c ; ' L = G cos d ; \ 3£= G cos e ; I (88) J^ = G cos/ J If a force and a couple, equal and opposite respectively tc the resultant force and resultant couple, be introduced into the system, there will be equilibrium, and H and G will both bo zero. Hence, for equilibrium, we have X=0; T=0; Z = 0; (89) Z = 0; Jf=0; ir=0. (90) 85. Discussion of equations (87) and (88). 1. Supjyose that the hody is perfectly free to move in any manner. a. If the forces concur and are in eqnilibrium, equations (87) only are necessary, and are the same as equations (60) ; hence, we will have x=o, r = o, Z=0. 1). If y? = and G is linite, equations (88) only are necessary. c. If R and G are both finite, then all of equations (87) and (88) may be necessary. 2 If one point of the body 'r&flxed^ there can be no trans- lation, and equations (88) will be sufficient. 3. If an axis jparallel to x is fixed in the body, there may bo translation along that axis, and rotation about it ; hence, the 1st of (87) and the 1st of (88) are sufiicient. [86.1 PROBLEMS. 115 4. If tioo 2)oints arejixed, it cannot translate, but may rotate ; and bv taking x so as to pass throngli the two points, the equa- tion Z = is sufficient. 5. If one point onltj is confined to the ])lane xy, the body will have every degree of freedom except moving parallel to s, and hence, all of erpiations (S7) and (SS) are necessary except the 3d of (ST). 6. If three points, not in the same straight line, are confined to th.Q plune xy, it may rotate about 2, but cannot move parallel tos; hence, the 1st and 2d of (ST) and the 3d of (SSj are necessary and sufficient. 7. If two axes parallel to x are fixed, the body can move only parallel to x, and the 1st of (ST) is suflicient. 8. If the fo'rces are parallel to the axis of y, there can be translation pai'allel to y only, and rotation about x and z. 0. If the forces are in the plane xy, the equations for eq^li■ librium become X = 2F cosa = B cos « = ; I Y= i:F cos ^= B cosh = ; I (91) Yx - Xy = X {Fx cos ^ -Fy cos a) ^- 0. ) [Obs. In a mechanical sense, whatever holds a body is a force. Hence, when we say '" a pomt is fixed," or, '' an axis is fixed," it is equivalent to in- troducing an indefinitely large resisting force. Instead of finding the value of the resistance, it has, in the preceding discussion, been eliminated. When we say "the body cannot translate," it is equivalent to saying that finite active forces cannot overcome an infinite resistance.] 88. Applicatioxs of equations (91). a. problems in which the tension of a string is involved. 1. A hody AB, %ohose weight is W, rests at its lower end upon a perfectly smooth horizontal plane, and at its upper end against a perfectly smooth vertical plane : the lower end is pre- vented from sliding by a string OB. Betermrne the tension on the string, and the 'pressure upon the horizontal ami vertical planes. 116 , EXAMPLES. [8G.] Take the origin of coordinates at C^ the axis of x coinciding with CB^ and 2/ with AC, x being positive to the right, and y positive npwards. Let TF = the weight of the body whose centre of gravity is at G\ R = the reaction of the vertical wall, and, since there is no friction, its direction will be perpendicular \oAO; iV= the reaction of the horizontal plane, which will be perpendicular to CB ; I — the horizontal distance from C to the vertical through the centre of gravity ; t — the tension of the string. The forces may be considered positive, and the sign of the component of the force will be that of the trigonometrical function. To determine the angle between the axis oi + x and the force, conceive a line drawn from the origin of co- ordinates parallel to and in the direction of the force, then will the angle be that swept over by a line from + x turning left-handed to the line thus drawn. The origin of coordi- nates may be at any point, and the origin of moments at any other point. Taking the origin of coordinates, and the origin of moments both at O, we have X=--B cos 0° + t cos 180° -f W cos 270° -1- i^Tcos 90° = ; r = /^sinO° + t sin 180° + JFsin 270° + irsin90° = 0; Moments = - B.AC + t.O - W.CD ^ I^.CB ^ 0. Keducing gives E-t = 0, - R.AC- W.CD ■{■ W.CB=0\ or, R = t, TF=iV^, ij=^'^^-/^,r= =^i'--- [S6.] EXAMPLES. 117 We see from this that the horizontal plane supports the entire wei^-ht of the piece, and that the pressure against the wall equals the tension of the string. "We also notice that the forces J^ and W being equal, paral- lel and opposite, constitute a couple whose arm is DB ; and this must be in equilibrium with the couple t, CA^R\ the arm being CA^ hence we have W.DB = LCA^ as before. 2. A ladder rests on a smooth horizontal plane and against a vertical wall, the lower end being held by a horizontal string ; a person ascends the ladder, required the pressure against the wall for any position on the ladder. 3. A uniform beam, whose length is AB and weight TF, is held in a horizontal position by the inclined string CD, and carries a weight P at the extremity ; required the tension of the strincr. Fig. 5ij. AD AC ^ ^ - Atis. t = 4. A prismatic piece AB is per- mitted to turn freely about the lower end A, and is held by a string CE\ given the position of the centre of gravity, the weight Wof the piece, the inclination of the piece and string, and the point of attachment E\ required the tension of the string, and the pressure against the lower end of the beam at ji. 5. A heavy piece AB is supported hy tv^o cords which pass over pulleys G and D, and have weights jPj and P attached to them; required the inclination to the horizontal of the line AB joining the points of attachmeM of the cord. (Consider the pulleys as reduced to tlie points C and D, ) F'G. 57. lis EXAMPLES. Let G, the centre of gravity of AB^ be on tlie line join ing the points of attachment A and B \ a = AG] b = BG; i = tlie angle DCM\ S = the inclination of BA to DC; a = BOA; and^ = CBB. Resolving horizontallv and vertically, we have X=P, cos (180° - MCA) + Pcos JVBB + TFcos 270°= 0; = - Pi cos (a -i) + P (!(.s (/3 + ?') = ; {a) Y = I\ sin (a - i) + F sin (/3 + /) - W = 0. {h) Taking the origin of niouients at G, wc have - 1\ X Gp + P X Gjh + TF X = 0; or, - Pi . « sin (a + S) + y\ Z- sin (/3 - S) = 0. (c) The angle * is given by the conditions of the problem ; hence the three eqnations («), {h), and (c) are snfficient to determine the angles a, /3, and 8, when the nnnierical vahies of the given quantities are known. Tlie inclination will be h + i. C. Snppose, in Fig. 58, that the strings are fastened at C awA B. and that BO, AC, and BB are given, reqnired the incli- nation of AB. [The solution of this problem involves an equation of the 8th degree]. C ^A 7. A heavy piece AB, Fig. 59, is free to swing ahont one end A, and is supported by a string BC which passes over a pulley at 0, and is attached to a weight P ; Und the angle ACB when they are in equilibi'ium. 8. A VN'cight W rests on a plane whose inclination to the horizontal is i, and is held by a string wliose in- clination to the plane is 6 ; reqnired the relation between the tension F and the weight, and the value of the normal pressure npon the plane. Fig. CO Ans. P ^ sm t cos 6 W\ Iformal pressure cos{6 + i) .y. cos 6 [86.] EXAMPLES. 119 Fia. 01. h. EQmLIBBTUM OF PERFECTLY SMOOTH BODIES IN CONTACT AVITII EACH OTHER. 9. A heavy heam rests on two smooth inclined planes, as in Fig. 61 ; required the inclination of the heam to the horizontal, and the reactions of the resj>ective planes. Let AC and CB be the in- clined planes; AB the beam whose centre of gravity is at G. When it rests, the reactions of the planes must be normal to the planes, for otherwise they would have a component parallel to the planes which wonld produce mo- tion. Let ay = AG; a^— GB ; Ji = the reaction at A ; B'= " " "B; W= the weig-ht of the beam ; a = the inclination of AC to the horizon ; Take the origin of coordinates at the centre of gravity G of the body, x horizontal and y vertical. The forces resolved horizontally give X=B cos (90° -a) + 12' cos (90° + ^) 4- TT'cos 270° = ; {a) and vertically, Y=B sin (90=- a) + 12' sin (90° + yS) + W sin 270° = 0. (//) -i- B sin a X ^16^^ sin 6. + B' shi /3 X GB sin 6. 0. — B cos a X AG cos 6. + B' cos /3 X GB cos 6. Hence Xy — Yx = Ba^ sin a sin d + B'a^ sin /3 sin 6 — Ba^ cos a cos 9 + B'a2 cos yS cos ^ = - Bai cos (a + ^) + B'a^ cos (13-6)= 0. (c) The moment of J? sin a is, . " " " B' sin yS is, . " " " TFcos 90° is, . « « " B cos a is, . " " " B' cos y8 is, . 120 EXAMPLES. [86.1 It is generally better to deduce the values of the moments directly from the definitions ; (see Articles 51 to 57). To do this in the present case, let fall from G the perpendiculars aO and LG upon the action-lines of the respective forces ; then hG = a, sin (90° - {/3 - d)) --= a^ cos i^-d); aG = «i sin (90° — [a + d)) = a^ cos (a + ^) ; and we have the moments = — R.aG + W - hG = — Ea^ cos (a + ^) + R'a2 cos O — ^) = ; as given above. Solving equations (a), (^), and (c), we find si n^ ^, si ng ^ ~ sin (a + /S)^*^ ' ~ sin (a + yS) ' ^ (2, cos a sin 8 — a^ sin a cos /3 tan e^ = -, ■ r^ : — 5 . («i 4- cii) sni a sin p HB = R\ then sin y8 = sin a ; which are the conditions necessary to make the normal reactions equal to each other. Tlie reactions prolonged will meet the vertical through the eenti-e -of gravity at a common point D, and if the beam be Bu«.pended at D by means of tlie two cords DA and DB it will retain its position when the planes AC and CB are removed. If /S = 90°, the plane CB will be vertical, and w^e find R = IF sec a; R' = ^^^ir= Tf tan a; ' cos a tan = r — cot a. «, + «2 If ai = ^25 then sin (/3 — a) tan d = 2 sill a sin ^ [86.] EXAMPLES. 121 If /? = 90° and a ■= 0°, then i?' = 0, ^ = 90°, and R=W. A special case is that in which the beam coincides with one of tlie planes. The formulas do not apply to this case. 10. Two equal, smooth cy- linders rest on two smooth planes whose inclinations are a and /3 respectively ; required the inclination, ^, of the line joining their centres. Fig. 62. Ans. Tan d = ^(cot a — cot y3). 11. A heavy, uniform, smooth beam rests on one edge of a box at C, and against the vertical side opposite; required its inclination to the vertical. Let (J be the centre of gravity. Ans. Sin 6 J BE Fig. 64. 12. Three equal, smooth cylinders are placed in a box, the two lower ones being tangent to the sides of the box and to each other, and the other placed above them and tangent to both ; required the pressure against the bottom and sides of the box. Ans. Pressure on the hottom — total weigJtt of the cylin- ders. Pressure on one side = i weight of one cylinder x tan 30°. 13. Two homogeneous, smooth, prisma- tic bars rest on a horizontal plane, and are preveiite.d from sliding upon it ; required their position of equilibrium when leaning az, etc., be the weights on the successive apices; which we will suppose are equal to each other ; hence i>=i>i=i?2 = etc.; Np = the total load ; V= the reaction at A\ and pr_ u a « J^ 1st. There will he equilibrium among the external forces. All the forces being vertical, their horizontal components will be zero, hence X=0; Y= V+Vi-Sj) = V+Vi-JVp = 0; (a) and taking the origin of moments at B, observing that the moment of the load is the total load multiplied by the horizontal distance of its centre of gravity from B, we have -. - -V.AB + ]Vp.^AB=0; or, V.L- N]).\L =0; 126 STRESS ON which in {a) gives V^ also equal to ^-iV/; ; hence the sni)])ortg sustain equal amounts, as they should, since the load is sjinc- trical in reference to tliem, and is independent of the form of trussing. 2d. To determine the internal forces. — Conceive that tht, truss is cut by a vertical plane and either jpart removed while we consider the remaining 'part. To the jpicces in the plane section^ apply forces acting in such a manner as to jproducc the same strains as existed before they were severed. Consider the forces thus introduced as external, and the problem is reduced to that of determining their value so that there shall he equilibrium among the new system of external forces. E A Let CD, Fig. 69, be a vertical section, and suppose that the right-hand part is removed. Introduce the ex- ternal forces in place of the strains, as shown in Fig. 70. Pio. 70. Let R = the compressive strain in the upper chord ; ZTi = the tensile strain in the lower chord; F — the pull in the inclined piece ; 6 — the inclination of i'^to the vertical ; n = the number of the bay, bD, counting from A (which in the figure is the 3d bay) ; and D = CD = the depth of the frame. The origin of coordinates may be tahen at any point. Take it at A^ X being horizontal and y vertical. Resolving the forces, A':=Z7,cos0°4-/ZcoslS0' + i^cos (270''+e'')+ Fcos 90' + 2p cos 270^ = ; r=i?isin 0°+i/sm ISO'+i^sin (270°+fl°)+ Fsin 90'+2p Bia270 =0; or, H^-II-\- F&in 6=0; V-iip - Fco&d --= 0; and the moments, - l-7i'pl + IIB -F.nl cos ^ = 0. («) 86.] FRAMES. 127 Eliminating i*^ between equations (h) and {c), substituting the value of V =^ i^^P, f^iic^ reducing, give n^^ni^N-n)', {d) that is, the strains on the hays of the upper chord vary as the product of the segments into which the lower chord is divided hy the joint directly under the hay considered. From {Jj) we have i^cos e = V- np = i(iY - 2n)p ; (e) and since 6 is constant, the stress on the inclined pieces decreases uniformly from the end to the middle. At the middle n = ^N', and F= ; hence, />/■ a tmiform load, there is no stress on the centred hraces. If F were considei-ed as a p)ush, equation ( =p, =i>2 = etc. = 1,000 lbs., AB = 56 feet and Z> = 4 feet ; required the stress on each piece of the frame. 22. In Fig. 69, if ^J»i and^o ^^"e removed, and_/>3 = _?>, — p-^ — 1,000 lbs., iind the stress on the bay 2 — 3, and the tie 2 — h. 23. If all the joints of the lower chord arc equally loaded, and no load is on the upper chord, required tlie stress on the n^" pair of braces, counting from li, Fig. 69. • Ans, \{N — 2;i + V) p sec 0. 128 STRESS ON A [86.1 24. A roof truss ADB is loaded with equal weights at the equidistant joints I, 2, 3, etc. ; required the stress on any of its memhers. [Obs. a load composed of equal weights on all the joints will produce the same stress as that of a load uniformly distributed, except that the latter would produce cross strains upon the rafters, which it is not our purpose to discuss in this place.] Let the tie AB be divided into equal parts, Aa, ah, etc., and the johits connected as shown in the figure. The joints are 1 y \ n \ f 1 ^ I A a i nt n c Fig. 71. assumed to be perfectly flexible. The right half of Fig. 71 may be trussed in any manner by means of ties or braces, or both, and yet not affect the analy- sis applied to the left half. Conceive a vertical section nm and the right-hand part removed. In- troduce the forces H, Hi and F as previously explained, and the condi- tions of the pi-oblem will be repre- fiq. 72. scnted by Fig. 72. The letters of reference given below involve both figures. Let N = the number of equal divisions (bays) in AB ', n = the number of the bay ho counting from A ; I = Aa = ah, etc. ; jp = the weight on any one of the joints of the rafter; V = the vertical reaction at A ov B ; /> = DC, the depth at the vertex; e = h2c; andi = DAC. Then {N — l)i? = the total load ; y^\{N-\)p. [86.3 ROOF TRUSS. 129 Take the origin of coordinates at A, and the origin of mo- ments at the joint marked 2, Eesohing the forces shown in Fig. 72 horizontally and vertically, we have X= //cos (180° + aAV)-\-H^ cos 0°+i^sin (- h=lc) = ; Y= V-{n-l)2) + JI&m{lS0°-{-aAl) + IIi&m0° + Foos{- 1'2g) = 0; or, — // cos i-\- Ri — i^sin ^ = ; Y— {n-l)p-E:^m i + /'cos ^ = ; also the m.oments^ IIJj2 - V.Ah+ {n-l)j>A{n -2)1=0. But from Fiof. 71 we have b2 CD AC (n - 1) I Substituting in the equation of moments the value of J2 found above, of F = i(iV"— l)j>, of Ai = {n — 1) ?, and reducing, give By means of the other two equations, and {n — 1) tan i tan 6 =1, we find H = i(iV — 7i)j) cosec i ; F = iln ~1) J) sec d. e. STRESS IX A LOADED BEAM. 25. Suppose that a beam is fii^Aj fixed in a 7call at one end, and that the projecting end is loaded with a weight P ,' required the forces in a vertical section mn., Fig. 73. Take the origin of coordi- nates at A, X horizontal and y verticalT" Take the plane sec- tion perpendicular to the axis of a;. Without assuraina: to know the directions in which the Fio. 73. 130 LOADED CORD. [86.] forces in the section act, we may conceive them to be resolved into horizontal and vertical components. Let F' be the typical horizontal force, then will X= %F^ 0; hence, some of the F-forccs will be positive, and the others negative. Neglecting the weight of the beam, and letting Y^ be the Bum of the vertical components in nni^ we have Z= Ti - P = .-. 1^1 = P; as shown in the figure. The forces, + P and — P, constitute a couple whose arm is Aa; and since the F-forces are the only remaining ones, the resultant of tlie + F''s and the — Fh must constitute a couple whose moment equals P.Aa witli a contrary sign. [Obs. Investisrations in regard to the distribution of the forces over the plane section belong to the Resistance of Materials. ] f. LOADED COED. 26. Suppose that a 'perfectly flexible, inextensihle cord is fleoed at two points and loaded continuously, according to any law / it is required to find the equation of the curve and the tension of the cord. Assuming that equilibrium has become established, we may treat the problem as if the cord were rigid, by consider- ins: the curve which it assumes as the locus of the point of ap- " Fig. 74. plication of the resultant. The resultant at any point will be in the direction of a tangent to the curve at that point; for otherwise it would have a normal component which would tend to change "the form of the curve. Take tlie origin of coordinates at the lowest point of the curve. Let a be any point whose coordinates are x and y; JT = the sum of the x-co^rnqyonents of all the external forces between the origin and a ; y = the sum of the y-components ; [86.1 LOADED CORD. 131 t = tiie tension of the cord at a ; ti = the tension at the origin. Eesolving the tension (t) by multiplying it by the direction- cosine, ^ve ha\e (a) dx i --- = the £5-component of t, and as t ~ = the y-component. cis For the part Ca^ eqiuations (91) become as Xy — Yx = 0. [Obs. In the problems which we shall consider, the third of these equations will be unnecessary, since the other two furnish all the conditions necessary for sohlng them. ] Let all iJie ajpjylied forces he vertical. Then X = 0, and the first two of equations {a) become dx I (J) as From the first of these we have t-j- = t^ = a constant ; hence, the horizontal component of the tension will le C07isfant throughout the length for any law of vertical loading. Frorathe second of (5), we have ^ d^ ^' 132 PARABOLIC CORD. [86.] Jieiice, the vertical component of the tension at any jpoint equals the total load hetioeen the lowest j)oint and the j?oint considered. 27. Zet the load he uniformly dis- tributed over the horizontal. (This is approxiipately the condition of the ordinary suspension bridge.) Let w = the load per unit of length, then Y =: — wx\ A ^ J, and {l>) l)eccmes -t.^ t dx = ds ~wx + t -^ = 0. Eliminating t gives ds t^dy — wxdx; and integrating gives i^y = iic:(P +((7=0); . 2^0 w y E Fig. 75. (P) id) hence, the cnrve is a parabola wliose axis is vertical, and whose parameter is — ^. The parameter will be constant when f^-^ w w is constant ; hence the tension at the lowest j>oint will he the same for all jparaholas having the same jparaineter and the same load per unit along the horizontal, and is indepemlent of the length of the curve. To find the tension at the loioest ])oint, substitute in equa- tion {cC) the value of the coordinates of some known point. Let the coordinates of the point A ])e a^i and y-^ ; then (i) gives ^0 = WXx 2yx' (^) [86.] THE CATEXART. 133 To find the tension at amj ])oint we have from the first of equations (c) and the Theory of Curves ds _ Vd^m^ _ i/rr^ ^-^'~dx- *' — di -^'^ ^ "^ ^ To find the tension at the highest jpoint A, from {d) find dx x^ ' ^' substitute in (/), and we obtain (To find U by the Theory of Moments, take the origin, at A. The load on Xi will be wr.ri, and its arm the horizontal distance to the centre of gravity of the load, or |.Ti ; hence, its moment will be iicxi"^. The moment of the tension wiU be ic^i ; hence, ^cxr to^i = iicx"-! or ^0 = o — 1 ^ before.) The slojye (or inclination of the curve to the liorizontal) may be found from equation (^) ; which gives tan ^ = ^^ . 28. The Catenary. A catenary is the curve assumed by a perfectly flexible string of uniform section and density, when suspended at two points nut in the same vertical. Mechanically speaking the load is uniformly distributed over the arc, and hence varies directly as the arc. To find the equation, let w = the weight of the cord per unit of length ; .♦. Y =^ — ws (s being the length of the arc) ; and equations {h) become -^0 + ^^ = 0; -ws-{-t^^=0. ds (h) 134: THE CATENARY. 186.1 Transposing and dividing the second by the first, gives dy w dx to ' and differentiating, substituting the vahie of ds and reducing. ,: T dx h Integrating gives ^^(1) \/^-i 10 or, passing to exponentials, gives /"_ ^ + x/rr^' dj/ d^^ [—,2 - « _ !^ 6'° ^^ J ' or, . 1 + from which we find which integrated gives jives 1 ^0 {k) (0 which is the equation of the Catenary. (y it Eliminating ~ between equations (*) and {j), we find ds 1 r /-« ~r* I 2i=*L' +' J' [88, j THE CATENARY. 135 the integral of which is s=h\\ c^^" -e '""' I +((7 = 0); (m) which gives the length of the curve. The following equations may also be found a; = - logg \ ■ — + V^^fh to cly s = -' ~ w dx U 9 — the iuclinatiou of the curve to the vertical, then X = s tan loge cot i;9. The tensions, t and fg, are so involved that they can be de- termined only by a series of approximations. The full devel- opment of these equations for practical purposes belongs to Applied Mechanics. The catenary possesses many interesting geometrical and mechanical prop- erties, among which we mention the following: — The centre of gravity of the catenary is lower than for any other curve of the same length joining two fixed points. If a common parabola be rolled along a straight line, the locus of the focus ■will be a catenary. According to Eq. (k) it appears that if the origin of coordinates be taken directly below the vertex at a distance equal to to -f- w, the constant of integra- tion will be zero. (This distance equals such a length of the cord forming the catenary as that its weight will equal the tension at the lowest point of the curve). A horizontal line through this point is the directrix of the catenary. The radius of curvature at any point of the catenary equals the normal at that point, limited by the directrix. The tension at any point equals the weight of the cord forming the cate- nary whose length equals the ordinate of the point from the directrix. If an indefinite number of strings (without weight) be suspended from a catenary and terminated by a horizontal line, and the catenary be then drawn out to a straight line, the lower ends of the vertical lines will be in the arc of a parabola. If the weight of tlie cord varies continuously according to any known law the curve is called Catenarian. 136 LAW OF LOADING. I8G.] 29. To determine the equation of the Catenarian curve of uniform density in which the section varies directly as the tension. Let 7c = the variable section ; B = the weight of a unit of vohime of the cord ; G = tlie ratio of the section to the tension ; then Y= -fhMs; h = ct] .:T=- Bcftds; which substituted in (b) and reduced, gives Bey = loge sec cBx, for the required equation. ^• LAW OF LOADING. 30, It is required to find the law of loading so that the action-line of the resultant of the forces cot any j)oint shall he tangent to a given curve. Assume the loading to be of uniform density, and the variations in the load- ing to be due to a variable depth. In X Fig. 70, let O be the origin of coordi- nates ; Z = «5 = the depth of loading over a point whose abscissa is a; ; J = the depth of the loading over the ori- gin, and B = the weight per unit of volume of the loading, then Y= -fBZdx', which in Eq. {b) gives dx ^ -&fZdx + t'^^=0, ds Transposing, and dividing the latter by the former, gives l = |A'^= [86.] LAW OF LOADING 131 which, differentiated, gives But, from the Theory of Curves, we have (i + '^-i\ d^y \ dxv sec' i dx^ ~ p p in wliich p is the radius of curvature, and i is the angle between a tano-ent to the curve and the axis of x. From these we read- ily find y — '1 fo sec'* At the origin p -— po, i = 0, and Z — d] which vahies sub- stituted in the preceding equation give = dpQ .: Z = dpo . (n) Discussion. For all curves which have a vertical tangent, we have at those points i = 90° ; .*. sec ^ = oo , and, if p is finite Z = ^ ', hence, it is practically impossible to load such a curve through- out its entire length in such a manner that the resultant shall be in the direction of the taiigent to the curve. A portion of the curve, however, may be made to fulfil the required con- dition. Zei the given curve he the arc of a circle ; then p = p^, and equation {n) becomes Z = d sec' i, from which the upper limit of the loading may be found. For 138 EXAMPLES OF A [86.] small angles sec** will not greatly exceed unity, and hence, the np])er limit of the load will be nearly parallel to the arc of the circle for a short distance each side of the highest point. At the extremities of the semicircle, i = 90°, and Z =^ cc . If the given curve he a jpcvrabola^ we find Z = d, that is, the depth of loading will be constant ; or, in other words, uni- formly distributed over the horizontal. This is the reverse of Prob."27. (The principles of this topic may be used in the construction and loading of arches.) 31. Let the tension of the cord he uniform. "We observe in this case that the loading must act normally to the curve at every point, for if it were inclined to it, the tangential component would increase or decrease the tension. Let J? = the normal pressure per unit of length of the arc ; ihen^ds = the pressure on an element of length, and this mul- tiplied by the direction-cosine which it makes with the axis of X, and the expression integrated, give I pds i-j ) = / J>dx — the aj-component, and / jpdy = the y-component of the pressures, hence, equations (a), p. 131, become -to + jpdx + ^ ^ = ^ ; fl>dy+t§L=0; differentiating which, give [86.] NORIVIALLY PRESSED ARC. 139 Transposing, squarinp;, addiuf^ and extracting the square root, give that is, the normal ^yressure varies inversely as the radius of .curvature. 1. If a string be stretched upon a perfectly smooth curved surface by pulling upon its two ends the normal pressure upon the surface will vary inversely as the radius of curvature of the surface, the curvature being taken in tlie plane of the string at that point. 2. If p be constant p will be constant ; hence, if a circular cylinder be immersed in a fluid, its axis being vertical, the nor- mal pressure on a horizontal arc being uniform throughout its circumference, the compression in the arc will also be constant. h. THE LAW or LOADING ON A NORMALLY PRESSED ARC BEING GIVEN, BEQDLRED THE EQUATION OF THE ARC 32. The ties of a susjpensioii ^ -g hridge heing normal to the curve of the cahle, and the load uni- / / /^J~Y^ form along the span, req;uired j) e the equation of the curve of fiq. tt. the cable. the origin being at C, x horizontal and y vertical. If tan i = -T-^ and Oq = the radius of curvature at the vertex, dxy ' then X = ^po (1 + cos^ i) sin i, y = ipo sin~ i cos i. (See solution by Prof. S. W. Robinson, Journal of the Franklin Institute, 1863, vol. 4G, p. 145 ; and its application to bridges and arches, vol. 47, p. 152 and p. 361.) 140 NORIIALLY PRESSED ARC. [86.] 33. A lyerfectly flexible^ inextensihle trough of imlefinite length is filled with a fluid, the edges of the trough heing i^ar- allel and supported in a horizontal plane ; required the egua tion of a cross section. The lenf^th is assumed to be iiKlefinitelj long, so as to elimin- ate the effect of the end pieces. The pressure of a fluid against a surface is always normal to the surface, and varies directly with" the depth of the Huid. The actual pressure equals the weight of a prism of water whose base equals the surface pressed, and whose height equals the depth of the centre of gravity of the said surface below the surface of the fluid. The problem may therefore be stated as follows : — Reguired the equation of the curve assumed hy a cord fixed at two points in the same hori- zontal., and pressed normally hy forces which vary as the verti- cal distance of the p>oint of ap>plication lelow the said hori- zontal. Let A and B be the fixed points. Take the origin of the coordinates at D, midway be- tween A and B, and y positive downwards. Let S be the weight of a unit of volume ; then p z=hy, which in equation {o) gives t = hyp, and for the lowest point t = SBpo ; in which D is the depth of the lowest point and po the radius of curvature at that point ; .: hyp = 8I)po, or -f^ = ^. But from the Theory of Curves we have p~\'^dx') dj?' which substituted above, and both sides multiplied by dy, may be put under the form [S6.] HYDKOSTATIC TROUGH. 141 the integral of wliicli is But -y- = 0, f or y = -Z> ; .'. 6^ =^7 ; which substituted and the equation reduced gives {ax + aif) - as - ^^^2jfip2_^2j)p-D'i+,ff] Squaring and reducing, gives Tliese may be integrated by means of EllijptiG Functions. Making y = D co?> i>, and c = -j—, they may be reduced to known forms. Using Legendreh notation, we have (See Article by the Author in the Jownal of the FranTdin Imtiiute, 1864, vol. 47, p. 289.) CHAPTER y. RELATION BETWEEN THE INTENSITIES OF FOKCES ON DIFFEKENT PLANES WHICH CUT AN ELEMENT. 87, Distributed Foeces are those whose points of applica- tion are distributed over a surface or throughout a mass. The attraction of one mass for another is an example of the latter, some of the properties of which have been discussed in the Chapter on Parallel Forces ; similarly, when one part of a body is subjected to a pull or push, the forces are transmitted through the body to some other part, and are there resisted by other forces. If the body be intersected by a plane, the forces which pass through it wnll be distributed over its surface. Planes having different inclinations heing passed through an element, it is jprojposed to find the relation between the intensi- ties of the forces on the different jylanes. 88, Definitions. Stresses are forces distributed over a sur- face. In the previous chapters we have assumed that forces are applied at points, but in practice they are always distrib- uted. A strain is the distortion of a body caused by a stress. Stresses tend to change the form or the dimensions of a body. Thus, ^jpull elongates, a j9wsA compresses, a twist produces tor- sion, etc. (See Resistance of Materials^ A simple stress is a pull or thrust. Stresses may be com- pound, as a combination of a twist and a pull. A dikect simple stress is a pull or thrust which is normal to the plane on which it acts. A ^w^^ is considered j9(95ii^*v«?, and 21. push, negative. The intensity of a stress is the force on a unit of area, if it be constant ; but, if it be variable, it is the ratio of the stress on an elementary area to the area. To form a clear conception of the forces to which an element is subjected, conceive it to be removed from the body and then [89,90.] RESOLVED STRESSES. 14^ subjected to such forces fts will produce the same strain that it had while in the body. 89. Formulas for the intensity of a stress. Let 7^ be a direct simple stress acting on a surface whose area is A^ and _p the measure of the intensity, then i^=l- F A' (92) when the stress is uniform, and 7) — -TTi when it is variable. •^ dA If the stress be variable we will assume that the section is so small that the stress may be considered uniform over its sur- face. 90. Direct stress resolved. Let the prismatic element AB, Fig. 79, be cut by an oblique plane DJS". Let the stress F be simple and direct on the surface CB, and iV" = the normal component of i^on D£; T = the component of F along the plane DE, which is called the tangential component ; = FON = the angle between the action-line of the force and a normal to the plane DE^ and is called the ohliquity of tlie plane / A = the area of CB, and A' that of BE. Then, according to equations (62), we have ir= i^cos6'; T = Fsm e. From the figure we have A' = A sec 9y hence, on the plane BE, we have Fio. 79. 144 SHEARING STRESS. [Ol.j J^ormaliniensity, 2>n^^ -j-/ = "i a — P *^*^s^ ^5 Jx j^ sec (/ Tangential intensity, j)t — ~a7 — ~a a = j? sin ^ cos 6. (93) Pass aiiotlier plane perpendicular to DE, having an obli- quity of 90° — 6] then, accenting the letters, we have I>'7i = jp sin=^ 6 ; a - a i (94) I> t = i? cos ^ sin d. ) ^ This result is the same as if a direct stress acting upon a plane perpendicular to GB, having an obliquity of 90° — 6 in reference to J)E, be resolved normally and tangentially to the latter. Combining equations (93) and (94) we readily tind . . (95) that is, when an element {or hody) under a direct simple stress is intersected by two planes the sum of whose obliquities is 90 degrees, the sum of the intensities of the normal components of the stress equals the intensity of the direct simple stress, and the intensities of the tangential stresses are equal to each other. 91. Shearing stress. The tangential stress is commonly called a shearing stress. It tends to draw a body side wise along its plane of action, or along anotlier plane parallel to its plane of action. Its action may be illustrated as follows : — Suppose that a pile composed of thin sheets or horizontal layers of paper, boards, iron, slate, or other substance, having friction between the several layers, be acted upon by a horizontal force applied at the top of the pile, tending to move it sidewise. It will tend to draw each layer upon the one immediately beneath it, and the total force exerted between each layer will equal the applied force, and the resistance at the bottom of the pile will be equal and opposite to that of the applied force. If other horizontal forces are apj^lied at different points along the ver- tical face of the pile, the total tangential force at the base of the pile will equal the algebraic sura of all the applied forces. A shearing stress and the resisting force constitute a couple, [91.] SHEARING STRESS. 145 ■>^ and as a single couple cannot exist alone, so a pair of shearing stresses necessitate another pair for equilibrium. When the direct simple stresses on the faces of a rectangular j)arallelojpipedon are of equal inte^isity, the shearing stresses will he of equal intensity. Let Fig. SO represent a paralellopipedon with direct and shearing stresses applied to its several faces. At pre- sent suppose that all the forces are parallel to the plane of one of the faces, as abfe, and call it a plane of the forces ; then will the planes of action, which, in this case, will be four of the faces of the parallelopipe- don, be perpendicular to a plane of the forces. If the direct stress + F = — F, and ■\- F' =■ — F', they will equilibrate each other. The moment of the tangential force T, will be j)c y. areafc x ah ; and of T' jy't X area ac x If. The couple T.ah tends to turn the element to the right and T'.hf to the left, hence, for equilibrium, we have pt X areafc x ah = p't x area ac x hf; but area fc x ah ~ area ac x ^Z" = the volume of the ele- ment, hence I>. = I>\- (96) The effect of a pair of shearing stresses is to distort the element, changing a rectangular one into a rhomboid, as shoum in Fig. 81. Direct stresses are directly opposed to each other in the same plane or on opposite surfaces ; shearing stresses fio. 81. act on parallel planes not coincident. 10 U6 NOTATION. [92, 93.1 92. Notation. A very good notation was devised by Co- riolis, wliicli has been used since 1837, and is now commonly employed for the general investigations on this subject. It is as follows : — Let J9 be a tyjncal letter to denote the intensity of a stress of some kind ; p,. the intensity of a stress on a plane normal to a? ; 2^xx tbe intensity of a stress on a plane normal to x and in a direction parallel to x, and hence indicates the intensity of a clireGt sim,j>le stress ; and f^ the intensity of a stress on a plane normal to x but in the direction of y, and hence indi- cates the intensity of a shearing stress. Or, generally, the first sub-letter indicates a normal to the flane of action and the second one the direction of action. Hence we have INTENSITIES OF TUE FOKCES 'parallel to X y 2 X\ I^yx J?vv !Pyz f on a plane normal to -| y ; If direct stresses only are considered, one sub-letter is suffi- cient ; iisj):„j),j, or^,. 93. Tangential stress eesolved. Let T be the tangential stress on the right sec- tion AB = A, the section being normal to y, then iV = ^ -^ ^• Let CD be an oblique section, normal to the axis y' ; x' and x being in the plane of the axes y and y' ; then will the angle be- tween y and y' be the obliquity of the plane CD. This we will denote by {yy'). lAit the tangential force be parallel to the axis of x. Resolving this force, we have Normal component on CD = T sin (yy') ; Tangential component on CD ~ T cos (yy'). Fig. S2. [94.1 RESOLVED STRESSES. 147 Dividing each of these by area CD = AB -^ cos {ijy), we have Tsin («y') cos (?/y') • , is ,• '^ 1 Normal inteiuity = j)y'y' ^ U ^v, =Vm^-^ ^VV ) cos (^^ ); ' area AB Tcoii"-(yu') „ , ,, Tancjential intensity = pyfx' = '"^'ab' ~ ^^"^ M97) J and for a tangential stress on a plane normal to x, resolved upon the same oblique plane CD, we have (98) i^Vj/ = JP^y cos W) sin {yy') ; P'y'^ = P^y sin2 {yy'). If the tangential stresses on both planes (one normal to y, and the other normal to x) are alike, and the obliquity of the plane CD less than 90°, the resultant of their tangential components will be the difference of the two components, as given by equations (97) and (98); that is, it will be J9^v — jy'y'^; bat the normal intensity will be the sum of the components as given by the same equations. The reverse will be true in re- gard to the direct stresses. 9^. Let a hody he suljected to a direct sirn,ple stress ; it is required to find the stresses on any two jplanes perpendicular to one another and to the plane of the forces ; also the intensity of the stress on a third plane perpendicular to the plane of the forces ; and the normal and tangential components on that plane. Let the forces be parallel to tlie plane of the paper ; AG and OB., planes perpendicuhxr to one another and to the plane of the paper, having any obliquity with the forces. Let the axis of x coincide with OB, and y witli A 0. Let AB be a third plane, also parpendicular to the plane of the paper, cutting the other planes at any angle. Take y' perpendicular to AB and ai parallel to it and to the plane of tlie paper. Fia. &3. 148 KESOLVED STRESSES. S95.1 The oblique forces may be resolved normally and tangen- tially to the planes AO and 0£, by means of equations (93) and (94). .The problem will then be changed to that shown in Fig. 84, in which one set of stresses is simple and direct, and the other set tangential ; and, according to Article 91, the intensity of the shearing stress ^°- ^^- on the two planes will be the game ; hence, for this case The intensity of the total normal stress on the plane AB will be the sum of the normal components given by equations (93), (94), (97) and (98), and the total tangential stress will be the sum of the components of the tangential stress given by the same equations ; hence P v'y = P^^ sin'' (&^') + Pyv cos' {yy') + 2piy sin {yy') cos (yy) ; \ ^99^ Pyx ^\Px^- Pyy \ sin {yy) cos {yy) + p^y \ cos" {yy) - sin= {yy) \ • ) Tlie resultant stress on AB will be, according to equation (46), being 90°, and the inclination of the resultant stress to the normal, y\ will be tan(ry')=-^-^. (101) 95. Discussion of equations (99). A. Find the inclination of the ^lane on which there is no tangential stress. In the 2d of equations (99) makej^y/^/ = 0, and representing this particular angle by (yy'), we find ^ ^ cos^ {yy') - sm^ {yy ) ^J^ry Pyy Pxx , (102) [95.1 PRINCIPAL STRESSES. liO which gives two angles differing from each other by 90°, or, the planes will be perpendicular to one another. Hence, in every case of a direct sim.pU stress upon a jpair of jflanes jperpendicular to one another and to a jplane of the stresses^ tJiere are two jplanes^ also x)erpendicidar to one another ami to thejylane of the stresseSjOn which tJiere is no tangential ■stress. These two directions are c&Wcd j^rincipal axes of stress. Frlncijpcd axes of stress are the normals to two planes per- pendicular to one another on which there is no tangential stress. Princiiml stresses are such as are parallel to the principal axes of stress. (In some cases there is a third principal stress perpendicular to the plane of the other two.) The formulas for the stresses become most simple by refer- ring them directly to the principal axes. a. Let one of the direct stresses Ite zero. Equation 102 gives tan 2(yy') = ^ (103) JPxx h. Let 07ie of the direct stresses he a jpull^ and the other a push. Then tan 2(yy') = ^^J (104) c. .Let them act in opposite senses and equal to each other. Then tan2(YY')=-^. (105) d. Let there he no tangential stress on the original planes^ orp^ = 0. Then7 tan 2(yy') = ; .-. (yy') = or 90° ; and the original planes 2sq princi'pal planes. 150 RESOLVED STRESSES. [95.] e. Let there he no direct stresses. Then, tan 2(ty') = oo ; or- (yy') = 45° or 135° ; (106) that is, if on twoj)lanes, 2:>er])endiGulaT to one another and to the plane of the stresses, there are no direct stresses, then will the stress on two planes, ferjpendicidar to one another and to the 'plane of the st7'esses, whose inclination with the original j>lanes is 45°, he simple and direct. f Let the direct stresses he equal to one another and act in the same sense, and let there he no shearing on the original planes. Then tan2(YY')=^; and (yy') is indeterminate; hence, in this case every plane perpendicular to a plane of the stress will be a principal plane. Examples. 1. A rongh cube, whose weight is 550 pounds, rests on a horizontal plane. A stress of 150 pounds applied at the upper face pulls vertically upward, and another direct stress of 125 pounds, applied at one of the lateral faces, tends to draw it to the right, while another direct stress of 50 pounds, tends to draw it to tlie left ; required the position of the planes on which there are no tangential Presses. If the cube is of Unite size it will be necessary to modify the problem, in order to make it agree with the hypothesis under which the formulas have been established. The force of gravity being distributed throughout the mass, would canse a variable stress, and the surface of no shear would be curved instead of plane. We will thei-efore assume that the cube is ivithout weight, and the 550 pounds is applied directly to the loiver surface. Then the vertical stress will be 150 pounds, the remaining 400 pounds being resisted directly by the plane on which it rests, an.d so far as the present problem is concerned, only produces fi-iction for resistim^ the shearing stress. The direct horizontal [95.1 EXAJIPLES. 151 Stress will be 50 pounds, the remaining 75 pounds producing- a shearina: on the horizontal ulane. The former force tends to turn the cube right-handed by rotating it about the lower right- hand corner, thus producing a reaction or vertical tangential stress of 75 pounds. Let the area of each face of the cube be unity, then we have J>x!/ = 75 pounds ; j?^ = 50 pounds ; j>^ = 150 pounds ; and these in (102) give tan2(yY) = ^— -— = 1.d; .-. (tt') = + 2S" 9' IS", or - 61° 50' 42". If the body be divided along either of these planes, the forces will tend to lift one part directly from the other without producing sliding upon the plane of division. 2. A rouijh body, whose weight is 100 pounds, rests on an inclined plane ; required the normal and tangential components on the plane. (Use Eq. (03).) 3. A block without weight is secured to a horizontal plane and thrust downward by a stress whose intensity is 150 pounds, and pulled towards the right by a stress whose intensity is 150 pounds, and to the left with an intensity of 100 jiounds ; re- quired the plane of no shear. 4. A cube rests on a horizontal plane, and one of its vertical faces is forced against a vertical plane by a stress of 200 pounds applied at the opposite face, and on one of the other vertical faces is a direct pulling stress of 75 pounds, which is directly opposed by a stress of 50 pounds on the opposite vertical face ; required the position of the plane of no shear. In this case the weight of the cube would be a third princi- pal stress, but it is eliminated by the conditions of the problem. The shearing stress is 25 pounds; and because the direct stresses-are unlike, we use Eq. (104). 5. A rectangular parallelopipedon stands on a horizontal plane, and on the opposite pairs of vertical faces tangential 152 PLANES OF [Q5.\ stresses of eqnal intensities are applied ; required the position of the plane of no shear. (See Eq. (106).) 6. In the preceding problem find the intensity of the direct stress on the plane of no shear. (Substitute the proper quanti- ties in the 1st of (99).) £. To find the planes of action for maxiniuin andminimum normal stresses^ and the values of the stresses. Equate to zero the first differential coefficient of the 1st of Equations (99), and we have 2^^ cos {yy') sin {yy') - 2^^^ sin {yy') cos {yy') | - 2i>xj, sin2 iijy') + 2p^ cos^ {yif) = ; I •• tan ^yy') = --^^ Pxx Pin/ which, being the same as (102), shows that on those planes which have no shearing stress, the du-ect stress will be either a maximum or a minimum. Testing this value by the second differential coefticient, we find that one of the values of (yy') gives a maximum and the other a minimum. Comparing (107) with the second of (99), shows that the first differential coefficient of the value of the direct stress on any plane equals the shearing stress on that plane. From (107), observing that cos {yy') — -/l — sin^ {!/l/')> we find sin^ (yy') = i I 1 T /=^^==r^ [ ; (108) and these values in the 1st of (99), and the maximum and minimum values designated by p^', give i?Y' = i(i?^ +i>J ± V\Ail>^ -Pwf +P'x^\ ; (109) in which the upper sign gives the maximum, and the lower the minimum stress. These are jprincipal stresses, and we denote them by one sub-letter. [95.] MAXIMUM STRESS. 153 a Ifx>xii — ^) we have (yy') = 0° or 90°, as we should. h. Ifjpyy = 0, we have a3 . xy 5 (110) maximuTn, p^f = ^Pxx + VijP^xx + P^. minimum,])^, = ^j>^ - ^/i^^a^Tpy^ hence, the niaximnm normal stress will be of the same kind as the prmcipal direct stress, j!?a^ ; that is, if the latter is a^t^Z)?, the former will also be a 7:>?i//, and the minimum principal stress will be of the opposite kind. c. If there are no direct stresses jp^ will also he zero, and we have (yt') = 45° or 135° ; and inaximum ])^> =^2^'-cy — ~ ]?\' foi' rainhmirrh / that is, the principal stresses will have the same intensity as the shearing stresses, and act on planes perpendicular to one another, and inclined 45° to the original planes.' EXAISIPLES. 1. Suppose that a rectangular box rests on one end, and that one pair of opposite vertical sides press upon the contents of the box with an intensity of 20 pounds, and the other pair of vertical faces press with an intensity of 40 pounds, and that horizontal tangential stresses, whose intensities are 10 pounds, are applied to the vertical faces, one pair tending to turn it to the right, and the other to the left ; required the position of the vertical planes of no shearing, and the maximum and minimum values of the direct stresses. 2. For an application of Equations (103) and (110) to the stresses in a beam, see the Author's Resistance of Materials^ 2d edition, pp. 236-240. C. To find the position of the planes of maximum and Tninirmi'in shearing. 154 PRINCIPAL STRESSES. [95. j Equate to zero the first differential coefficient of the second of (99) and reduce, denoting the angles sought by {YY'), and we find, - cot2(rr') = tan2(YY'); .-. 2rJ":-2(YY') + 90°; or, ' YY' = yy' + 45° ; that is, the plaiies of Qnaximum and minimum shear make angles of 4:5 degrees ^vith the pkincipal planes. D. Let the^planes he pkincipal sections. Then the stresses will be j^rincijpal stresses, and p^ = 0. Using a single subscript for the direct stresses, equations (99) become 2\' = Vx shr {yij) + j^y cos- {yy) ; (^ ^ I>y'x = {px - I>y) sii» iuu') cos {yy'). \ a. Let J9a: =Tyi t^^®^^ Tv' =I>x'-, and ^v it' = 0; that is, when two princij)al stresses a^e alike and equal on a pair of planes perpendicular to the plane of the stresses, the normal intensity on every plane perpendicular to the plane of the stresses will he equal to that on the principal planes, and there loill he no shearing on any plane. This condition is realized in a perfect fluid, and hence very nearly so in gases and- liquids, since they offer only a very slight resistance to a tangential stress. If a vessel of any liquid be intersected by two vertical planes perpendicular to one another, the pressure per square inch will be the same on both, and will be normal to the planes ; hence, according to the above, it will he. the same upon all planes traversing the same point. This is only another way of stating the fact that fluids press equally in all directions. h. To find the planes on which there will he no normal pres- sure. For thispj^ in (111) will be zero ; [90.] PROBLEM 155 .•.tanQ//)=\/f I/-ZT1 which, being imaginary, shows that it is impossible when the stresses are alike ; but if tliej are unlike^ we have If j9a. — — Py, then {yy') = 45", and the 2d of (111) gives wliich shows that when the direct stresses are luiliJce and of equal intensity on planes perpendicular to one another, the shearing stress on a plane cutting both the others at an angle of 45 degrees, will be of the same intensity. Let {ijy) — 45'', or 135°, then (111) become (112) v^/^' = ±*(i>^-i>j.); in the latter of which the upper sign gives a maximum, and the lower a minimum value. Using the upper sign, we find 96. Problem. Find the jplane on tohich the obliquity of the stress is greatest, the intensity of that stress, and the angle of its obliquity. Let the stresses be jprincipal stresses and of the same hind, and ^ the angle of obliquity of the required plane to the stress ; then sin (^ — r^~Vy .^ the intensity ■= \^{,VvJPv)\ ^^^^ ^^^® angle be- tween the principal phxne x and the required plane = 45°— \^. 156 CONJUGATE STRESSES. [97. | If the principal stresses are unlike^ then gin (f)' =fcti^; the intensity = V'^?^r)y, and the angle be- tween the principal plane x, and the oblique plane = 45° — -I*/)'. Example. If a body of sand is retained by a vertical wall and the intensity of the horizontal push is 25 pounds, and of the ver- tical pressure is 75 pounds ; required the plane on which the resultant has the greatest obliquity, and the intensity of the stress on that plane. CONJUGATE STKESSES. 97. A pair of stresses, each acting parallel to the plane of action of the other, and whose action-lines are parallel to a plane which is perpendicular to the line of 'intersection of the planes of action, are called conjugate stresses. Thus, in Fig. 85, one set of stresses acts on the plane YY, parallel to the plane XX^ and the other set on XX, parallel to YY. In a rigid body the intensities of these sets of stresses are independent of each other; -for each set equilibrates itself. Principal stresses are also conjugate. There may be three conjugate stresses in a body, and only three. For, in Fig. 85, there may be a third stress on the plane of the paper, which may be parallel to the line of intersection of the planes XX and YY, and each stress will be parallel to the plane of the other two. A fourth stress cannot be introduced which will be conjugate to the other three. Conjugate stresses may be resolved into normal and tangen- tial components on their planes of action, and treated according to the preceding articles. The fact that the stresses have the same obliquity, being the complement of the angle made by the planes, simplifies some of the more general problems of stresses. ■OS.] GENEKAL PROBLEM. 157 Fig. 86. GENEKAL PROBLEM. 98- Given the stresses on the three rectangular coordinate f lanes ; required the stresses on any ohUgite plane in any re- quired direction. As before, the element is supposed to be indefinitely small. Let ahc be the oblique plane, the normal to which designate by n. The projection of a unit of area of tin's plane on each of the coordinate plains, gives respectively cos {nx\ cos (7iy), cos {iiz). The direct stress parallel to x acting on the area cos {7\x) Avill give a stress of jpxx cos {nx), and the tangential stress normal to y and parallel to x will produce a stress _^ya: cos {ny), and similarly the tangential stress normal to s and parallel to x gives ^?j^ cos ((m) ; hence the total stress on the unit normal to n and parallel to x will be 2)nx = J>xx cos {nx) + 2yyx COS (???/) + ^^x COS {jiz) ; - similarly, , ^^^^x _^„y = ^^^ cos (nx) +jpyy cos {ny) + j[>,y cos {ns) ; j ^^ —2)x^ co^inx) +j>y^ cos (ny) + j)zz cos (nz). J Let these be resolved in any arbitrary direction parallel to s. To do this multiply the first of the preceding equations by cos (sx), the second by cos (.sy), and the third by cos {sz), and add the results. For the purpose of abridging the formulas, let cos (7ix) be written Cnx, and similarly for the others. Then we have Pns =p>xxCnxCsx +j)yyCnyCsy + p),,CnzCsz | 4-^yc {C/vjCsz + OnzQsy) + j),JS^nzGsx V (115) + CnxGsz) + 2?xy (CnxCsy + CfiyCsx). ) This expression being tyjncal, we substitute x for n and s, and -i>hus obtain an expression for the intensity on a surface normal to x' and parallel to x'. Or generally, substitute suc- cessively x', y\ z' for n and s, and we obtain the following formulas : 158 GENERAL FORMULAS. [98.] DIRECT STRESSES. + 'ij?iJdsx' Cxx' + 'ilp^Cxx'Cyx ; j>y'y> =2):^xG\rj/+ pyyC'yy' + p,S^'zy' + ^2Jy,Cyy'Czy' + 2j)j:^zy' Cxy'+ ^j):^Cxy'Ctjy' ; i?^^ =I>J^'xz' +PyyChjz' +p.S^'zz' + 2j>yS^yz'Czz' + 2j>zxGzz'Cxz + 'ijy^Cxz'Cijz' ; TANGENTIAL STRESSES. i?vv =i^^^C.z'y'C^'3' +PyyCyy'Oyz' + jy^zGzy'Gzz' + iyy^ {Gyy'Ozz' + Q/s'C^y') + i>- (<"-/Ca;3' + CzzCxy') + p^y{Gxy'Cyz' + C^/C^y') ; _p^^ z= :^a,^Cj33'C«*' + jpyyQyz'Cyx' + _^e2 Cs^'Cg-r' + iV.(6/s'Csa?' + C33'C2/2;')+i^^:«(C33'C^,:c'+C3^'C*3') +Pxy {Cxz'Cyx' + Ca%c'C2/3') ; j)^j^, = ^^^^C.zja?'C^?/' + p^>jCyx'Cijij' + p,,Czx'Czy' +Py,{CyxVzy' + Cyy'Czx') +p,^{Czx'Cxy' + C3^'Ca;.z;') + p^y{Cxx'Cyy' + Cxy'Cyx'). It may be shown that for every state of stress in a hody there are three jplaines jperjpeiidicular to each other ^ on vjhich the stress is entirely normal. [These equations are useful in discussing the general Theory of the Elasticity of Bodies?)^ These formulas apply to oblique axes as -well as right, only it should be observed when they are oblique that jpy>z> is not a stress on a plane normal to ?/', parallel to z\ but on a plane nor- mal to x' resolved in the proper dh-ection. CHAPTER YI. VIRTUAL VELOCITIES. 99. Def. If the point of application of a force be moved in the most arbitrary manner an indefinitely small amount, the projection of the path thns described on the oriirinal action-line of the force is called a h virtual velocity. The product of the force f \ ^ into the virtual velocity is called tlie virtual ^^^ g^_ moment. Thus, in Fig. 87, if a be the point of application of the force F, and ah the arbitrary displacement, ac will be the virtual velocity, and F.ac the virtual moment. The path of the displacement must be so short that it may be considered a straight line ; but in some cases its length may be finite. If the projection falls upon the action-line, as in Fig. 87, the virtual velocity will be considered positive, but if on the line prolonged, it will be negative. 100. Prop. If several concurring forces are in equilibrium, the algehraic sum of their virtual moments will he zero. Using the notation of Article (47), and in addition thereto let I be the length of the displacement ; and jp, q, and r the angles which it makes with the respective coordinate axes; then will the projections of I on the axes be I cos^, I cos q, I cos r, respectively. Multiplying equations (50), by these respect- ively, we liave Fx cos ai I cos^ -+- 7^ cos Oo ^ cos_p + etc. = ; Fx cos yS) I cos q + Fi CO?, /3^ I cos q r etc. = ; Fx cos 7i I cos r + /% cos 70 I cos r -f- etc. = 0. 160 VIRTUAL VELOCITIES. [101.] Adding these together term by term, obsei-ving that cos a cosp + cos /3 cos q + cos e will be the virtual velo- city of P, and we have — PJje + W.ae ~ ; and ae .-.P.CB (S Fig. 93. AB BO, eh W.BA; or, the weight is to the hoinzontal force as the hase of the tri- angle is to its altitude 6. In Fig. 27 show that Pdr = Wdy. 7. One end of a beam rests on a horizontal jDlane, and the other on an inclined plane ; re- quired the horizontal pressure against the inclined plane. This involves the principle of the. -wedge ; for the block AB O may represent one-half of a wedffB beins: forced a2:ainst the resistance TF. Conceive the ■plane to be moved a distance AA', and that the beam turns £ B' Fig. 94. 164 VIRTUAL VELOCITIES. [lOl.J about tlie end D^ but is prevented from sliding on the plane ; then will the virtual velocity of the horizontal pressure be JL^', and that of the weight will be Eg\ hence, for equilibrium wo have W.Eg - P.AA' = 0. {a) We now find the relation between Eo and AA' , Let I = DF^ the length of the beam ; a = DE^ the distance from D to the centre of gravity of the beam ; a = CAB\ ^ = ADF, The end at F will describe an arc FF' about Z^ as a centre. From F' draw F'd parallel to AA' ^ and from F drop a per- pendicular Fe upon clF' . Then, from similar triangles, we have Fe = - Ec, a FF' will be perpendicular to DF^ and Fe perpendicular tc dF\ hence eFF' = ADF= yS; dFe = 90°— a; .-. dFF' = 90°-~a + /3; and FF' = Fe&ecB =:^ - Eg sec B. a The triangle FdF' gives FF' sin a hence, AA! = dF' sin (90^ - a. + fif Eg a sin a cos /3 AA! ~ I cos(a — /3)' whichj substituted in equation («; above, gives „ _ -„ « sin a cos /3 ~ I cos (a — ^)' 8. Deduce the formula for the triangle of forces from the principle of Virtual Velocities. CHAPTEE 711. MOMENT OF INERTIA. (This chapter may be omitted until its principles are needed hereafter (see Ch. X.) Although the expression given below, called the Moment of Inertia, comes directly from the solution of certain mechanical problems, yet its prin- ciples may be discussed without involving the idea of force^ the same as any other mathematical expression. The term probably originated from the idea that inertia was considered a force, and in most mechanical problems which give rise to the expression the moment of a force is involved. But the expres- sion is not in the form of a simple moment. If we consider a moment as the product of a quantity by an arm, it is of the form of a moment of a moment. Thus, dA being the quantity, ydA would be a moment, then considering this as a new quantity, multiplying it by y gives y-dA, which would be a moment of the moment. Since ice do not consider inertia as a force, and since all these problems may be reduced to the consideration of geometrical magnitudes, it appears that .some other term might be more appropriate. It being, how- ever, universally used, a change is undesirable unless a new and better one be universally adopted.) DEFINITIONS. 102. The expression, yy^fi? J., in which dA represents an ele- ment of a. body, and y its ordinate from an axis, occurs fre- quently in the analysis of a certain class of problems, and hence it has been found convenient to give it a special name. It is called the moment of inertia. THE MOMENT OF INEKTIA OF A BODY is the sum of the jproducts ohtained hy multijplying each element of the hody hy the square of its distance from an axis. The axis is any straight line in space from which the ordinate is measured. The_quantity dA may represent an element of a line (straight or curved), a surface (plane or curved), a volume, weight, or mass ; and hence the above definition answers for all these quantities. 166 MOSIEXT OF INEKTIA. [103.] The moment of inertia of a plane surface, wlien the axis liea in it, is called a rectangular moment; but when the axis is ]-)erpendicular to the surface it is called ^])olar moment. 103. Examples. 1, Find the moment of inertia of a rect- angle in reference to one end as an axis. Let J = the breadth, and d = the depth of the rectangle. Take the origin of coor- dinates at 0. We have dA = dydx; ; if dydx = ^ / "^fdy — ^hd^. and ^0 Pig. ho. 2. What is the moment of inertia of a rectangle in reference to an axis through the centre and parallel to one end % X Ans. -^M. 3. "Wliat is the moment of inertia of a straight line in reference to an axis through one end and perpendicular to it, the section of the line being -.considered unity? Ans. \1?. 4. Find the moment of inertia of a circle in reference to aa axis through its centre and perpendicular to its surface. ■We represent the jpolar moment of inertia by I^. Let r = the radius of the circle ; p = the radius vector ; Q = the variable angle ; then dp = one side of an elementary rectangle ; pdO = the other side ; and dA = pdpdd ; and, according to tlie definition, we have ^ = p^dpdd = ^r*. [104.] EXAMPLES. 167 5. What is the moment of inertia of a circle in refereneo to a diameter as an axis ? (SeeArticle 105.) Ans. ^Trr^. 6. AVhatis the moment of inertia of an ellipse in reference to its major axis ; a being its semi- major axis and ^, its semi-minor'^ fig. 97. Ans. ^7ra¥. 7. Find the moment of inertia of a triangle in reference to an axis through its vertex and parallel to its base. Let J be the base of the triangle, d its altitnde, and x any width parallel to the base at a point whose ordinate is ■// ; then dA = dxdy, and we have ^ = / 1^ fdijdx =--^ I fdy = lhd\ S. What is the inoment of inertia of a triangle in reference to an axis passing through its centre and parallel to the base ? Ans. -^jM^. 9. AVhat is the moment of inertia of an isosceles triangle in reference to its axis of symmetry ? Ans. ^^b^d. 10. Find the moment of inertia of a sphere in reference to a diameter as an axis. The ecpiation of the sphere will be x^ + 'if + z- = If. The moment of inertia of any section perpendicular to the axis of x will be ^tt/ ; hence for the sphere we have r= iTTf/klx = TT / {li'- arf dx = A^i?'. FORMULA OF KEDUCTION. 104. The moment of inertia of a hody, in reference to any XuiSj. equals the moment of inertia in reference to a jparallel axis passing through the centre of the hody jplus the product of the area {or volume or mass) hy the square of the distance between the axes. IGS FORMULA OF REDUCTION. [104. J This proposition for plane areas was proved in Article SO. To pi'ove it generally, let Fig. OS re- present the projection of a body npon the plane of the papei', B the projection of an axis passing through the centre of the body, A any axis parallel to it, C the projection of any element; AC — r^ BG = 7\, the angle CBB = 6, and V = the Yolnnie of the body. Then I^ = fr^cl V will be the moment of inertia of the volnme in reference to the axis through the centre ; and I —fi^dV^ the moment in reference to the axis through A. .,2 snr Let AB ^ />, then AE = B + o\ cos 6, and 1^ = + (i> + Tx COS 6f ; ... f,M V =fi\^d V + 2Bfj\ COS ed V + B'fd F. But //'i COS 6dV = 0, since it is the statical moment of the body in reference to a plane perpendicular to AB passing throngh the centre of the body and perpendicular to the plane of the paper, therefore the preceding eqnation becomes /=/i+ VB'; (119) which is called the formida of reductio7i. From this, we have I, = I- VB\ (120) Examples. 1. The moment of inertia of a rectangle in reference to one end as an axis being ^hd^, required the moment in reference to a parallel axis through the centre. Equation (120) gives I, = ^M' - bd iidf = j\M\ ■ 2. Given the moment of inertia of a triangle in reference to an axis through its vertex and parallel to the base, to find the moment relative to a parallel axis through its centre. [105. J EXAMPLES. 169 Example 7 of the preceding Article gives /= ^hd^ ; heuce equation (120) gives I, = ihd' - ¥^d {^df = ^\hd' 3. Find the moment of the same triangle in reference to the base as an axis. Equation (119) gives /= ^\M' + ^7jd i^df = -^^hd\ 105, To FIND THE RELATION BETWEEN THE MOMENTS OF IN- ERTIA IN REFEKENCE TO DIFFERENT PAIRS OF RECTANGULAR AXES HAVING THE SAME ORIGIN. Let X and y be rectangular axes, Xi and yi, also rectangular, having the same origin ; a = the angle between x and x^ ; ia = the moment of inertia relatively to the axis X, similarly for /i,,ix, and /y^; B = JxydA ; and For the transformation of coordinates we have a'l = y sin a + x cos a ; yi=^y cos a — X sin a ; K^ + 7/1^ = 3^ + y^. Also dA = dxdy = dx^dy^. Hence, /j. :=fyi dA = ^cos^ a + ly sin^ a —^B cos a sin a ; /^^ = I^ sin^ a + ly cos^ a + 27/ cos a sin a ; B-i^ = {Tx -~ J^y) '^'os a sin a + B (cos^ a — sin~ a) ; .'. 7a;j~+ ^y, = 7 X + /y =^ Vp I Fia. 09. (121; the last value of which is found fi-om the expression J'y^dA + fx^dA = f{y^ + x^) dA =jydA = 1^; which shows that the 170 MOMENT OF INERTIA, [105.] 2)olar moment equals the sum of two rectangular moments, the origin being the same. If the rectangular moments equal one another, we have Ip — 2/a: • Thus, in the circle, Ip = ^irr^. (See Ex. 4, Article 103), hence I^ — ^r*. The last of equations (121) is an isotroinc function; since the sum of the moments relatively to a pair of rectangular axeSjCquals the sum of the moments relatively to any other pair of rectangular axes having the same origin ; or, in other words, the sum of the moments of inertia relatively to a pair of rect- angular axes, is constant. To find the maximum or minimum moments we have, from the preceding equations, dio. da and z=z — {Ix — ly) COS a sin a — ^ (cos^ a — sin^ a) = ; — J^ = + (Z: — Z/ ) cos a sin a + ^ (cos~ a — sin^ a) = ; da ^ ' ■ .'. By = 0. From the first or second of these we have — 2i^ 2 cos a sin a Z — Iv cos^ a — sin^ a — tan 2a. It ma}' be shown by the ordinary tests that when /g^ is a maximum, ly^ will be a minimum, and the reverse ; hence tliere is always a pair of rectangular axes in reference to one of which the moment of inertia is greater than for any other axis, and for the other it is less. These are called j9H?^c^j?a^ axes. Thus, in the case of a rectangle, if tlie axes are parallel to the sides ajid j^ass through the centre, we find B= xydA = ; ^' J ^\d hence ■x and y are the axes for maximum and minimum moments ; and if (i > Z*, -^^^d^ is the maximum, and -^i^lj^d a minimum moment of inertia for all axes passing through the origin. In a similar way we .find that if the origin be at any [105.] EXAMPLES. 171 other point the axes must be parallel to the sides for maxirauni and minimum moments. , The preceding analysis gives the position of the axes for maximum and minimum moments, when the moments arc known in reference to any pair of rectangular axes. But if the axes for maximum and minimum moments are known as I^ and ly , then B = 0; and calling these 7^' and I^/, Eqs. (121) become J^^ z= 1^1 cos^ a + lyf sin^ a ; \ ly^ = I^, sin- a + Ly eos^ a ; >- (122) B = [Ix' — ly') COS a sin a. ) In the case of a scpiare when the axes pass through the centre I^' — ly' \ .'. I^^ — I^, (cos*^ a + sin^ a) — I^>\ ly^ — Iy>, and ^. = 0; hence the moment of inertia of a square is the same in refer- ence to all axes passing through its centre. The same is true for all regular polygons, and hence for the circle. Examples. 1. To find the moment of inertia of a rect- angle in reference to an axis through its cen- tre and inclined at an angle a to one side, we have I^ = -^KlxP and ly = Tj^^d .-. Jx^ = -i\hd {d^ cos^ a + 1)^ sin^ a) ; ly^ = -^KM {d^ sin^ a + P cos^ a). Fia^m. 2. To find the moment of inertia of an isosceles triangle in reference to an axis through its centre and inclined at an angle a to its axis of synnnetry. We have Zc = i^M^ and ly = -i-^^^d, in which h is the base and d the altitude ; .'. Ix = ^-ghd {d"^ cos^ a -f ^1/ sin^ a) ly — J-gJc? {d'^ sin^ a -f f J^ cos^ a). The moment of inertia of a regular polygon about an axia 172 MOMENT OF INERTIA. [10G.J through its centre may be found by divicliDg it into triangles having their vertices at the centre of the polygon, and for bases the sides of the j)olygon ; then finding the moments of the triangles about an axis through their centre and parallel to the given axis and reducing them to the given axis by the fvr- inula of reduction. It R be the radius of the circumscribed circle, r that of the inscribed circle, and A the ai'ea of the polygon ; then, for a regular polygon, we would find that For the circle R = r, as before found. For the square, t = ^h, R = ^hV'2, and A = ¥; 1 = as before found. -1 I* Fig. 102. 106. Examples of the moment of inertia of solids. (The following results are taken from Moslem's MccJianics and Engineering.) 1. The moment of inertia of a solid cylinder about its axis of synnnetry, r being its radius and h its heiglit, is ^ttIip*. 2. If the cylinder is hollow, c the thickness of the solid part and R the mean radius (equal to one-half the sum of the external and internal radii) then /= 27rhcR {R'' + Jc^). 3. The moment of inertia of a cylinder in reference to an axis passing tlirough its centre and perpendicular to its axis of sj'mmetry is ^irhr'^ {r^ + -JA^). 4. The moment of inertia of a rectangular paral- lelopipedon about an axis passing through its cen- tre and parallel to one of its edges. Let a be the length of the edge parallel to the axis, and b and c the lengths of the other edges, then I = -^^ aba Q}^ + 0^) = yV of the volume multiplied by the Fio. 103. square of the diagonal of the base. [107.] RADIUS OF GYRATION. 173 5. The moment of inertia of an upright triangular prism liaving an isosceles triangle for its base, in reference to a vertical axis passing through its centre of gi'avity. Let the base of the triangle be «, its altitude 5, and the altitude of the prism be A, then Fig. 10-1. G. The moment of inertia of a cone in reference to an axis of synnnetry is ^jjirr^h. {r being the radius of the base and h the altitude.) Fia. 106. Fiii. lUT. 7. The moment of inertia of a cone in reference to an axis through its centre and perpendicular to its axis of synnnetry is ^\7rrVi {.^ + y^. S. The moment of inertia of a sphere about one of its diam- eters is -^jirli,^. 9. The moment of inertia of a segment of a sphere about a diameter parallel to the plane of section. Let H be the radius of the sphere, and h the distance of the plane section from the centre, then / = -^VtT (16i?^ + li)E'h + lOIiW - W). FIG. 1.8. EADIUS OF GTKATION. 107, ^6 may conceive the mass to be concentrated at such a point that the moment of inertia in reference to any axis will be the same as for the distributed mass in reference to the same axit The radius of gyration is the distance from the moment axis to a point in which, if the entire mass be concentrated, the moment of inertia will be the same as for the distributed mass 174 EXAMPLES. [107.] The princvpal radius of gyration is the radius of gyration in reference to a moment axis through the centre of the mass. Let h = the i-adius of gyration ; Z\ = tlie principal radius of gyration ; M = the mass of the body ; and D = the distance between parallel axes ; then, according to the definitions and ecpiation (119), we have Ml^ = Xim^ .-. 7.^= h^ + i>2. (123) from which it appears that k is a minimum, for Z> = 0, in which case Jc — ki\ that is, the princiiMl radius of gyration is the minimum radius for jjarallel axes. We have . /. 2 _ ^1 . '■ ^' -TV Iience, the square of the principal radius of gyration equals the moment of inertia in reference to a moment axis through the centre of the body divided by the mass. Examples. 1. Find the principal radius of gyration of a circle in refer- ence to a rectangular axis. Example 5 of Article 103 gives, Ix = Jtt;'*, which is the moment of an area, hence, we use tt/^ for J/, and have 2. For a circle in reference to a polar axis, k^ = ^7^. 3. For a straight line in reference to a moment axis perpen- dicular to it, k^ = -^^P. 4. For a sphere, k^^ — p^^. 5. For a rectangle whose sides are respectively a and 5, in reference to an axis perpendicular to its plane, k\^=^;^ (a^+l^). 6. Find the principal radius of gyration of a cone when the moment axis is the axis of symmetry. CHAPTER YIII. MOnON OF A PAKTICLE FKEE TO MOVE IN ANT DIEECTION. 108. A free, material particle, acted upon by a system of forces which are not in equilibrium among themselves, will describe a path which will be a straight or a curved line. The direction of motion at any point of the path will coincide with that of the action-line of the resultant of all the forces which have been impressed upon the particle prior to reaching the point, which will also coincide with the tangent to the path at that point. Let ds be an element of the path described by the particle in an element of time dt ; R the resultant of the impressed forces, and 771 the mass of the particle; then, according to Article 21, we have It — m -y^ = 0. dt Let a be the angle between the action-line of the resultant R (or of the arc ds) and the axis of cc; multiplying by cos a, we have dj^s R cos a — 771 -j-^ cos a = ; in which R cos a is the x-convponent of the resultant, and according to equation (51) equals X; or, in other words, it is the projection of the line representing the resultant on the axis of X ; d^s cos a is the projection of d^s on the axis of x, and is d'x. Hence, the equation becomes, and similarly, X- d^x dt^ 0; Y- d?y 0; Z- d^z ''' df = 0; (124) 176 MOTION OF A [109.1 whicli are tlie equations for the motion of a particle along the coordinate axes ; and are also the equations for the motion of a body of finite size when the action-line of the resultant passes through the centre of the mass. They are also the equations of translation of the centre of any free mass when the forces produce both rotation and translation ; in which case m should be changed to M to represent the total mass. See Article 38. VELOCITY AND LIVING FORCE. 109. Multiplying the first of equations (12-i) by fZ.c, the second by dy^ and the third by dz, adding and reducing, give Xdx + Ydy + Zdz = ^;^^('i^-±-^|J-±^) = 1 ^^^^ and inteirrating sives / {Xdx + Ydy + Zdz) = hn -^^ = ^mv' + C. The first member is the work done by the impressed forces ; for if Ji be the resultant, and s the path, then, according to Article 25, equation (2G), the work will be /lids, and by pro- jecting this on the coordinate axes and taking their sum, vre have tlie above expression. The second member is the stored energy plus a constant. Let X, Y, Z be known functions of x, y, s, and that the terms are integrable. (It may be shown that they are always integrable when the forces act towards or from fixed centres.) Performing the integration between the limits x^, y^, Sq, and a^i, Vi, ^u we have i'Vo, Vo^ ^o) - «/> (■»!' 2/1, ^i) = hn {v\ - v\) ; (125) hence, the w^ork done by the impressed forces upon a body in passing from one point to another equals the difference of the living forces at those points. It also appears that the velocity at two points will be independent of the path described ; also, that, when the body arrives at the initial point, it will have the same velocity and the same energy that it previously had at that point. [109.1 FEEE PARTICLE. 177 Examples. 1. If a tody is projected into spaee^ and acted iipon only hy gravity and the impulse ; required the curve described hy the projectile. Take the coordinate plane xy \\\ the plane of the forces, x horizontal and y vertical, the origin being at the jwint from which the body is projected. Let W — the weiglit of the body ; V = the velocity of projection ; and a = BAx^=^ the angle of elevation at which the projection is made. "We have, X=0; Y=-mg\ Z=0; and equations (124) become di' EX Fig. lOy. 2 = 0; Integrating, observing that v cos a will be the initial velocity along the axis of x, and v sin a that along y, we have, dx dt dy dt = V cos a : = V sin a — gt', (a) and integrating again, observing that the initial spaces are zero, we have, X = vt cos a ; y ■= vt sin a — ^gf. Eliminating t from these equations, gives 11 ^ X tan a — ■—- — 5— ; (J) -^ 2 r COS'' a' ^ ' which is the equation of the common parabola, whose axis is parallel to the axis of y. 12 178 PROJECTILES. fl09.] Let A be the height through which a body must fall to acquire a velocity v, then ty^ = 2gh, and the equation (5) becomes, 7/ = X tan a — —; 5—. (c) •^ 4A cos^a ^ To find the range AE, make ?/ — in equation {J)), and we find a? = 0, and X = AE = 4/i cos a sin a = 2/i sin 2a ; (r//-. Since F is a function of r, the integral of this equation gives V in terms of r, or the velocity depends directly upon the [116,117.] FORCES. 185 distance of the body from the centre. Hence, the velocity at any two points in the orbit is independent of the path between them, the law of the force remaining the some. 116. To determine the time of descrihing any jportion of the orhit. To find it in terms of r, eliminate dd between equations (129) and (133), reduce and find r^^ dr which integrated gives the time. To find it in terms of the angle, we have from (129) t^^J^^dO; C from which r must be eliminated by means of the equation of the orbit, and the integration j)erformed in reference to 6. 117. To find the cojnponents of the force along the tangefiit and normal. Let T = the tangential component ; N = the normal component ; and resolving them parallel to x and y, we have dt^ ds ds ' Eliminating iVgives d^x ^ d^y , ^, m-j^ dx^ m -j-^ dy — Tds ; or r = x^ + r% also = 7/1^. da ds ' dt^ 186 CENTRAL FORCES. [117.] Eliminating T gives ,^ dx dhi dy d^x as dr ds dr _ mds^ idxdhi — dyd^x\ ~ d¥ds \ ^^ ■/ = m-: (141) P hence the comjponent of the force in the direction of the nor- incd is dependent entirely ttpon the velocity and radius of cur- vature. This is called the centrifugal force. It is the measure of the force which deflects the l)ody from the tangent. The force directed towards the centre is called centripetal. If w = the angular velocity described by the radius of cur- vature, then, V — po), and equation (141) becomes JV = viay'p. (142) Examples. 1. If a body on a smooth, horizontal plane is fastened to a point in the plane by means of a string, what will be the num- ber of revolutions per minute, that the tension of the string may be twice the weight of the body. 2. A body whose weight is ten pounds, revolves in a liori- zontal circle whose radius is five feet, with a velocity of forty feet per second ; required the tension of the string which holds it. (Use equation (141).) 3. Required the velocity and periodic time of a body re- volving in a circle at a distance of n radii from the earth's centre. The weight of the body on the surface being mg, at the dis- (r \ mo — I = — 2", and this is a nri n [117.] EXAMPLES. 187 measure of the force at that distance, (Use equation (141) or (139).) ^,„._(^)^, = 3.(f)' (This is substantially the problem which Sir Isaac Newton used to prove the law of Universal Gravitation. See Whewell's Inductive Sciences.) 4. A particle is projected from a given point in a given direction with a given velocity^ and moves under the action of Oj force which varies inversely as the square of the distance from the centre ; required the orhit. Let [x = the force at a unit's distance, then F =^ IJiu^, and equation (132) becomes dho Ji_ — f\ or, the first integral of which becomes by reduction de = v/^'-(»-^r in which A is an arbitrary constant, and the negative vahie of the radical is used. Integrating again, making Oq the arbitrary constant, we have - 1 -< O' 6 — 6q = cos I vvliich by reduction gives 1 (i / AO^ \ 188 ORBITS OF [117.) wliicb is the general polar equation of a conic section, the oriofin bein^ at the focus. As this is the law of Universal Gravitation, it follows that the orbits of the planets and comets are conic sections having the centre of the sun for the focus. In equal ion {a), Oq is the angle between the major axis and a line drawn throuo-h the centre of the force, and is the ° ' fi eccentricity = e', hence the equation may be written u = -^ (l + . cos (^ - ^o)). (5) The magnitude and position of the orbit will be determined from the constants which enter the equation, and these are determined by knowing the position, velocity, and direction of motion at some point in the orbit. Draw a figure to represent the orbit, and make a tangent to the curve at a point which we will consider the initial point. Let /3 be the angle between the path and the radius vector at the initial point, ^o the initial radius vector, and Vq the initial velocity; then at the initial point ^* = -5 '^ = ^' ^^^"^ = ::np= - .tt^' (^) r. * = «' -"3 = ,-S = du tion (h) e cos Uq = 1, id) du lie . f. which, combined with equation (c), gives (72 ^.Qt ^ From equation (137) = — (S sin ^0- ( — , e > 1, and the orbit is a hyperbola, 9, T^o^ ^ — , e = 1, and the orbit is a parabola. Fo^ < — , (a?o,2/o) ~g^ i^,y) ', hence, for such a system, the velocity loill he independent of the path described^ and will he dejpendent only upon the coordinates of the points ; also, the velocity toill he independent of the nor- mal pressure. 119. To find the normal pressure multiply the first of equations (144) by dy, the second by dx^ subtract, observing that d,^ + dif = d^, and we find \dxd?ii dyd'^x) ^dx ^dy ^-j- or in ds dt^ ds d(^ C ds ds ds^ \ dxdhj — dyd^x \ _ ^ dx y-^^V j_ -kt. If] '-d? \-^ ds~^ds'^^^' ^ ( iV^^xt'-rf +m^; (146) ds ds p in which p is the radius of curvature at the point. The first and second terms of the second member are the normal com- ponents of the impressed forces. The total normal pressure il20,131.J A PARTICLE. 103 loill, therefor'e, he that due to the imjpressed forces plus that due to the force necessary to deflect the lody from the tangent. The last term is called the centrifugal force, as stated in Article 117. If the body moves on the convex side of the cnrve, the last term should be subtracted from the others ; hence it might be written ± m — ; in which + belongs to movement on the r concave are, and — on the convex. 120. To find the time of movement, from equation (145), we have ds" S-'.'i7r/'-">+'"*> f J Sr, Vm ds Vy\Xdji + Ydy) + mvf (^^^) 121. To find where the particle will leave the constrain- ing curve. At that point ir= 0, which gives m — — X -§- — Y-j- : (148) p ds ds which, combined with the equation of the curve, makes known the point. If a body is subjected only to the force of gravity, we have X= in all the preceding equations. Examples. 1. A hody slides down a smooth inclined plane under the foreer of gravity ; required the formulas for the motion. Take the origin at the upper end and let the equation of the ]ilane be y = ax\ 13 194: EXAMPLES. [131. J y being positive downward. Tiien we have ^ = 0, Y = rag, dy = adx, %\ = 0, and equation (145) becomes %» = "Ifgadx - 2gax = ^gy ; (a) hence, the velocity is the same as if it fell vertically through the same height. To tind the time, equation (l-iT) gives that is, if the altitude of the jplane (y) is constant the time varies directly as the length, 8. We may also find s = t V^gy = ^g^ sin a. (c) 2. Prove that the times of descent down all chords of a ver- tical circle which pass through either extremity of a vertical diameter are the same. 3. Find the straight line from a given point to a given in- clined plane, down which a body will descend in the least time. 4. The time of descent down an inclined plane is twice that down its height ; required the inclination of the plane to the horizon. 5. At the instant a body begins to descend an inclined plane, another body is prr>jected up it with a velocity equal to the velocity which the fii-st body will have when it reaches the foot of the plane ; required the point where they will meet. 6. Two bodies slide dowi\ two inclined lines from two given points in the same vertical line to any point in a curve in the same time, the lines all being in one vertical plane ; required the equation of the curve. 7. A given weight, P, draws another weight, TT, up an in- clined plane, by means of a cord parallel to the plane ; througli what distance must P act so that the weight, TT, will move s feet after P is separated from it. risi.i SBIPLE PE^T»ULOI, 195 S. Beqidred a curve such that if it revolve with a uniform angular velocity about a vertical diameter, and a, smooth ring of infinitesimal diameter he placed upon it at any pointy it wiU iwt slide on the curve. Let CO be the angular velocity, then we have Y = — mg, X = mi» A Fig. 11-3. V = i'2g {h - y) = The equation of the arc is ds dt (a) hence drC^ dy". 196 EXAMPLES OF riBl.l But ds^ = dx^ + dy^, r dy ds = V^yy — y Considering this as negative, since for the descent the arc is a decreasing function of the lime, we have from («) r r'*' dy Tliis may be put in a form for integration by Elliptic Fnnc- tions ; but by developing it into a series, each term may be easily integrated. In this way we find by means of which the time may be approximated to, with any degree of accuracy. When the arc is very small, all the terms containing ^ will be small, and by neglecting them, we have for a complete oscillation (letting I be the length of the pendulum), T=^t = 'rrJl; (5) that is,, for very smcdl arcs the oscillations may he regarded as isochronal, or performed in the same time. . For the same place the times of vibration are directly as the square roots of the lengths of the jpendulums. For any pendulum the times of vibration vary inversely as the square roots of the force of gravity at different places. If t is constant Icr. g. 11. What is the length of a pendulum which will vibrate three times in a second ? 12. Prove that the lengths of pendulums vibrating during the [121.] CONSTRAINED MOTION 197 same time at the same place, are inversely as the square of the iiuinber of vibrations. 13. Find the time of descent of a particle down the arc of a cycloid. The differential equation of the curve referred to the vertex as an orio;in. x being horizontal and y vertical (;• being the radius of the generating circle), is 2/' — y dx = . - dy. V Iry - if Ans. IT < HL. V g The time will be the same from whatever point of the cui'\'c the motion begins, and hence, it is called tautochronal. 14:. In the simple pendulum, find the point where the tensicai of the string equals the weight of the particle. 15. A ])article is placed in a smooth tube which revolves horizontally about an axis through one end of it ; required the equation of the curve traced by the particle. The only force to impel the particle along the tube is the centrifugal force due to rotation. Letting r = the radius vector of the curve ; ro = the initial radius vector ; w = the uniform angular velocity ; we have d'r which, integrated, gives 1 / (at , — ^££-£^^>- B, = R cos L. fei=f^^^ of transla- tion. If the body be unconstrained, the motion of transla- tion of the body will be that of the centre of tlie mass. If the axis of rotation is rigid, it may be located anywhere in the body, or even without the body by considering it as rigidly connected with the body, in which case the motion of translation will be that of some point of the axis. In either case the motion of translation may be considered as resulting from a force acting directly upon the axis of i-otation, and the rotation, by a force acting at some other point. The two motions may then be considered as existing independently o£ each other. 131. Formulas for the movement of a body involving both translation and rotation. The general equations for this case are (164) and (165) in the next chapter. In this Article let the rotation be about an axis parallel tos, and the centre of the mass move in the plane xy. Resolve the forces into couples and forces applied at the origin of coordinates, as in Article 83 ; then will the third of equations (S6) be the impressed forces which produce rotation. Let R be the resultant of the forces at the origin at any instant, (see Article 84), and s be measured along the path described by the point of intersection of the axis of rotation with the plane xy. Then will the principles of Article 21, give dKv (l^y^ (156) The expression 5* {Xy — Yx) is the sum of the moments of the impressed forces = :SFa (Article 60). Transforming the second term of the last equation into polar coordinates having the same oi-igin, we have ^(^ny^,-mx-£^=^mr-^^', 204: REDUCED JIASS. [132-1 hence, the equations become (If ''^17^= ^; fPd ZFa \ (157) I I J df Xmr^' 132. Eeduced mass. A given mass may be concentrated at such a point, or in a thin annuhis, that tlie force or impulse will have the same effect upon it as if it were distributed. To accomplish this it is only necessary that Xmi^ in the second of (157) should have an equivalent value. Let Mhe the mass of the body, h the distance from the axis to the required point, then in which k is the radius of gyration, as defined in Article 107. But any other point may be assumed, and a mass determined such that the effect shall be the same. Let k be the distance to that point (or radius of the annulus), and J/i the required m-ass, then we have .-. J/1 = J/|; (158) which is called the reduced mass. Examples. 1. A prismatic lav AB^ falls through a height h^ retaining its horizontal jyosition until one end strikes a fixed ohstacle C ; required the angular velocity of the piece and the linear velo- city of the centre immediately after the imjyidse. Let M be the mass of the bar, I its ___^ length, V the velocity of the centre >a |!t W 'S. . \B at the instant of impact, and t\ the velocity of the centre immediately after impact. Consider the bodies as ! S^ perfectly non-elastic ; then will the "^^p^ effect of the impact be simply to sud- ' ^^^ ^^„ denly arrest the end A. The bar will rotate about a horizontal axis through the centre, as shown by Article 3S ; and, as shown by Articles 27 aud 38, the (132.] EXAMPLES. 205 impulse will be ^ = 21 {v — v^ ; that is, it is the cJiange of velocity at the centre multiplied hy the ?na-ss. The impact will entirely an-est the motion of the end, A, at the instant of tho impact, and hence at that instant the angular velocity of A in reference to G will be the same as G in reference to A. Equation (155) gives moTnent of impulse moment of inertia _ M {v - V,) jl 0) = 6 V — ^'l I ' But at the instant of the impact Vi = ^1(0, golving these give « = l^» ^'i = f V. We now readily find Q = iMv. To find the velocity of any point in a vertical direction at the instant of the impact, we observe that it may be considered as composed of two parts ; a linear velocity v^ downward, and a right-handed rotation. The actual velocity at A due to rota- tion will be ¥(0 = iv, which will be upward, and the linear velocity downward will be Vi = fv, hence the result will be no velocity. Similarly, the velocity at B will be |^ -f |^ = Iv. Also, for any point dis- tant X from Gy we have at the left of G ^v — (OX = fv|l — 2 v-J; and-to the right of G we have When the bar comes into a vertical position, we easily find 206 ANGULAR MOVEMENT. [132.] that A has passed below a liorizontal through C. Every point, therefore, has a progressive velocity, except the point A, at the instant of imj)act. After the impact the centre will move in the same vertical and with an accelerated velocity, while the angular velocity will remain constant. 2. Suppose that impact takes place at one-quarter the length from A, required the angular velocity. 3. At what point must the impulse be made so that the velocity of the extremity JS will be doubled at tlie instant of impact ? 4. An inextensible string is wound around a cylinder, and has its free end attached to a fixed point. The cylinder falls through a cer- tain height (not exceeding the length of the free part of the string), and at the instant of the im- pact the cord is vertical and tangent to the cylinder; all the forces being in a plane; re- Fia. 118. quired tlie angular velocity produced by the impulse, and the momentum. A71S. f - ; Q = iMv. 5. In the preceding problem, let the body be a homogeneous sphere, the string being wound around the arc of a great circle. A Q=M v: Q. A Jiomogeneous prismatio har AB, in a hoi'izontal posi- tion constrained to revolve about a vertical fixed axis A, receives a direct imjndse from a sphere whose momentum is Mv ; re- quired the angular velocity of the har. The momentum imparted to the bar will depend upon the elasticities of the two bodies. Consider them perfectly elastic. The effect of the impulse will be the same as if the mass of the bar were concentrated at the extremity of the radius of gyration ; hence an equivalent mass at the point C may be determined. Fio. 119. [133.] EXAMPLES. 307 Let J/"i = the mass of AB ; J/y = the reduced mass ; v^ = the velocity of the reduced mass after impact ; a = AC. Then, by equation (15S), the mass of the bar reduced to the point O, "will be by equation (■iO) the velocity of Jl^ after impact will be hence, the momentum imparted will be and the moment will be Mit.\ — ^^ o . \r-,^ V ; According to equation (155), we have _ tnoment of the impulse Tnoment of inertia Ma? + J/;F 2J/tj i/«2 + MJc v. This result is the same as that found by dividing equation {a) by «, as it should be. T.Suppose, in the preceding prohlem, that there is no fixed axis, hut that the hody is free to translate ; find where the im- pact mttst he made that the initial velocity at the end A shall .he zero. 208 ANGULAR MOVEMENT. [132.] Let Mv be the impulse iinjparted to the body; M]c^= the principal moment of inertia; h = the distance from the centre of the bar to ihe required point ; then moment of the impulse tnoment of inertia _Mvh_vh^ , ~ Mk^ ~ h^' ■ ^ and the movement at A in the circular arc will be vlh ^ and the initial linear movement will be vlh ^~ W' which, by the conditions of the problem, will be zero ; hence vlh _ or. A = ?^. (b) I ' The distance from A will be The bar being prismatic, l\^ = -^V^ ; .-. h + il = ^l. The result is independent of the magnitude of the impulse. From {h) we have h{il) = h'', hence, h and U are convertible, and we infer that if the im- pulse be applied at A the point of no initial motion will be at fl33.] AXIS OF SPONTAJ!fEOUS ROTATION. 209 the point given by equation (^), where the impact was previ- ously applied. 8. In the preceding j)rollem find where the impulse tnust he applied so that the point of no initial velocity shall he at a distance h' from the centre. The initial linear velocity due to the rotary movement found from {a) of the preceding example, Avill be A'oints a and h are convertihle. AXIS OF SrONT^\Js'EOUS KOTATION. 133. In the preceding problem the initial motion would have been precisely the same if there had been a fixed axis through a perpendicular to the plane of motion, and hence the initial motion may be considered as a I'otation about that axis. If a fixed axis were there it evidently would not receive any shock from the impulse. The axis ahout which a quiescent hody tends to turn at the instant that it receives an impulse is called the axis of spon- taneous rotation. 14 210 INSTANTANEOUS AXIS. ri34,135.j CENTRE OF PERCUSSION. 134. AVlaen there is a fixed axis and the body is so struck that there is no impulse on the axis, any j[>oint in the action- Une of the force is called the centre of percussion. Thus in Fig. 120, if a is the fixed axis, h will be the centre of percus- sion. It is also evident tliat, if J be a fixed object, and it be struck by the body AC, rotating about a, the axis will not veceive an impulse. AXIS OF INSTANTANEOUS ROTATION. 135. An axis through the centre of the mass, parallel to the axis of spontaneous rotation, is called the axis of instantaneous rotation. A free body rotates about this axis. In regard to the spontaneous axis, we consider that as fixed in space for the instant ; but at the same time the body really rotates about the instantaneous axis which moves in space witli the body. EXAMPLES UNDER THE PRECEDING EQUATIONS CONTINUED. 9. A horizontal uniform disc is free to revolve ahout a ver- tical axis through its centre. A man walks around on the outer edge ; required the angular distance passed over hj the man and disc when he has walked once around the circmnfer- ence. Let TF= the weight of the man ; w = the weight of the disc ; r = the radius of the disc : &>! = the angular velocity of the man in reference to a fixed line ; 0) = the angular velocity of the disc in reference to the same fixed line ; Q = the force exerted by the man against the disc ; The result will be the same whether the effort be exerted suddenly, or with a uniform acceleration, or irregularly. We will, therefore, treat it as if it were an impulse. The weights arc here used instead of the masses, for they are directly pro- 1135.1 EXAMPLES. 211 portional to each other, and it is more natural to speak of the weio-ht of a man than the mass of a man. We have inoment of the impulse moment of inertia Qr _ Wvr Tf, in a nnit of time the man arrives at the initial point of the disc, v.-e have « -|- ft)^ = Stt ; which, combined with the preceding equation, gives "^^ ~ w + 2W' If TF = 10, we have 0)1 = f TT, for the angular space passed over by the man, and 0) = I- TT, for the distance passed over by the disc. 10. In Fig. 115 let the force F he constant; required the nuniber of complete turns which the hody C will make about the axis DE in the time t. Let r = the radius of the circle passed over by F; 7\ == the distance of the centre of the body from the axis of revolution ; ki = tlie principal radius of gyration of the body in refer- ence to a moment axis parallel to J)F; k = the radius of gyration of the body in reference to the axis DF; then, according to equation (123), k^ = r,' + k,'; 212 ROTATION OF fl»5. f\nd, according to equation (152), cPd moment of forces df' Tnoment of inertia _ Fr ~ 3W ' Multiply by dt and integrate, and we have de _ Fr dt~jsm the constant being zero, for the initial quantities are zero. Multiplying again by dt, we find whieli is the angular space passed over in time t ; and the num- ber of complete rotations will be j9_ _ _Frf 277 ~ 47rJf7? ■ 11. If the ])ody were a sphere 2 feet in diameter, weighing 100 pounds, the centre of which was 5 feet from the axis ; F, a force of 25 pounds, acting at the end of a lever S feet long ; required the number of turns which it will make about the axis in 5 minutes. 12. If the data be the same as in the preceding example ; required the time necessary to make one complete turn about the axis, 13. Su2:)pose that an indefinitely thin body, whose weight is TF, rests upon the rim of a horizon- tal pulley which is perfectly fi-ee to move. A string is wound around the pulley, and passes over another pulley and has a weight, P, attached to its lower end. Supposing that there is no resistance by the pulleys or the string, required the distance passed over by P in time t. FiQ. 121 [135.1 SOLID BODIES. According to equation (152), we Lave cPd _ Pr Pg 213 W + P ^~ {W + P)r' from which it may be solved. (This is equivalent to applying tlie weight P directly to the weight IF, aa in Fig, 10, and hence we have, according to equation (21), W+P (Ps^ a de = P; but referring it to polar coordinates, we have r -— = — ^, which substituted dt- di^ ' reduces the equation directly to that in the text. ) 14-. A disc whose weight is TFis free to revolve about a horizontal axis pass- ing through its centre and perpendicular to its plane, A cord is wound around its circumference and lias a weight, P, attached to its lower end ; required the distance tlirough which P will descend in t seconds. We have Frs. 12-2 U ROTATION OP [135.1 Fig. 123. 16. Two weights, P and TF, aro suspended on two pulleys by jneans of cords, as shown in Fig. 123, the pulleys being attached to the same axis 6'. ISTo resistance being allowed for the pulleys, axle, or cords ; required the circumstances of motion. "We have (if moment of the forces moments of inertia P.AC -W.BC F{AOf + W^BUf^disGACk^-^discBCki g'-> in which disc A C, etc., are used for the weights of the discs. Let the right-hand member be represented by M. then we have d = Ulf. 17. In the pi'eceding example let the discs be of uniform density, J. 6^ = 8 inches, ^C' = 3 inches ; the weight of AC —-^ pounds, of CB = 2 pounds, of P = 25 pounds, and of TF = GO pounds ; if they start from rest, required the space passed over bv P in 10 seconds, and the tension of the cords. IS. A homogeneous^ hollow cylinder rolls doion an inclined flane hy the force of gravity ; required the time. The weight of the cylinder may be resolved into .two components, one parallel to the plane, which impels the body down it, the other nor- mal, which induces friction. Fio. 12-4. The friction acts parallel to the plane and tends to prevent the movement down it, and is assumed to be sufficient to prevent sliding. Let W = the weight of the cylinder ; i = the inclination of the plane to the horizontal ; ]V = TFcos i = the normal component ; ni — the mass of a unit : the altitude = 1 ; f 135.1 SOLID BODIES. 215 (^ = the coefficient of friction ; T = (^N = the tangential component ; T' = W sin i = tlic component of the weight parallel to the plane ; 7\ = the internal radius of the cylinder ; r = the external radius ; = the angular space passed over by the radius ; s = A C\ the space. This is a case of translation and rotation combined, and ec| nations (157) give Wdh ^ - T=: TFsin i - T: g dt' dW _ T.r _ '^gTr df ~ ^mir (/•* - i\') ~ Wi:,^ + ri') ' and from the problem s = rd. Eliminating .■? and Tfrom these equations, we get d'^d 2f/rsmi dl ^ S/M^n"" Integrating and making the initial spaces zero, we have _ gr sin ?• ^ : t =\/ — ^-^^ s. V r//-~ sm ^ 9' If ri = 0, the cylinder will be solid, and V g sin % and hence, the time is independent of the diameter of the cylinder. If /»! = r, the cylinder will be a thin annulus, and -v; g sm % hence, the time of descent will be \/% times as long as 216 THE C03IP0UND [135.1 when the cj'lincler is solid ; the weight being tlie same in both eases. If it slide down a smooth plane of the same slope, we have / 2.V V y sill 1. which is less than either of the two preceding times. THE PENDULUM. 19. Let a hody he suspended on a horizori- tal axis and moved by the force of gravity ; required the circumstances of motion. We have d^6 m^oment of forces dt^ m,o'ment of inertia Wh sin e Pio. 125 ^^"^ in which h = Oa, tlie distance from the axis of suspension to the centre of gravity a of the body ; TF = the weight of the body ; 6 = bOa; and let ^'i = the principal radius of gyration; then the preceding equation becomes gh d^d df '~ Ir + k^ sni This equation cannot be completely integrated in finite terms, but by developing sin 6 and neglecting all powers above the first, we find for a complete oscillation r = 7r n/ h^ + li^ gh (161) which gives the time in seconds when A, Jc^ and g are given in eeet. To find the length of a simple pendulum which will vibrate [130, 137.] PENDULUM. 217 ill the same time, we make equations (5), page 196, and (101) equal to one another, and have I = ^L±R = Od. (1C2) Ih Let ad = Ai, then ; — A = Ai = -T- ; 136. Definitions. A Lody of any form oscillating about a iBxed axis is called a compound j)endidum. A material particle suspended from a string without weight, oscillating about a fixed axis, is called a sim.ple pendulum. The point d is called the centre of oscillation. It is the point at which, if a particle be placed and suspended from the axis by a string without weight, it will oscillate in the same time as the body Od. Or, it is the point at which, if the entire mass be concentrated, it will oscillate about the axis in the same time as when it is distributed. The point 0, where the axis pierces the plane xy, is called the centre of susjpeixsion. 137. Results. The centres of oscillation and of percussion wincide. (See Article 134.) According to equation (163), the centres of oscillation and of suspension are convertible. According to the same equation the principal radius of gyra- tion is a mean proportional between the distances of oscillation and of suspension from tlie centre of gravity. Equation (161) indicates a practical mode of determining the principal radius of gyration. To find it, let the body oscillate, and thus find T, then attach a pair of spring balances to the lower end and bring the body to a horizontal position, and find how much the scales indicate ; knowing which, the weight of the body and the distance between the point of attachment and the centre of suspension 6>, the value of A may easily be computed. The value of g being known, all the quantities in equation (161) become known except h^, which is readily found by a solution of the equation. 218 CAPTAIN KATER'S [138.1 Examples. 1. A prismatic bar oscillates about an axis passing through one end, and perpendicular to its length ; required the length of an equivalent simple pendulum. 2. A homogeneous sphere is suspended from a point b}' means of a fine thread, find the length of a simple pendulum which will oscillate in the same time. 138. Captain Kater used the principle of the convertibility of the centres of suspension and ' W oscillation for determining the length of a simple Fig. 126. sccouds pendulum, and hence the acceleration due to gravity. — Phil. Trans., 1818. Let a body, furnished M-ith a movable weight, be provided with a point of suspension C (figure not shown), and another point on which it may vibrate, fixed as nearly as can be esti- mated in the centre of oscillation O, and in a line with the point of suspension and the centre of gravity. The oscillations of the body must now be observed when suspended from C and also when suspended from O. If the vibrarions in each posi- tion should not be equal in equal times, they may readily be made so by shifting the movable weight. AVhen this is done, the distance between the two points C'and is the length of the simple equivalent pendulum. Thus the length Z and the corresponding time T oi vibration will be found uninfluenced by any irregularity of density or figure. h\ these experiments, after numerous trials of spheres, etc., knife edges were pre- ferred as a means of sujjport. At the centres of suspension and oscillation there were two triangular apeilures to admit the knife edges on which the body rested while making its oscil- lations. Having thus the means of measuring the lengtli L with accuracy, it remains to determine the time T. This is effected by comparing the vibrations of the body with those of a clock. The time of a single vibration or of any small arbitrary number of vibrations cannot be observed directly, because this would requii-e the fraction of a second of time, as shown b}' the clock, to be estimated either by the eye or ear. The vibrations of the [138.] EXPERIMENTS. 219 body may be counted, and here there is no fraction to be esti- mated, but these vibrations "will not probably fit in with the oscillations of the clock pendulum, and the dift'erejices must be estimated. This defect is overcome by " the method of coinci- dences." Supposing the time of vibration of the clock to be a little less than that of the body, the pendulum of the clock will gain on the body, and at length at a certain vibration the two will for au instant coincide. The two pendulums will now be seen to separate, and after a time will again approach each other, when the same phenomenon will take place. If the two pendulums continue to vibrate with perfect uniformity, the number of oscillations of the pendulum of the clock in this in- terval will be an integer, and the number of oscillations of the body in the same interval will be less by one complete oscilla- tion than that of the pendulum of the clock. Hence by a simple proportion the time of a complete oscillation may be found. The coincidences were determined in the following manner : Certain marks made on the two pendulums were observed by a telescope at the lowest point of their arcs of vibration. The field of view was limited by a diaphragm to a narrow aperture across which the marks were seen to pass. At each succeeding vibration the clock pendulum follows the other more closely, and at last the clock-mark completely covers the other during their passage across the field of view of the telescope. After a few vibrations it appears again preceding the other. The time of disappearance was generally considered as the time of coincidence of the vibrations, though in strictness the mean of the times of disappearance and reappearance ought to have been taken, but the error thus produced is very small. {Encyc. 2L:'t. Figure of the Eaith.) In the experiments made in Hartan coal-pit in 1854, the Astron.omer Royal used Kater's method of observing the pendulum. {Pldl. Tra?iS.,lS56.) The value of T thus found will require several coi-rections. These are called '' Reductions." If the centre of oscillation does not describe a cycloid, allowance must be made for the altera- tion of time depending on the arc described. This is called '' the reduction to infinitely small arcs." If the point of sup- port be not absolutely fixed, another correction is required 220 A STANDARD OF LENGTH. [188. J {Phil. Trans., 1831). The effect of tlie bucijancy and the resistance of the air must also be allowed for. This is the ''reduction to a vacuum." The length X must also be cor- rected for changes of temperature. The time of an oscillation thus corrected enables us to find the value of gravity at the place of observation. A correction is now required to reduce this result to what it would have been at the level of the sea. The attraction of the intervening land must be allowed for by Dr. Young's rule {Phil. Trans..,lSld). We thus obtain the force of gravity at the level of the sea, supposing all the land above this level were cut off and the sea constrained to keep its present level. As the sea would tend in such a case to change its level, further corrections are still necessary if we wish to reduce the result to the surface of that spheroid which most nearly represents the earth. (See Camb. I^hil. Trans., vol. x.) There is another use to which the experimental determina- tion of the length of a simple equivalent pendulum may be applied. It has been adojDted as a standard of length on account of being invariable and capable at any time of recov- er}'. An Act of Parliament, 5 Geo. lY., defines the yard to contain thirty-six such parts, of which parts there are 39.1393 in the length of the pendulum vibrating seconds of mean time in the latitude of London, in vacuo, at the level of the sea, at temperature 62° F. The Commissioners, howevei*, appointed to consider the mode of restoring the standards of weight and measure which were lost by fire in 183-i, report that several elements of reduction of pendulum experiments are yet doubt- ful or erroneous, so that the results of a convertible pendulum are not so trustworthy as to serve for supplying a standard for length ; and they recounnend a material standard, the distance, namely, between two marks on a certain bar of metal under given circumstances, in preference to any standard derived from measuring phenomena in nature. {Report, 1841.) All nations, practically, use this simple mode of determining the length of tlie standard of measure, that of placing two marks on a bar, and by a legal enactment declaring it to be a certain leno;th. [109, 140.] ELLIPTICITY OF THE EARTH. 221 139. T'oEM OF THE Eaktii. The pendulum furnishes one of the l)est means for determining the form of the earth. Let a ■= the equatorial radius of the earth ; Tj = the semi-axis ; e = the ellijiticity of the earth ; then a — h € = . a Let 7?i = ratio of the centrifugal force at the equator to the force of gravity at the same j^lace ; /q = length of a second's pendulum at the equator ; Zgg = the length of a second's pendulum at the poles ; then, from the Mecanique Celeste, tome II., Xo. 34, we have ^ _ 5 -- ^90 ~ ^^ ra — The value of vi is ^j^- The formula for the length of the second's pendulum when the length at Paris is taken as unity, is ?= 0.996S23 + 0.00549745 sin^iJ, when ^is the latitude of the place. See Puissant's Traiie de Geodesic, page 461. By this means it has been found that e is about -^\j. P>ow- dich, in his translation of the Mecanique Celeste, p. 485, re- marks, " It appears that the oblateness (e) does not differ much from 3^, and may possibly be a little more, though some results give a little less." 140. Torsion Pendulum. If an elastic bar, CD, be fixed at one end, and at the other end have in/mm/'/m/. two weights, A^ and A-^, rigidly fixed to it by means of the cross arm, A^ A^, then if the arm be turned into the position J^^B-i, the elastic resistance of the bar 5,.-:% z;-.^ Z^C' will cause the weights to move i)ack "■-••. to yli^2) 3-nd by virtue of the. energy of ' ...---'c'--.,. the weights at that point, they will pass 2?''--' ^■■■'B that position, and move on until their ' fio. i2t. motion is arrested by the action of the elastic resistance of the 223 VIBRATIONS OF A [111.] bar ; after which the}' will return to their former position, thus having a motion similar to that of the common pendulum. This arrangement is called a torsion iJCRdiilum. The motion will be the same for one weight as for two, but when the bar DC is vertical, the arms 6Vii and CA^ should equal each other, and the weight ^1^ equal A-^. 141. To find the force neceasary to twist the 7vd DC through a given angle. Let F=^ the force at A^ perpendicular to the arm CA„ ; a -^ CA^= CA,; l = DC; a = AiCDi ; /= the polar moment of inertia of a trans- verse section of the bar DC; and Q = the coefficient of the elastic resistance to torsion. The moment of the twisting force, K, will be M ; and the moment of the elastic resistance will be (see Mesistance of Materials, 2d Ed., p. 206), Gl^^; hence, we have Fa = GIp .: F= Gl\ at The weights A^ and A^ are not involved in this problem. If the ano-le be measured from some fixed line making an angle ^ with the neutral position of A^A^^ then instead of a we would have a^ — (/>, and the last equation becomes Fa G f , , If the force be reversed, it will twist the bar in the opposite [142. J TORSION PENDULU3I. 223 diroction, making an angle a^ with the fixed line of reference, and we would have Fa G , , . Adding these equations gives 2 -^ = J (oi — oo). 142. To find the time of an oscillation. The bar CD having been twisted by moving the bar from its normal position A^An,/^-mz,^) = ^Zy,-XYz, Dropping X before X, Y, Z, and letting those letters repre- sent the total axial components upon tlie entire body, and Zy-i^ — Yz^,^ etc., the resultant moments of the applied forces, we have the six following equations : iff = X; ^^ df ^ ' M^ — 7 ■ (167) ^Lnx, -Ji - my, -^) = Yx, - Xy,. (168) '\^ di' de Equations (167) do not contain the coordinates of the point of application of the forces, hence, the motion of translation of the centre of a m-ass is independent of the point of apj>lica- tion of the force or forces / or^ iii other words, it is indejpen- dent of the rotation of the mass. Equations (168) do not contain the coordinates of the centre of the mass, and being the equations for rotation, show that the rotation of a mass is indejpendent of the translation of itn, centre. [149, 150.] MOTION. 23D These equations are sufficient for determining all tlie circum- stances of motion of a free solid. In their further use the dashes and subscripts will be omitted. 149. If X, Y, Z, are zero, we have and similarly for the others. Transposing, squaring, adding and extracting the square root, give ^ = V ±2 = VC\'+ a + 6'^; (169) which, being constant, shows that the motion of the centre of the mass is rectilinear and uniform. This is the general principle of the conservation of the CENTKE or GRAVITY. conservation of areas. 150. The expression, xdy — ydx, is, according to Article 112, twice the sectoral area passed over by the radius vector of the body in an instant of time. Hence, if ^{rnxdy — Tnydx) = dAi ; differentiating, we iind ^/ d^i/ d-x\ d-Ax 2{r,ix ^, - my -^,) = ^ • cl J* If there are no accelerating forces m -y-^ = ; and similarly for the others ; hence " ^^-0 ^2-0. ^^3-0. dt' " ' df ~ ' ' dir ~ ' .-. Ai = Cit ; A^ = c^t ; A3 — c^t ; (170) 1.5 236 COXSERVATION OF [151.; the initial values being zero. These are the projections on the coordhiate planes of the areas swept over by the raclins vector of the body. They establish the principle of the conservation OF AREAS. That is, In amj system of hodies, moving without accelerating forces and having only mutual actions upon each other, the jprojections on any plane of the areas swept over hy the radius vector are proportional to the times. CONSERVATION OF ENERGY. 151. Multiply equations (167) by dx, dy, dz, respectively, add and integrate, and we have Jfy2 _ j/y^2 ^ ^f{Xdx + Ydy + Zdz) ; and for a system of bodies, we have k-Xi^Mv") - iX(JAv) = Xf{Xdx + Ydy + Zdz). (171) The second member is integrable when the forces are directed towards fixed centres and is a function of the distances between them. Let «, 5, G be the coordinates of one centre, «!, ^1, Ci, of another, etc. ; X, y, s, the coordinates of the particle m ; r, ?'!, etc., be the distances of the particle from the res- pective centres ; 1\ Fi, etc., be the forces directed towards the respective centres ; then, resolving the forces parallel to the axes, we have X^= F cos a ■\- Fi cos a + etc. = F + F^ + etc. ; r r r=^^^ + /;^^^ + etc.-, r r Z = F^^^ + F,^-^^^ + QtQ. [151.] ENERGY. 237 Multiplying by dx^ dy, dz, respectively, and adding, we have Xdx + Ydy + Zdz = f\ ^^-=^dx + ^-^^dy + '-^^dz \ + Fi ] dx + '-dy + dz \ + etc. But r' = {a- xf + {h -yf + {c - zf ; (172) and by differentiating, we find _ a — x 'b -y -. c — 3 dr = — dx — dy — ■ -dz similarly, «i - X Ji - ?/ 7 ^ 1 -g ^ dr\ = — dx -d'j dz ; etc., etc., etc. These substituted above, give Xdx + Ydy + Zdz = - Fdr - F,dt\ - F,d)\ - etc. Therefore, if F^ etc., is a function of r^ etc., and /i, yu-i, etc., the intensities of the respective forces at a distance unity from the respective centres, or F=i.^{r); F, = fj^i(f>{r,) ; etc., etc. ; the second member, and hence the first, will be integrable. In nature, if a particle ??i attracts a particle nii, the particle mi will attract w^, each l)eing a. centre of force in reference to the other, and hofh centres will be movable in reference to a fixed origin. But one centre may be considered fixed in ref- erence to the other, and consequently the proposition remains true for this case. The second member of equation (171) being integrated be- tween the limits «, ?/, z and a^g, ?/(,, Sq, we have i^(J/r)- i^(i/vo-) = ^}^9 ('^, y, 5) ~^}^V O^o, 2/0, ^-o)- (173) Hence, the gain or loss of energy of a, system, suhject to forces directed towards fixed centres and lohich are functions of the distances from those centres^ is independent of the jpath 288 CONSERVATION OF [151.] desd'ibed hi/ the hodiea, and dejpends only upon the2)osition left ami arrived at hy the hodles, and the intensities of the forces at a unifs distance from the respective centres. Therefore, when the system returns to the initial position, or to a condition equivalent to the original one, the vis viva will be the same. From equation (171) we have The gain or loss of energy in passing from one j^ositlon to another equals the loork done hy the impressed forces. Let TFq = the work done bj the impressed forces in passing from some definite point (.^05 ^o? ■^'oO ^o an- other definite point (.rj, _?/i, ,ci,); and W = the work done by the same forces in passing from tlie first point to any point {x, y, 2,) ; then from equation (171), we have i^illvo') - i^iJIv') = W; and subtracting the latter from the former, gives miM) - i^iMv,') = TTo - TF; or, by transposing, i2{Mv') + W= i:^( JA'i^) + TTo ; in which the second member is constant. The first term of the first member is the kinetic energy which the system has at any point of the path, and the second term is the work which has been done by the forces upon the body and has become latent, or potential ; hence, in such a system the stem of the kinetic and potenticd energies is con- stant. This is the principle of the conservation of energy in theoretical mechanics. This term has been extended so as to include the principle of the transmxitation of energy as estab- lished by physical science. If a portion of the universe, as the Solar System for in- stance, be separated from all external forces, the sum of the kinetic and potential energies will remain constant, so that if the kinetic energy diminishes, the potential increases, and the [151.] ENERGY. 239 converse. If external forces act, the potential and kinetic energies may both be increased. To be more specific, suppose that the earth and snn consti- tute the system, the sun being considered the centre of the force. The velocity of the earth will be greatest when nearest the sun, and will diminish as it recedes from it. While reced- ing, the amount of work done against the attractive force of the sun will be \Mv' - ^Mv^ = - TF; in which J/ is the mass of the earth, Vi the maximum velocity, and V the velocity at any point. The second member is nega- tive because the first member is. When the earth is approaching the sun the velocity is in- creased and the living force is restored, and the kinetic added to the potential energy is constant. Again, if a body whose weight is W be raised a height h, the work which has been done to raise it to that point is TFA, and in that position its potential energy is Wh. If it falls freely through that height it will acquire a velocity v= V 2yA, .-. W/i = ^v' = ^Mv" 2c/ which is the kinetic energy. If the same body fall through a portion of the height, say Ai, its kinetic energy will be TF%i = i-3^^i^, and the work which is still due to its position is W{h—hi), which, at that instant, is inert or potential. It is found, however, that in the use of machines or other devices, by which work is transmitted from one body to another, all the work stored in a moving body cannot be utilized. Thus, in the impact of non-elastic bodies there is always a loss of living force. (See Article 32.) This, so far as theoretical mechanics is concerned, is a loss, and is treated as such, and until modern Physical Science established the correlation of forces it was supposed to be entirely lost. But we know that in the case of impact heat is developed, and Joule determined a definite relation between the quantity of heat and the work necessary to produce it, and called the result the mechanical equivalent of heat. Further 240 ROTARY MOTIONS. [152.1 investigations show that in every ease of a supposed loss of energy, it may be accounted for in a general ^vay by the appearance of energy in some other form. It is impossible to trace the transformation of energy as it appears in mechanical action, friction, heat, light, electricity, megnetism, etc., and prove by direct measurements that the sum total at e\ery instant and with every transformation remains rigidly con- stant; but by means of careful observations and measurements nearly all the energy in a variety of cases has been traced from one mode of action to another, and the small fraction which was apparently lost could be accounted for by the imperfec- tions of the apparatus, or in some other satisfactory manner ; until at last the principle of the conservation of energy is recognized to be a law as universal as that of the law of gravi- tation. The exact nature of molecular energy which manifests itself in heat, chemical affinity, etc., are unknown, but, accord- ing to the general law, all energy whether molecular or of finite masses, is either kinetic ox potential, COMPOSITION OF KOTAKY MOTIONS. 152. Take the origin of coordinates at the fixed point when there is one, or at any point of the axis of rotation when the body is free. Conceive three cones having a com- mon vertex at the origin of co- ordinates and each taugent re- spectively to the coordinate planes; then will the angular velocity of that element of the cone which for the instant is in contact with the plane xy, considered as rotating about the axis of z, be defined as the anguhir velocity of tlie hody about that axis ; and similarh" foi- '2/5 &7^, be the angular velocities of the hody about the respective axes x, _?/, s. If ca, considered as infinitesimal, be the actual velocity of a particle in the plane xy, positive from x towards y, p = Oc, ca = po^^', then cd = poj^ cos (180° — acd) = — cog • p sin J^Oc = — ojgy, da = poog sin (180° — acd) = cjg • p cos XOe = cogX. Fig. 127. the other axes. Let go^, cj^ dt~-"^ a,.2/; o^a; — &)a^3; |- , (174) [152.] ROTARY MOTION. 24-3 Similarly in regard to the axis of y, we have — WyX, WyZ, and in regard to x, — oi^, (^xV- AVhen all these rotations take place at the same time, we have, by adding the corresponding velocities, the several velo- cities along the axes dx 1 'dt dy_ dt dz — = w^y- o3yX. The particles on the instantaneous axis have no velocity in reference to the movable origin, hence dt ' dt ' dt ' .-. onyZ — (Ozy = 0, w^x — Q)xZ = 0, o)j,y — (OyX = ; (175) which are the equations of a straight liiie through the origin, and are the equations of the instantaneous axis. Let a, /3, 7, be the angles which it makes with the axes x, y, 2, respectively, then {Anal. Geom.), COS a = / 2 , 2 I ~i ■> ^^^ ^ cos 7 ,2 * To determine the angular velocity of the body, take any point in a plane perpendicular to the instantaneous axis. Let the point -^ on the axis of x, and from it erect a perpendicular to the instantaneous axis, and we have / 5 — / <«>?/+<»/ p = x&ma = xV 1 — cos'' a = \ / — 2~. !'■ — 2^- 24^2 ROTARY MOTIONS. [153, 154.] For this point y=0 and s = in equations (174), and \ve find for the actual velocity, V — -4 = « r wj,~ + w^ ; and hence (176) which represents the diagonal of a rectangular parallelopipe- don, of wliich the sides are tu^., 03y, w^. 153. Moments of rotation of the centre of the mass ahout the fixed axes. ' Multiply the second of equations (167) by s, the third by y, subtract the former from the latter, and we have J^(yf d^2 _ cry\ ^_ df-'d¥) = ^^ df " dt^ Treating the equations two and two in this manner, dropping the dashes, and substituting Z^, Mi, iYi, for the second mem- bers, we have dh d^y^ ^,/ dh d^y\ J- ^\y-di^-'w=^^ (177) These equations are of the ^^meform as equations (168). MOTION OF A BODY DUEING ESIPACT. 154. Ifotion of the centre of the 7nasses. The second mem- bers of equations (167) are the accelerating forces. If any two of the bodies collide, they being free in other respects, the action of one body upon the other is equal and opposite to that of the latter upon the former; hence, in regard to the system they neutralize each other, and the motion of the centre of the masses will be unaffected by the collision. If there are no [155, 156.] CONSTRAINED MOTION. 2^3 accelerating forces the velocity of the centre will be uniform and in a straight line, as shown in Article 140. To find the velocity of the bodies after impact requires a knowledge of their physical constitution. See Articles 28 and 29. 155. The motion of rotation of the centre of the entire mass about the origin will also be unaffected by the collision, when the bodies are acted upon by accelerating forces; for, the mo- ments of the forces due to the collision will neutralize each other, and the second members of equations (177) will contain only the applied forces. This would be illustrated by the impact of two asteroids, or in the bursting of a primary planet. But the rotation produced in each body about the centre of its mass depends upon the moments of the forces applied to the body, and hence, upon the moment of the momentum produced by the impact, CONSTEAINED MOTION. 156. General equations of rotation about a fixed point. Take the origin of coordinates at the fixed point. For this case equations (164) vanish. Substitute in (1G5), the values of -3-3, etc., obtained from (174), and reduce. We have d?x dwy dwz df ^ ^ ~df ~ ydt '^ '"i'Ky - «y«) - <»e(«.« - «.~^) ; and similarly for the others. Let Z, J/, iT, be substituted for the last term respectively of equations (165), and substituting the above values in tJie last of these equations, we find ~ ^m{x^ + y') + w^wy ^m{x^ — if) - (o),* ~ «/) Xmxy - i^ -f ^,0),) ^myz )- = iVi (178) 244 PRINCIPAL • [157. J Tlie other two equations may be treated in the same manner. But they are too complicated to be of nse. Since the position of the axes is arbitrary, let them be so chosen that "Zmxy = 0, ^mxs = 0, ^myz = ; (-1-79) in which case tlie axes are called principal axes / and we will show in the next article, that, for every point of a Ijody, there are at least thr(3e principal axes, each of which is perpendicular to the plane of the other two. Let a?!, 2/i, Sj, be the principal axes, having the same origin as the fixed axes, and A = Sm{i/c + Zi^), the moment of inertia of the body about Xi; B = Zmiz{ + x^), moment about y^ ; C = I^m{x^-\-yx), moment about % ; also let &>i, 0)2, 6)3, be the angular velocities about the respective axes Xx, y^-, and ^i, and substituting these several quantities in (17S), we have B^ +(^- C) 0,30,1 = J/; (7^' + (J?-^)o,,o,3= iV^. (ISO) These are called Euler's Equations. The origin of coordinates may be taken at the centre of the mass, and as tlie I'otation about that point is the same whether that point be at rest or in motion, as shown at the bottom of jDage 23-lr, equations (180) are applicable to the rotation of a free body when acted upon by forces in any manner. PKINCIPAL AXES. 157. At every point of a hody there are at least three prin- cipal axes perpendicular to each other. Wlien three axes meeting at a point in a body are perpen- dicular to each other, and so taken that Xmxy = 0, Xmyz = 0, Smsx = ; they are called Principal Axes. 1157.] AXES. 245 Fig. 12S. The planes containing the principal axes are called Principal Planes. 'The moments of inertia in reference to the principal axes at any point are called the Principal Moments of Inertia for that ]>oint. Let OI^hQ any line drawn through the origin, making angles a, /3, 7, with the respective coordinate axes. First find the moment of inertia about the line OJy. From any point of the line OJV. erect a perpendicular, JVF. The coordi- nates of P will be X, fj, z. Hence we have 6»p2 = a;M-2/'+s'; ON = X cos a + y cos yS -f 2 cos 7 ; , 1 = (;os*a. + cos^/3 + cos^. The moment of inertia of the body in reference to ON^ will be I z= -2.711. PlSf^ = 2m (OP- - ON-) = 2m ■{ {x- + p'^ + z'^ — {X cos a + ycos /3 + 2 cos yf }■ = 2m -( {x"+y- + 2-) (cos-a + cos-/3 + 003-7)— (a; cos o + y cos &+z cos 7)- }■ = 2m (g- + 2-) cos'^o + 2m {x^^ + 2-) cos-/3 + 2m (a;^ + f-) 003-7 — 22m?/2 cos /3 cos 7 — 22,mzx cos 7 cos a — 22mxi/ cos a cos ^ =.1 cos-o+Z?cos-/3 + COS-7— 2Z)cos /3 cos 7— 3£'cos 7 cos 0—2^^^003 a cos ^ ; in which A,B, C,hiiye the values given on page 24i, and B, E,F, are written for the corresponding factors of the preceding equation. This may be illustrated geometrically. Conceive a radius vector, r, to move about in space in such a manner that for all angles a, /3, 7, corresponding to those of the line ON, the square of the leno;th shall be inversely proportional to the moment of inertia of the bodv. Then /: T in which c is a constant. Hence, the polar equation of the locus is - =i4 cDs'-a + Z?cos-/3-(- Ccos'7— 32? cos /3 cos 7-2£'cos 7COS a— 2Fcos a cos p. 246 PRINCIPAL [158.] Multiplying by t^, we have c = Ax" + Bf + Oz- - 2D(/2 - "lEzx - 2Fxy ; which is the equation of the locus referred to rectangular coor- dinates, and is a quadric. Since ^,^, and C' are essentially positive, it is the equation of an ellipsoid, and is called the momental ellipsoid. Therefore, the moment of inertia about every line which passes through any point of a body may be represented by the radius vector of a certain ellipsoid. But every ellipsoid has at least three principal diameters, hence eveiy material system has, at evei-y point of it, at least three principal axes. If the ellipsoid be referred to its principal diameters the coefficients of ys^ zx, xy^ vanish, and the equation of the ellipsoid becomes c = ^ar^ + Bf + Cz\ In many cases the principal diameters may be determined by inspection. Thus, in a sphere every diameter is a principal axis. In an ellipsoid the three axes are principal axes. In all surfaces of revolution, the axis of revolution is a principal axis, and any two lines perpendicular to each other and to the axis of revolution are the other two principal axes. 158. Ifct hody revolve ahout one of the prinovpal axesjxfss- ing through the centre of gramty of the hody, that axis will silver no strain from the centrifagal force. Let s be a principal axis, about which the body rotates. The centrifugal force of any particle will be maP'p = mfi? Vx^ + y^ ; which, resolved parallel to x and y, gives moy^x, m^xy, ^mv?yx ; [159, 160.] AXES. 247 but these, according to equations (179), are zero. If the body be free and revolves about this axis it will continue to revolve about it. For this reason it is called an axis of permanent ro- tation. If the hody he free, and the initial rotation be not about a principal axis, the centrifugal force will cause the instantane- ous axis to change constantly, and it will never rotate about the jperinanent axis. If, therefore, we observe that a free body revolves about an axis for a short time, we infer that it revolved about it from the beginning of the motion. RELATION BETWEEN" THE AXES .T, 7/, Z.^ FIXED IN SPACE AND THE PRINCIPAL AXES X\. ?/i, ^-j, FIXED IN THE BODV. 159, If the body be free, take the origin of coordinates 0, Fig. 129, at the centre of the mass ; but if there be a fixed point about which rotation is forced to take place, take the origin O at that point. Conceive a sphere, radius unitj-, hav- ing its centre also at 0. The line PP' will be tlie intersec- tion of the planes xy and x^yx^ and P one of the points where it pierces the surface of the sphere. Positive angles will be deterniined in substantially the same manner as positive moments* described in Article 5-i, page 80; thus, positive rotation will be from + x towards + y, from + y towards 4- 2, and from + z towards + x\ and the opposite directions will be negative. In the following iigure the rota- tions are all positive, and the angles or amounts of rotation are represented as less than 90°. In passing from one system of rectangular axes to another, * It may be observed that the relations of the axes a;, y, and 2, in the follow- ing figures are not the same as on the preceding pages. Thus, for instance, on pages 70 and 111 the axis of y is vertically upward, while in the following figures z will be in that'position. While the author prefers the former arrange- ment, for the reason that the axes x and y would then retain the same rela- tive position in the printed page as is most common when only two axes are used ; yet for the sake of conforming with the usage of most writers, and for greater e, Eliminating a?', y', z, x", y", 2", be- tween these equations gives X = (cos (p cos ^' — sin cp sin ^ cos 6) Xi + (— sin cp cos rp — cos 9? sin tp cos ^) ?/i + sin rp sin ^ . z^ y = (cos q) sin ^' + sin q) cos ^ cos ^ Xi + (— sin

cos ip cos ^) yi — cos ^' sin /^. Zi z = sin ^ sin ^ . a'l + cos dt ^ dt ^ CO, = ^ co&ZOY, + '^cosZ,Or^ ^ dt dt 4- -r cos POYi + cjj cos X^O Yi + cjg cos Z^OY^ at dip . n dd . = -J- cos (p sm t> y- sill ^ eft at CJg = -z— cos ZOZi -j- -j^- cos ZiOZi dt it rk- -^ cos POZy + a?! cos XxOZy + ci^a cos Y^OZ^^ dt d4> n dq) .(186) 252 SPONTANEOUS AXIS. [161.] In a similar manner the values of -7-, -rr, -ir 5 Ji^f^Y be found, (It (It ^ (It -' ' but they may be as readily found by elimination. Eliminat- ing among (186) gives : dt = Gi?ji cos

d^ &?! sin q} + GJ2 cos

than r, according as v is <, =, or > than rco. Only one line, equa- tions (194), fulfils the condition of a spontaneous axis. KELATIOXS BETWEEN THE SPONTAKEOUS AXIS OF ROTATION, THE CENTRAL AXIS AND THE LINE OF ACTION OF THE RESULTANT. 163. Observing that x, y, 2, equations (174), are coordinates in reference to the centre of a free body as an origin (the sub- scripts having been dropped), and hence are the same as Xi, 3/1, 01, in equations (192), we have, by substituting the values of the former in the latter, dx " dy GOyXl ^.y^ dz It (195) which are the equations to the axis of spontaneous rotation^ the origin of coordinates being at the centre of the mass. The 256 RELATIONS BETWEEN THE [163.] equations to the axis of iiistantaneous rotation or central axis, in the same notation are, equations (175), (restoring the sub- scripts), oD.Xi = Go^si [ . (19G) A comparison of equations (195) and (196) shows that tliese lines are parallel ; hence. The spontaneous axis of rotation is parallel to t/ie instanta- neous {or central) axis. Letting V be the velocity of the centre of the mass, a, 5, f, the angles between the line of motion and the axes a?, y, s, re- spectively, we have ~y- = Vcosa, ~ = 1 cosi, -T, — y cos c. (197) at at at ^ Substituting these in (195), we find for the distance be- tween the central axis (196) and the axis of spontaneous rota- tion (195), {Qoordinite Geometry, Appendix III.), omitting numeral subscripts. , _ F / (c£ix-\-ciay)cQ^ a +2ooxGOz cos a cose + (oo^ + a>z)cos^c f" cox + a)y + GO, Let the axis of y be the axis of instantaneous rotation, then will oox = 0, oog = 0, and we have, dropping the subscript, h, = — , (199) GO * This may also be written _ , . (ttj^ + Gol) cos" a + 2oox: QOy cos a cos 6 + {ooy + ool) cos " & ^zy - 00^ + 0!>y-{- GO V /C^x + ^«) cos'^ft + 2oOy 00^ cos 6 cose + (o^^ -+- t»|) cos^ e [163.] SPONTANEOUS AND INSTANTANEOUS AXES. 257 a result readily deduced from Fig. 133, since ultimately we ^vould have CC' = V = Ca.co; lience Ca = h^ =V -^ co. If the angular rotation and velocity of the centre both be imiform, hi (198) will be constant ; and it is also evident that the linear and rotary velocities may vary proportionately in such a way as to make h-^ constant. This is readily seen in the more simple case of equation (199), and illustrated by ex- amples 3 and 4, p. 20G. If any number of forces, I\, K, Fz, etc., act upon a body producing both a translation of the centre of the mass and ro- tation about that centre, they will be equivalent to a single force R applied at some point of the body ; for we have (Eqs. (85), (87), and (86)), '2F cos «r — X = 7? cos « "] JSi^cos /5 = Y= Rzo?>l \\ (200) I 'S.F cos X = Z = It cos c J .-. i?= VX' + Y- + Z'] (201) from which R becomes known, and hence «, I, e, equations (200), also become known. To find the line of action of Ji; since the moments of the separate forces are known, 2F cos y .iji— 2F cos ft .z^, = L (say) becomes known, and similarly for the others ; hence Zyy - J\ = Z A\ - Zxi = M (202) where Xi, yi, Zi, are the coordinates of the line of action of i? in reference to the centre of the mass as an origin ; hence (202) are the equations to the line of action of the resultant. The third of (202) is a consequence of the other two. Since any point in the line of action of a force may be taken as its point of application, the point (ccj, ?/i, z-^) may be considered as 17 258 SPONTANEOUS AND [163.] the point of application of the resultant. To find where the re- sultant pierces the plane 3/2, make iCi = in (202), and we have, ]^ M yi = - 2, 2:, = ^ ; (203) and similarly for the other planes. If the forces reduce to a couple, we have i? = 0, and there will be no motion of the centre produced by this system of forces, although the centre may have a motion due to initial conditions. In this case the left members of (202) all reduce to zero, and the point of application will be found to be infi- nitely distant. The right members for statical equilibrium would also be zero, but if Z, J!/, iV, one or all, have finite values, they will produce rotation, and must be placed equal to the left members of (168). The inclination, ^, between the line of the resultant (202) and the central axis (190), will be {^Coordinate Geometry^ Eq. (6), Art. 199), dropping the numerical subscripts, XoOj. + Yo^y + Zco, cos d = ^ ' (204) li 12 = 0, Equation (201), cos ^ = — , which is indetermi- nate. If the forces are parallel to the axis of a?, then Y = 0, Z — 0, oj^^O, and cos ^ =. ; that is, the action line of the resultant will be perpendicular to the central axis, and hence also perpendicular to the axis of spontaneous rotation. The shortest distance between the ac- tion line of the force, equations (202), and the spontaneous axis, equations (195) and (197), will be (Appendix III., Coor- dinate Geometry)^ dropping the numerical subscripts, )\y z%)^\z - A ^\ J (205) -y/ {Xoo^ — Zo^xf + ( Yoog — Zooyf: -f- {Xooy — F(»s)2 [163.] IXSTAXTAXEOUS AXES. 259 As a special case, let there be a single force (or forces having a single resultant) acting parallel to the axis of a? and at such a point as to produce rotation about the axis of y only, and let this be a principal axis, then we have : x= ^, r = 0, ^ =z 0, Z = 0, a = 0, I c = 90", cjj. = 0, cjy = ctj, cj. = ; and (205) becomes, omitting subscripts, 90°, B Fig. 134. I M R + and (204:) gives cos z? = ; F . 6" = 90*^ (206) hence the spontaneous axis will be perpendicular to the action line of the force, will lie in the plane yz^ and be at a distance from the action line equal to the value given by equation (206). The first three conditions immediately preceding equation (206), in equations (167) give, after integrating once, R being considered constant. in-^, =1/1^= Rt at dy _ dt ~ dz ^ ; (206a) where ?/?, the mass, is used to distinguish it from J/, the mo* ment, in (206) above. The velocities i\ and c^ are constant, and are unaffected by the force or impulse R. If c'l and c-.. are not zero, Fwill not be the actual velocity of the centre, but will be the velocity produced by the resultant it. From the first of (206a) and (199) we have i? = 711 V mhxGO t t (207) 260 SPONTANEOUS AXIS. [163.] In equations (180) we will have G?! ~ 0, cl>3 = 0, B = 2m. (x^ + 2") = mh^ ; and, considering the moment of the forces as coiistant^we liave by integrating — J— = M; (208) and hence (206) beeonraes In Fig. 134, I) being the point of action of the force in reference to the body AB, and a the projection of the axis of spontaneous rotation, then I = ab, h^ = Oa. Let Ob — h^ then from (209) we Imve .\hji^= Af ; (210) and since ^'i is constant, it follows that in the plane containing the action line of the force and the centre of the mass, the spontaneous axis and foint of aiyplication of the force are con- vertible. In other words, if a be the point where the spon- taneous axis pierces the principal plane xz when the force is applied at b in the same plane, then if the force be applied at a the spontaneous axis will pass through h (See also Prob- lem 8, page 209). Although, in establishing equation (206), we have as- sumed a constant force applied at some point in one of the principal planes of the body, yet these conditions are not always practicable. Examj)les 4 and 5, page 206, illustrate the case when the forces are constant and parallel. Eve nin these cases it is necessary also to assume that the body has no initial velocity, or if it has it must be the same as might have [163.] MOMENTS OF MOMENTUM. 261 been produced by the action of the force, or forces, under the conditions above imposed. But all these equations are applicable to the case of instan- taneous forces, or, in other words, of impact, the initial veloci- ties either being zero or abstracted from the above conditions. The equation to the line passing through the origin of coordinates — which is at the centre of the mass — and the point where the action line pierces the plane ?/.?, equations (203), will be found by dividing one equation of (203) by the other, and is — restoring the subscripts — (211) Let ff be the angle between this line and the axis of spon- taneous rotation, and we have (195), (211), (See Appendix III. of Coordinate Geometry), cos V =■ CO.^Mi — G7j/^iVi Vo^xl + (^h + ^H ■ V.MI + ^\"l (212) For the case of impact this may be reduced to a form which will lead to an interesting general result. Since this condi- tion excludes accelerating forces, X = 0, etc., in (107) and (168), and integrating once, we have (213) '"'^'dt -'"'' Tt d,i\ dz. mz, ^ - mx, ^^^ = A ^=MA\ (21^) ,/ dy dxA where Zj, J/"i, iVi are the moments of the momentum, as shown by the left members of the equations. Substituting in (214), 262 ROTATION. [163.] for -jT^ etc., their values from (174) (since (1T4) are equally true for the origin at the centre of the mass), and making the axes principal ones by the conditions of equations (179), we finally have _ A _ A 1 COv, = ^ il/l COs, =^ iVi - ^ 1 -2m{xi +y\) a J (215) Substituting in (212) the values of a?:^, oj^^, GOy^, f^'o™ (215) gives B cos 0-' i/5^-j/(My-(^ST-- (216) If, in (216), B = C, cos 0' = ; tliat is if two of the prin- cipal moments of inertia equal each other, the spontaneous axis of rotation will be perpendicular to the line joining the centre of the mass and the point where the action line pierces the plane of those moment axes. If the line of impact be parallel to the axis of x, Zi, equa- tions (215), will be zero, which gives oo^^ = 0, and since ^ and— will also be zero, these in (195) show that the axis (U ' at of spontaneous rotation will be in the plane yz. If now B = C, the axis of spontaneous rotation will lie in the plane of the moment axes of B and C, and be perpendicular to the line drawn from the centre of the mass normal to and intersecting the line of the impulse where it pierces the plane of these axes. Further, if the line of the impulse be in the plane xy, making any angle with those axes, then Mi = 0, and ]\\ = 0, and the [164.] CENTRE OF PERCUSSION". 263 denominator becomes infinite, and hence S' = 90° ; hence the axis of spontaneous rotation will be perpendicular to the plane xy, or parallel to the axis of z ; and since this result is inde- pendent of the angle which the line of the impulse makes with the axis of a?, it will be true when it is parallel to that axis, and in the plane xy^ in which case the spontaneous axis will lie in the plane yz and still be parallel to the axis of z. This investigation beginning at equation (211) may be still further generalized by drawing from the centre of the mass a line normal to and intersecting the line of the impulse and finding the angle 6" between this line and the axis of spon- taneous rotation, but the results so found will be but little, if any, more general than those given above. CENTRE OF PERCUSSION. 164. It appears from Article 162 and Fig. 133, that the point a in the axis of spontaneous rotation may be considered at rest at an instant, and hence if the elements on that axis were held rigidly in space when the body is struck at h, Fig. 13-i, the axis would suffer no shock. Such an axis generally exists, as shown by equation (205), or more simply by equa- tions (206), (209) and (210). Any point in the line of im- pact is called the centre of perciission in reference to the axis of spontaneous rotation, and the centre of percussion in ref- erence to any axis is any point which may receive a blow without imparting a shock to that axis. Equation (210) enables one to find the centre of percussion in reference to any assumed axis. In the case of a compound pendulum acted upon by gravity only, the resultant foi'ce passes through the centre of gravity of the oscillating body, and the centre of percussion is farther from the axis of suspension than is the centre of gravity. In the ballistic pendulum, the blow should be at the centre of percussion, if possible, to avoid shock upon the axis of suspen- sion. Queries. 1. Can the centre of percussion be at the centre of gravity of a body? 2. Can the axis of spoutaneous rotation ever pass through the centre of gravity of the body 1 264 CONSERVATION. [165. 166.] CONSERVATION OF THE MOTION OF THE CENTRE OF THE MASS. 165. Any condition that will render the second members of (167) zero, gives 0, lan€. In the solar system, knowing the positions and motions of the planets at any time, the position of the invariable plane may be found. [168.] ATTRACTION. 267 MUTUAL ACTION BETWEEN TARTICLES. 168. Tlie mutual action between particles, or bodies, is of the nature of attraction, or of repulsion, or of both these forces. We will Urst consider attraction. Newton's law of universal gravitation is — Two ])artides attract each other with a stress directly propor- tional to the product of their masses, and inversely as the square of the distance hetwecn them. Thus if m and iii are two masses considered concentrated at mere points — in other words, the masses respectively of two particles — 77 the stress due to the mutual attraction of two units of mass, tlie one upon the other, at distance unity be- tween them, i^ the stress between the masses in and iii due to their mutual attractions when the distance between them is ic, then F^Il—4-. (224) X- where F^ m, and m are in the same nnits as 77, Taking the origin of coordinates on the line joining ^ ^^ -^ the particles, od the abscissa of m, x" * — .x- — ^< — -— x > of m, and x the distance between them at "* ^ time tj then for m' we have Fig. 136. ^■ = -n'h (225) dt? X' and for m. or also' VI -77.— = i^ cos 0° — 77 — ^— , di- X- 'M- = U%, (226) dt' x^ ^ ^ differentiating which, gives 268 ATTRACTION OF d?x' — d'x" = d^x, in which substituting (225) and (226). we have fJ^T 1 [169.] d(. Integrating, y2 = ^^ = 277 {m + 7u') "' ~ '" df ax (227) (22S) where a is tlie distance between the particles when their velocity is zero. Integrating again, t = + a COS" 1 fX \-i -—-^ y\(ax-x')\ (229) In the preceding investigation the mass lias been conceived to be concentrated in a mere point ; it will now be extended to that of a mass of finite dimensions. 169. Tojimd the attraction of a homogeneous sjphere ujJon a pai'ticle exterior to it. Let ABD be a spherical shell, centre C, P tlie position of an external particle. Conceive two consecutive radial lines drawn from P, cutting the shell in the points A and i>, and join P and C. Let B — APC, y = perpen- dicular from A on PC, ds = an element of lengtli of the circle at A. PA = r, PC = c, CA = CB = p, dp = thickness of the shell, d = density, a = the radius of the sphere, and m the mass of the particle at P. The revolution of the semicircle about PC as an axis will generate the surface of the shell. The area of a section of the shell at A whose leno;th is ds, will be Fig 137. dp . ds. [169.] SPHERES. 269 Let this ai-ea be revolved about PC as an axis through an angle dcp ; it will generate a solid whose altitude is ydq), and the volume so generated will be ydq) . dp . ds, and its mass will be Sydcp dpds. The attractive stress between this element and the element at P will, equation (224), be proportional to m . dydcpdpds ^ the component of which along the axis PC will be ^^dpdsy^of.dcp and the attraction between the entire sphere and the particle will be 7n6 r-'^p t -'' dpdsycosddcp IT "^0 ^0 = ^nm8 '"' p"'' dpdsyQmB ^230) Jo r"- In order to integrate this, y, B and t must be found in terms of ds ; or, including rf.s, all must be given in terms of a single variable. Drop the perpendicular CE—p^ Fig. 137, from the centre, (:^, upon PB^ then p =: G sin^; differentiating dj:> = ccosdde', (231) also from the triangle PCA, r^ — 2f'c cos6 + (T = p^ ; 270 ATTRACTION OF [169.] and since p will be constant for any particular shell, we have for that shell, by differentiating, dr re smd dd r — c cos^ ' From the Theory of Curves, ds^ = di^ + 'i^d^, which, combined with the two preceding equations, gives ds _ pr dO r — c co&O * and this with (231) gives c cos^ ds = r — CCOS& From the figure we have r — c cos^ = Vp' —p't also or y _p P y = r-» ^ c These in (230) give ^TtmS ['' r pdppdp "^ Jo Jo VW^ where tlie former result has been multiplied by 2, since the same perpendicular CE corresponds to two elements A and 3. Integrating gives c and integrating again. 4:7rmd f" - , 169a.] SPHERES. 271 ^ o — = wi X voliwie of the sphere x -r- tnass of the sphere = m X ^ ^^ . (232) That is, the attraction between any homogeneous sphere and a particle exterior thereto varies directly as the mass of the sphere, and inversely as the square of the distance between the particle and the centre of the sphere. It is hidejyendent of the volume of the sphere^ and is the same as if the entire mass he coyisidered as at the centre of the sphere. For the same reason, if the mass 7n be also a homogeneous sphere, it may be considered as concentrated at its centre of gravity; hence the attraction between any two spheres varies as the product of their masses conjointly and inversely as the square of the distance between their centres. This result is the same as that given by Newton in his Prindpia.^ 169a. Mass and stress are often referred to by the common name pounds ,' but when necessary they are distinguished as pounds of mass, or pou7ids of force. If two homogeneous spheres of equal size and known masses be placed at a known distance d, from each other, and under such circumstances that they are free to approach each other under their mutual attractions ; then if the equal opposite forces P, applied to each sphere just sufficient to prevent their approach, be ac- curately measured in pounds by an accurate balance, we have from the preceding article and equation (221) ■■n=F'^,. (2.33) But this method of finding 77 is impracticable, chiefly on ac- count of the difficulty of measuring the exceedingly small * Principia, B. 1, Prop. LXXYI. Cor, 3. 272 ATTRACTION OF SPHERES. [169a.] valne of 77 which would result from manufactured spheres of manageable size. The usual method of finding 77 and one suffi- ciently accurate in practice, is to consider the earth as a homo- geneous sphere whose radius is the mean radius of the earth. The stress due to the attraction between the earth and any body at its surface equals the weight of the body, or m! x g = F^ which in equation (224) gives TI = ^g, (234) where R is the radius of the earth at the place where the body is weighed. The same result follows from equation (227) by neglecting the mass m of the small body when compared with that of the earth, and substituting g for the acceleration, and R for a?, the distance. Let R — 20,850,000 feet, the mean radius of the earth; E = the mass of the earth, being about 5^ times that of an equal A'olume of water, page 227 ; ff = 32.16, the mean value of the acceleration of a free body on the earth, p. 144 of Elementary Me- chanics ; then from (234) we have 3 X 32^ X 2 _ 67 ,, . ~ 4 X 11 X 3,1416 X 20,850,000 ~ 1,000,000,000 "^''^^ •' ' if the density of the unit sphere be the average density of the earth and a foot radius. From this i-esult 11 may be found for any assumed material, since the attractive stress will vary directly as the mass, for the same volume. Equation (227) thus becomes completely determined. The unit of mass for the solar system is sometimes taken as that of the earth con- sidered as 5^ times that of an equal volume of water. Equation (227) is for motion in the line of centres of the bodies. If [169b, 170.] LEAST ACTION. 273 their initial motions be not in this line, the bodies will describe orbits, j>«/'^/erfect fluid in which the repulsive forces always exceed the attractive ones. Such a substance would expand indefinitely if not restrained. Air is usually taken as a type^ though hydrogen is a more perfect gas. A heavy fluid \?, one in which its weight is considered. 173. It was formerly supposed that water was incompress- ible, while it was known that air could be easily compressed, and for this reason fluids were divided into compressible and incompressiUe, or elastic and non-elastic ; the former of which were called gases and the latter liquids. Although it has long been known that liquids are compressible, yet since the com- pression will be very small for pressures to which they will ordinarily be subjected, we still consider a perfect liquid as incompressible. LAWS OF PRESSURE. 174. The pressure upon any particle of a p)e7fect fluid at rest is equal in all directions, for if it were not, there would be a resultant pressure which would produce motion of the particle, since it is assumed to be perfectly free to move. The force here considered is Unite. The weight of the particle acted upon by gravity is infinitesimal, and hence if [175, 176.] LAWS OF PRESSURE. 277 it be at rest the upward pressure against the particle must ex- ceed the downward by an infinitesimal amount, the amount being equal to the infinitesimal weight of the particle. 175. The pressure of a perfect fluid at rest, upon the sur- face of the vessel containing it, will be normal to that surface at every point of it. For otherwise there would be a tangential component, and this would produce motion, which is contrary to the hypothesis. This proposition is also true in regard to any surface exposed to fluid pressure ; hence, if a body be immersed in a fluid, the pressure upon it will be normal at every point of its surface. The discussion in regard to the stress in a fluid might be founded on Article D, p. 154. 176. Eoery external 'pressure upon a 2>eff&ct fluid at rest will he transmitted with equal intensity to every part of the fluid in the vessel. For if there were any unequal pressures thus transmitted, motion would result. This is called The Law of Equal Trans- mission. It is independent of the form of the vessel. The pressures due to the loelght of any particle of the fluid will also be transmitted equally to all parts of the fluid in the vessel helow the point where the particle is located. It cannot be transmitted above that point, for its weight is equilibrated by the upward pressure at that point. This proposition is illustrated by means of a vessel having closely fitting pistons at different parts of it, and the vessel filled with water, when it is found that a pressure on any one of the pistons [the pressure being j) pounds per square inch] produces tiie same pressure per square inch on each of the other pistons. Tli« pressure per unit area is called the intensity of the pressure. The ^^^^^^^^ proposition is not rigidly proved by ^iq- i38. this experiment, for the pistons cannot work without friction. 278 LAWS OF PRESSURE. [177, 178.] 177. The pressure upon the base of a vertical prismatic vessel containing a perfect heavy fluid equals the weight of the fluid ^Zms the pressure upon the upper surface of the fluid. For, according to the preceding article, the pressure upon the upper base is transmitted to the base with undiminished intensity ; and by the same article the entire pressure due to the weights of the particles is transmitted to the base, and must then be supported by the base. 178. The pressure upon the base of a7iy vessel containing a heavy perfect fluid equals the weight of a prism of the fluid having for its base the base of the vessel, and for its altitude tlie height of the fluid ; j:>Zws the pressure per unit area on the upper base into the area of the lower base. It is independent of the form of the vessel. For, if over any element of the base a vertical prism of the fluid be conceived of the same weight as that of the liquid, and the upper surface be subjected to a pressure of the same in- tensity as that of the given fluid, the press- ure on the element will equal the pressure on its upper base ^^^^s the weight of the vertical prism, Article 177. But this 1 c pressure will be transmitted equally in all Fig. 139. directions, Article 176, and hence pro- duces an equal pressure on every element of the base, and thus balance the real pressures. The intensity of the transmitted pressures is the same throughout the containing vessel, Article 176. If, therefore, the vessel be oblique, so that no real verti- cal prism can be erected on any element, the pressure wnll re- main the same, depending upon the vertical height. The ideal vertical prism being suppressed the proposition will be estab- lished. If ^ = the area of the base of the vessel, a = the depth of the fluid in tlie vessel, 6 = the weight of a unit of volume of the fluid, p = the pressure per unit of area upon the upper base of the fluid, P = the total pressure upon the base of the vessel ; [179, 180.] LAWS OF PRESSURE. 279 then we have I> = daS+pS= {6a + 2))S. (236) 179. Static Head. — In the preceding Article, conceive a prismatic vessel having the same base and filled with the same fluid to such a height, h, as to produce the pressure P upon the base, then P = ShS. (237) which compared with equation (236) gives h=a+^' (238) o This value of h is called the head due to the pressure, or the reduced head. It is a height of the fluid which would produce the actual pressure upon the base. The pressure varies directly as the head. In the case of gases confined in small vessels, the weight of the fluid, compared with the external pressures, may generally be neglected, in which case we have ^ = 0, and (236) becomes P=2^^- (239) Gases, confined in a vessel, are always subjected to a press- ure at the upper surface ; but in the case of liquids, p, equa- tion (236), may be zero, in which case a becomes the same as h in (237), and we have ^ = S.=f' ^'"'^ where j9' is the pressure upon a unit of area of the base of the vessel. 180. Free Surface. — If the upper surface of a fluid is not subjected to a pressure, it is called a, free surface; and is someti,mes so called in the case of liquids when the atmosphere causes the only pressure. Liquids may have a free surface ; but, according to the definition, a gas cannot have a free sur- face, since it may expand indefinitely. When the surface is 280 MECHANICS [181.] free the pressure on the base of the vessel is given by equa- tion (237). 181. Pkessure on a Submerged Surface. — The sub- merged surface may be the interior of the containing vessel, or the surface of any body within the fluid, and of any form. The normal pressure upon any element of the surface will be, Articles 176 and 179, ^ = dxdAy where x is the reduced head, dA the element, and S the weight of a unit of volume of the fluid at the place of the element. The entire normal pressure upon the surface will be = 1 SxdA, (241) and if the fluid be incompressible, or if its density be uniform throughout the surface considered, S will be constant, and we have P=d\jdA. (242) If X be the reduced head over the centre of gravity of the surface considered, then, equations (79), taking the origin in the surface of the fluid vertically over the centre of gravity of the surface, we have csA xdA ; .-. P = 6^A, (243) that is : The normal pressure of a fluid against any sur- face suljinwrged in it, equals the loeight of a prism of the fluid whose hase equals the area pressed and whose altitude is the reduced head over the centre of gravity of the area pressed. If the fluid be an incompressible liquid having a free sur- [182, 183.J OF FLUIDS. 281 face, the reduced head will be the actual head over the centre of gravity of the area pressed. 182. Resolved Pressures. — The pressure iu any fixed di- rection will be the sum of the components of the normal press- ures in that direction. If d be the angle between the normal and required direction at any point of the surface, we have resolved jpressure = d xdA cos 6. dA cos 6 is the projection of the element on a plane normal to the required direction, which call dB ; then resolved pressure = 6 :cdB — dxB, (244) where x is the reduced head above the centre of gravity of the projected surface. Hence, when the projections of the ele- ments are not superimposed The component of pressure of a heavy perfect fluid upon any snhnerged surface in any direction, equals the iveight of a prism of the fluid having a lase equal to the projection of the surface on a plane normal to the direction, and whose alti- tude is the reduced head alove the centre of gravity of the pro- jected elements, each considered to he at the depth of the corre- ponding surface elements. 183. Resultant Pressures.— If a body be submerged in a perfect, heavy fluid, the resultant of the horizontal pressures will be zero ; for the projection of all the ele- ments upon parallel planes will be equal, and hence the opposing pressures, according to the preceding article, will be equal, and their re- sultant zero. For the same reason the residt- ant horizontal pressures upon the interior sur- llG. 140. face of such a vessel will also be zero. The resultant vertical pressures Avill also be zero, except that due to the weight of the displaced fluid, Article 176, hence 282 MECHANICS [184.J the resultant pressure against a submerged surface will be upward and equal to the weight of the displaced fluid. 184. Centke of Peessure. — The point through which the resultant of all the pressures upon a surface passes, is called the centre of pressure. Let AB be Xa submerged surface, CD the line of inter- section of its plane with the free surface, -|G or the upper surface of the fluid for a re- i duced head. Take the origin at any point c /~^^^\ ®^ ^^^® ^"•'^ ^^' ^/''^^ong ED, and x along ^\^^^ EF. FG = the head on the element dA = dxdy at F. Let d = FEG = the incli- nation of the surface pressed to the hori- zontal, then GF — X sin 6 ; and the pressure on the element, w being the weight of a unit of volume, will be wdA . X sin 6*, and its moment in reference to the line CD will be w dA . X" sin 0, and the moment for the entire surface will be w sin B \^dA. Denoting the distance to the centre of gravity of the area A by X, and the distance to the centre of pressure by Z, we have for the reduced head over the centre of gravity : X sin B, and hence for the total pressure on the surface wAx sin B, [184] OF FLUIDS. 283 and for the moment of the pressm-e, wAx sin .1', hence we have, by equating the preceding values, loAxl sin<9 = 10 sin^ ofdA; ••• ' = St- ' (245) that is, Articles 137 and 134, The centre of pressure coincides with the centre of percussion^ the axis of rotation heing in the free surface / or wliat is the same, it is tlie moment of inertia of tlie surface divided by its statical moment. It is inde- pendent of the inclination of the plane, and of the density of the fluid. EXAMTLES. 1. Required the entire piressure upon the interior of a cone filled with toater, and standing on its hase. Let r = the radius of the base of the cone, and h = its altitude. The weight of a cubic foot of water is 622- pounds. The area of the base will be jtr- ; hence the pressure upon the base will be 62^ . Ttr^ h. The normal pressure on the concave part will be 62i.27rr.Wr^ + h\^h, or I X G2i X Ttrh Vr^ + h% which added to the preceding gives 2S4 MECHANICS [184.] 2. Required the normal pressure upon the interior of the cone in the preceding example when inverted. 3. Kequired the normal pressure upon the interior of a sphere tilled with water, and compare the result with the weight of the water. 4. Find the normal pressure upon the interior of a cylindric- al vessel including its base, when tilled with water. 5. Find the pressure upon the interior of a cone filled with water, the axis being horizontal ; the radius of the base being 1 foot and the altitude 4 feet. 6. Required the centre of pressure of a plane triangu- lar surface immersed in a fluid, the base being in the free surface. The moment of inertia of a triangle in reference to its base as an axis is. Article 104, Example 3, Its area will be i M, and the distance to its centre of gravity ^ d ; hence, Equation (245), we have 7. Required the centre of pressure of a rectangle hav- ing one end in the free surface, a being the breadth and d the depth. 8. Find the centre of pressure of a rectangle immersed ver- tically in a fluid, its upper end being a distance h below the free surface, a its breadth and d its depth. 2 d' - ¥ 171S. 3 d--b^' 9. A cone standing on its base is tilled with water ; required the vertical pressure upon the concave part, the radius of the base beins: r and the altitude h. [185.] OF FLUIDS. 285 10. In Example 9 show that the pressure upon the base minus the vertical pressure upon the concave part equals the weight of the water. 11. The concave surface of a cylinder filled with a liquid is divided by horizontal sections into n annuli in such a manner that the pressure upon each annulus equals the pressure on the base ; the radius of the base being r, required the altitude and breadth of the mtli annulus. Ans. Depth h = 7i)\ Breadth of mtli anniihis = '\/rh \_Vm —y^rn — Ij, 12. A rectangle, breadth 14 feet, depth 20 feet, is immersed vertically in a liquid with one end in the free surface ; re- quired the distance below the free surface of a line which divides the pressures equally. Ans. 21.213 feet. 13. A vessel, in the form of a paraboloid of revolution, stand- ing on its l)ase, is filled with water ; required the normal press- ure on the concave part, and the vertical upward pressure on the same, the radius of the base being \\ feet, and altitude 4 feet. Flotation. 185. Consider the case of a body in an incompressible liquid. Let Y be the volume of the body, D its density ; Y' the volume of the liquid displaced, and d its density. Then, ac- cording to Ai'ticle 183, the pressure vertically npward will be hence, if there be equilibrium, we have gDY^gdY\ or 286 MECHANICS [186.] that is, The volume of the lody will he to that of the displaced liquid as the densitij of the liquid is to that of the hody. If then d = D, V= V; or the body will be entirely submerged, but if then 6>D, F>7'; or only a part of the body will be submerged, and the body is said to float. The intersection of the plane of the free surface with the floating body is called i\\e plane of flotation. The line joining the centre of gravity of the solid, G, and the centre of gravity G of the displaced liquid is called the axis of flotation, and if this line be vertical when the body is in equi- librium it is also called the litie of rest. If the body be displaced from its line of rest, the vertical through the centre of gravity C of the dis- placed Kquid is called the line of Fig. 142. suppovt ; and the point J/ where this line intersects the line of rest is called the metacentre. For the EQUiLiBRirM of a floating body it is necessary that the line of support shall coincide with the line of rest, and the equilibrium will be stable if the metacentre for an indefi- nitely small displacement is above the centre of gravity of the solid ; for in this case the reaction of the liquid along the line of support tends to turn the body toward the line of rest. If the centre of gravity of the body is below the centre of gravity of the displaced liquid, there will also be equilibrium. 186. The depth of flotation may be found by means of [186.] OF FLUIDS. 287 equation (246) when the density and form of the body and density of the liquid are known. For example, to find the depth of flotation of a paraboloid of revolntion with the vertex downward : Let h = the radius of the base, h the altitude, and x = the depth of immersion. From the equation of the meridian section we have y- = ^px, also, • • y- = r^' The volume of the solid will l)e and of the displaced liquid, ^Tty^x ; and these substituted in Equation (246) give ic = ... 6 h E XAMPLES 1. Find the depth of flotation of a solid sphere whose radius is 6 inches and density | that of the liquid in which it floats. Ans. 8.084 inches. 288 MECHANICS [187.] 2. Find the depth of flotation of a cone whose altitude is 5 feet, radius of the base 8 inclies, and whose density is one-half that of the liquid in which it floats, the axis being vertical and apex upward. 3. Eeqnired the depth of flotation of a solid paraboloid of revolution, base downward, radius of base r, altitude A, and density § that of the liquid in which it floats. 4. Kequircd the diameter of a spherical cavity in a uniform spherical shell of iron so that the depth of flotation shall be equal to tiie external radius of the shell, the external radius being ?', and density 7 times that of the liquid in which it is submerged. Ans. r — r 5. Required the pressure necessary to just submerge a cu- bical block of wood each of whose edges is a feet, and whose density is i that of the liquid in which it is submerged. 6. In a uniform spherical shell, external radius ?■, density 7, required the radius of the cavity that the plane of flotation shall be tangent to the top of the cavity. Ans. r = 0.95r + . SPECIFIC GRAVITY. 187. If '"^n external pressure act upon the body, either forc- ing it up or down, thereby producing equilibrium, we have, wiien the force i^acts vertically down on the body, J^+gJ)V = rj6V'', (247) for, the weight of the body, gB V, added to the downward pressure will equal the vertically upward pressure of the liquid. If the force i^'acts upward, then the upward pressure of the liquid and the force i^ will equal the weight of the body ; hence gDV=i/dV' + F. (248) [188.] OF FLUIDS. 289 By means of these formulas, the weight of a body com- pared with the weight of an equal volume of the liquid may be determined. If the liquid used be selected as a standard, the relative weight thus found is called the sj?ccijic weight, or specific gravity. The body weighed in a vacuum gives directly, W=gDY; (2^9) then immersing it in the standard liquid and ascertaining the value of F necessary to produce equilibrium, we have from the preceding equations, W=gdV' ±F', or W gdV 7=1 ± F g6 V ' (250) where + i^is used when the body is heavier than the liquid, and — i^when it is lighter. AVater at a fixed temperature (usually 60^ F.) and pressure (about 29.92 in. of the barometer) is usually taken as the standard. For a further development of the subject see the Author's Elementary Meclianics. 188. When a mass of liquid is in motion under such con- ditions that its form becomes permanent, certain problems pertaining thereto may be solved by the principles of Statics. SVe notice the two following Problems. 1. A heavy perfect liquid having a free surface, is moved in a given direction icith a con- stant acceleration, required the character of the free sur- face. Let the vessel move hori- zontaTIy under the action of a constant force F, produc- ing an acceleration/, the weight of the liquid being TF— the weight of the vessel being neglected. 19 290 MECHANICS [188.] Then we have, Equations (20) and (21), W=JI(?/v/i«w€?«^. Let *Si be the section of the tube at A, S that at B, v^ the velocity of the flow at A, v that at B ; then SiVi = Sv, which is the volume of the liquid flowing m a unit of time. JjQt j)i be the intensity of the pressure exerted at A, and ji? that at B, then will [190.] OF FLUIDS. 293 be the entire pressure on the section Si, and PiSiVi will be the work done hj j)^ in a unit of time, since the veloc- ity nia^y be considered constant for an element of time, and may represent the space passed over in a unit of time. Sim- ilarly, the work done by j? in the opposite direction in a unit of time will be Take any datum plane, above which are the ordinates s^ to A and s to B, then the work done by gravity wliile the liquid is passed from the height Si to that of s, will be, w being the weight of a unit of volume, wSii\{si - z). The difference in the kinetic energies at A and B will be wSiVi , „ 2\ -w ^'■" " '^- Since, according to the assumed condition, there is no re- sistance between A and B, the liquid between these points may be discarded — or, what is better, we may conceive that the liquid just before and behind the respective elements at A and B serve as a pair of perfectly flexible pistons which yield just enough to keep the tube full at all points passed by this elementary mass. Then v/ill the entire work done upon this mass in passing from A to B equal the difference in the kinetic energies at those points, or PiSjVi -pSv + wS,v, {,1 - z) = "^ (^ - Ti) ; (2.51) or pAv, + loSiv^, + "^v\ =pSv + ivSiv.z + ^^ v\ (252) 294 Bernoulli's theorem. [191.] In this equation, piSit\ is the potential energy of the ini- tial pressure (Articles 26 and 151) ; wSiViSi is the initial po- tential energy due to gravity in reference to any arbitrarily assumed horizontal plane ; and —^vl is the initial kinetic 2 g energy of the mass lo ~ g. The sum of these will be con- stant for steady motion. The second member of the equation represents corresponding quantities for any point of the stream ; hence For steady tnotion without resistances^ the sum of the ])oten- tial and kinetic energies is constant. This is JBernoidWs Theorem. It may be expressed in another form, for dividing Equation (252) through by loSyVx we have Z. + ,. + «5=J? + , + ^; (253) w %j w zg where ^^1 -^ ly is the head due to the initial pressure, 2^ the ini- tial head in reference to the datum, v[ — 2g the head due to the initial kinetic energy ; and similar general expressions apply to the second member. The sum of the heads in each member is called the total head; hence For steady 'tnotion vnthout resistances, the total head in ref- erence to any horizontal jylane is constant. 191. Discussion. — If the extremities of a stream in steady motion be in the atmosphere, the pressure at the ends, jp-^ and ^, will be that of the atmosphere, and in most cases will be practically equal in magnitude, but opposite in direction ; for which condition (253) becomes ,.^ = « + J^'. (254) The initial velocity is often so small that it may be neglected, for which case v^, = 0, and (254) becomes v^ =2g{zr-z). (255) L192.] TORRICELLl'S THEOREM. 295 and if the datum plane passes through the lower end of the stream, we have z.= 0, and (255) becomes V = V'2gzi, (256) wliich is called TorricelWs Theorem, Comparing (255) or (256) with the first of (16), we have In steady 'motion without resistances, the head due to the velocity equals the height through xohich a hody must fall to acquire that velocity. If the datum plane yjasses through the lower point consid- ered, we have » = 0, and (254) gives V = Vt'i + 2^2i. (257) Fig. 146. 192. To REPRESENT EqTJATION (253) GRAPHICALLY, let AB be a stream having a steady motion subject to different press- ures along its path, and as- sume AHi equal to % and draw the horizontal line JHiH. Let AG represent the head due to the pressure at A, which may be that due to the atmosphere, oi' the at- mosphere and any other extraneous pressure, and CU the head due to the actual velocity at A. Then will AU be the total head above A, and if HxH be the datum, then will HiU be the total head above the datum. At H^VG will have z = HB\ ^>-^^/j= i?i^; v''~1g = BF. If a vertical tube be inserted in the stream, having an open- ing up stream, the liquid should rise to the height ED\ but if it be turned so as to be open sidewise, it would rise only to the height GF. The latter is called the hydraulic head. 296 MECHANICS [193.] If the initial velocity be neglected, the horizontal line ED will pass through C. If at any point the pressure in the stream is zero, the line CF will be depressed and touch the stream at that point. Should it fall below the stream, the pressure would be nega- tive, but as liquids have no tensile strength, this condition would destroy the "steady " motion, and the equations would not be applicable. 193. If ^ vessel of varying sections be left free to discharge itself, or generally if a fluid has a " steady " flow through a pipe of varying sections, the pressure of the fluid in the small sections will be less than that due to the statical head, trictional re- sistances being abstracted. Let Si be the section at D, S that at B, then vS = ViSi', o o -" J 1 — -tJ o .-. V- -vi = -^^— vi ; Fig. 147. and (253) becomes ^=-a+,^_, + ^5izL^ ^0 2(/S\ (258) Avherc z, — z = BD, ^ = head due to the pressure on I), = hi w (sav), and -^ the head due to the pressure at £. w If Si = S, we have w = hi + DB ; (259) or the pressure will be exactly that due to the head. If S > Si, the last term of (258) will be positive, and hence [193.] OF FLUIDS. 297 p-^ w will exceed the head given by (259), and we may write for this case ^ = 7^1 4- BN. w Let the section at A be less than at D ov S < aS^, then henceJ. M will be less than the height of the free surface above A. Also ^ = hi + AM. If S is so much less than Si that is negative, then p -^ w will be negative, and there will be a tendency to a vacuum. Let C be such a section. Then if a bent tube CEG be inserted at C, having its outer end below a liquid, the fluid from F will rise in the pipe a height FG, so that FG = -^. (260) Examples. 1. A surface elementary stream of water having a velocity of 16 ft. per sec. undergoes changes in its sectional area as it passes a vessel which are proportional to the numbers 4, 6, 4, 3, 4, 6, 4 ; in what way can the head remain constant ? Draw a vertical contour of the stream with figured dimensions or distances. We have, since v^ varies inversely as the square of the sec- tion, nil r\r 1111111 ^ ^ T¥5 ■51') To? "0 5 T6^» i'^y TF* 298 EXAMPLES. [193.] Since the stream is in the surface^; will be constant, being the pressure of the atmospliere, therefore z or the height must vary, and since the head remains absolutely the same the in- crease of z must be the same as the decrease of ^. v = 16 on entrance, and ^ = l?o^ "lf > = ^ ^^' Hence, as tlie successive 2^ 32 X 2 values of pr ^i'^ .1__X or the vertical contour will be of wave form. FiG.148. 2. Water flows without loss of head through a horizontal pipe of a diameter varying uniformly from 3 in. to 1 in. at smaJlest section, and then gradually enlarges. The velocity on entrance being 7 feet per second, what will be the mini- mum pressure at entrance, in order that the pipe may run full- and what may be the maximum diameter of exit into the at- mosphere ? Since the pipe is horizontal, 2 = Zi in Equation (253), and according to the conditions of the example, ^j> = at the small- est section. We also have V at entrance = 1 feet ; .'. V at smallest section = T x 9 = 63 ; . j^i 49_(G3y. "to +2^- 2^ ' .-.^1 = 26.57 lbs. For maximum diameter of exit into the air we have in the same equation p^ = 0, p = 14.7 lbs. per square inch, v^ = 63, and V = 63 -f-c?; [193.] EXAMPLES. 299 . 147 X 144 _^/7x9y 1 _iG3f_ .*. d =^ 1.22 inches. 3. "Water flows from a tank through a pipe, the lower end of which is 15 feet below the entrance, the sectional area of the pipe at the tank end being twice that of the lower end. !N^ear the tank the pipe is perforated or broken; tind the head of water in the tank necessary to prevent air leaking in, or water out, through the fi-acture. The pipe must be full, and the motion "steady." At tlio ends j3 = 2^i — ^^ ^^^^' 0^^' ^^-^ ■^^^)- If V be the velocity at the joint, then 2y will be that at exit, and we will have 34 + ^ + 15 = ^+34; .•. pr- = 5 ft. = head in vessel. ^ r be- low the surface. 1 //'\2 5 (r\^ Ans. 71 f' v'-Igh ^-wAv-Tm\h) +"''■ 6. To determine the time in which a vessel will empty itself of a jpeifect liquid through an orifice in its hase. Take the origin at the orifice, z vertical, a the area of the orifice, K i\\Q area of the free surface ; then Kdz will be the elementary volume discharged in an element of time, acvdt = ciG 'y/^gs .dt= — Kdz, the quantity passing through the orifice in an element of time, and is negative, since t and z are inverse functions, hence t =. ^ — r^^_:^«^ (2665) acy'^g Jo Vz [196.] EXAMPLES. 305 If the section K be variable, its value must be found in terms of z before intetjratinir. 7. To find the time in which a prismatic vessel filled with a perfect fluid will discharge itself through a mere orifice, «, in its base. Arts. ■ ; ■ . 8. A vessel, formed by the revolution of the semi-cubical parabola, lif = ^, about its axis z, which is vertical, is filled with a liquid to the height h ; to find the time in which the liquid will be discharged through a small orifice, section a, at the vertex. Here K= 7ty'= -^; and the limits are for t = Q, z = h, and for t = t^, z = ^. Ans. t = -s ~. 9. Find the time in which a paraboloid of revolution whose altitude is h and parameter p, full of liquid, will empty itself through a small orifice at its vertex, its axis being vertical. Here Tty- = ttjjx. Am. t = 7^==. dacy 2^ 10. A conical vessel, the radius of whose base is ?•, and alti- tude A, is filled with a liquid ; required the time in which the surface of the liquid will descend through half its altitude, tlie orifice being at the vertex, and the axis vertical. Here z- and the limits are z = h, and z = \h. '^'''' ^ - 20ac.Vy ao 306 EXAMPLES. [196.] 11. Find the time in M'liich a liquid contained in a parabo- loidal vessel, i^ = hz, will descend equal distances h, the flow being through a small orifice whose section is a, at its vertex. . , TthVb Atis. t = ^^ Tlie times are equal for equal heights taken anywhere along its axis. This would form a water clock in which equal times would be indicated by equal spaces. 12. A tank whose height FD is 100 feet above the level of the ground is supplied by a 1 inch pipe which communicates with a Ih inch horizontal pipe at the level of the ground, and is fed by a third pipe 2 inches in diameter proceeding from an accumu- lator 3 feet in diameter, with piston, A^ 10 feet from the ground, loaded with 30 tons. Neglecting all resistances, find velocity with which the Fig. 154. Water enters the tank. Equating heads at G and D we have ,.=^=^-^^-=^ F E D A ' 1 i^ B C 30 X 2,240 9 X 3.1416 from which X 62.5 -[- 10 + 34 = 100 + ^ + 34 •y = 64 feet per second. 13. "Water is discharged from a vertical rectangular orifice of height d and area A. Show that approxi- mately Q = c 's/2gh X a(1 —w^i where Jt is the depth of centre of or- ifice below the water level, the orifice being fully immersed. 1^ 3/(V' Fig. 155. [107.] DIVERGENT TUBES. ' 307 From the figure, with the origin at the upper edge of the orifice we have ,2 = ,.£''-^j/2y(A-'|+y>, 2<; A _r/, rl XJ-]'' or Developing (h +0" and U ""^)' ^^ ^""^^^^ terms, and per- forming the operations indicated, we have 14. Find the time in which the liquid in two prismatic ves- sels will come to the same height, one discharging itself into the other, through a short pipe connecting them at their bases. Let l> be the area of each base, h their height, A the section of the pipe ; and initially, let one of the vessels be filled, and the other empty. 197. If a conically convergent tube BE oi the form of the vena contrada be attached to the orifice B, and to the small end Eq. tube slightly divergent be attached, it is found by ex- periment that the amount of flow is in- creased, and is even greater than if the discharge be through a single orifice, except when the flow is into a vacuum. It appears that the liquid adheres to the sides of the tube, carrying away the particles of air from within the tube, tending to make a partial vacuum at E, or at least to diminish the internal pressure, thereby making more Fig. 156. 308 PRESSURE OF [197.J effectual the head AJ^ and the pressure of the air on the upper surface. Tlie tube being entirely filled, it is the case of steady mo- tion, and Bernoulli's Theorem applies. If J) be the pressure per unit of the atmosphere, jh the pressure at J^, at first unknown, and consid ring the velocity of the surface i^ = 0, equation (253), we have for the head at A (potential) and for the head at F^ z being zero, and for the head at E, z also zero, ^ Vi. but these heads are all equal, hence A + £ = ^-+^-i = 5l + i^; (267) %0 l(J 10 l(j 10 .'. v" = Igh ; which is the velocity which it would have through a mere ori- fice at E\ but as the section at F\& larger than at E, the flow has been increased, and hence the velocity at ^has also been increased. This becomes apparent from the equation vS\ — vS ; .'.v,=^v, (268) where Vi and Si apply at E, and S at ^ wliich is larger than S^. [107] FLUIDS IN MOTION. Wo also have from (2G7) 2^ w or the hydraulic head exceeds 7i. To find the pressure ^^i, we have from (207) and (2GS) 309 (269) 2h = 2^ + u^^i ( 1 — --^ and since Si < S, \\c havc^^i < i^- If ^>i = 0, we have 'S V __ l) + wh \Si J ~~ w/i (270) (271) and if the liquid be water, p -f- 7a = 34 feet, nearly, the head due to the pressure of the air, and the exj)ression becomes S_ ^\ Since pi cannot be negative for steady motion, equation (270) gives the limiting ratio of the sections, but this limit cannot be quite reached in practice. Eytelwein found that when the mouthpiece BJE was shaped like the contracted vein, followed by a divergent tube whose length was 8|| inches long and angle of divergence 5° 9', that 2.5 times as much water was discharged as through a simple orifice of the size of the section at £J, and 1.9 as much as through a short tube of the same section as at ^. If two vessels be connected by a tube as shown, and filled to the same height with the same liquid, ^'"^- ^^'^"• and a stream be established in any manner, it will continue to flow across the space when a small portion of the tube is .L^^ 310 REACTIOX [198, 199.] removed, provided the velocity be sufficiently great, and will cease only when overcome by friction, or by the difference in heads in the vessels. REACTION OF FLUIDS. 198. Newton's Third Law of Motion, being universal in its application, includes the action and reaction of fluids. If a heavy fluid discharges itself through an orilice in the side of a vessel suspended by a cord, the vessel will be forced away from the stream and the cord held in an inclined position. The pressure which would be exerted against the side of the vessel if the orifice be closed, is removed when the orifice is open, and the pressure con- tinuing on the side of the vessel directly opposite the orifice forces the vessel in that direction. Fig. 1566. There is also a pressure exerted in the same direc- tion due to the deflection of the fluid veins from their course, as will be shown hereafter. The latter is called a reaction. 199. Centrifugal Action. — The force wh^'ch deflects a body from a tangent to a curve is called a centripetal force^ and the equal opposite action of the body upon the curve is called centrifugal force. Strictly speaking, we should say that a force is developed between the body and the curve, which, acting one way against the body, forces it away from the curve, and in the opposite direction produces an equal pressure against the curve. If the body be attached to a cen- tral point by means of a cord, the centrifugal action would be exerted upon the fastenings at the centre ; or, if there be no rigid connection, as in the case of the planets moving about the sun, the centrifugal force would be the same as if the planet moved on the concave surface of a solid coinciding in curvature with the orbit. If the motion be in a circular arc, let m be the mass of the body, V its velocity, r the radius of the path, and (p the cen- trifugal force. Since the centripetal force simply changes the direction of motion, the velocity of the body will be constant. [200.] OF FLUIDS. 311 and q) also constant. At A the body will be moving in the direction of the tangent AB, and the centripetal force will act in the direction ^0 ; so that the body will ^ reach B on account of the motion in the same time that its centripetal force will draw it to D, the points B and D being con- secutive to A. Let AB = dx, which ultimately =■ AG = DG) then dx = vdt. From the circle we have CW = AD (2r - AD), d'j? = 2rds, or ultimately , where ds = AD, and AD compared with 2r disappears. From equation (24) or intesratino; twice = 3f or, ultimately, and substituting ds from above ^r'^ds = M and, substituting dx from above dx- (272) which is the value given in equation (141), and also in Article 119. 200. Resolved pkessures. — If a particle be shot into a 312 REACTION [200.] Fig. 158. small perfectly smooth tube having a circular bend AEB, it will exert a uniform radial press- ure upon the tube ; the compo- nents of which in a given direc- tion as (7X will depend upon the position of the particle. The sum X of these components for a given length of arc will be the same as if that length were full of such particles all moving with the same uniform velocity ; and if such a stream of particles be contin- uous, the pressure will be constant. Suppose then that a steady stream of fluid passes through the tube, and let 10 = the weight of a unit of volume of the fluid. k = the section of the stream, r = tlie radius of the centre line of the tube, V = the velocity of flow, s = AD = any portion of the path from the initial point of the curve, e = ACD, CX parallel to the tangent at the initial point of the curve, and OT" normal to it. Friction being discarded, the velocity will be uniform throughout the tube. It is required to find tlie pressure in tlie directions 6'.Xand CY. For any element of length, we have from the figure, and from (272) Integrating ds = rdB^ dm =z — kds — g r w J V' (p = ~ks — , (J r (273) = 31 sv (273a) where J/ is the mass flowino^ into the tube in one second. [200.] OF FLUIDS. 313 Resolving dcp, equation (273), parallel to CX and CY xq. spectively, and integrating gives X, = - W f sin eae = - ytzr' (l - COS e\ (2U) i/ Jo g Y, = — kir fc'os BdB = - kv" sin ^. (275) H Jo ^ If the angular deviation be 90% then 6 = \n, and (274) and (275) become X. =^A'.S (276) which are identical. For motion through a semicircle 8 t=z n^ and we have X=2^Z•^;^ (2TS) 'J Y. = ; (279) the last of which shows that the pressures normal to the line of the stream balance each other. For the entire circum- ference 6 = 27r, hence ^^'^ == ^ I . (280) We observe that equations (274) to (280) are independent of the radius of the path ; hence we infer that the path may have a variable radius, but cannot be zero, since equation (272) will be infinite for r = and v finite. If h be the height due to the velocity v, then v^ = 2(/h, and (276), (277), (278), become respectively Xi. = 2wM, (281) Yi.=2wkk; (282) X„ = 4:wkh. (283) 314 REACTION [201, 202.] Hence : Dcjiecting a continuous stream of frictionless suhstance through an anM (286) i\. = JA; ; (2S7) X„ = 23Iv. (288) Equations (284) to (288) show that the resultant pressure due to deflecting a fluid from a rectilinear course varies frst as the momentum of the fluid per second, and second, as a function of the angle through which it is deviated. APPLICATIOXS. 201. In the following applications all resistmices due to friction, contractions, enlargements, or whorls and eddies in the stream will be discarded. 202. Discharge from the side of a vessel.— In Fig. 156^', considering that the fluid filaments have their origin in [203, 204.] OF FLUIDS. 315 the free surface, their initial direction will be vertically down- ward, and in order to issue from the orifice horizontally must be deflected through an angle of 90° ; hence, equations (2S1) and (282), the pressure on the opposite side of the vessel and on the base, due to the discharge of the liquid, will equal a prism of the liquid whose base is that of the contracted section, and whose height is twice the head above the orifice. The correctness of this conclusion in regard to the horizon- tal pressure has been proved by one Peter Ewart, an English experimenter, who determined the pressure by direct measure- ment {Memoir's of the Manchester Phil. Soc, Vol. IV.). In regard to the increase in the pressure vertically down- ward, we know of no experiments. If the vessel were pushed horizontally in the direction of the issuing stream, with an acceleration equal to ^ = 32^ leet per second, the discharge would cease, and it will be readily seen that the pressure on the inside the vessel opposite the orifice would be doubled. 203. Discharge vertically upward. — In this case the change from the initial direction of the motion will be 180°, and equations (283) and (279) are applicable, hence the pressure vertically downward will equal the weight of a prism of the liquid whose base is the section of the stream, and whose altitude is four times the head due to the velocity. 204. If the discharge be from an orifice in the fi»- i^^- base of the vessel, the fluid veins will issue in a direction par- allel to that of their initial motions, and B, equations (274) and (275) will be zero, hence A-o-O, To^O; or there will be no reaction due to the flow, and the apparent weight of the liquid will be Fig. 160. diminished only by a column of the liquid hav- ino" for its base the contracted section, and for its height the . A P ffivi 1 E? 5? 1 j:i-il yin %^^m 316 REACTION- [205, a06.] head due to the velocity.* That such a column may not have for its base the area of the entire oriiice, is evident from the fact that if the filaments could interfere to such an extent as to prevent the discharge, as they might in the ease of semi- fiuids, the orifice would not affect the apparent weight. If a vessel containing a liquid falls with the acceleration of gravity, the liquid will not flow out of an orifice in its base. 205. If a stream of liquid impinge normally against a plane surface, in the immediate vicinity of the intersection of the axis of the stream and the plane an eddy or whorl of h'quid will be formed, over which the stream will flow as along a — P curve and he discharged tangentially to the plane M-ith its initial velocity. The direction of motion beinw chano-ed 90°, Fig. 161."" equations (281) or (286) will determine the pressure exerted by the plane, and we have - E P^'^wTiJi^Mv, (289) that is : Thepress^ire exerted hy a liquid stream flowing nor- mially against afixedjplane equals (nuwteHcally ) the weight of a prism of the liquid whose hose is the section of the stream^ and whose altitude equals twice the head due to the velocity^ or eq uals (mimerically) the momentum (j}er second^ of the tnass imjjing- ing against the surface. The experiments of Michelotti, Weisbach, and others show that this conclusion is very nearly realized when the impinged surface is at least six times that of the section of the stream, and placed at a distance of not less than twice the diameter of the stream from the orifice. 206. Cup vane. — If the axis of a stream coincides with that * Weisbach states that the apparent loss of weight will be twice this amount. Mechanics, Vol. I., p. 1006. But he errs in his analysis in §496, p. 1005, where he finds V = 2hFy sin a, instead of F = 2hFy (1 — sin a) . 207.] OF FLUIDS. 317 Fig. 162. of the axis of revolution of a surface, and impinges against the concave surface, thej will be deflected as before, and flow- ing along tlie eddj as along a curve, the fila- ments will leave the surface tangent! ally, and equation (284) will be applicable. If the tangents to the surface at C and D are parallel to the axis of the stream AB, we have e= Tt, and (283) or (288) will be applica- ble, and we have P = AwU = 2Mv. (290) The resultant jpresswe due to the imjndse of a liqxdd stremn against a concave heimsiyhere equals the weight of a prism of water whose hase is the cross section of the stream, and lohose height is four times the head due to the velocity. Weisbach found by experiments Avith air impinging against a concave surface that the pressure M-as about 0.88 of the theo- retical value. c / 207. Bent pipe. — If the tube through which the liquid flows be bent through an angle B at the point B, an eddy, or whorl, will be formed at the angle, so that practi- callv the flow will be along a curve, and equations (28-i) and (285) give the resultant pressures parallel and normal to the initial direction of the stream, which being from A toward B will be Fig. 16.3. X = Mv (1 - cos d) Y = Mv sin e (291) The resultant R will be B = VX^ + Y^ = M'vV2 (1 - cos 6) ; (292) which alone will ^^I'event the bodily movement of the tube 318 REACTION [207.] were no other external forces acting. The direction of It will bisect the angle ABC. In practice the quantity of liqnid discharged through a bent pipe will be less than through a straight one of the same sec- tion, on account of the contraction of the stream at the bend. The mass J[f will be the quantity actually discharged. If two forces, each equal T, Avere acting along the branches of the tube and away from the angle, and of sufticient magni- tude to produce the resultant R, we would have 2""- + T' -2T-cose = Ji" = 2M-v" (1 - cos 6), .'.T^Mv, (293) which gives the required value of the force T. If there be two bends, B and C, in the pipe, let two forces T = J/v, one at A and the other at C, act away from the angle B\ they will hold the part ABC. Similarly, one at D and another at ^, each equal to T and acting away from C will hold BCD ; but the equal and opposite tensions along BC neutralize each other, leaving the tensions at A and D. Hence, if frictional Fig. 164. resistances be neglected, A peifedly flexible tule of uniform section having lends of any curvature, and its ends fixed in any 2^osition^ will not change its curvature on account of the jyressure due to the flow- ing of a fluid through it. The effect of the weight of the fluid, which is not included in the above inference, would cause the tube to conform with the plane, or other surface on which it rests. The pressure in a pipe being normal to the curve at all points will be, from Equation {o), page 139, T r [208, 209.] OF FLUIDS. 319 but from (272) this becomes r as before shown by (293). 208. Impinged surface inclined. — If the plane receiving the impulse be inclined an angle d to the axis of the stream, and the stream be confined be- tween guide plates so as to flow along the plane in the line of greatest inclination, the case will be essentially the same as that of a bent tube, and hence the press- Fio. les. ure directly opposed to the stream, and also normal thereto, will be given by equations (284) and (285) respectively ; or P = Mv (1 - cos 0) ) ^294) Ye = Mv sin 6 \ 209. Remark. — Liquids act by impulse when suddenly changed in direction ; and we have seen that the measure of this action is the same as when the stream flov»'s along a curve of finite radius. In practice, however, certain resistances fol- low an impulse, due to various causes, such as the contraction of the stream, eddies, or so-called ivhorls, which make the effi- ciency of the fluid less than when it acts by simple pressure. Some writers improperly use the term impact in this con- nection, as if the action were the same as that of the impact ol inelastic bodies ; but the impact between liquids and solids is only infinitesimal in amount, and hence eludes measurement. The true action is not an impact but a pressure — an action and a reaction of finite magnitude. The stress between finite solids during impact is rarely sought.;.. and, indeed cannot generally be found, for tlie law of action is generally unknown. To find it, the stress as a func- tion of the time must be known, so that the value of fFdt may be found. Also the law of the distribution of the stress 320 EXAMPLES. [209.] tliroiigliout the section in contact must be known, from wliicli it appears that the stress may be variable, and the finite value sought will be the sum of the infinitesimal stresses acting upon the several elements of the section of contact. Inelastic bodies have a common velocity after impact ; and if a small inelastic body impinge against an indefinitely large one at rest, the motion is destroyed ; but if a liquid impinge against such a body, the direction of motion is simply changed. In the eases above considered, both the momentum and the kinetic energy of the liquid are the same after the impulse as before. Exam p les. ■ 1. The nozzle at the end of a flexible pipe of a fire engine is directed at 45° to the horizon, the pipe lying along the ground. What is the apparent increase of weight of nozzle when 150 gallons of water per minute are discharged with a velocity of 80 feet per second ? Also, wdiat is the tension of the pipe ? For the tension we have T = Mv = -iO lbs. ; the vertical component of which will be rsin-t5° = 28 1bs. ; ' which is the apparent increase of weight. 2. A jet of water of sectional area A impinges beneath a horizontal plane of weight W. Find the energy of the jet re- quired per second to support the plane? Let Til be the mass of a unit of volume of water. Then the momentum acquired per second must equal W, - Av'- = W. (1) The energy of the jet is wAv X |- , (2) [209.] but from (1), EXAMPLES. i / ^^ . 321 and substituting in (2), Enerqii = wA A / — r ^^ y mA mA 2^ 2 y mA 3. A vessel containing water and weighing 1 ton, within which the pressure is 9 atmospheres, is supported by discharg- ing water downward ; what is the diameter of the jet ? Tlie avaihible head, supposing the discharge against the at- mosphere, is 8 atmospheres = 8 x 34 feet of water. The momentum of the jet, and the consequent reaction is to — Av' = 2,240 ; or ^'x — X 2r/ x8x 34 = 2,240; .*. d = 3.474 inches. 4. A jet of water strikes a fixed vane at an angle of 30" and then glances off at an angle of 60° rela- tively to the tangent through the initial point of impulse; required the resultant pressure on the surface, the jet deliver- ing 600 gallons per minute, with a veloc- ity of 10 feet per second. The entire deflection of the jet will be 90° ; hence Fig. 166. li = ViMvf + {Mvf = MvV^, 600 X 6.4 60 X 324- 10 X v^2 = 28.2 lbs. 21 322 EXAMPLES. rSlO.] 5. A jet of water 4 inches in diameter, velocity 25 feet per second, impinges on a fixed cone at its vertex, tlieir axes coin- ciding and the apex angle being 30° ; find the pressure tending to move the cone. F = — Q .{1 - cosl a) = 2^ Q sin^ 7° 30'. 9 ^ ^us the work imparted to the vane ; or Fig. 107 D "'"" E iJ/y- = Pu + ^MV\ (305) where Y is the velocity of discharge. Let FGhe, the stream having a velocity v, GP the direction of motion of the vane having a velocity u in the same direc- tion as the stream ; then will v — u be the velocity of the stream relatively to the vane at entrance, and since there are [213, 214.] WORK OF IMPULSE. 325 no resistances, it will quit the vane tangentially with the ve- locity BD = V — ic, and at the same time it will move for- ward with the velocity BC = u ; hence the actual velocity will be BE= V= V{v - iif + w^ + 2w {v - u) cos e ; (306) which in equation (305) gives iMv- = Fu + UI[{v - tif + u^ + 2u {v - n) cos 0] ; (307) from M^iich we find, Fu = 3Iu (y - v) (1 - cos 6), which is the same as equation (300). Dividing by u gives equation (295). From (305) we have Pii=UIv"-i3IV% (308) or, The energy im/parted to the vane equals the loss of energy of the fluid. 213. Efficiency. — The efficiency of a siremn in imjparting work to a vane^ is the ratio of the energy so imparted to the actual energy of the stream. Let e be the efficiency ; then, for the j^receding case, we have, equation (308), 214. Tub Resultant Pressure in Fig. 167 will be E - ^Xl + Yl= M{v - w) 2 sin ^d ; (310) and -'- • tan EGP = ^' = cot id = tan (90° - ^0) ; .-. BGP = K180° -0) = iFGB. (311) 326 WORK OF IMPULSE. [215.] The line of the 7'esultcmt pressui^e hisects the angle hetiveen the approaching stream FG and the pilane AB. 215. General Case. — Let the vane move in any direction. Let AB be the vane, CI) the initial direction of the stream, having a velocity v = DF^ u = DF= tlie velocity and direc- tion of motion of the vane ; v^ = BG 3= the velocity of discharge relatively to the vane, \A-hich will be tangentially p - s^^^^gj. a. \ ^_ r r to the vane: Y =z BII — the actual velocity of discharge, and P the press- ure exerted in the direction of motion F>E. There being no frictional re- ^H sistance, we have, as before, iifr- = Pu + mV\ (312) Fig. 168. To find F, let 8 = FPJ= the angle between the stream prolonged and a tangent to the vane at the point of impulse; cp = FBF, /3 = the angle between i>i^ and the tangent BG. The line BF joining B and F, represents the magnitude and direction of the stream relatively to the vane at Z>, and since there is no loss of velocity relatively to the vane while passing along it, we have BG = EF =vi= Vv' + iC- - Ivu cos (p. (313) Drawing {riZ equal and parallel to BE = ?/, the line BH = V, will represent the actual velocity of discharge, and we have V"- = vl + 10- — 1v{>i COS BGH, — v\-\- u' + ^v^u COS {/3 + qj); (314) which after substituting Vi from equation (313) gives Y^ = v^ — 2ic {v cos cp — ti) + 2u cos ( /i +cp) \/v- + u^ — 2vu cos cp } ; (315) [216.] AND PRESSURE. hence, equations (312) and (315) give Pu = 3fu[v cos q) — u 327 — cos {fi + (p) 's/i? + •. P = M{v cos cp — u 2v2* COS cp\ \ (316) .(317) — COS (/? + (p) Vv^ + 10^ — 'ivu cos qj) = ^/[ycos cp — %!, — v^ COS (y3 + ^)] If ;& be the loss of energy compared with the total energy of the stream, the efficiency in imparting work to the vane will be e = 1 - yl- = Pu IMv'-UIY^ mv" \Mv^ = 1 cos qi — ^ ; (318) - - cos (/? + 9>) a/i 4-^-2 -cos cp V y v V from which the condition for maximum efficiency, u being variable, may be found, but the result will be too complex to be of practical value. These equations include not only all the results of the preceding cases, but also several others. They are independent of the initial slope of the vane, rela- tively to the stream, but are dependent upon the relative di- rection yS with which it quits the vane. 216. Discussion.— Case I. Let v cos q)—u— -v^ cos (/? + cp). Then equations (317), (31G), (315), (318) become, respect- ively, P = 23I(v cos (p — u) ,^._. Pu = 231 {v cos

and N of the triangle EDN equal the external angle KEN^ or KEN =q, + 13', .-. DEK = ISO^ - KEK = 180" - (r^ + y5) ; (321) hence (320), (321), I)EK= KEF, as it should, since, as shown above, "o^ may be laid oif on EF or EN\ and if the angles be all measured from 2)^ prolonged, we have, equation (321) : The angle hetioeen the direction of motion of the vane^ and that of the stream relatively to the vatie at the point of impulse, inust equal the stipplement of the angle between the former line and that of the direction of the stream relatively to the vane where it quits it. The line GB produced will not generally pass through E, but in any case the relation of the angles will be that given above, since a line may be drawn through ^parallel to BG. The speed of maximum efficiency will be found by making the second member of the last of equations (319) a maximum in reference to -w as a variable, which requires that u = \v cos cp ; (322) [216.] AXD PRESSURE. whicli reduces (319) to P = 3Iv cos cp Pu = illv- COS" (p yz= V sin cp e = cos^ (p 329 (323) and P, Pu, and e will be a maximum for (p=0, when V will be zero. Case II. Zet the vane he flat and oblique to the stream. For this case yS becomes (9, and by substituting the latter for /? in (315), (316), 317), (318), the required results will be found. The equations will be of the same form as for the general case. Case III. Let the vane he flat and move normally to the vane. For this case d — (i^ and ^ + /? = 90° ; and equations (315), (316), (317), (318) become P = M{v cos cp — u) Pu = M{v cos cp — u)u Y^ =. V- — 2u (ycos qi — ii) __ 2(v cos cp — u)u (324) (325) The speed for maximum efficiency will be when n = ^v cos q)f (326) which reduce (325) to 330 EFFICIENCY P = ^Mv COS (p Pu = ^Mv"' cos^ cp V^ = (1 - i cos^ cpy [216.] (327) Vv^ + u- — 2vwcos q})i V'^ — V- — 2(v cos q}~- u)u — 2u sin q) ^Jv^ + %^ — "Iva cos q) . (328) Case Y. Let the vane he flat and normal to the stream, and inove in the direction of the stream. In this case /? = 90°, 9> = 0, which in (315) to (318), or (325), (326), (327), give P ^M{v-u) Pu= M{v — u)^i F' = tr - ^uv + 2r'^ (32^) which will be theoretically a maxhiium when the circumfer- ence of the wheel has one-half the velocity of the jet ; in which case (339) becomes ■^-^"32i X 2,200 ^""'^ n early, 1,132 - ~ Q^ (310) where A is the head due to the velocity v. If S be the area in feet of the section of greatest contraction, we have .:Jir = 0.4:558 Shi (341) The theoretical maximum efficiency in this case will be (5th of equations (330)) but this considerably exceeds the value realized in practice. The several losses due to the contraction of the vein, the clear- ances^ about the wheel through which the water escapes with- out acting upon the wheel, the lack of normal impulse, tlie imperfect action of a thick vein, and other causes, combine to reduce the theoretical efficiency. Experiments show that a 336 WATER WHEELS. [223.] well-made wlieel will realize about 60 per cent, of the theoret- ical efficiency, giving for fair practice, equation (339), For a practical maximum efficiency we have €x = from 0.30 to 0.3S. (343) H\\Q j>ower oi the wheel is independent of its size ; and hence may be so proportioned as to make a desired number of revo- lutions per minute. 222. The PoJsrcELET "Wheel. — M. Poncelet, a celebrated French scientist, improved the undershot wheel by making the vanes so curved that the water upon leaving them would ilow backward rela- tively to the vane. For this case, equations (331) and (332) are applicable, and we have theoretically, Fig. 171. TT-n n ^0 X 62i ^ , -, (3i4) and for a theoretical maximum IIP lie "''^^^^^• (345) If the velocity of the circumference of the wheel be |- that of the jet, and the buckets be so curved as to completely re- verse the direction of motion of the water, the wheel should be perfect, or e=l, (346) but these conditions not being realized, combined with the [323, 224.] WATER WHEELS. 337 losses noticed in the preceding Article, make the inactical ef- ficiency in good wheels about 0.60. (347) 223. -A. BREAST WHEEL is one in which the water is admitted at some point opposite the face of the wheel, and the water retained in the wheel to the lowest point by means of a curved trouo-h, or passage way. The jet will enter the wheel with a velocity, but if the buckets be properly curved and the velocity of the wheel be properly regulated, there will be, theoretically, no loss from this cause. After the water has entered ig the wheel it will act by its weiffht throuo;h the remain- ing height. Let h be the head due to the velocity, and lix — CD — the height through which the water acts by its weight ; then Fig. 173. will the maximum theoretical power be (348) If II = EF be the entire fall of the water, the theoretical power will be 621^77, and hence the ilieoreticdl eQiciency will be li + llx (349) Experiments with the best wheels of this class have given an actual efficiency of --1 ei =.75 to .80. (350) 224. Overshot wheels receive the water at or near their highest part, and retain it in buckets during its descent. If 23 338 EXAMPLES. [224.] retained only by buckets, the water spills out before reaching the bottom of the wheel, and thus produces loss of efficiency ; and if the water be retained by a curved trough, this class of wheels will not dif- fer essentially from the breast wheel in theory. If the water acts by its weight through the height CD = h then will Fig. 173. the mechanical power be Pu =AM{v - u)u + 62|^Ai, (351) where ^ is a coefficient whose value in the ease of flat vanes will be unity, as shown by the 2d of Equations (329), and for a complete reversal of direction of the jet will be 2, as shown by the 2d of equations (331) ; hence the real value of A will be more than 1 and less than 2. If the velocity of the circumference, ^l^ of the wheel be half that of the water when it enters the wheel, and there be no loss from the impulse, then — ^—^ — will equal AB = h ; and we have Pu=e^Q{h+ h,), (352) the same as (342), and the efficiency of the wheel would be perfect. But there will necessarily be some loss from impulse, and an additional amount from clearance in the curved trough, and still more from the imperfection of action due to the thickness of the jet, all of which combine to make the practi- cal efficiency of M^heels well constructed and properly oper- ated, of e, = 0.75 to 0.80. (353) [234.] EXAMPLES. 339 Examples. 1. Water approaches an overshot wheel with a velocity of 12 feet per second, the buckets moving with a velocity of 6 feet per second wliere the fall is 20 feet, but on account of empty- ing the buckets 4 feet are wasted ; find the efiiciency, and if the supply be 600 cubic feet per minute, find the horse power. The relative velocity of 6 feet pei* second in the race is equiv- alent to a head 7i = r— . The height through which the water 2,) = v^ + uK (355) The actual velocity of the jet will be V-u; which. will be less than v for any finite value of ?^, and the pressure due to the reaction will be, equation (297), p = 31 {V- v) = 31 (Vv- + u^ - u), (356) 844 REACTION MOTORS. [227.] ■which may be compared with the first of equations (333). Tlie mechanical power will be Pii = M{^ir + V' - u) u, (357) which has no theoretical maximum. Developing, we have Pu =. 4 M ( V' -:r-^ + 7r-i - —-, + etc. ) , (358) which is \Mv^ for u ~ oo, for which case the motor would be perfect. But it is impossible to realize this condition, and it is found in practice that the efficiency is a maximum for u equal nearly to v. Making -m = v, in (357), we have Fu = Jf ( V2 - 1) u' =0A14:2Mif. (359) The potential energy of the water having the head AB will be 621Q ' AB = yVv" ; (360) hence the theoretical efficiency of the motor will be _ 31 (Vt^^ + v^ — ^0 ■* = 9 = .r^; (361) Vn' + V' + u ^ + ^ in which, if v = u, we have 6 = 0.82+. (362) A vessel floating on water might be propelled according to this principle. A pump, drawing water from the bow of a vessel and forcing it out at the stern, would propel the vessel ; and for velocities u > v would be theoretically quite efficient. 227. Barker's (or Seguin's) Mill consists of a vertical hollow shaft communicatino- with hollow transverse arms. [227.] REACTION MOTORS. 345 The water is admitted into the upper end of the hollow shaft, and passing downward, flows horizontally into the hollow arms, escaping horizontally through orifices near their extrem- ities, one orifice being on one side of one arm, and the other on the opposite side of the other arm. The deflection of the water from the vertical to the horizon- tal direction will cause both a hori- zontal and vertical pressure, as shown in Article 21 ; but the horizontal pressures will neutralize each other, while the vertical pressure will be resisted by the support, and the water will flow into the arms with a velocity ^^^^||||C unaffected by these pressures, and, if -^?^.^^^KSt|P the arms are stationary, the water will ""-^^^^lLJ^Z^^^K^^ be discharged at the orifices with a yk. it5. velocity due to the head in the vertical shaft. But wdien the arms are rotating, the water in them will be forced radially outward on account of the centrifugal force developed, and thus the head of discharge relatively to the orifice will be increased. Or, instead of considering it in the light of cen- trifugal force, we may consider that the water has a velocity common with the arm, which alone would be equivalent to a head producing a velocity. As the water approaches the ends of the arms it will be deflected through an angle of 90^, thus producing a pressure radially outward, and also trans- versely to the arms, the former of which will be resisted by the solid parts of the machine, and the latter Avill produce the rotary motion. The two reactions — one at each end of the outer extremities of the arms — produce a couple, the moment of which will equal the moment of the resistances overcome by the machine. Let v = the velocity due to the head h — C 'B, a = the arm to the orifice, CO = the angular velocity of the arms, u = the actual velocity of the orifices G and 7^= aco^ Y= the actual velocity of discharge, Aj = the head due to the velocity u. 34G REACTION MOTORS. [227.] Then, as shown in the preceding Article, and the actual velocity of discharge will be Y—n— \/a'co^ + V- — acj ; (364) and the pressure due to the reaction Avill be F = 31 (a/«'c^ + t;2 _ aco) ; (305) and the mechanical power resulting, will be Pu = 3f{Va^co^ + v' - acj) ao) ; (366) and the theoretical efficiency will be _ 3f{Va-Gj- + V- — aoo)ao3 2«ca V + aoa^ (367) '\rhich has no maximum in reference to co as a variable, but approaches unity as a limit as co is increased indefinitely. In practice it is found that the efficiency is a maximum when rcD is about equal to v ; and making roo = v, and neglecting all losses, except that due to a loss of actual velocity of dis- charge, we find for the efficiency, ^ = 0.82+. (368) If there were several orifices at distances «', a", a'", etc., from the axis of the shaft, producing reactions P', P", P"\ etc., giving expressions similar to those in equation (365), and li the resistance overcome wdth a uniform velocity w at a dis- tance 1) from the axis of the same shaft, then w'ould the mo- [22"-J REACTION MOTORS. 347 ments of the reactions equal the moment of the resistance overcome, and hence lih - P'a' + P"a" +, etc., = 31' {V<^('^'^ + t^ — a 00) a + J/" {Va^^o^^ + v^ - a" go) a" + , etc. ; (369) and the mechanical power will be Pdco = M' {V'-i''co^ + v^ — aoo) a' GO +, etc. (370) An analysis of the second member of the last efjnation gives the following : — the quantity Vo-'^go^ + v^ — a go is tlie actual velocity of discharge as shown by equation (364) ; M' {^a'^Gi? -{■ v^ — a' go) is the momentum of the mass If', as shown by the definition in Article 27; and JI'iVd'Go^ + u^ — a'co) a' is the moment of the momentum, as shown in Ar- ticle 166, or in Article 168 of the Elementary 2fechanics. Hence the entire exj^ression, M' {V<-i^'^(i^ + v^ — (i'go) cl'go^is, the moment of the momentum of the mass M' inulti]?lie(l hy the angular 'velocity, and similarly for the other terms. Separat- ing tlie expression into other terms, we have 3T{V(t'^Gj^ + '('^)a\ which is the moment of the momentum of the mass M' havino: the velocity 's/cl'^go^ + v^ in reference to the orifice ; and Ma GO • a', which is the moment of. the momentum of the mass 31' having an actual velocity a'co, equal to the actual velocity of the orifice, wJiich, being in a direction opposite to that of the discharge, will be negative. Each of these mul- tiplied b}' the angular velocity, go, will give tlie corresponding mechanical power developed ; and similarly for the other terms containing 31 ', 31", etc. In,.this case the water enters the arm without angular velocity, still the principle shown in Article 166 is general, for the effect produced is tlie result of tlio mutual action and reaction between the water and arm. Hence, 3J:8 REACTION MOTORS. [228.] W/i£?i iDciter hj Us reaction jy^'ochices a Totary motion, the momenz of momentwni into the angular velocity c<2iials the work done. Tliis niacliiiio lias been the subject of many improvements. To remove as mucli as possible the resistance of the water in passing through the arms, sudden turns, eddying, etc., the arms have been made large where they join the body of the shaft, and gradually contracted and curved as they pass outward toward the openings. Instead of arms, there may ^_B be a disc liaving openings which permit the water to escape tangentially, wdiich form is yf^ ^ ^_^ ^ ^^ known as Whitelaw's Turbine. In the form shown in Fig. 175, the pivot at the lower end of the shaft supports the mill, and all the w\ater running through it, tlius producing Fig. iro. much friction. This objection has been in part removed by introducing the water into the mill from the under side, thus producing an upward pressure, and counter- balancing, in part, the weight of the machine. But notwith- standing all these improvements, the mill has been superseded by other more efficient motors. 228. To determine the pressure due to the centrifugal force. — Let A be the uniform cross-section of the arms, co the angular velocity, j9 the pressure per unit, m the mass of a unit of volume, p any distance from the axis of rotation, tlio initial and terminal values of which are 7\ and ra respectively, then will the mass of a transverse section be mAdp, and the centrifugal force will be inAdp'Ou-p, wliich will equal Adjj ; Adj^ = niAdr pdp\ [328.] KEACTIOX MOTORS. 849 or i^ —pQ = \mcir{n — r% (371) wliich is the pressure required. If 10 be the weight of a unit of vohmie, we have ^^ (I - w) = ^^ (^^ - ^''^ = ^' ^'' - '''^' ^^ ' ^) wliere h is tlie head due to the velocity r^oo, and h^ tliat due to ;'i(i7. The velocity due to the difference of these heads will be u = ^%j Qi — //(,) = oo\/rl — rl ; (373) in which if i\ = 0, and r^ = a, we have u = acj, (374) which is the same as u in equation (363). But both the velocity of the orifice and centrifugal fo7'ce are not involved in determining the velocity of discharge. Centrifugal force — or energy — may he discarded in determining the efficiency of a motor in all cases where the head due to the centrifugal force has heen induced by the rotation of the wheel. For this head tvill have heen secured at the expeiise of the energy of the outflowing jet, and, in turn, this head will impart to the outfloioing jet exactly the same amount of energy as that tvhich it received. Or, hriefly, in a reaction ivheel, or turbine, centrifugal force is self neutralized. 1. Find the horse power, number of revolutions per min- ute, gallons of water discharged, and efficiency of a Barker's mill, neglecting friction. Area of nozzles 1 sq. ft., radius 3 feet, total head 30 feet, speed of orifices |tlis that due to the head. ^i =z ^\/2g X 30 = 32.88 feet per second ; 27Tn X 3 = 32.88 x 60, .-. 71 = 104.6 ; 350 TURBINi:S. [229.] Y = ^ti' + 2(/k, or V= 6LS ; «, = 5Miil^8i^O= 24,594. JijficiencT/ = jy = 0. 1 o. Horse power = 140.7. 2. A rotating wheel receives 2 tons O- water per minute at a radius of 2 feet, and delivers it at a radius of 3|- feet ; on entrance, the water is rotating with a velocity of 14 feet per second, and on delivery it is rotating in the opposite direction with a velocity of 4 feet per second. Find the couple exerted on the wdieel, and if the wheel makes 200 revolutions per minute, find the HP developed. 9 X 2240 Moment of momentum, on entry — ^ "^"^ x 14 x 2. 2 X 2240 Moment of momentum on deliveini = -tk — "*!,," x 4 x 3i. •^ -^ 60 X 3ii ^ ^ Sum = Cou2)le = 98. rrj^ 98 X 2;r X 200 „ _ ^^ = - 60 550 = ^'^' TURBINES. 229. The general definition of a turbine is a water motor rotating about a vertical axis ; but it is usually restricted to those horizontal wheels in which the water is conducted to the vanes by curved guide-plates. Tliey may be divided into four classes : 1. Outioard flow tu7'hines, in which the water flows hori- zontally outward from the central shaft, and is discharged at the outer circumference of the wheeh 2. Inward floio tui^bines^ in which the water is admitted at [230. J TURBINES. 351 the outer circumference mid flows horizontally inward toward the central shaft. 3. Parallel flow turbines, in which the water flows down- ward (or upward) through the wheel. 4. Mixed flow turbines, in which the flow may be com- pounded of the 1st or 2d with the 3d above. 230. Fourneyron's Turbine, invented about the year 1827, is a g-ood type of an outward flow turbine. The central shaft Fig. irr. (7 of the wheel, to which the driving mechanism is attaclied, passes down through the supply chamber Gil, from which it is separated by the tube LL. 352 TURBINES. [230.] The water passes downward between the guide plates oo, and outward between the same, througli the gates MM, thence against and along the vanes BB, finally escaping at the outer circumference of the wheel. The vanes curve backward so as to cause the water to be discharged in a direction opposite to that of the wheel, relatively to the vanes. The vanes are attached to the sliaft Ghy means of the rigid armsj^. To find the mechanical power expended upon the wheel, let, in Fig. 177, V = the velocity of the water as it enters the vanes, y = PQB = the angle between the tangent 7^^ to the guide plate, and QB, the tangent to the inner rim of the wheel, Vt = the tangential component of the velocity, Vr = tlie radial component of the velocity, r^ = OQ — the internal radius of the wheel, a = angle between the direction of tlie water as it escapes from the wheel, and XS' the tangent to the outer rim of the wheel, v', i;/, V;.', 7'3 = respectively, the velocity of discharge, its tan- gential and radial components, and the radius of the outer rim. Then, Vt = V cos Yf V,. = V sin y ; {^^^) Vt = v'cos a , Vr'= v'sin a . (376) The moment of the momentum of the radial velocities will be zero ; hence, the moment of the momentum of the water as it enters the wheel will be 3IriVt , and as it quits it, it M'ill be Mr,v: ; hence, Article 227, the work imparted to the wheel, neglect- ing friction, contractions, eddying, etc., will be Fu = 3I{riVi - 7\v;) CO ; (377) [230.] TURBINES. 353 and if h be the head in the supply chamber above the point of discharge, the efHciencv will be ^^{'\v,-r,v;)c^^ (378) g/i Equation (377) is a fundamental one in the theory of tur- bines. The velocities v and v' will generally be independent of each other, depending upon the form of the guide plates, the vanes, and the speed of the wheel. For the highest efficiency, the water must quit the wheel with no velocity. There will, however, practically, always be a radial velocity, but when this is reduced to a minimum by the proper construction of the vanes, a Tnaximxiin efficiency will be secured by running the wheel at such a speed as to make the tangential velocity of discharge zero, in which case we have vl = 0, Fic = Jlr.vcco, (379) e = 'j^. (380) (j/t The coefficients of 31, equations (377) and (379), divided by g, are the effective heads for the respective cases. If to the efiective head there be added the head due to the loss of velocity, there being no frictional resistances, the sum will be the head in the supply chamber. The only head lost in the case of maximum efficiency will be that due to the radial velocity of discharge. For this case, «, equation (376), will 1)0 90°, and Vr = v. But if %o be the velocity of the water along and relatively to the vane, and ii\ and i^v the relative tangen- tial and radial velocities, respectively, ft = the angle SXT, Fig. 177, between the tangents respectively to the outer rim of the wheel and the vane ; then 10 1 = 10 cos ft ; v.\ = w sin ^ ; ) ^ Wr = u't tan ft; j 23 354 TURBINES. [230.] the head corresponding to which will be w^t tan^ /? -7- 2(7 ; hence, for the case of maximum ethcienc}', h = {2riVtGo + 7^jf tan^ /5) -r- 2^ ; (382) from which we find that the velocity with which the water enters the wheel is V = Vt sec r = i-^ — ^-^-^ — '-) sec y. (883) Hence the velocity of entering the wheel varies inversely as the velocity of the inner rim. As limiting cases assume that ;/ = 0, and /3 = 0, and we have v=-^. (384) If rico = V, then v'- = ffh, (385) or the velocity will be that due to one-half the entire head. If rioj — ^Vj then v' = 2fjh, (386) or the velocity will be that due to the head. If ViCJ — Iv, V- = 4:(/h, or the velocity will be that due to twice the head, which is an hypothetical condition. The above analysis is equally applicable to the other classes of turbines ; but the construction must be different for each class in order to realize the conditions here imposed. Thus it will be found that /3 must be larger for a parallel-flow turbine than for an outward flow, and still larger for an inward flow, if the transverse sections of the passages through the wheel arc uniform. Its value may also depend upou Ti and ?'i, as will be shown. [-•^1-] TURBESTES. boo 231. Special Cases. — 1. In. the Fonrneyron turbine the initial elements of the guide plates are radial, and passing out- ward from the axis of the supply chamber, are gradually curved until they finally terminate as nearly tangent to the inner rim of the wheel as practicable, allowing sufficient space between the plates for the passage of the required quantity of water. The least value of y, tan r = — , (387) gives the greatest possible tangential velocity to the water when it enters the wheel. The initial elements of the vanes are also radial, but in passing outward they are curved backward, and at their terminus make a small angle, /?, with the outer rim. These conditions give tan /5 = ^ . (3SS) In. order that the water shall enter without impulse, the velocity of the inner rim of the wheel must equal the tangen- tial velocity of the water, or, Vt = nco ; (389) in which case the initial velocity of the water relatively to the wheel will be the radial velocity v,.. It will then pass along the curved vanes and be finally discharged backward relatively to the vane with a tangential velocity Wt, but the vane has a forward motion of r»Gj = n}\Go ; hence the actual tangential velocity of the water will be ^'/ = '?'-'! — n7\co, (39v0) which for the best resnlt must be zero, for which condition we iind, combined with equations (3SS), (389), and (387), w, = nr,cj = -^=:nv,= -^^^'^, (391) tan /J tan y ^ ^ ^ .'. tan r=^^~ tan /? ; (392) to*. 356 TURBINES. [231.] which establishes the relation between the outer angles of the blades and vanes relatively to the I'espective rims of the wheel. The radial velocity through the wheel may be made const^int when the Mdieel passages are full, centrifugal force being neglected, by making the transverse sections of these passages uniform; and since their breadths vary as the radii ?'i and /'o, the depths must vary inversely as the same numbers, thus making the axial sections hyperbolas. This being done, Vy. = iv,., and we have tan Y = n tan A (393) which is the condition commonly assumed. These conditions being realized, we have for the Fourneyron type — frietional losses being abstracted — equations (380), (382), (389), and (390), ^""2 +n^tanV^' ^^^"^^ 2.yA (395) ^"' 2 + ;r tan^ /i and from (379), P„=||e.|=ir/,^_^-l^-^^; (390) from which it appears that the efficiency is greatest when /5 is least. If /5 = ; then e = 1 and v = Vgh ; and the head due to the velocity with which the water enters the wheel is one-half that due to the head in the supply chamber. Since the water enters the Avheel with a small velocity relatively to the vanes, Vi tan y, and is discharged with a greater relative velocity, w = nviOJ sec ^, it follows that from the point of entrance the velocity of tiie M'ater relatively to the vane is accelerated to the point of discharge, while the actual velocity diminishes along the same path. 2. In the cvp vane turhine the guide plates are essentially the same as the Fourneyron type ; but the water issues from them [231.] TUEBINES. 357 with a velocity due to the head in the supply chamber, and hence enters the wheel as a jet. The vanes are vertical, and their surfaces are as nearly semi-cylindrical as the circum- stances of construction will permit. To avoid loss from shock, the initial elements of the vanes should be tangential to the direction of the water relatively to the vanes as it enters them. The velocity of the inner rim of the wheel should, according to Article (211), equal one-half the tangential velocity of the water as it enters the vanes ; hence r^Go = ^v cos y. (397) The radial velocity of the water as it quits the guide-plates will be Vr = v sin y ; (398) hence, if y' be the angle between the vanes and the inner rim of the wheel at their intersection, we have tan y' = II^^LJL = 2 tan y. (399) \v cos y If ;/ = 0, y' will also be zero, or the terminal elements of the guide plates and the initial elements of the vanes will both be tangent to the rim of the wheel. The radial velocity of the water being unaffected by the motion of the wheel, and the tangential velocity relatively to tlie wheel being one-half that due to the head, the velocity relatively to the vane will be v' = fyVcos" y + -^ sin- ;/, (400) and if the motion were rectilinear, the velocity relatively to the vane would be uniform, and the water would quit the ' wheel with a velocity v' = w\ but since the velocity of the elements of the vanes varies directly as their distance from the centre of the wheel, the relative velocity will vary. But, in any case, we have, for best efficiency, n}\co = w cos /5 ; (■^^1) and the velocity lost will be w sin (3. 858 TUEBINES. [231.] The energy imparted to the wheel will be, equations (379), (397), Pu = IIvt'Ti 00 = Mv COS y-lv cos ;/, = Jil/^'^ cos- y. (402) The kinetic energy of the water as it leaves the gates will be, W being the weight of water flowing per second. Wh = lllv" ; hence the theoretical efficiency will be iify^ cos- y „ (403) (404) 3. In the inward flow turhine the initial elements of the guide curve are at the outer circumference of the supply chamber, and curve so as to form a small angle with the outer rim of the wheel. The outer elements of the vanes are radial, and as they approach the inner rim of the wheel, they curve in a direction opposite to that of the guide plates. Since n, in equation (393), will be fractional, the angle will exceed y. The water should enter the wheel with a tangential velocity equal to the velocity of the outer rim. The remainder of the analysis is the same as for the Fourneyron turbine. If the vanes are cup-shaped, the analysis will be the same as the second case above. 4. In the jparaUel flow turbine^ the initial elements of the guide plates are vertical, and are curved as they pass downward so as to terminate as nearly horizontal as possible. The initial elements of the vanes are usually vertical, and the vanes curve in an opposite direc- tion from that of the guide plates, termi- nating at the same angle with the horizontal. This last is a result of making w = 1 in equation (393), giving [232.] TURBINES. 359 5. If, in an outward flow turhinie^ the vanes make an angle with the inner rim, exceeding 90° and less than 180°, which will be an angle gi-eater than for the Fourneyron type, and less than for the cup vane, we shall find rco> Vyh and < ^/'ighy for a speed of maximnm efficiency. 6. An unlimited number of turbines having different forms of vanes may be devised which will give, theoretically, high efficiencies ; but if friction, whirls, contractions, etc., be in- volved, those forms will be, practically, most efficient, which for the same theoretical efficiency, have the least of the above named resistances. 232. GrENERA-L CASE. — In any hydraulic motor, let H be the total available head, h the loss of head due to location and form of wheel, or that portion of the head below the wheel (and possibly above) not utilized, h' the loss by shock, h" the head lost on account of the velocity of the water at quitting the wheel, h'" the head lost from contractions, whirls, friction, etc., then will the mechanical power be e,^Q[H-h- h' - h" - A'"] ; (405) and the gross efficiency of the motor will be H-h-h' - h" - h'" H (406) The efficiency of the motor as a machine may exceed this, for it may be so constructed and placed that its entire efficiency will be developed with a less head than H\ as, for instance, in the case of an overshot wheel, if there were 4 or 5 feet fall below tlte lowest point of the wheel, it would in no way affect the efficiency of the wheel, but the wheel would fail to utilize the power of the water. In determining the quality of the wheel as a motoi", only so much of the head should be 360 TURBINES. [232.] considered as is actually necessary for operating the wheel in the place where it is established. EXAMPLES. 1. A stream delivers 10,000 gallons of water per minute to a Fourneyron turbine. If the fall be 100 ft., tan /3= ^,u= 1.4, and revolutions per minute i= 240 ; find the internal and external radii for the most efficient speed, the HP, and the depth of the casing at the outer periphery. From equation (395) we find o , /2 X 321 X 100 2 + {lAf X i ' or n = 2.14 feet. r., = n/'i = 2.996 feet. From (394), efficiency = 2 + (1.4)2 ^ ^: .90. ^ 10,000 X 231 X 62^ x 100 x .9 ~ 1,728 X 33,000 * = 227.8. From (391), Wj. = Vr = ni\Go tan /5, or V,. = 25.1328 feet per second, and , 10,00 X 231 X 12 ~ 1,728 X 25.133 x 67r x 60 = .56 inches. 2. An outward flow turbine is supplied with water having a head of 350 feet. The internal diameter is 15 inches, and the external diameter is 21 inches. Angle of guiding plates, [232.] TURBINES. 361 ;/ = 16°, and tlie vanes of the wheel are placed at the corre- sponding angle. Depth of wheel at the inner periphery, 2 inches. Find the number of revolutions per minute required for the best efficiency, neglecting frictional resistances. Taking account of frictional resistances, and supposing the true efficiency to be fths, find the best number of revolutions and HP developed. From equation (395), = 104 feet per second. y 2.0822 104 X 60 27r X 1,589 revolutions per minute. If we take frictional resistances into account, and suppose the wheel to be running at its greatest efficiency, we will have for the total energy, mgQh = inQv\ + -$ v\n'' tan" /? +''I^:2Fv\{\ + nHanV^). Therefore the efficiency is e = or 2 + n- tun^ (i + :2F{X + n"" tan^ fi) ' •yf =r \ forces are functions of the distance between the centres of force, Article (151), a condi- tion not realized in this case. The particles act upon each other according to a law not definitely known. It may be partly of the nature of friction, but chiefly it results in the production of whirls at the angles of the tube. It is also known that the reduction of the kinetic energy develops heat, and this, being retained in the tube, partially maintains the head, so that the loss is not as great as it otherwise would be. The action is not exactly of the nature of the impact of finite bodies, but resembles it in regard to the non-elasticity of the watew and the comparatively sudden reduction of the velocity ; and for these reasons we analyze the case as for inelastic im- pact. In all cases, however, when the law of action is unknown, the result should be checked by direct experiment. We may assume that the mass m passing out of the small tube in a unit of time impinges upon the mass 31 in the enlarged [234. J RESISTANCES. 863 part of the tube ; then, according to equation (42), the loss of energy will be pi{v-v,)\ (407) and it is found that this expression represents with sufficient accuracy the results of experiment. This amount of energj^ transformed into heat would raise a pound of water a number of degrees in temperature given by the expression _ m{v-v,y ^ - 2x772 ' ^^^^^ where 772 is Joule's mechanical equivalent of heat. The head lost will be 234. Resistances in long pipes. — The principles of work and energy on which Bernoulli's theorem is founded may be extended to the case in which there is resistance alono- the stream between A and JB, as is the case in actual tubes. The lai'j of the resistance can be determined only by experiment. It has been found that, with sufficient accuracy, the resistance varies as the perimeter of contact between the liquid and pipe in a transverse section, called the loetted perimeter^ also as the square of tlie velocity, and a factor/ dependent upon the con- dition of the pipe. Hence, if B, be the resistance at any sec- tion, s^ 10 the weight of a unit of volume of the fluid, and c the wetted perimeter, we have i 2t = fiocr-— . 2g If the velocity and sections be uniform, then will the loss of head ibr a length I be 364 RESISTANCES. [385.1 The value of y for rivers, as given by Eytelwein, is where a = 0.007164, I = 0.007409. (411) 235. Geneeal case. — Let AB be a stream flowing from a reservoir whose upper surface is at C\ and whose exit is at B. Or, to generalize it, let B be any paint in the pipe. Through B draw a horizontal GB, then will OG = BD be the total Fis. 180. head on 5, neglecting the pressure of the air as we may do in most practical cases. If B be the exit, let BL be the head necessary to overcome the pressure at the orifice. If the opening be the full size of the pipe, BL will be zero ; but if it be contracted, it will be of finite value. Lay off on BD^ tlie followino; heads : » I) II = ^h = o~ ~ ^^^^ head due to the velocit}' of exit, ^I = 7^2 = the head lost due to the resistance of entering the pipe at A^ IM — /ig = the head necessary to overcome the resistances along the straight portions of the pipe, MN — 7/4 — head lost by bends in the pipe, NO = 7^5 = head lost by sharp angles, [235. J EESISTANCES. 365 OL —Jif!, — head lost by enlargements and contractions, LB — h-j = head due to pressure at B, DB = H= total head at B. Then B-=h^ + h, + h, + h, + h + h + K (412) The values of these several heads, with the exception of the first, are determined by experiment. They are usually de- termined as multiples of hy = — , and hence may be written with the use of corresponding subscripts, thus Zr=(l+/,+/3+/4+/5+/c+/7)|. (413) The following are some of the principal values of the coefficients as determined by experiment : Orifice in a thin plate, which would be applicable when the length of the pipe is zero, /a = 0.054. Straight, short cylindrical mouth-piece perpendicular to the side of the vessel, /o = 0.505. For the resistance along the pipe whose length is Z, section Sj wetted perimeter c, "Where/" is dependent upon the dimensions of the pipe, and according to Darcy, its value is 866 RESISTANCES, [235.] Where d is the internal diameter of tlie pipe in inches, and k for clean cast-iron pipe, is Z: = 0.005; and for pipe very rough from sediment, k = 0.01. According to Weisbach, for clean cast-iron pipe, 0.0043 / = 0.0036 + V'? where v is the mean velocity. The larger computed value of f would ordinarily be used. For a bend through an angle V, the radius of the curve being r, diameter of the pipe Fig. 181. /4 = ^|0.131 + 1.847(|;)^'|. n\ For a knee, or sharp angle, «', /g = 0.946 sin= | + 2.05 sin* ^ , and if the knee is 90°, this becomes /s = 1 nearly. The value of /"« may be computed approximately, and f^ is dependent upon the condition of the opening at exit. These values in (413) enable one to determine v; or -y and the other quantities being known, H, the necessary head, may be determined. Indeed any one of the many quantities may be found in terms of the others. To find the pressure on the pipe at any point, as J, lay down from CD the head dh due to the velocity at h, hi the head lost at entrance A, im the head lost by the resistance of the straight part of the pipe from A to h, mn the head lost by [235.] RESISTANCES. 867 bending between A and h due to bends, no that lost by angles, ol all other heads lost ; then will l^ = pressure 011 pipe at h. The head hg is potential in reference to BG. The line FL limits the upper ends of the heads due to pressure on the pipe ; and if the pipe Lay along this line there woukl be no pressure except on the under side due to the weight of the fluid. If an orifice be made at J, the liquid M-ould flow out in a stream rising to the height hi, less the resistance at exit from said orifice and the resistance of the air. But if the pipe lay along FL, the liquid would not flow out of an orifice at I. E XA3*IPLES. 1. Let it be required to deliver water with a given hydraulic head Jiq, such, for instance, as is necessary to drive an engine. Then where ^Fiq the sum of the coefiicients due to valves, bends, knees, sharp-edged entrance, etc., and / tliat for surface fric- tion. Therefore we have IT H~h ^^ l + ^^i^+4/V -^ d from which the velocity is obtained. If Q be the required quantity, and S the section, then 2. Water flows from a tank through a 1 in. vertical pipe. Find the head in the tank so that the velocity of discharge may be the same for every length, taking into account the resistance of sharp-edged entrance as well as of sm'face fric- tion. 368 RESISTANCES. [235. J Let / be the length of the pipe, and H the head above the orifice ; then i...= (i./..f)l' 2^' or ^=(i+/4,-(rJ-0'- But the conditions require that the term containing I shall disappear ; hence ^ yb _ and i?=(i+/-)J. where v is independent of I as required. Then and d 4 X .005 ( 1 + ^) (1 + i) = 3.7 feet. 3. Water flows from a tank throngli a uniform sloping pipe of diameter d. Taking into account the resistance at entrance, show that the water will flow with the same velocity what- ever be the length of pipe, if the pipe slopes at an angle, sin (9 = — (^^¥^) ^^0, Ih being the head in the tank above the entrance. [236.] GASES. 369 We have, as before, We will suppose 6 so small that the resistance to sharp- edged entrance may be taken as F'— |. We have/" = .005 ^ d ' and m ~ — . Hence we have 4 which may be regarded as an equation of identity in I. There- fore equating like terms, or Also 2g 1.5 --3-— ^i— ^u- GASES. 236. The density of an ideal, incompressible fluid is inde- pendent of the pressure to which it is subjected, and depend- ent only upon its constitution ; but the density of a compress- ible fluid is dependent upon both its constitution and external pressure." Thus, if water were strictly non-compressible, its density at all depths in the ocean would be the same as at the surface ; but the atmosphere, being a gas, diminishes in den- sity as we ascend, since the weight of the atmosphere above 24 870 GASES. [237.] any point is less the higher the point. The laws of the pressnre of fluids given in Articles 174 to 178 inclusive are applicable to gases. But if the i:)ressui'e varies, the density and temperature both varj'- according to laws which are deter- mined by experiment. 237. Boyle's (or Mariotte's) Law. — According to the experiments of Boyle and Mariotte, the volume of a gas at uniform temjjerature varies inversely as the pressure to which it is subjected. Hence, if v^ be a known initial volume of gas, andj!?o the initial pressure per unit of area to which the gas is subjected — being the pressure upon its bounding sur- face, or at any point within it, for which latter reason it is also called the tension of the gas — and v and j) any other con- temporaneous volume and pressure ; then we have 2W =2)i)^\ — constant = m, (414) which, if j> and v vary simultaneously, is the equation of an equilateral hyperbola referred to its asymptotes. In testing this law experimentally, it is necessary, after compressing or dilating the gas, to make the temperature of the gas the same as the initial temperature ; in the former case cooling, and in the latter heating it. The "q — X law is found to be very nearly, but not exactly, correct for air and other known gases, the agreement being nearer the more perfect the gas, from which it is inferred that it cor- rectly represents the law for the ideally perfect gas. Since the density, S, varies inversely as the volume, we have, for uniform temperature^ 6v = SqVo = constant, (415) and 6p,= 6,p. (416) [338.] GASES. 871 238. To FIND THE TENSION AT ANY POINT OF A COLUMN OF GAS OF UNIFORM TEMPEKATUKE. — On account of the Weight of the gas and its compressibility, its density will vary as some function of the height. Conceive a prismatic column of gas whose base is unity, and at the lower base jh ^^^^ tension, d^ the density, w^ the weight of a unit of volume, and at height z let the corresponding quantities be 2>i ^^ '^' Then will the pressure on the lower base equal the pressure at tlie height z added to the weight of the prism of gas of that height ; or, and differentiating, wdz. dj) = — wdz, (417) which is independent of the law of ])ressure and of the total head. According to Mariotte's law, 8\S^'.\-\-^..])'.ip,, (418) and considering gravity as uniform, ^ = ^> (419) Substituting and reducing, integrating between the limits^;* and^j'o? ~ -iid 0, gives 1 P ^^'o . log ^—= -Z' .-. j9=:j9oe ■''" . (420) 872 GASES. [239, 240.] Similarly, w = Woe ^'°"; S=doe ^'^ . (421) The weight of a prism of gas of height h, will be W = xcdz (422) If /<- = CO , 7F =^->o, as it should. 239. N^uMEEicAL VALUES. — The mean pressure of the atmosphere at the level of the sea is ^j>o = 14.7 pounds per square inch, or 2116.8 pounds per square foot. The weight of a cubic foot of pure dry air under the press- ure of 14.7 pounds per square inch, and at the temperature of melting ice (32° F.), Wq = 0.0S0728 j>ound avoirdupois. (423) If the atmosphere were pure, dry, and uniform, and of the density as at present at the level of the sea where the tem- perature is 32° F., the height would be H^ 2116.8 0.080728 26,221 feet (424) or the height of one atmosphere of uniform density is nearly 5 miles; but as the atmosphere does not fullil these condi- tions, and being a little heavier, the height is a little less than this value; At the level of the sea the barometer stands at 29.92 inches nearly. At 5,000 feet above " 24.7 At 10,000 feet (Mt. ..Etna) 20.5 At 15,000 feet (Mt. Blanc) 16.9 At 3 miles 16.4 At 5 miles 8.9 240. To find the tension at extreme heights when the variation of gravity is considered. [240.] GASES. 373 From equation (418) we have ^ = !^.£^. (425) Let R be the radius of the earth, z any distance above tlie earth, then and (417), (425), (42oa), give ^ ^ _ !^ ^ f/. . integrating between the limits of j) and ^oj ^ and 0, Jog - = ^-- — , w„ liz and similarly, .•.i^=i^o^ ^"'^^^; d=(yo^~^^''^"\* (425?') Assuming the density of air at the earth d^ = 4^0 of a pound per cubic foot, as it is very nearly, and R = 20,860,000 feet, and equation (4255) reduces to -aiSyr^— 6^-,U^0) '''\ .(425.) If 2 = GO , we have ^ = 400 xW ' ^^'^'"^^ M-liich is the limit of the density. If gravity be considered uniform, we would have for the height equal to the radius of * A more cumbersome demonstration of this formula may be made by- means of Prop, xxii., B. II., of Newton's Principia. 374 GASES. [241.] the earth, 2 = 20,860,000, and (421) and (418) reduce to the same as (425d). 241. Heights may be determined approximately by tlie pressure of the atmosphere. If its temperature were uniform and free from currents, we would have from equation (420), ^^£o^ £0^26,221 log. 1% where 5o and h are the readings of a barometer at the lower and higher stations respectively. Adapting this to the use of common logarithms, we have oq qo z = 26,221 X 2.30258 log ^^ = 60,735 log ?^^. But this coefficient is known to be too large, and practice shows that 2= 60,345 log ?^ (426) gives better results. If «i be the height of another station, then 2, = 60,345 log ?^; from which subtracting the former, we have z^-z = h = 60,345 log \ . (427) This result, however, requires to be corrected for the effect of the temperature on the mercurial column, the effect of the variation of gravity due to great heights, change of gravity due to latitude, the reduction of observations when not simul- taneous, the effect upon the mercurial column due to the [242.] GASES. 875 attraction of the mountain on which observations are made, all of which are well discussed by Poisson in his Traite de Mecanique. \i T — the temp'ture of attached thermometer at lower station, nr" u « ■ (( upper a t = " detached li lower u t = " " u upper u q> = the latitude of the place ; then, ._..o..l + 0.00102 (^ 4- t' -64) X logio 0.002551 cos '2cp 1 h 1 + 0.0001001 (r- T')_\ feet. (427«) Observations should be taken at both stations simultane- ously, in calm weather; but when this cannot be done, two observations may be made at one station at different hours, and one at the other at a time midway between, and the mean of the former be used as one observation. 242. GrAY LussAc's (or Chakles') Law. — It was found by these men by direct experiment, that the increase of tension in a fixed volume of gas is directly proportional to the increase of temperature. If i->o? ^05 "^05 ^^ respectively the initial pressure, temperature, and volume of a gas, and p, t, v, any other corresponding contemporaneous values, and /3 a coefficient of expansion, then we have, the pressure being constant, v = v,{l + ^{t-to)), (428) the volume being constant, p=Po{l+/^{i-to))' (429) From equation (428), 376 GASES. [243.] In practice, the initial temperature is taken as that of melt- ing ice, 32° F., or 0° C, and the initial pressure that of one atmosphere, or 147 lbs, per square inch, or 29.92 inches of mercury (760 millimeters). The initial volume may be of any convenient value. For these values it has been found that for carbonic acid, /? = 0.0037099 for 1° C, for air, /? = 0.0036706 for hydrogen, /? = 0.0036613 " From these values it appears that the more perfect the gas, the less will be the coefficient of expansion ; and Kankine concluded, in accordance with tlje theory of molecular vortices, that the limiting value is j3 = .000365 = ^4-j for 1° C. = 0.0020275 = — ^ for 1° F., (431) which values are now used for the ideally perfect gas, and are sufficiently accurate for air and several other of the more perfect gases. 243. Absolute zero. — Equation (129) becomes, for ^o = 0°C., i'=i^o(l+2^4)' (^^2) in which, if if = — 274°, p becomes zero, and a perfect gas would be destitute of tension. Similarly, according to (428), the vohune would vanish. In Fahrenheit's scale ^o is 32°, and equation (429) becomes in which, \i t — — 461.2, p becomes zero as before. Accord- ing to the modern theory of heat, at this temperature the molecules of the gas would be at perfect rest. It is the point of absolute deprivation of heat, and is called the zero of also- [244.] GASES. 377 lute temjperature. ISTo such temperature can even be approxi- mated by any known process, tlie lowest recorded temperature obtained by artificial means being — 140" C. (— 284:° F.) ; but an extension of the law of uniform expansion leads to this result. Temperatures reckoned from the absolute zero are called ahsolute temperatures^ and are especially useful in simplifying many formulas in regard to heat. If t be the temperature on Fahrenheit's scale, and r the same tempera- ture on the absolute scale, Tq = 493.2, the temperature of melting ice, then r = 461°.2 + ^ = To - 32° + t, and equation (433) becomes !> ^Ih- . (434) '0 If w be the weight of a cubic foot of dry air at pressure /> pounds per square inch, and at absolute temperature r, then xo = w^^ -°: (435) and if w be the weight at temperature r' and tension _p', then y,'=^o^'L' s=dj-^.'^; (436) and since for gravity uniform, the volume varies inversely as the weight, ^; = ro^'•-. (437) Since the mechanical properties of gases, whether at rest or in motion, involve the property of heat, we now consider some of the abstract properties of the latter. 244. Heat is a form of energy. It is not force, since force is only one of the elements producing energy. See Articles 25, 26, 151. It is believed to be a certain manifesta- 378 GASES. [245, 246.] tion of the motion of the particles of a bod}^* The following principles, the result of observation and experience, have become established. 1. Heat may be transformed into external work ; and con- versely, work, under certain conditions, may be transformed into heat. 2. Heat cannot be transferred from a body of a lower to one of a higher temperature except by the aid of a machine and the expenditure of mechanical energy. 245. The thermal unit is the amount of heat energy necessary to raise a imit of weight of ice-cold water one degree on the thermometric scale. The English thermal unit is the amount of heat energy necessary to raise one pound of water from the temperature of °32 F. to 33° F. ; the value of which, as found by Joule, is the heat produced by friction in bringing a body weighing one pouiul to rest after falling Y72 feet in a vacuum.t Or, more briefly, the English thermal unit equals 772 foot-pounds of work. This is called Joule's equivalent, and is usually represented by J. The equivalent in French units is the heat energy neces- sary to raise one kilogramme of water from 0° C. to 1° C, and equals 424 kilogramme-metres. The amount of heat energy necessary to raise the tempera- ture 1° is not the same from all temperatures, although for ordinary mechanical purposes it is considered constant for the same substance. 246. Specific heat is the heat necessary to raise a unit of weight of any substance one degree on the thermometric scale, the thermal unit being unity. Strictly speaking, specijic heat is a ratio, being the quotient obtained by dividing the quantity of heat required to raise the temperature of one pound of the substance one degree by the quantity required to raise the temperature of one pound of water the same amount. * Elementary Mechanics, p. 69. Stewart on Heat. f Elementary Mechanics, pp. 71, 72. [247, 348.] GASES. 379 The specific heat under constant 2^i^€ssure of gases, is the specific lieat determined when the gas is permitted to expand under the constant pressure. "We denote it by e^. The specific heat under constant volume is the specific heat determined when the vohime of the gas remains fixed. We denote it bv c^. The former always exceeds the latter, for in addition to increasing the molecuhir motion of the gas, external work is done against the pressure of the air, expanding it about 0.0020275 of its volume for each degree, as shown in equation (431). If the same quantity of heat will raise one pound of water m degrees, and one pound of any other substance n degrees, both under constant pressure, then 7n c„ n •-P The following are the values of Cp, c\,, and Cj, -^ c,, for a few gases : GrAS. Cp Cv Cp — Cv Air 0.238 0.169 1.408 Oxygen 0.218 0.156 1.400 Hydrogen 3.405 2.410 1.412 Steam 0.480 0.370 1.207 247. The dynamical specific heat is the specific lieat ex- pressed in foot-pounds of work. If K„ be the dynamical specific heat under constant volume, and Kp that under con- stant pressure, then K^ = Jc^., Kp — Jcp ; (438) where J= 772 foot-pounds, or 424 kilogramme-metres. 248. Adiabatic curve. — Let q be tiie heat energy neces- sary to change the temperature of a unit of weight of a gas t degrees, the pressure being ^;, and density S. Then q=f{p-S-r). (439) Differentiating, j) being constant, and reducing by the aid of (436), gives i 380 GASES. [248.] dq\ _ dS dq _ d dq and differentiating, considering S as constant, e, = m='k/h^P.(?); (445) wherey is an arbitrary function, that part of which depend- do I mg upon -ji is implicitly a function oip and d, as shown by [248.] GASES. 381 1-v (444), and the other part, 6 -r- 2> y , is explicitly a function of the same quantities. The product of the two parts is some function of the same quantities expressed in terms of the ratio J9 -~ S ; the entire transformation being made so that an integral may be obtained, though it be in a functional form, jp" represents some other arbitrary function, and q) the inverse of it. In the particular case where q remaijis constant^ as it would when a gas is compressed or dilated in a close vessel imper- vious to the transmission of heat through its walls, although the indicated temperature would change,' we would have for another pressure j9o, and density 6q, and eliminating cp{q) between these equations, we have ^^l =J)o^^\ (446) and since the volumes will be inversely as the densities, we also have 2)v^ = 2^o'v^ = consta7it, i'^'^'^) and from (436), (446), (447), we have v-l Fig. 183. Equation (447) is the sim- plest form of the equation of the Adiabatic curve^ or "curve of no transmission" (from the Greek a-, not, and Sza/3aiy£iv^ to pass through), and is represented by the curve AB. If an isothermal line pass through the point A, where the volume of the gas is 1 and pressure j?^, it 382 GASES. [249.] will pass above the adiabatic for values of v > 1, and under it for values of v < 1. Examples. 1. If a given volume of air has a temperature of 85° F., what will be its temperature when dilated to double the vol- ume, performing work, but receiving no heat ? We have, from equation (-MS), 2^\ 0.403 V J ~ 461.2+ 85, 461.2 + t ' .'. i = - 50° F. 2. If a prism of air having a tension of 1| atmospheres at a temperature of 88° F. expands adiabatieally to a tension of one atmosphere, required the final temperature. We have /I Y-29o^__ 461.2 + t .,, 549.2 .-. t = 24°.35 F. 3. Eequired the ratio of the volumes before and after ex- pansion in the preceding example. Ans. 1.335. 4. If air be compressed adiabatically from a tension of 15 pounds at 50° F. to that of 90 pounds, required the final tem- perature. 5. If air at 60° F. and six atmospheres expands adiabati- cally to one atmosphere, required the final temperature. 249. Velocity of a wave m an elastic mediuim.* Assume * The general problem of wave propagation has received the attention of several of the most eminent mathematicians since the days of Newton, and many problems have been solved in a satisfactory manner. The simple method of Newton, Principia, Prob's XLIII.-L, B. II., has not been excelled, and the definite theoretical result obtained is quoted to the present day, although the effect of heat upon the velocity of sound was not then known. La Place, in the Ilecaniquc Celeste, tomes II. and V., has treated of the oscil- lations of the sea and atmosphere ; Lagrange, in the Mecanique Analylique, [249.] GASES. 883 that the medium is confined in a prismatic tube of section unity, ^the coefficient of elasticity for compression, p a force which will produce a compression chj in a length dx, then from definition we have ■^ dx The lamina dx will be urged forward — or backward — by the difference of the elastic forces on opposite sides of it, and as the quantities are infinitesimal, this difference will be cZj? ; or Let D be the density of the lamina, then its mass will be M— Ddx, and we have from equation (21), page 18, ^"^""de -^ dx' or, df~Dd7?' ^^ ^ which is a partial differential equation of the motion of any lamina. Let^-f-Z> = a", and adding a ^-4- to both members, ' ^ dxdt we have dt \dt dx) dx \di dx. tome II., has discussed the problem of the movement of a heavy liquid in a very long canal ; M. Navier published a Memoire on the flow of elastic fluids in pipes, in the Acndemie des Sciences, tome IX. ; and M. Poisson wrote several Memoirs on the propagation of wave movements in an elastic medium, and the theory of sound, for which see Journal de VEcole Poly- iechniqite, 14tYi chapter, and of the Academic of Sciences, tomes II. and X. These eminent mathematicians established the basis of all the analysis for the solution of the problem. More recently we have M. Lame's Lemons sur I'fjlasficife des Corps solides, and Lord Rayleigh's Treatise on Sound, both of which are works of great merit. 884 GASES. [249.] Let F-^+a'^^ (450) at ax ^ then where the parenthesis indicates a partial differential coefficient and \dx J d,t -«5|; (^52) and equations (449), (451), (452), give dt J~ \dx The total differential of V =f{x, /) is by substituting (452), and integrating, , d(x + at), dx ' F=^(. + .0 = |+«|, («3) where i^is any arbitrary function. Similarly, subtracting a -^^ from (449), '^'=/(--«0 = |-«|. (^51) [249.J . GASES. 885 Adding and subtracting (453) and (454), and we have the respective equations ^^ = iF{x + at) + if i-i- -«/■), :l=4^(-+''')-^y<--«')- But and substituting from above, gives dy = — F{x + at)d{x + at) -^ — f(x - at)d{x — at) ; ^a Aa integrating, y = 'p (x + at) + (p {x — at), (455) ■where rp and q) are any arbitrary functions whatever. Their character and initial values must be determined from the con- ditions of the problem. The equation represents a wave both from and towards the origin. If the wave be from the origin only, the ip function may be suppressed, and. we have y = cp{x- at), (456) and differentiating, CI) = '''(■" - "')■ which is the rate of dilation (the expansion or contraction of a prism of the air), and (1) = " ■ *''(*' - "'^ ' 386 GASES. [249.] which is the velocity of be the pressure per square foot, w the weiglit of a cubic foot, and H the height of a homogeneous atmosphere, equation (i24), then u=A/-ni=VgII', (459) y|' = vpr; hence the velocity of sound should equal the velocity of a body falling through a height equal to one-half the height of a uniform atmosphere. This principle is applicable also to the vibration of elastic cords, and it is found that The velocity of vibration of an elastic cord equals the velocity of a body falling freely through a height equal to half the length of the same cord ivhose lueight icould equal the tension. Similarly, in water neither too shallow nor too deep. The velocity of waves on the sea equals the velocity of a body falling freely throiigh a height equal to half the depth of the sea. It has been assumed that ^and D remain constant in wave motion ; but it Avas long since known that the results given by equation (458) for gases did not agree with those found by experiment, and La Place showed that the elasticity was increased by the action of the wave due to compression. It is necessary, therefore, to consider that the expression is cor- rect only for ultimate values ; or « = |/5-- («») dD' [250, 251.] GASES. 387 Since EOZ wy, clE— '^'-^dv^ — ^^^^dD, w w f /^= Vg^. (461) 250. To FIND THK VALUE OF y, \ve luivc froiii equatioD (461), w y^fj,n^\- (^62) "by means of which it may be fonnd when the velocity of sound in a gas of given weight and tension is known. In this way the vahies of y have been fonnd for a variety of sub- stances, a few of which are given in Article 246, It is con- sidered as constant for any given gas, and nearly constant for all the more perfect gases. 251. Remarks. It is difficult to find the specific heat of a gas under constant volume by direct experiment, bnt by means of equations (462) and (442) it may be readily com- puted when the specific heat at constant pressure is known. In this way the values of Cp, given in Article 246, were deter- mined. If the temperature of air in its quiescent state is uniform, the tension will vary as the density, equation (416) ; hence P ~ w will be constant, and the velocity of sound in any gas will be the same at all temperatures or densities, bearing in mind that the factor y appears on account of the condensation resulting from the transmission of the wave. But if the tem- perature varies on account of external causes, we have, equa- tions (435) and (461), Y' ^y^^ (^^^> in which the velocity is made to depend upon the absolute temperature. If r,, = 493.2, jh -^ Wq = 26,221 feet, equation 388 GASES. [252.] (424), and g — 32| ; the velocity in dry air at any tempera- ture r, will be ,,^ / 32i X 1.408 X 26,22 _1 ^^^Qggg . /Ifeet (464) Examples. 1. Required the velocity of sound in air at the temperature of 0° F. 2. Required the velocity of sound in air at 95° F. 3. If the weight of a cubic foot of hydrogen at the tension of the air, 14.T lbs. and 32° F. be 0.005592, required the velocity of sound in it at 80° F., y being 1.4. 4. Required the velocity of sound along a steel bar, the coefficient of elasticity, p, being 29,000,000 pounds per square inch, and w = 486 pounds. u = |/^ 000,000 X 144 X 321 ^ ^^^^^5 feet per second. 486 The velocity of sound in solids depends much upon their homo2:eneity. Experiment shows that its velocity is from four to sixteen times as great as in air. The above equations- are no more than approximately correct for solids. 252. Velocity of discharge of gases through orifices. If the density and temperature remained uniform during- flow, the law of discharge would be the same as for liquids^ and the same formula would apply, and we would have for the velocity, v^ = 2gh, where h would be the reduced head, being the height of a prism of the gas of uniform density equal to the density at the orifice. If the temperature be uniform, the density will vary according to Mariotte's law, or p — Po^^ Let u be the velocity, then will the mass passing a transverse section of area unity in a unit of time be Su, and the difference in pressures on opposite sides of a section will [252] GASES. 389 be djp, which will equal the momentum of the elementary mass, or ,,,^^. dp= - dudu ; (465) in which the quantities have contrary signs, since the greater the velocity, the less the difference of the pressures. Divid- ing by^?, integrating, where p is the tension of the gas in the reservoir, jf^j the ex- ternal pressure, and if the gas flows into the atmosphere it will be 14.7 lbs., w^ the velocity of exit, and u the initial velocity in the receiver. If the receiver be large, and the orifice small, p may be considered constant and u zero, in which case we have in which, if 2h= 0? ^/l = go , or the velocity of a gas into a vacuum would be inflnite according to this hypothesis. If the temperature and density both vary, the effective head will vary. Letting z be positive downwards, we have, from equation (417), dj? Ulp "J VJ Following the method of Joule and Thomson, and consider- ing gravity as constant, lo will be a function of p) and ;/ ; and substituting and reducing by means of equations (418), we have !^-2,3 = f'iioi.^^^^.Z_o.I^ri_^1; (468) in which p^ is the tension within the reservoir, and p^ t^^^t just outside ; then if t, = 0, the flow will be into a vacuum. 390 GASES. [253, 254.] and according to (-iGS) the velocity will be a maximum, the value of which will be (469) which is i/ z, times the velocity of sound; and for air y r-i becomes V = 2,413 a/ — feet per second ; and if the air be 32° F., then V = 2,413 feet per second, or about 2.2 times the velocity of sound in air. 253. The volume of gas flowing out per second will he Q = sv, (470) where s is the area of the contracted section. 254. The wkight of gas flowing out per second will be, equations (448), (468), 470), ^^,^ = '.!^.l!i(l!^y . (471) i^o n \lh- which is a maximum for W ~ Pi~ V + 1^ ^^1^ ' which for air becomes 1^=0.8306, ^-0.527, J^ = 0.6345. APPENDIX I. SOLUTIONS OF PEOBLBMS. PEOBLEMS. CONSTRAINED MOTION. 1. A particle is placed at the extremity of the vertical minor axis of a smooth ellipse, and is just disturbed; show that if it quit the ellipse at the end of the latus rectum the eccentricity must satisfy the equation e^ + be* + "Se^ = 5. Let V be the velocity of the particle at the end of the latus rectum, then v^ = ^h=2g(h-\y (1) Also in equation (148), p. 193, we have p-a(l-0(l+Os X=0, Y=-mg. From the equation to the ellipse, dx 1 „ -— = lor X = as, ds Vl + e' hence, v^ = ga{l- e-) (1 + r)i ^^7= = ^« (1 - e^ (3) Equating (1) and (2), 2ya { Vl - e' - (1 - e') ) = ga (1 - e*), or. 2VI - e- = (3 + e-) (1 - e') ; hence, "'"' e^ + 5e' + 3r = 5. 2. A particle slides doiun the upper surface of a frictionless 393 394 PROBLEMS. wedge, the wedge leiiig free to slide on a friciionless horizontal plane; determine the motio7is. Let W = the weight of tlie body, W'= the weight of the wedge, 7^ = AC, b = AB, 6 = ACB, ]V= normal action between the body and wedge, E = the vertical reaction of the plane AB on which the wedge slides, a = DB the distance moved by the wedge, b — a = AD, ACD = a. Take the origin of coordinates at C, x positive to the right, and y positive vertically down. The forces acting on the body are N and W, and on the wedge iV, TF', and R. The angle <9„ equations (143), is — 6, and B^ — 90° + Q ; and the equations of motion become — '^ = IT COS 90° + iYcos (- 6) = JV cos d, g df - ^= TFsin 90° + iVsin (- 8) = TF- J\^sin 6 ; g dt- 0) (2) (3) W dhi' iL-'ll. = w' sin 90° + iV^sin (180 + 6) + B sin 270 = ; (4) g dt which is zero, since the plane is fixed, and hence there will be no motion in that direction. g df = W cos 90° + N cos (180 + 6) + R cos 270= — iVcos 6. CONSTRAINED MOTION. 395 From equations (1) and (3) we have dt"' ^^ df' integrating, w '^^ - * dt ~ ^ dt ' and again, Wx = - W'x ; or. and Wil-a) = - W'a; . Wb W- W Let s be the distance from C down the plane, then y = s cos 6, differentiating, reducing: X = s sn\ tf — — y d-y= cos d^s, d''x= sin 8 d's — j d^y, sin |9 — — cos d)d^s', a which, in (1) and (2), gives dri fiX cos 6 W- N sin dt' ~ .,rf ■ n « ^\ "~ W cos 6 ^' ^^^ \y ( sni V — Y cos 8 Solving, TFTf'sin^ ~ TF' sin^ 6^ + ( TF + TF') cos^ ^ ' ^°^ and iVis constant. 396 PROBLEMS. Integrating (1), (3), and (3), gives dc W -ir = Wv^ — a jY cos 6 • t, at W^ = Wv^ = g {W - N ?\n e)t, W dx' dt W'vJ= -gNeosd . t; and integrating again, gives Wx = \gN cos e • t\ Wy =^jj{W- N&m d)r; W'x'= -^gWcos e . t\ (9) (10) (11) (12) (13) (14) Making x = h — a in (12), x = a in (14), and we find equa- tions (5) and (6) as before. Also make y = h in (13), and find iV^as before. Then equations (8) and (13) give if == \' 2h g cos cos- a + W W+ W -,sm' 0) (14') Integrating (1) and (2) in reference to x and y, and we have for the velocity with which the particle reaches the foot of the plane. = V2gh From (11) find WW ( W+ W')[{ W+ W) cos^^+ W sm'6]i_ (15) W W sin* d \ . (16) {W + }V){W + W) cos'^ 6 + W' sin" ^J The velocity with whi«h the particle reaches the horizontal CONSTRAINED MOTION. 897 plane is readily found by the principles of energy. Thus the work done by gravity equals Wh. Hence we have Wh = i^v' + i^v''; (17) which, reduced by the aid of (16), gives (15), If pf' = 00 , we have » i\^= Tf siD e, (18) ^ =^/^e> ^''^ V = V2P, (20) t;' = ; (31) which are the formulas for the motion of a particle down a ^m.ooi\i fixed plane. For the inclination of the path of the particle, we have I -a W , n tan«r = -^=-^^^p^tan^. A comparison of (15) and (19) shows that the particle acquires a less velocity when the wedge is free than when it is fixed. Questions.— 1. Show that whatever be the relative weights of the body and wedge, 'the common centre of gravity of the two masses will remain in the same vertical line during the motion down the wedge. (See Art. 149.) 2. Is the principle of the conservation of energy, Article 151, illustrated by this example ? 3. What is the relation between W and W, that will give a minimum velocity for the body IT? (Eq. (15).) 4. Given W, equation (15), is it possible to assign such a value to W or to d, or to both, so that vwill be \V^gh iV^^ or - ^/'igh ? 398 PROBLEMS. 5. Is there an algebraic minimum to the normal pressure, equation (8) — or an algebraic maximum ? 3. A homogeneous cyli7ider rolls doimi tJie upper surface of a wedge ivitliout slip>ping, the wedge heing free to move on a fric- tionless horizontal plane ; required the motion. The problem may be generalized, providing the successive elements of contact of the body and plane shall be in the succes- cive elements of a cir- cular cylinder circum- scribing the body. The centre of gravity will move in a vertical plane, and the body may be a cylinder, a screw, a coil of wire, a sphere, an ellipsoid of revolution, a volume of revolution having two circles of contact with the centre of gravity between, and other forms. Eolling may be secured by a string (or flexible band) wound around the body, one end being secured at the upper end of the wedge, at C, and the other to the bod}^ Let W = weight of the body, h its radius of gyration, W = weight of the wedge, N= normal reaction be- tween the body and plane, T the tangential action producing rotation, DB~z, AB = i, AD = h-z, AC = Ji, ACB = 6 ; origin at C ; x positive to the right, and g positive downward, r the radius of the circle of contact. The forces acting on the body are gravity, W, acting vertically downward, the normal reaction, iV, acting directly away from the plane, and friction, or tension of the string T— whichever produces the rotation — acting upward along the plane. For the equations of the motion of the centre of the body we have (167), Wc^x y df = Wcos 90° + .Vcos (- 6) + Tcos (270° - 6), (1) W d-y _ 9 df TTsin 90° + iV^sin (- 0) + Tsin (270° - 6). (2) CONSTRAINED MOTION. 899 The forces acting upon the wedge are its weight, W, acting vertically downward, the normal reaction, uV', of the horizontal plane acting vertically upward, the normal reaction, JV, between the body and wedge acting downward, and the pull, T, acting along the upper surface of the plane downward. Hence we have iLf^ z= W' cos 90° + J\^' cos 270° + iYcos (180° - 6) g dt' ^ ' + Tcos (90°- 8). (3) The origin of moments being taken at the centre of the body, we have, equation (lo2),for the moment of rotation of the body. at- (4) These equations may be reduced to T sin d TV drx „ . — = JV cos (9 N&m d— TcosS 9 df ^—~= _ j\rcos^+ Tsin 6 9 dt' a dt- (5) Integrating twice, assuming N and T constant, as they are. W — X 9 W — y 9 J -9 L^{h-z)= l{Nco^ d - Tsin 8) f, W li = |-(Tr- .Ysin d - Tcos d) t\ — 2 = i (- -^'cos ^ + Tsin B) r- — 1c"-cp = l-Trf, 9 * (') (8) (10) 400 PROBLEMS, When the body rolls completely down the wedge, we have r(p=zCB = \^¥T¥ = a (say) (11) which in the preceding equation gives 2 Wh^a T = which reduces the preceding equations to W ¥a sin d F(6 -z)= \gN cos e . f Wh = ig{W - N&in e)f Wh'a cos 8 W'z = yNcos 6 ' f — Subtracting (15) from (13), gives Wb Wk^a sin z = i~z = W+ W" W'l ir+ w" (12) (13) (14) (15) (16) (17) From (13) and (17), 7 + -F- W -vW A _ g cos ifN N (say) fronj (14), Wh + B Wlc"a cos 6' .-. N=z AW B -{■ A sm 6 r' J^(lf-iV^sin^)' (say) ; CONSTRAINED MOTION. 401 t = \/ T=- B + A sin 8 W 2 W'k'a Wk"- cos ft rqiB + A sin 6) „ . ., . W ^ r sin- u -—. —- W + H 7 + r- cos' 6 + F y= r*IF(^ + ^ sin 6^) V. = i/ TF i^ + ^ sin ^ 2 ITX-'g - r'ffA r' W' The entire work done by gravity is TT7/, and this equals the energy of the body due to the velocity of the centre, plus the energy due to the rotation of the body, jjIus the energy due to the motion of the •\vedge ; or Wh = iMv- + iMk'co"- + iM'v'\ 4. A particle slides down a fridionless arc of a fixed vertical circle ; required the motion. The forces acting on the particle will be gi-avity, W, and the normal action, iV", between the body and arc acting normally outward, and will be the difference between the nor- mal component of the weight and centrifugal force. Using polar co- ordinates, origin at the centre, the angle between r, the radius of the arc, and a vertical diameter, and resolving normally and tangentiallv, we have ■ — r ^7T = IT sm 6, (J dt- (1) |V(L';)=..-eos.-.V. (2) 402 PROBLEMS. Integrating (1) gives = 2g (1 - cos 6), (3) which is independent of the mass. In (2) this gives i\^= (3 cos (9 -2)IP''. (4) At the point where the body leaves the arc JV = 0, for which condition (4) gives cos ^ = § ; (5) .-. (9 = 48° 11' +. The velocity at this point is, equation (3), r§=VWr- (6) The subsequent motion will be that of a projectile, having the initial velocity of (6) and angle of depression of 48° 11' +. The normal pressure, JV, varies inversely as 6, as shown by (4). To find the time of the contact, we have from (3), f "^ •' cos dft a/1 — cos & losr tan iff The superior limit makes f =z —cc , which shows that the particle will not start from rest at the highest point. It will be necessary either to give it an initial velocity, or place it at a finite distance from the highest point. 5. A lochj rolls down the arc of a fixed circle without sliding, starting at a point indefinitely near the highest 2)oint of the arc; required the motion. CONSTRAINED MOTION. 403 Let be the centre of the arc GB, C the centre of the rolling body, W = weight of the body, R = OD, r — DC, the radius of the circle of contact, h = the principal radius of gyra- tion in reference to a hori- zontal axis through C, 6 = BOD. Eesolving tangen- tially and normally, we have for the motion of the cen- tre C, E(R + ,.) 'l^ = r sin d + .Ysin 180'' + 2^ sin 270°, g ^ df- E(E + r)f'^) = Weos 6 + iV^cos 180° + ^cos 270°, W{E + 0(^)'= "> cos d - Ng. or And for rotation we have (152), g dt- (1) (») (3) where q) is the angle which some fixed line CE of the body makes with the vertical ; thus cp will be zero when is zero. Since arc ED = arc BD, and DCW = d, DCE -^ cp - 6, we have r{cp-8) = Rd; .'. cp E + r r Solving these equations gives T k- + r W sin 6. (4) (5) 404 PROBLEMS. ^■. 9P 3 COS ^ + ,, , (1 - cos 8) -2 k' 4- r^ W. (6) When equation (G) becomes zero the body leaves tlie arc. Making iV"= 0, Ave find If the body be a sphere we liave h"^ = |r^, and cos 8 = {^; .'. 6 = 53° 58'. To find the velocity of the centre of the sphere when it leaves the arc, substitute 7" from (5) in (1) and we find {R + r)^ = y sine, (8) and integrating and reducing, we find (^ + r)^= V\'g (1 - cos d) {R + r), (9) which is the velocity at any distance d from the vertical. Making cos 6 :=\^, we have for the required velocity, ^^g{R + r). (10) To find the angular velocity of the sphere at the point where it leaves the are, substitute dd found from (4) in (9) and make cos 6 =. \^ ', Ave thus find From this point the sphere will continue with the uniform angular velocity given in (11), (the body having rolled by fric- tion, or at that point being freed from the string), and with the initial A^elocity giA^en by (10), the centre moving as a projectile. The sphere Avill strike the horizontal plane tangent at the lowest CONSTRAINED MOTION. 405 point of the complete circular arc, when its centre is at a dis- tance r above that plane ; hence the sphere will be a projectile through the height h = Rcos cos-^ 1} + R —r = \\R-r. (12) The time of movement as a projectile will be given by equa- tions (a), p. 177, and will be given by the equation f + 0.2254 ^R + r ' t=z 0.1049372 - 0.062177r. During contact it will rotate on its axis i2 , 10 COS"^ -n 27rr 17 times, and as a projectile it will rotate — times. 7t If the arc be not infinitely rough, let jj. be the coeflBcient of friction, rotation being caused entirely by friction ; required the motion for this condition. The friction will be / ; . •. dy'=^—{R + r) sin i^^tp, (5) B{cp-e-(3) =z r{d'-cp + ft) ; R{dcp-dff)^r{dd' -dcp). (6) Eliminating JV, J", and T' from equations (1) and (2), gives d'e .d'X d-x' d'd' Integrating once, the initial yalues being 0, gives ^Ax dx „ dd' „^r-r,dd dt dt dt ' (7) (8) from which we see that ivTien -^ = -^y^ • —^ , we have MV = dt MR dt ' — mvj or the moments are 7iumerically equal, and the centres CONSTRAINED MOTION. 409 are moving in opposite directions, {V and v being respectiYely the "velocities of the centres). Eliminating dO, dd', and dx by means of (3), (4), and (6), gives l{M+m)'^ = m{R + r)(3,-co^cp)^, (9) and |(i¥+ w) ^J- = [m (f - cos q)) + |(i>/+ m) cos (p]{R + r)-£. (10) From (9) it appears that tJie motion of the centre of the lower sphere loill be positive so long as cos cp is less than f , or ^ < 66° 26'. Should the spheres separate at a less angle, they will con- tinue to roll in the same direction. From (9) it appears that when cos = f, the centre of the lower sphere will be at rest, in which case (3) shows that there will be no rotary motion ; in short, the lower sphere will then be at rest. Equation (10) shows that the motion of the upper sphere will be continually positive. By means of the equations given above, a complete solution may be found, giving the nor- mal reactions and angular velocities. 11. A imiform rod of length a, capable of making comjjlete revolutio7is iti a vertical plane about one extremity, is placed in a vertical position tvith its free end upiuard, and being slightly displaced, moves from rest ; find the time of revolving from an angle /? to an angle 6. The equation of motion is, Art. 126, a'^cVd a . a ^^ q" Ji^ =" ^'^3 Sin 6, or 3 cW 2 a (I'd Multiplying both members by 2dd and integrating between the limits 6 and 0, we have 410 PROBLEMS. (If 1 /2a\ i , ^ or (U 1 /2a\ i Hence the time fi-om ft io 6 is /2a\* ^ tan J-^ S^r/ ^ tan {ft' 12. ^ ro^ rests with one extremity on a smooth plane and the other against a smooth vertical ivall at an inclinatimi a to the horizon. If it then slips doivn, show that it will leave the wall ivhen its inclination is sin "'(-§- sin a). Let the mass of the rod be m, its length 21, its inclination to the horizon d, and the coordinates of its centre of gravity x and y ; the origin being such that for the time, t, considered, a; = Z cos 6 and y = I sin 6. Let the horizontal and vertical reactions at the ends of the rod be -ffand F respectively. Then the equations of motion are fZV ,ff (sin ff) Tr /,N d^x -, d^ ("cos ff) r7-ff ^ml'jj = Rl sin 6 - VI cos 8. (3) Multiply (1) by cos 6, (2) by - sin 6, and (3) by (1 -f- 1), and add the products. There results COXSTEAINED MOTION. 411 -J a whence, since 77 = *^ '•^'^^ 6 = a when «" = 0, at — = —%- Tsm a — sm d). (It- 2 1' (5) The rod will leave the vertical wall when //= - ml i sm Oj^ + cos 6^^ 1 = 0. Substituting in this the values of -^ and j- given hj (4j and (5) we have = sin"^ (f sin a). i^TJie Analyst, 1882, p. 193.) 13. An angular velocity having ieen impressed on a Jietero- geyieous spliere, aUut an axis perpendicular to tJie vertical plane wliich contains its centre of gravity G, and geometrical centre O, andpiassing through G, it is then placed on a smooth horizontal plane. Find the mag?iitude of the impressed angular velocity that G may rise into a point in the vertical line SCK through C, and there rest ; the angle GCS being a at the beginning of the motion, a the radius, and cp the required angular velocity. Draw the radius CGA, and from G drop the perpendicular GM to the plane. Let m = the mass of the sphere, k the radius of gyration of the sphere about an axis through G perpendicular to the plane containing C and G, R the mutual reaction of the sphere and the plane, SM = X, GM = y, CS =- a, angle A GM z= angle ACS= (p, and CG = c. Since there is no friction, we have for the motion of the centre of the sphere d'x - (1) 412 PROBLEMS, resolving forces vertically, m-^=R-mg, (2) and taking moments about G, 'm¥ -^? = -Re sin cp, (3) q) being the angular motion of the sphere. We have y = a — c cos qj, whence d-y . d'cp wdqP' Substituting in (2), R =1 mic sm cp-^ + c cos (p-j^^ + ffj- This in (3) gives by reduction, (c- sm- cp + k-) -Tij + c- sin qj cos cp-^-= — eg sm q). (4) Integrating, {& sin' qj + ¥) -^Y = C +2cg cos — cos a). (6) Now if the initial value of ^ = o', the terminal value — a + n, hei and wlien also ~ = ; then the left member of (6) becomes zero. Acg cos a & sin'- ix v Tc'^ ' (The Analyst, July, 1882.) kinetic energy. 413 Kinetic Energy. 14. A hall, mass tn, radius r, is shot with a velocity v into a perfectly liard, smooth tuhe of length a, radius r', mass m', free to turn about its middle point, which is fixed, imparting to the tuhe a rotary motion ; if the hall just reaches the centre of the tuhe, required the angular velocity of the latter. In this case the kinetic energy of the ball and tube due to the rotary motion, will equal the kinetic energy of the ball before it enters the tube. Let Jc and k' be the radii of gyration of the ball and tube respectively, and oj the required angular velocity, then, equation (153), Imv" = ^mh'oo- + hn'lc'oo^ ; mh" + m'h'- ' If the tube be considered slender, we will have k'"' = yV«^ ; also k^ = |r% hence, 60mv^ CO = . 24:nir- + 5m'a^ 15. A cone, mass m and vertical angle 'ia, is perfectly free to move about its axis, and has a fine, perfectly smooth groove cut in its surface, snaking constant angle fd ivith the elements of the cone. A heavy particle, mass M, moves along the groove under the action of gravity, starting at a distance c from the vertex ; required the angle through which the C07ie has turned when the particle is at a distance r from the vertex. Let p be any variable distance from the ver- tex, (p the angle through which the particle has moved, k the radius of gyration of the cone, and 6 the required angle. The geometrical relations give p sin « = radius of the cone at the place of the particle at time t, p sin adqi = the horizontal arc through which the particle moves in time dt, dp tan /? = the same arc ; 414 PROBLEMS. .-. dp = p sin a cot ^dqj. (1) The monicut of the momentum imparted to the cone in an element of time, will be at The angular advance of the particle will be dq) — dd, and the horizontal component of the momentum of the particle will be ,^ . /dcp - dO Mp Sin « ' dt and the moment of the momentum will be ^dcp — d& Mp" sin^ a dt Since gravity has no horizontal component, the horizontal motions will be due to the action and reaction between the bodies, and these moments must be equal ; hence ,^ „ . , /dcp-de\ ..-.dd . Eliminating dcp by means of (1) and (2), we have gJfsin^ apdp mk- + MfX sin- a 2 sin a cot (5dd ; fi— ^ *^^ /^-i ^ '?»^'''' + ^^^'' sin" a ~ ^ sin « '^ mk"^ + Mr sin" a ' 16. An elastic ring, mass m, natural radius a, modulus of elasticity e, is stretched around a cylinder; the cylinder sud- denly vanishes ; find the time in which the ring will collapse to its natural length. _ /nam 17. A prismatic lar in a vertical position rests on a invot at its loiver extremity ; it is slightly disturbed, required its kinetic energy when it loill have rotated 180°. KINETIC ENERGY. We have, Prob. 19, p. 216, and Art. 107, 3^ . and integrating 1 (W ^ a - 2 at- 2 1 -3^. 415 which in equation (153) gives 1 ' r '^^' Wl that is, tJie e^iergy is the same as if the iar had fallen freely through the vertical descent of the centre of gravity of the har. 18. If a sphere, "pivoted on a horizontal dia- z meter as an axis 'without friction, oscillates al)out an external axis ; required the hinetic energy ivhen vertically under the support. Let tlie body rotate about the axis of y, then will equations (177) be applicable, and we shall have AB = l, L, = 0,M, = W'l sin 6, iVj = 0,Mg = W,x = l sin 8, z = I cos 6 ; .-. d-x = — / sin 8d6- + I cos 8d'd, (Pz = - Z cos Odd- - I sin dd'd ; and these, in equations (177), give dW g . n which, integrated, gives 1 d8- a 2df=l''' .r= |- (1 - cos 6). The velocity of the centre of the sphere at the lowest point will be 416 PROBLEMS. ,dd and the kinetic energy of the mass when it rotates through 180° •will be iM.r~ = 2Mffl = 2Wl; hence in this case also, the kinetic energy is the same as if tlie sphere had fallen freely through the height equal to the descent of the centre of gravity. 19. Stqjpose, in the preceding example, that the axis of the sphere he rigidly connected with the rod AB ; required the velocity. Here we have, as in example 19, p. 216, d^e df ~ ' V gi ;sin 1 dd" 2 df- ~ f gi i(l- cos 6) ; which, compared with the preceding problem, shows that the angular velocity loill he less at the loivest point when the spthere is rigidly connected loith the rod AB, than when it is free to roll on its 01071 axis. The kinetic energy in this case, when it will have rotated from the highest to the lowest point, will be, equation (153), page 202, dd'' i.'m(r + |r^).|^ = 21f7; which is the same as in the preceding case. The time of vibra- tion will be greater in this case than in the i^receding — equation (161). Queries. — 1. If the sphere in example 18 be free to rotate on its horizon- tal axis as a diameter while the entire mass rotates about an external axis, and it rotates through 180", starting with no velocity from a point vertically over the fixed axis ; if, at the lowest point in its path its axis instantly becomes rigid with the bar AB, will it rise to the highest point ? MOMENT OF THE MOMENTUM. 417 2. In tlie preceding example, will the time of describing the second part of the arc be the same as that of describing the first 180' ? 3. In example 18, if the sphere gradually melts away, will the velocity or time of vibration be thereby affected ? 4. In example 19, if the sphere gradually melts away, will the time or velocity be thereby affected ? 5. In examples 18 and 19, which -will produce the greater stress on the axis of suspension, the masses and arcs of vibration being the same in both cases ? G. At what points of the rotating masses must they strike a fixed obstacle so as to produce no shock on the fixed axis ? 7. If a spherical shell B be rigidly connected to the bar AB and filled with a frictionless fluid, would the time of vibration and the velocity at the lowest point remain the same if the fluid should suddenly freeze ? 8. If a spherical shell rigidly connected to a bar and filled with a friction- less fluid, be rotating about a vertical axis with a uniform velocity under the action of no forces ; should the fluid suddenly freeze, will the velocity of rotation remain the same ? \V'ill the kinetic energy remain the same ? Conservation of Aeeas (Art. 150). Moment of the Mo- mentum (Art. 166). 20. A cylinder of ice, radius r, length I, revolves with a tmi- form angular velociiij co ; if it he subject to no external force and melts, required the angular velocity of the resulting sphere. Neglecting the contraction due to melting, and the spheroidal form due to rotation, and letting go' = the required angular velocity, F = |i2' = the principal radius of gyration, we have, Article 166, m ' jtr-oo = ni • ^R'oo', , 5 r"- To find R we have i7tB'= 7ir-l', .-. R"- = (|r=03 ; 5 /4 r\i5 ^=4V3-r 27 418 PROBLEMS. 21. If n spherical shells of infnitesimal thickness, mass m of each, radius r, move in contact without friction, having their axes of rotation all in 07ie pla)ie, the angles between the axes and an assumed line being ft-t, f3„, ^3, ..../i„, and having angular velocities (i?„ go.,, cd>,, a)^, respectively, suddenly become solid ; required the position of the resultant axis, and the resultant angular velocity. Let B be the required angle and &? the required angular velocity ; ^•' = \r-, the principal radius of gyration before and after becoming solid ; then, since the moment of the momentum will be constant, we have, resolving parallel and perpendicular to the line of reference, nmlc- cos ^•CL>=«m/l'-(fi7iC0S /?, + CL>2COsy52+ &?„cosy?„)=?w^--^ ; mlc^ sin 6 • oj=:nih^{ooi sin ft^ + oo„ sin /Jo + . . . . &?„ sin /?„) = ml = -— TT OJ = \00. 28. A cone revolves about its axis loitli an angular velocity go; the altitude contracts, the volume remaining constant, required the resultant angular velocity. Let h be the original altitude, x any subsequent altitude, r the original radius, y the radius when the altitude is x ; then, the volume being constant, i^7tr-h = i7ry% the moment of inertia will be mk" = ^^Ttr^h, m]c\ = ^o7ty*x ; but the moment of the momentum being constant, mk'i GO = mJc'oo^ ; ^'i r" X .'. Ot?, = ^, GJ 1=: — ti? = 7- GO, k- y h or the angular velocity will vary directly as the altitude of the cone. 29. 07ie end of a fine, inextensiUe string is attached to a fixed ^ point, and the other end to a j^oint in the swface of a homo- geneous sphere, and the ends brought togetlier, the centre of the sphere being in a horizontal through the ends of the string, and the slack string hanging vertically. The sphere is let fall and an angular velocity imparted to it at the same instaiit, the sphere winding up the string on the circumference of a great circle until it tvinds up all the slack, when it suddenly begins to ascend, winding up the string, the sphere returning just to the 422 PEOBLEMS. starting point. Required the initial angular velocity, the ten- sion of the string during the ascent of the sphere, the initial iqnmrd velocity of the centre of the sphere, and the time of move- ment. Let I = the length of the string, r = radius of the sphere, m =.its mass, k = its radius of gyration, x = the length of the unwound portion of string at the end of descent, v — Telocity at the end of descent, v' the velocity with which the sphere begins to ascend, 00= the initial angular Telocity, co' = the angular velocity with which the sphere begins to ascend, t^ = the time of descent, t^ the time of ascent, T= tension of the string during the ascent, and / the impulse communicated by the string to the sphere at the end of descent. The length of string wound up at end of descent is /9 TGDti — rooA / '^^, (J t, being the time of falling freely through the height z, whence ■^ = ^^i/i- <^) For the impulsive motion, / V + V = — . 7n Eliminating /, — (co — co) = V + V. r But Jc- = |r, and v = V^gx, .-. go' = &)-^ {V^gx + v'). (2) MOMEXT OF THE ilOMENTUlL 423 The upward motion, the origin being at the point where the centre begins to ascend, and the axis of y positive upwards, ^'^^ = ^-^"y. (3) For the angular acceleration, dt or W -T7T = — ^7— . (4) dt- 2r ^ ' Also y = rO, or dhj _ rd'd dt:' dt' ' (5) Eliminating -rf and -j- from (3), (4), (5), T = f/ng = f the pa weight of the sphere. Eliminating T and -r^ from (3), (4), (5), -jp- = — ^g. Integrating, observing that when t — 0, dt ~^' % = V-^9L (6) When ^^=0,t = f,; .'.v'=^gt^ (7) Integrating (6), observing that when ^ = 0, y =0, y = v' ■ t - -ijgt\ When y = X, f = t, X = v't, - ^\gtl (8) J4 PROBLEMS. From (7), (8), v' = W-^-'^, (9) '-|/f- (10) From (5), dy dd dt ~^dt' 1 v' = rco', ca' = — '^^.^-gx. (11) r Substituting these yalues of v' and cd' in (2), a;= ?(G- V35). Substituting this yalue of ic in (1) and (9), (V35 - 5)\/gl 00 — := mz • *'V[2(6- VSS)] v' = V[-t5'?(6 - V35)], From (10), The whole time is (Problem by the Author in Matliematicdl Visitor, Jan., 1879.) 30. If n concentric iiniform spherical shells of infinitesimal ihicTcness moving zvithout friction about axes whose inclinations to three rectangular axes are a-^, /5„ y^; a,, /Jj, y^, etc., toith angular velocities co^, 002, etc., respectively, suddenly become one MOMENT OF THE MOMENTUM. 425 solid; required the rcsuUa7it angular velocity and resultant axis. Let a', J3, y, be the direction-angles, oo the resultant angular velocity, h the radius of gyration, aad m the mass of each ; then, Art. 16G, nm1c-0D cos a = mh'^{GOi cos a^ + a?, cos a^ +, etc.) = A, (say) nnik-cj cos /? = 7nk\GJi cos /3i + cj^ cos /^a +, etc.) = B, mnk'^GJ cos ;/ = mk-{coi cos 7/1 + cj^ cos ;^2 +> etc.) = C ; and from Coordinate Geometry, Art. 198, Eq. (3), cos* a + cos" /? + cos- y = 1. Substituting from the preceding, we have A' + B- + C'^ oi'm'Tc'oj- ; ^A' + B' -I- C* nmk- This value in each of the preceding values gives A cos a = cos /3 = ^A"" + B' B + C" f VA^ + B^ C + c ) cos V = — , . . ^ ^A' + B' + 6'^ If the motion be in the plane xij, we will have y^ = 90° - y, =iys, etc.; hence y = 90°, and the third of the preceding equa- tions vanishes, and the case reduces to the problem given by the author in the Mathematical Visitor, Jan., 1882, p. 14. 31. A screw of Archimedes is free to turn about its axis placed vertically ; a particle placed at the upper end of the tube runs down through it ; determine the resultant angular velocity im- parted to the tube. 426 PROBLEMS. Let m = the mass of the particle, nm = the mass of the screw, a = radius of the screw, z = the vertical distance the particle has descended, and, and (p — the angles through which the screw and particle have respectively revolved about the axis in the time t. The energy of both bodies equals that imparted by gravity, or Prom the principle of Conservation of Areas, we have (2) ,dcp' ,dd' We also have the geometrical equation z = a {(p + 6) tan a. (3) From (3), — = a^ tan^ a{-^^^ + 2- . - + ^^j. (4) Substituting in (1), / 1 dcp\jXn'a dq> dd sm' a dd" dd'X .. From (3), dcp dd dt ~ dt' Substituting in (5), / tf 2Msin^a' sin* « , Xdd"^ _ \cos^ a cos^ a cos^ a J dv or, dt flfp a\n + l)(w + sin* ^) ^j^ = ^9^ ^os* a. MOMENT OF THE MOMENTUM. 427 When -jj = GO, z = li, y 'Zgh cos'^ a {n + l){n + sin^ a)' 32. A cube slides doivn an inclined plane ivith four of its edges liorizontal. The middle imint of its loivest edge comes in contact with a small fixed obstacle ; determine the limiting veloc- ity that the cube may be on the point of overturning. Let V be the velocity of the cube at the instant of impact, 25 the length of one edge of the cube, h the radius of gyra- tion in reference to the edge at P, and Tcy the principal radius of gyration in reference to a parallel axis, oo the initial angular Telocity, and m the mass of the cube. The moment of the momentum just before impact will be mv -b. (1) The initial moment of the momentum after impact, will be, equations (155) and (123), and example 4, page 172, Qa = mlfoD = m[W + ]cX)co = m{2b- + p=')&? = ^m¥oD. (2) Hence, Article 1G6, mvb = fmb'^GJ ; (3) The cube will be on the point of overturning when the energy due to rotary velocity is just sufficient to raise the centre to a point vertically over P. The energy will be, equations (153) and (123), ^2mr • d)^ = ^mk^ • a?* = lmoo'^¥ = ^^m¥oar. (4) The work of raising the centre to its highest point will be 428 PROBLEMS. mg-h\/'2[l -cos (Xtt -/?)], wliicli being equal to (4), we have ^^ = |V2|-[l-cosa.-r->5)], and this substituted in (3) gives (5) (6) tr' = -VV2 • Ml - cos a^r - ^)], (7) which is the required result. In this problem we may consider two imjiulses, one that of the momentum before impact ; the other that destroyed by the impact. It is now required to find the magnitude and direction of a single imjDulse, which applied at P, will produce the same effect. At first it seems that this impulse will be parallel to the plane and ojjposed to the direction of motion, but if the body were free, it would, in this case, rotate about its centre, equa- tions (168), in which case the corner at P Avould move perjjen- dicularly to the diagonal through the centre, whereas, in the problem, this corner becomes instantaneously fixed. The result- ant impulse may be considered as the resultant of an impulse acting along the plane, and another acting at P perpendicular to the diagonal of the centre through that point. The angular velocity is, equation (3), 3 V (8) lience the actual velocity, w, of tlie centre is ^/2h ' 00 = u ■= \'\/~Zv ; which is also the initial velocity with which the point P would move normally to the diagonal if the body were free. The plane and obstacle then, impose the two component velocities v and w, the angle between which is |;r \ hence the resultant velocity V will be MOMENT OF THE MOMENTUM. 429 ' V =z v' + u' + 2VU cos fTT, — v' + -^^v' — Iv' ; .-. V=W^v. (9) To find the angle between V and v, we have u"^ = V^ + v'^ — 'ZvV cos

gm sin i — F, gm sin { < 2F', 432 PBOBLEMS. but F=gm}.i cos i ; therefore tan i < 2//. 36. A sphere having a rotary motion ahout a horizontal axis is 2Jlciced gently on a rough plane; deternmie its motion. Let W = the weight of the sphere, pi the coefficient of friction, I' the tangen- tial action due to friction. There may be two cases : 1st, if there be slipping ; and 2d, if there be no slip- ping. In the former case T = /xW. For the motion of the centre we have or cPx >\ (1) and for the rotary motion. or dv f-iWr, \, (3) Integrating, we have dx dt = l^gt + Ci dcp _ dt f.igrt + c' K. (3) Ic^cp = — li-igrf^ + c't + c" > For ^ = 0, we have a: = 0, and 9? = .'. c^ = 0, and c" = ; also iov t = o/-~ = (i?o, the initial angular velocity, .-. c'=k^coo ; ctt FKICTION. 433 and — = 0, the initial velocity of the centre, and the corrected dt equations become dx . di = ^9^ dcp lAgrt 1 f-tgrl'' (4) The Telocity of the point of contact will be plus ^ , and dq) ^ dx di minus r -jr ; hence dx dq) }^grH -u-r-^ = l^gt-ooor + -^j^', dt dt (5) and so long as this is finite the preceding equations will hold true, but when it reduces to zero the conditions change. To find when they change, make each member of (5) equal to zero, and solve for t, which call t^ ; hence h = ooqI'Ic^ f.ig{]c' + r')' The left member of (5) becomes integrating, differentiating. dx _ dq) ^ di~'''~di' X = rq) ; dt' ~~ ^ df- ' (6) (8) (9) which are the equations of motion when there is no slipping. The value of T is not equal to fiW after the time A, and to find its value combine the first of (1) and (2) with (9), giving 28 434 PROBLEMS. T{^) .-. T=Q; (10) hence it requires no friction to cause the point of contact to have no progressive motion. The motion is the same as if the body were in void space under the action of no forces, having a uniform motion both as to the translation of the centre, and of rotation about the centre. The total amount of slipping will be 2/^- -J- St*' roD^^ti — rep — \ixg — ^ — t\ — o^^rU. (11) 37. If a sphere, radius 3 feet, weight 20 pounds, rotating ten times per second, be placed on a horizontal plane whose coeffi- cient of friction is y^ ; how long will it be in coming to a uni- form velocity, how far will it have traveled, how much will it have slipped, what will be the uniform velocity of the centre, and the uniform angular velocity ? 38. If a cylinder have the same amount of material, diameter, and rate of rotation, as the sphere in the preceding example, and placed on the same plane, which will first attain a uniform motion, which will have the greater uniform velocity, and which will have slipped most ? 39. If a cylinder whose altitude equals the diameter of the sphere of example 37, the same amount of material and rate of rotation, be placed on the same plane, which will finally attain the greater uniform rotation ? 40. What must be the coefficient of friction that there be no slipping at the point of contact at the beginning of motion ? 41. Which will first attain a uniform motion, the sphere in example 37, or a sphere of the same material and twice the diameter ? 42. If the body gradually contracts, retaining a constant mass and same form, will it go farther or not before attaining a uni- form velocity ? 43. If a sphere preserves the same circle of contact, but grad- ually contracts laterally, changing to an oblate ellipsoid, will it FEICTION. 435 affect the time of attaining a uniform velocity ? What will he the time if the polar axis hecomes half the equatorial diameter ? 44. A rope is stretched roimd a rough cylindrical surface sub- tending an angle 6, the coefficient of friction ieing ju; required the force F acting in the direction of the tangent at one end in order that P will he in a state just bordering on motion- towards F. Let the tension at any point, a, be t, that adjacent, h, will be t + dt, p the normal pressure on the arc per unit of length if it were uniform ; hence, on an element of length, it will \>Qpds, }.i the coefficient of friction ; then dt = /.I • pds, also pds = tV + {t - dtf + %t{t + dt) cos (tt - d) = 3(1 + cos (tt — 6)), (ultimately) = t-2 cos -\{7T — 6) = t ■2s'ml8 = tdd (ultimately) ; ;. dt = l.it> dd. Integrating, But for and for log t = M^ + 0. e = o, t = F, e = AB, t:=F; .: P = Fe>-». From the relations and we find pds = tdO, ds = rdO, t the same as equation (o), p. 139. 436 PROBLEMS. 45, A shaft having a bearing the entire length is driven by a pulley at one end, the power being taken at the other. Find the diameter of the shaft at any point for uniform strength. Let IV = the weight of the shaft per unit volume, /x = the coefficient of friction, h = the diameter at the driving end, x the distance of any cross section from the driving end, and y the diameter of that cross section. Then wfXTty'^dx — friction of an element, wuTiy^dx =■ moment of friction. Then w}xny^dx + PR — cy^, [a) PR being the moment of the driving power. From {a) W)A7zy^dx — dcy'^dy, or, letting {'Adx^i"'-^, Jo J, y y y =z h£-^\ and ATTRACTION". 46. Assume that ttoo spheres of the same material as the earth, each one foot in diameter, are reduced in size to a mere point at their ce^itres, and placed one foot from each other, required the time it would take for them to come together hy their mutual attraction, they being uyiinfluenced hy any external force. Let E= the mass of the earth, m = the mass of one of the spheres, m' = the mass of the other sphere, R = the radius of the earth, r = the radius of one of the spheres, ATTRACTIOK 437 r' = the radius of the other sphere, g = the acceleration due to gravity on the earth, jx = the acceleration due to the attraction of a sphere of mass unity, upon another sphere of mass unity, the distance between their centres being unity, a = the original distance between the centres of m and m', and X = the distance between their centres at the end of time f. Then from (237) we have d'x df ,.E'g 1 which integrated (pp. 33, 34), observing that for t = 0, x = 0, R'-fi and V = 0, and that j.i in the reference equals {m + m') —~- in this case, gives t = Ea _2{m + m')R^g_ which for the limits gives t = ^Tta Ea _2{m + m')R'g_ If both spheres are of the same density, their masses will be as the cubes of their radii ; or m = -^3 E, rrt' = -j^.E', Ra and we have and if the spheres are equal, as in the problem, we have 438 PROBLEMS. If a = 1 foot, R = 20,850,000, r = i foot, (/ ^ 32^, we have \ O/i.loO / = 1,788 seconds, nearly, = 29.8 minutes, nearly. 47. ToJi7i(l the stress in pounds which would he exerted ly the mutual action of two such spheres as iii examjjle 46, at a dis- ia?ice of one foot hetiveen their centres, we have, from equations (224) and (235), siiice m = m', atid x=l, 64 W ' The mass of the earth is 5| times an equal mass of -water. The weight of a cubic foot of water is 62^ lbs. at the place where g = 32^, and the volume of the earth is i7rB\ hence B = 5| X 62| X i7tE\ which, substituted above, gives for the stress ^rr X 5|- X 62^g x 32^ 64 X 20,850,000 = .00003471+ lb. or nearly of a pound, a quantity inappreciably small. 48. If the density of the earth at the surface he unity, and at the centre is m, g the force of gravity at the surface, and f that at any distance x from the centre luithin the earth, what is the law for f, the density increasing uniformly toward the centre. The density at any distance, x, from the centre will be (m — 1) , X , -, ^^ ~ (r —x) +1. ATTRACTIOX. 439 The volume of a sphere of radius x is 4;r I x'dx. Hence we have, •{m - 1) (r — x) +1 Ix^dx, and If hence. Mass = Art . 4:7r r(m —1) /x^ x*\ x'' = c J; 4(m + r — l)x — 3(wi — l)x° x^r, f=g, 5^ = f ^ \_Km + r - l)r - d{m - l)r^ ; _ Mm + r — l)x — 3(771 — l)x' ■ -^ ~ 4 771 + r — !)/• — 3{m — l);-^ ' ^* 49. Considering the earth and moon as itniform bodies, the mass of the earth b\ ti77ies that of an equal volume of water, and its radius 3,956 miles, the ^)iass of the moon 3^ times that of an equal volut7ie of water, and its radius 1,080 miles, the mea7i dis- tance hetweeii the centres of the ea7-th and moon 60 times the radius of the earth, and the acceleration due to gravity at the surface of the earth 32^ feet per second ; required the time in which they would co7ne together hy their mutual attraction. "When they are in contact the distance between their centres will be 5,036 miles, and we have from Problem 46, \^{M + m)R^g_\ [ax — x-)' + a cos \IV / _| 237,360 But 7if B" 7 J/r' and t = Ba 2{E^ + -i\r')g_ T( ax — x")- + a COS"' x\i (2) 440 PROBLEMS. Substituting numerical values in equation (2), -3,956x237,360x5,280x3 .193pr956)- + ^(17080«J LV'"''''" "" 6,030 -(5,036)^ = 412,945 seconds. Loomis gives 415,600 seconds as tlie time required for a par- ticle to fall from the moon to the earth, t^^e distance to the moon being the same as that given in this example, which is 2,855 seconds, or nearly 48 minutes more than the time for the earth and moon to meet. 50. When the earth is in perihelion, suppose the sun's mass to he increased hy x times its present value. Required the change in the elements of the terrestrial orbit. The eccentricity of any planetary orbit is (Problem 4, page 189), FV sin /? V*r^ sin' /3 2- '- + M- in which V is the velocity of the planet at the point whose radius vector is r, /? the angle between the curve at that point and r, and /.i a measure of the attractive force. In this case p = 90°, and the equation becomes, by reduction. FV , M If now the mass of the sun is increased suddenly x times its present value, /i becomes x/j, and the eccentricity of the new orbit will be e^ (say), hence, V"-r ^ = l±e^; x/^ and ±e,= — ■ - 1, X/.1 = ^(1±^)-1, ATTRACTION. 441 in which ^'i must be used when the right member of the equa- tion is positive, and — e^ when it is negative. Itx < 1(1 ± e), Ci > 1, or the orbit will be a hyperbola. " x = l{l ±e), e^^l, " " " parabola. " X \ -^l^} ^ ^} [ ,e,< 1, " " " ^^^ ellipse. '' x=l ±e, ±c, = " " " -A circle. '' x>l±e, -e,> -I, " " " an ellipse. In all but the last the given and resulting perihelia coincide ; but in the last the given perihelion will coincide with the new aphelion, x = | and e — 0.01G784, the eccentricity of the earth's orbit, we find - Ci = —0.32314, or — e^ = 0.3445--3 ; hence the present perihelion would become the aphelion point of the new orbit, and the new orbit would be an ellipse with the eccentric- ity 0.32214 or 0.34452, The mean distance would become 3r = 0.7436 2(1 + r) of its present distance. The time of rotation about the sun Avould be T: y[( a + r X 3651 days = 191.25 days. 51. Suppose infinitesimal aerolites equally distrihuted tlirongh all space, everyivJiere moving equally in all directions tvith a uni- form and constant absolute velocity. The aggregate mass inter- cepted in a given time ly a given stationary spliere is supposed to be known. Determine the effect upon the eccentricity of a spherical planet of given mass and volume moving in an eccentric orbit all of whose elements are hnomi. {Math. Visitor, July, 1880. ) If the velocity of the planet be less than that of the aerolites, the same mass Avill be intercepted as if the planet was at rest. Consider this case. The change of the elements of the orbit will be due to two causes. 1st. The increase of the mass of the planet will increase 442 PROBLEMS. the attractive force between the sun and the planet. 2d. The aerolites will cause a direct resistance to the motion of the planet. Let M he the mass of the sun, w' the initial mass of the planet, s the distance between them, x a constant ; then will the acceleration of one body in reference to the other at the end of time t (the time in the problem being unity) be ]c{M + m' + mt) hence at distance unity, fx = ]c[M + m' -f mt)f and djx =1 Jcmdt, For an elliptical orbit we have, page 190, Vlrl = /S[(p-p')(p'-p/)(p"-A")] = (^+-^+^0>S'(p-p.r(p'-A)=0, PROBLEMS. 445 which proves that the apparent orbit is a plane. — {Codrdinate Geometry, p. 277, eq. (3).) {Math. Visitor, July, 1880.) 53. An elastic string, loithout weight a7id of given length, has one end fixed in a perfectly smooth horizontal plane, and the other to a point in the surface of a sphere, the stri^ig being un- loound. The sphere is projected on the plane from the fixed point with a linear velocity v, and an angular velocity oo, loindiiig the string on the circumference of a great circle ; required the elon- gation of the string ivhen fully stretched, and the subsequent motion of the sphere. Let r — the radius of the sphere, a — the original length of the string, cy = the initial angular velocity of the body, v — the initial velocity of the centre of the body, and t^ — the time of winding the slack. Then vt-i, 4- rootx = a ; .: t, = V + roo and the initial stretched part will be vti = = I (say). V + rcj ^ "^^ Immediately following this time the string will be stretched, and the tension at first diminishes both the linear and angular velocities. Take the origin at the remote end of I for the variable motion. Let m = the mass of the body, s = the space passed over by the centre during time t, 6 = the angular dis- tance passed by the initial radius in the same time, k = the radius of gyration of the body, e — the coefficient of elasticity of the string, A the cross section of the string, and A the elonga- tion produced by the tension T of the string. Then Mariotte's law gives r= — - (1) ^ l-rd' ^ ' Assume that I is so long compared to rd, that the latter can be neglected, and let B = eA -^ I, then T = B\. 4i6 PROBLEMS. The conditions of the problem give dX = ds + rcW ; (2) .-. cVX = d's + rd'O. Also, for motion of the centre, m^^=-T=-B\, (3) and for the rotary motion, mJc"-'^= -Tr=-BrX, (4) dr which two equations in the preceding give dt^ mk^^ Integrating, observing that for A = 0, ^ = 0, and dX -r- dt = t' + TOO, we have -. V + rcj . , . A. — — - — sm Bf. (5) The elongation X will be a maximum for sin Dt = 1, or t = TT ~ 2D, for which . V + rco Tcs/ml , . The time of producing the maximum stretch of the string is independent of the initial motions. When the string returns to Its original length X will again be zero, and sin Dt = 0, or Df = 7r; .-. t = ^. All the circumstances of the variable motion may be deter- mined by integrating equations (3) and (4). Integrating after ds substituting from equation (5), observing that for ^ = 0, j^ = v, s = 0, ~ — cj, and 6 = 0, we have, if we jmt i^for eA{v + roo) PROBLEMS. 447 ds dt = F\co& Z)^ - 1] + V, (6) s^^l&inDt-Dt'] + vt, (7) ~ = i^p[cosi)^f-l] + oa, (8) e = ^p [sin Dt - Dt] + cot. (9) For the maximum of (5) dX ~ dt = 0, which in (2) gives dt ~ '^ dt' which combined with (6) and (8) gives cos i)^ — 1 = — 1 ; .*, Dt = Irr as before found, and serves as a check upon the work. The relation ds = — rdO shows that the direction of one of the motions changes sign. At the point where the linear motion is reversed, y- = 0, and for this we have fl R and if the direction of rotation is reversed, -^ = 0, and (8) gives ^3 = ^cos^(^l--^-j; from which it appears that if v < ¥co -^ r the motion of the centre will be reversed, but otherwise the angular motion will be reversed. The value of t2 in the former case will be less than -y . Both motions will change at the instant of greatest elon- gation if rv = Pgj. If the values of 4 and 4 are both less than rr -^ D, one motion will change sign before the instant of greatest elongation and the other after ; otherwise only one will change sign. To find 448 PROBLEMS. tlie total variable movement, make Dt = rr, acd (7) and (9) give ^=(--4)5- K (6) reduces to zero when Dt = tt, the body would rest at the moment the string regains its original length, and F= Iv, but it would still have an angular velocit}^ of cj + {rv -^ k-), as shown by (8). Similarly, if the rotary motion is destroyed at that instant, the linear velocity will he v + (h-oo h- r), and will continue uniform. It may be shown that the kinetic energy of the moving body at the end of the variable motion is the same as at the beginnins^. It has not been attempted to solve the general case represented by equation (1). It is evidently intricate. — (Problem and solu- tion by the author in TJie Analyst, Jan., 1882.) 54. Find the minimum eccentricity of an ellipse capalle of resting iii equilibrium on a perfectly rough iJiclined plane, in clination fi. The centre of the ellipse will be vertically over the point of support, and since the plane is tangent to the ellipse, a vertical through the point of support and a parallel to the plane through the centre will make conjugate diameters. Hence the acute angle of the conjugate diameters is 90° — ji. At the point bordering on motion, the potential energy is a maximum, and the major axis will bisect the acute angle of the conjugate diameters ; hence the positive angles made by the conjugate diameters with the major axis will be ^ = 45° — ^/?, d' = 135° + \fi. The condition for conjugate diameters is a^ sin B sin 8' -f- IP' cos 6 cos B' = 0, which, by substituting the preceding values, gives a' sin^(45° - ^/3) - d' cos\4:5° - ^/5) = ; FLUIDS. 449 which, reduced, gives / (a- ~ ¥\_ _ / ( 2sin/g '^ {Math. Visitor, Jan., 1879.) FLUIDS. 55. A sphere 4 inches in diameter, specific gravity 0.2, is placed 10 feet under ivater. If left free to move, what ivill be its velocity at the surface of the ivater, and what ivill le the maxi- mum height it will attain. Let r = \— radius of the sjDhere, p = ^ = its specific gravity, h = 10 feet, V the velocity acquired in ascending a distance re, Fthe velocity at the surface of the water, h = the resistance. For the motion of the sphere, 'dx ~~^\p J Putting g(^^-lj = g', vdv dx — —r g' - hv'' p _ r vdv ]o^~ ],g' - lev" h = :-r log whence. 2^' ^ \q' -kV\ V'=\{l-e-^"^), fc — ^^ — nearly, The required height is V 1- p 2g 2kp Tcp ( - e''^% 2hp 29 - nearly. 450 PROBLEMS. The value of Tc reduced from Newton's Principia, Book 2, 3 Prop. 38, is —— , which giyes F = \}i-r(j{\ - p)]^ = ^ ^J^g nearly. 1 and v h = —- — 4| inches nearly. {Math. Visitor, Jan., 1879.) 56. The first of two casks contains a gallons of ivine, and the second b gallons of ivater ; c gallons ivere drawn from the second cash, and then c gallons were drawn from the first cash and poured into the seco?id, and the deficiency in the first supplied hj c gallons of ivater ; c gallons were then drawn from the first cash, and c gallons from the second, and poured into the first, and the deficiency in the second cash supplied hy c gallons of wine. Re- quired the quantity of ivine in each cash after ni such operations as that described above. Let w„ and v„ represent the wine in the first and second casks respectively at the end of the wth operation ; the quantities of wine in each cask at the successive stages of the {n + l)th opera- tion are u,„ v,r, ( 1 — ^ ) w«' ( 1 — rVu ; (1 — - ) «« (1 — rV« + T w„ a J " V b J "' \ a J "' V 57" ' b 1 ) + -y a J ab "Whence ^ A-. ^\ f . ^ \' ■ ^ A ^ ''•*bV- i) "■' {' - i) '• ■ + aV-i>«- + <-- y-^a-K--^)- (^) FLUIDS. r.^,= il-t\\+Ul-i)«. + c. Also «„ + 2 = 1 - - + -, a J ao_ «„+i + |(l-^)v.. + ij v„^, = [ 1 -~ Vz-'t+i + ^fl -rV^''+i+ ^• jy' Eliminating v„ from (1) and (2), i_'_W._»-„,„.=(i-iYfi-i,«.- Eliminating v„ + i from (3) and (5), li^n + 2 i-v.::,.(i-,;j «...+(i-^) (1 =?'^ 451 (3) (3). (4) (5) c (6) Let ri, r.,, be the roots of the equation c\' . r . /, ^\' L. , /, ^\7i_£.^ =0. i-:7^^n^-^) i^n^-. The solution of (6) is, (Hymer's i^iwiite Differences, pp. 54, 55), t<„=CVrr + Gr?+ - Let the last expression = S. Erom (1), «o = n, 452 whence PROBLEMS. C\ , _n,- S- r,(a - S) i\ — 1\ n _ ^i(^ — S) —Ui + 8 L/2 — ■ , n — r. and \ ri — r. J \ r, — r. J " in Eliminating u^ from (1) and (2), and w„ + i from this and (4), t-'n + « 1 - t\.i + 1 - lib l-^)r„ whence From (2), therefore ^'« = CsT^ + C^rl + S, . Vq = 0, Vi = c( 2 — ij' (J/«f//. l7s?7or, Jan., 1880.) 57. ^ servant drmvs off a gallon a day for 20 days, from a cash Jwlding 10 gallons of wine, adding each time a gallon of water to the cask. He then draios off 20 gallons more, adding as taken, a gallon of wine to the cask for each gaJloii drawn. How much water remains in the cask ? Put 10 gallons = a, 1 gallon = J, 20 = t, and let w^ = the number of gallons of water in the cask at the end of the icth operation. Then we have U, + I) = M. + i, (1) FLUIDS. 453 an equation in Finite Differences. Integrating (1), When X = 0, 11^ = ; .-. C = — a. V - (a — hy^ •. w^ = « ( ^ ) = c. by the first condition. Let Vi — the quantity of water in the casl? at the end of the tth. operation under the second condition ; then, Integrating, When a — i V, = v, + \. 'a - b^ ' Vt = C[ a t = 0, Vo = c ; ''a — h C = CiVi = c When a J \ a" x = t = 20, a -10, b = l, t;j=::10 (.9™ -.9'°). {The Wittenlerger, Jan., 1880.) 68. A piston, lueigJif w, is dropped into the end of a vertical cylinder filled tvith air, length I ; how far ivill the piston descend, assuminq no friction nor escape of air, nor heat from the com- pressed air 9 454 PROBLEMS. There being no escape of heat, the law of pressure will be ex- pressed by pv^ = constant = p'v''^, (1) where j?' = the initial pressure of the atmosphere = 15 lbs. nearly ; v' = volume of the cylinder = al, if a is the area of the base ; p = the pressure within the cylinder when the weight has descended a distance x ; a{l — x) = the volume when the press- ure is^ ; k = 1.408 ; then, from (1), and Integrating, observing that for x = 0, v — Q, \ 10 J ^ io{k — l)\ (l — x)^ "• / At the end of the downward movement v = ; therefore x{w + pa){l — xf-^ + Y~: (^ — ^Y'^ = ?^_ , , from which x may be found by trial after numerical values have been substituted for the known quantities. 59. If each ofn vessels closely cojinected in circuit contaijis a different liquid, each q gallons, and the liquids circulate liy flow- ing uniformly in one direction at the rate of a gallons per min- ute, mixing uniformly, how much of each liquid luill there he in any one of the vessels at the end of the time t. Let the vessels be numbered in the natural order 1, 2, 3 ... w ; let X denote any particular liquid and x^, x^, x^, . . . x„, the quantity of it in any vessel at the end of the time t. adt = the FLUIDS. 455 amount flowing out (and in) each instant, of which - adt, — adt . . . -^ adt, will be of the liquid denoted by x. In vessel 1, —adt flows out, and '-^ adt flows in from the ni\\ vessel, and q q the difference will be an element of the decrease of the x liquid ; («) Letting - = r, and dividing by dt, wo have •. dx, = — adt adt. q q rdxi or dt rdxi ~~dt Xi x„ Similarly for the others, rdx^ ~df rdx; ~dT rdx^ __ dt " ~ Differentiating n — 1 times, we have rd^x^ d"'^Xi _ d"~^x„ "* dt""- rd^'-'^x^ d^-^Xx d"~-x„ dt"-' rdx^ ~df Xy Xn ip) {c) 456 PROBLEMS. Similarly, Td"-^x^ _ d"-% _ d"'''x\ dt"-' dt"-" ~ ~~di''^' rdx, and finally. dt ^' ~ ~^' * ~dF' dt''-' ~ ~ ~~dF^ ' dt"-' dt^^~' ~ ~ di"-' ' rdx„_i _ — ^ - x„., - -.r„ Multiplying the successive equations in (c) by the successive terms of the development of the binomial [r —1)""^, and add- ing results, we have r"d"Xi wr""V7""'aTi 7i(n — 1) r""V7""'a:i nrdxi ~df" dt''-' "^ \2^ dt"-^ '-^ ~df "^ ^' _ _ r"-VZ"-X / _ , , r"-V7''-l'g„ _ (n — l){n — 2) r''-^d''-'x„ ~ dt"-' + (^ - 1) —^t^^ 12 W- The second member of this equation is one degree less than the first, and since this is true for all values of n, it will be FLUIDS. 457 true if we reduce the exponent and subscript still further by 1. This process ultimately gives for the second member. rdx. ~di hence we have ± — ^ =F a-2 = =F a;i ; r"c^% nr"-'^d''-^Xi n{n — 1) r" " d" "-x ^ mi- w^^~ "^ ^ dv'-^ ' ' • 7i{n — 1) rd'Xi nrdx _ ^ "^ " " [2 w ~dr ~ ' Adding =F 1 to both members, the characteristic equation becomes (Price's Infinitesimal Calculus, Vol. IL, p. 634), yn^« _ nr"-'/^"-' ± nr/3 :p 1 = T 1, or (r/i-l)"= + l; ., ^.^llv^^ which may be written ^^l+VtiT, (,) where the exponent n, of - 1, is simply for the purpose of determining the sign of this term. If n bo even (-1)" will be + 1, and there will be two real roots + 1 and — 1 ; if w be odd there will be one real root of V^^ = - 1. The other roots will be imaginary and in pairs. The number of values of /i will equal the degree of the equation, one value being zero, and letting these values be *„ b„ h . . . h,,.„ the integral becomes (Price's Infinitesimal Calculus, Vol. II., eq. (107), p. 037), X, = (7ie'^>< + Cx^"J + + C'„eV, {e) and similar expressions for .Ta, x,, etc. 458 PROBLEMS. To find the constants of integration we have for t = 0, Xi = q, = a, from (a) since, initially, x„ = 0, a dxi a dx„\~\ _ a^ a _ ^' t.o~\q dt g dtJ_\t=o~q ~ ~q d''xC~\ a^ -fjr = -> etc. .-. q= C\ + C,+ C, . . . . (7„, a = h,C\ + ^2^2 o . . . hJJ„, if) and finally from which the constants of integration can be found As a special case let 7i = 3, then from (d), ^ = 0, ^(3-V^, and^(3 + V=^). We then have from (e), a / — a / — ^ 5-(3-v/-3)< s;-(3fV-3)« and similarly. X2= Ai + AoB X3 =z Bi + B^e d I CL i~ r;(3-V-3)« s;;<3 + v-3)< 2q + A^e' d I — a / — s-(3-V-3)« 5r(3+v-3;< 29 + B^ 2s' \ {9) The sum of these, or Xx + x^ ■{■ x, ■= q ^t any time t. FLUIDS. 459 Also by the process sliown in (/), we have for t = 0, x^ - q, -dJ-"^' cW -q ' ' ' dt elf q dt " ' df q ' Hence, to find the constants in equation {g), we have, for Xi , q=C\+ C, + C„ 'I = -J^ rs -v^sra + £^. (3 - V^^ra ; q iq -^Q for Xi, = A^ + An + Az, _ « = f (3 - ^^^)A, + I; (3 -^ V="3)^3, 2q '^Q 2a; a^. _V~3AJ. + -[^ (3 +^^3)^43 ; q 4:q'^ ^ ^ " 4:q and for x^. = « (3 -V'=^d)B, + I- (3 + V^3)5„ g Aq ^'1 These give C, = C, = C^ = \q- B = \q, B. = - \q(l - V^^), ^3 = - ^^(1 + V=^). 460 These in (^7) give PKOBLEMS. -3 -"V- = \q[\ + 26 23 cos— — ^; ^2 = \q 3 — 2e 27 ( cos ~ 1- V3 sin ---. — H ^ JA h=\c^ Sat 2e"^ cos v3 sm 25 yj In a similar manner, if y is the liquid in the second vessel, the quantity of it in vessel 2 at the end of time t will be y^ = ^x above ; and similarly for the others. 60. To find the velocity of an ice hoat. Let AB represent the track of the boat, BD the position of the sail, 6 = DBA and WG the direction of the wind, which we will assume to be normal to the sail. When the boat ad- vances to C, the posi- tion of the sail will be CE. If V be the ve- locity of the boat, pro- portional to BC, and V the velocity of the wind relatively to the earth, then Avill ^c V —V sin 8 be the velocity relatively tp the sail, since the wind passing any point as B must travel a distance BH before coming in contact with the sail. The pressure of the wind is assumed to vary as the square of the velocity relatively to the surface pressed, and if M be the mass of the boat and sail, B a constant depending PROBLEMS. -i^l upon the size of the sail and the unit of velocity, and neglecting all resistances, we have m'^^ = B{v- V sin or-, .li'i \sin dt '^^' f,= Adt, where A = ^^ . Integrating, making F = for / = 0, dt from which it will be seen that V increases as t increases in- definitely, the limit of V being F = sin^' and this increases as 6 decreases, from which it appears that, according to the above hypothesis, the boat might attain an immense velocity for a small velocity of the wind The smaller the ancle of the sail with the track of the boat, the greater the ultimate velocity of the latter, there being no resistances. _ But the resistances may be considerable. The coefficient o friction on the ice may possibly be as low as 0.04. Should 1 o air move away from behind the sail with the velocity o the wind, or even with the velocity Fsin 6 no --t-ce .oald be offered bv the air ; but such will not be the case. The ma.t, !ai and other parts will be opposed by the resistance of the air as Ihev move hrough it, but it is difficult to determine the LcrLount. General Tower, in Van ^'ostrand's En,u^n^ Magazine, January, 1880, pp. 83, 84, according to an exam p^ of probable conditions, concluded that the maximum velocity of an ice boat might exceed twice that of the wind. 462 PEOBLEMS. Gl. Determine the j^f'th of a rotating tody projected into a resistitiff meditnn. A general solution of this problem has not been obtained, but certain qualitative results may be determined without finding the quantitative. Let the body be a sphere rotating about a vertical axis, the centre moving in a hori- zontal plane. Let go be the angu- lar velocity, and v the velocity of the centre ; then will OK, the distance of the spontaneous axis K from the centre be, equation (199), CO The combined motion of rotation and translation of the body at any instant being considered the same as that of the entire body rotating an infinitesimal amount about the axis K, the quadrants A and D will move Avith equal velocities exceeding those of B and C. The resistance of the medium to a body moving normally against it will vary as some power of the velocity of the body, and in this case may be considered as pro- portional to the same. The quadrant A moves against the medium with a greater velocity than B, hence the pressure on that quadrant will be greater, while the velocity of the quadrant D moving away from the medium, exceeds that of C. There- fore the pressure of the medium will be greatest on quadrant A, next B, then C, and least on D. The resultant of these press- ures will not be zero, and generally not parallel to OG. Let B be the resultant, the component of Avhich, parallel to OG, will be the pressure directly opposing the motion, and ab, normal to OG, the pressure which will deflect it from its initial direction. Neglecting friction, the path will be a curve convex towards the quadrant of greatest pressure, and will be more nearly a right line as K is more remote, or the more v exceeds go. Still, neglecting friction, the rotary motion will be constant, while the velocity of the centre will be diminished ; hence the curva- ture of the path will increase with the distance traveled. The friction between the medium and body tends directly to PROBLEMS. 463 diminish the rotation, but if the sum of the components of tlic frictional resistances resolved in reference to two rectangular planes be not zero, there will be a resultant. The friction will be greatest where the pressure is greatest, as at A, and act tan- gentially to the surface. Let F be the resultant ; it will be equal to a couple, and an equal parallel force at the centre, the former of which reduces the rotation; and of the latter, that component which is i^arallel to OG directly opposes the motion, and that which is normal to OG tends to deflect the path in the direction OB, opposite to that produced by pressure only. If the body be comparatively smooth and the medium rare, the friction •will be only a fractional part of the pressure, and the resultant friction will be only a very small part of the entire friction, in which case the direction of the curvature of the path will be de- termined by the resiiltawt jjressure, but the amount of curvature will be diminished by the friction. This case is illustrated by a rotating sphere projected into air. But if the body be rough and the medium dense, frictional resistance might exceed the pressure, in which case the direc- tion of curvature would be determined by the resultant friction, the amount being modified by pressure. This could be illus- trated by a wheel with flat vanes rotating about its axis, placed vertical and pushed along in water. Or still more strikingly, if a rough cylinder rotating about a vertical axis be pushed into a bank of earth, the tendency to a lateral motion might be almost entirely dependent upon the friction. APPENDIX II THE POTENTIAL. SO THE POTENTIAL. The term ''potential function," or simply the pote7itiaI, as used by Gauss and subsequent writers, is applied to a certain expression appearing in certain investigations involving forces depending upon some function of the distance between the bodies. It is of value in the higher investigations of the theory of attraction, hydromechanics, electricity, magnetism, and heat. Before defining it definitely, tal^e an example. Let m and 7)i' be the masses respectively of two particles, the place of m being X, y, z, oim', x', y' z', the distance between them r, and/(r) the law of the mutual attraction or repulsion. Then will the stress between them be P= =F 7nm'f{r), (a) — being attractive and + repulsive. The axial components will be /y' ^ q/' — f/ P cos « = JT = mm'/ (r) '' , Y = nim'f (r) —^^ , and for the distance between them, r' = {x - xy + iy' - yf + {z' - zj ; (c) from which we have for the partial differential coefficients, x, y, z, being considered fixed, rdr = (x' - x)dx', rdr = (y' - y)dy', rdr = (z' — z)dz. (d) Let/(r) be considered as the differential coefficient of some other function of r, so that '^^=/«; » 467 468 APPENDIX II. then will equations {d) and {e) reduce (b) to „ , dF{r) ^ , dF{r) „ , dF{r) X ■=. mm — ^\^ , Y = m m — ^— - , Z = 011 m — r4-^ . dx dy dz If m,, mz, etc., be the masses of other particles distant r^, r^, etc., from m', then we have, m being typical of W2„, m^, etc., ^ ,d2mF(r) „ ,d2mF(r) „ ,d'EmFir') , .. X = m ■ , ^ , 1—7)1 , , , Z = m YT^— '(f) dx dy dz ^•' ' Let 2mFir) = F, (g) then we have ,A?^\ XT rA^^'N 7 .A^^^ =-Ci^> ^='"'(:f)^ \dz'J' the parentheses indicating partial differential coefficients. The function Fis called tJte pote7itial, and may be defined by inter- preting equation (^). When found, the axial components of the stress appear as partial differential coefficients of F regarded as a function of the coordinates of the particle. It is rarely used in this general form, but is confined to the cases where the law of the force is that of the inverse squares — the law most common in nature. Let the system of particles m be continuous, form- ing a solid, and the force attractive, then we have and if d be the density at the place x, y, z, then dm = Sdxdydz ; _._ Y^ I r r ddxdydz ^ ^^ that is, The potential is sum of the quotients of all the element- ary masses divided by their dista7ice fro))i the attracted particle. THE POTENTIAL. 469 Again, if the mass of the attracted particle be unity, and that of the attracting particle be m, then at any distance r we have the stress, and hence an element of the work done upon the unit mass at that point in being moved over the space dr will be in dw = :;dr '. r- and similarly for any number of particles forming a continuous body ; W=:2w Sdxdydz _ I £^«^. Qc) which is the same form as that above, hence, also, The potential is the energy acquired hy a unit mass in fall- ing from infinity under the attraction of a given body to a dis- tance r. Again, m may represent any quantity of action, either attrac- tive or repulsive, as in magnetism, or electricity, in which the law of action is that of the inverse squares and product of their quantities. From equation (c) we have the partial derivative {dx ) X — X and from (0, considering Fas a function of /•, d V\ _ [[{ d{x' —x)dxdydz ^ dx j r 470 similarly, \dx"y APPENDIX II. ' 3{x' - x)' 1 (S^)-\\\[ dz "^YAW '3(y - yr 1 and adding drV d. -, + dif + dxdydz, dxdydz, dxdydz ; = 0; (0 ■which theorem was discovered by La Place. It is not general, however, for it is found to fail when the particle is one of the particles of the attracting mass ; hut it is correct when the particle attracted is external to the attracting body. Examples. '1. To find the potential of a slender uniform rod, length a, density d, and section s, upon an external particle m'. Take the origin at one end of the rod, x, along the rod, and x, y', the position of the particle m'. Then V=- V(a;' - xy + y" ^ -x + {x' ^- x'y Hence the attractive forces parallel and normal to the rod will be respectively. jr= — m l-j-T ) = 7n OS Wy'^ + {a - xj Vy" + ^'^- , (d V\ m'Ss Y — —m ,— = — — \dy J y a — X Wy' + (a-xj Vy' + a;'^J 2. To find the potential of a thin, homogeneous, spherical shell uj)on an external particle. THE POTENTIAL. 471 Let a = radius, d = density, da = thickness, p = distance of particle from the centre of the shell. Using polar coordinates, origin at the centre of the shell, (0, 0, p) the place of m', = polar distance of element of m from wliere p pierces the shell, qj = longitude, initial at any point, then dm = 6da • a sin Odq) • add, r" = a- — 2ap cos ^ + p- ; sin Sdddq) V = dda • d' (a" — 2a p cos 6 + p-)* ' or. {n) (P) For an external particle p > a, lience the last equation is the correct form for this case, and the former gives the potential for a particle internal to the shell. For the latter, -j.-_ ^Tcda'da _ m ~ P ~ p ' and for a concentric shell of finite thickness, ' p' v= 4:7T6 p J a« "7 4 a-da = - o P (al- •' Q ^ M P" which is the force along the line p ; hence, TTie atfracfion of a sjjlierical, homogeneous shell upon an 472 APPENDIX II. exterior partide is the same as if the mass of the shell were co7i- densed into ajiarticle at the centre. Equation {n) becomes V = 4:7r6ada ; hence, The attraction of a spherical homogeneous shell upon a par tick within it is zero. / J . L " ■ 'nj i n H I .uiwift^tnm-^"^ -Zt4-£._J;Ai-.i-s-a--,S.^ 2A.^V. \ i-4 u^.^ r^> V- ^\ C^' ^ It-ib '^ -^/l- t^i^-^ p ■at .> /jfTv^t.'!^ /^ cl^ 3~~>r?-^ in^C i AT V;ip\v :■#"' ■vf^* ^'•:>i$^:'' : 7 DAY USE RETURN TO ENGINEERING LIBRARY T©L No. 642-3339 This publication is due on the LAST DATE and HOUR stamped below. oubject to recall after FEB 2 1977 je eriCi of Sr.iiNG qucrifcf Subject to recalj after — MONOGRAPH OCT -8)1977 FOR DME DUE SEE LOAN SCHEDULE MONOGRAPH JAN\l 7 1978 FOR DAte DUE SPP )GRAPH APR l\4 1978 FOK DATE |RUE SEE lOA W SCHEPULE General Library University of California Berkeley W^^^ 7883C2 ^'^^^■l Library UNIVERSITY OF CAUFORNIA LIBRARY