ELEMENTARY DYNAMICS. A TREATISE ELEMENTARY DYNAMICS FOR THE USE OF WILLIAM GARNETT, M.A., \s HONOKAEY D.C.L. OF THE UNIVERSITY OP DURHAM; (LATE WHITWORTH SCHOLAR;) LATE FELLOW AND LECTURER OF ST. JOHN'S COLLEGE, CAMBRIDGE ; PRINCIPAL AND PROFESSOR OF PURE AND APPLIED MATHEMATICS IN THE DURHAM COLLEGE OF SCIENCE, NEWCASTLE-UPON-TYNE. FIFTH EDITION, REVISED. CAMBRIDGE : DEIGHTON, BELL AND CO. LONDON: GEOEGE BELL AND SONS. 1889. BUTLER & TANNER, THE SKI.WOOD PRINTING WORKS, FIIOME, AND LONDON. atask 00 CONTENTS. CHAPTER I. ON THE GEOMETRY OF A MOVING POINT, AND THE FUNDAMENTAL LAWS AND PRINCIPLES OF DYNAMICS. Time PAGE 1 2 6 7 9 12 15 17 20 21 21 22 Velocity ...,-...-. Unit of Length . Acceleration Composition of Velocities Belative Velocity Parallelogram of Accelerati Belative Acceleration . Angular Velocity Areal Velocity Matter ons Mass Density 23 Kinetic Energy defined 24 First Law of Motion . 25 Second Law of Motion 28 Unit of Force 29 Dyne 30 Impulse 31 Independence of Forces 34 Parallelogram of Forces . . . . . . .35 Weight 36 Value of " g " 36 Weighing . . . . . . . . . .37 Work . 38 Horse-Power and Energy . Third Law of Motion . Conservation of Energy Examples on the Laws of Motion Attwood's Machine Weight Independent of Velocity Weight Proportional to Mass Empirical Formula for " g " 5002128 41 43 48 51 54 57 59 60 VI CONTENTS. PAGE Evidence for Laws of Motion 62 Fundamental Units 63 Change of Units 63 Examples of Change of Units 68 Unit of Length 78 Unit of Area 78 Units of Volume and Time . 80 Units of Mass and Velocity ....... 81 Unit of Acceleration 82 Units of Momentum and Force . . . . . . .83 Unit of Work . 81 Units of Energy and Density 85 Unit of Impulse 86 Astronomical Unit of Mass 87 British Association Units and Nomenclature .... 88 Transmission of Power by Belts ...... 91 Transmission of Power by Shafting 93 Examples of Change of Units 94 Examination .......... 97 Examples 100 CHAPTEE II. ON UNIFOBM, AND UNIFORMLY ACCELERATE!*, MOTION. Uniform Velocity 107 Uniformly Accelerated Motion 107 Space described with Uniform Acceleration .... 108 Geometrical Representation Ill Space described during a Given Interval 114 Equations for Uniformly Accelerated Motion .... 115 Examples of Uniformly Accelerated Motion . . . . 117 Particle sliding down a Bough Plane 119 Pressure of Shot on Target 122 Motion of Particles connected by a String 123 Time of describing a Given Path 132 Time down Chords of a Vertical Circle 133 Straight Line of Quickest Descent 134 Kinetic Energy equivalent to Work done 136 Examples of Kinetic Energy 137 Examples of Horse-Power ........ 139 Convertibility of Energy 140 Conservation of Energy and Perpetual Motion . . . . 140 Examples on Units . 144 Examination 145 Examples ........... 147 CHAPTEE III. ON PROJECTILES. Motion of a Projectile . . . . . . . . 155 Path a Parabola . . . . . . . . . . 156 CONTENTS. Vll PAGE Range . . . .158 Latus Eectum and Position of Focus 159 Height of Directrix . . . 159 Velocity due to Fall from Directrix 162 Examples of Projectiles 163 Range on Inclined Plane 165 Geometrical Solution of Problems . . . . . . 167 Envelope of System of Trajectories 171 Fountain Jet 171 Change of Kinetic Energy and Work done 173 Problems on Projectiles . . . . . . . . 174 Examination 180 Examples 181 CHAPTEE IV. ON COLLISION. Nature of an Impulse . . . 188 Elasticity 189 Coefficient of Elasticity 189 Impact on a Moving Plane ........ 191 Oblique Impact on a Fixed Plane 192 Oblique Impact on a Rough Plane 193 Direct Impact of Smooth Spheres 194 Oblique Impact of Smooth Spheres 197 Action between Elastic Balls 199 " Forces of Compression and Restitution " 200 Examples on Collision 201 Continuous Impact 211 Falling Chain 212 Falling Chain of Variable Density 214 Kinetic Energy after Impact 218 Nature of Impact 221 True Ratio of Forces of Compression and Restitution . . 223 Ratio of Impulses 223 Energy dissipated by Sudden Changes of Motion . . . 224 Dissipation of Energy in Inelastic Strings . . . . 225 Effect of Shocks in Machinery 227 Examination 228 Examples 229 CHAPTER V. ON CUHVILINEAU MOTION, CYCLOIDAL MOTION, THE PENMJLUM, NOKMA ACCELERATION, CENTRIFUGAL FORCE, INITIAL TENSIONS, ETC. Motion of a Particle in a Smooth Tube Newton's Experiments on Impact Properties of the Cycloid . Motion of a Particle on a Cycloid Length of Arc of a Cycloid 243 246 249 251 253 Vlll CONTENTS. PAGE Cycloidal Pendulum 254 " Simple Equivalent Pendulum' 1 256 Determination of " g " by Pendulum 257 Normal Acceleration 258 Centrifugal Force 261 The Conical Pendulum . 261 The Parabolic Governor 264 Tension in a Kunning Belt 266 Velocity of a Transverse Wave in a Stretched String . . . 267 Harmonic Motion 268 Motion of certain Systems of Bodies 270 Initial Actions 273 Pressure of a Steam Hammer 277 Examination 279 Examples 280 CHAPTER VI. ON THE DYNAMICAL THEORY OF GASES. Constitution of a Simple Gas 286 Pressure due to a Stream of Particles 288 Mean value of cos 2 a 289 Velocity of Mean Square 293 Boyle's Law 295 Gay Lussac's Law 295 Velocity of Mean Square of Hydrogen 296 Absolute Zero of Temperature 297 Work done by Gas in Expanding 297 Cooling produced by Expansion 299 Expansion into Vacuum (Joule's Experiment) .... 300 Graham's Law of Diffusion 301 Gases act to one another as Vacua 301 On the Average Value of cos 2 a ... ... 302 Miscellaneous Examples 303 ANSWERS . 313 ELEMENTARY DYNAMICS. CHAPTER I. ON THE GEOMETRY OF A MOVING POINT, AND THE FUNDA- MENTAL LAWS AND PRINCIPLES OF DYNAMICS. 1. A POINT is said to be in motion, when it changes its position relative to surrounding objects. From this definition it will be seen that all cases of motion which will come under our consideration are essen- tially relative ; in fact, we have no means of measuring absolute motion, or of determining whether any given point is absolutely at rest in space, or not, and it may even be doubted whether the human mind is capable of form- ing a distinct conception of absolute motion. We shall not attempt to give a definition of space; our idea of it must be considered a primary conception. Time is defined by the metaphysician as " the suc- cession of ideas " ; the physicist treats time, like space, as a primary conception. Equal times are generally defined as those intervals during which the earth turns through equal angles rela- tive to the fixed stars, and any duration of time may then be measured by the angle turned through by the earth during the interval. The most obvious unit of time is therefore the sidereal day, or the period during which the earth makes a complete rotation on its axis relative to the fixed stars. The unit generally adopted is the second of mean solar time. That our fundamental conception of the measurement of time is not, however, based upon the rotation of the G. D. l B 2 VELOCITY. earth is apparent from the fact that we sometimes ask whether the length of the day has changed during the last two or three -thousand years. "Were the definition of equal times to which we have just alluded generally accepted, this question would be absurd, since all days would be equal by definition. The test of equality be- tween two intervals of time will be discussed in con- nection with the first law of motion, meanwhile the definition above given will be found sufficient for our purpose. 2. The velocity of a point is the rate at which it is changing its position relative to surrounding objects ; in other words, the degree of speed with which it is moving. A point is moving with uniform velocity when it passes over equal distances in equal intervals of time : under other circumstances its velocity is said to be variable. A distinction has been drawn between velocity and speed by restricting the use of the word speed to cases in which the direction of the motion is not considered while velocity implies direction as well as degree of quickness. If this distinction were generally adopted we could not speak of a body revolving in a circle with uniform velo- city but only with uniform speed, while the velocity would be undergoing a constant change. Uniform velo- city would imply uniformity in direction as well as in magnitude. 3. If we wish to convey the idea of speed in speaking of the motion of anything we say that it passed over a certain distance (say 60 miles) in a certain time (say an hour) ; while, if we wish to convey an idea of the slow- ness of the motion of anything we say that it took a certain time (say an hour) to traverse a certain distance (say two miles). Now the velocity of a point, being defined as the degree of speed with which it is moving, must always be expressed according to the former method ; viz. as so many units of length per unit of time. 4. The complete representation of any physical quan- tity consists of two factors, one of which is the unit in VELOCITY. 6 terms of which the quantity is measured and must be of the same kind as the quantity itself, while the other is a pure number indicating the ratio of the quantity to this unit arid called its measure. Velocity like all other quantities must be measured by its ratio to a unit of its own kind, that is, to a certain velocity selected as the standard, and the magnitude of this unit is all that is in our power to select. The velocity of a point which, mov- ing uniformly, passes over the unit of length in the unit of time is taken as the unit of velocity, and the velo- city of any other point is measured by its ratio to this unit. In this case the measure of the velocity of a point which is moving uniformly will be equal to the number of units of length traversed by it in the unit of time. 5. In order to measure the velocity of a point which is moving uniformly we have but to measure the space traversed in a given time, and dividing this distance by the measure of the time we have the space traversed during each second, or in other words, the velocity of the point. 6. When the velocity of a point changes continuously we may speak of its value at some particular instant, but a little care is required in order to determine exactly what is meant by the measure of a variable velocity at any particular instant. If the velocity be increasing, then, during the second succeeding the instant in ques- tion the point will move over a greater space than if the velocity were not on the increase but remained the same as at the commencement of the second. On the other hand, during the preceding second the space traversed is less than if the velocity were constant throughout the second and the same as at the end of it. We shall obtain a better result if we observe the space traversed during the second which contains the instant in question and half of which precedes while the other half succeeds the instant, but this will not give an accurate result unless the velocity changes uniformly. If however we take a very short interval including the proposed instant, the mean or average velocity during that interval will be ob- tained by dividing the measure of the distance traversed 4 MEASURE OF VELOCITY. by that of the interval. Now if the interval taken be exceedingly short the velocity has no opportunity of changing sensibly during it, and the mean velocity dur- ing the interval cannot sensibly differ from the velocity at any instant contained therein. Hence, by making the interval sufficiently short, we can obtain a result which differs by as little as we please from the velocity at the proposed instant. If we could make the interval indefinitely short and still perform the above operation we should realize our conception of the velocity of the point at a particular instant. 7. It is usual to state that the velocity of a point at a particular instant is measured when variable by the space which would be passed over in the unit of time supposing the velocity constant during the unit and the same as at the proposed instant. The words in italics take us back to the original difficulty, so that we appear to gain very little by the definition. 8. If the unit of length be increased or decreased, the unit of time remaining the same, the space passed over in the unit of time by a point moving with unit velocity is increased or decreased, and therefore the unit of velo- city is changed, in that same ratio. If, on the other hand, the unit of time be increased or decreased, the unit of length remaining the same, the time required by a point moving with unit velocity to pass over the unit of length is increased or decreased accordingly. Now the longer the time occupied by a point in moving over the same distance, the less must be its velocity, and the shorter the time the greater the velocity. Hence the unit of velo- city must vary inversely as the unit of time, if the unit of length remain constant. Also we have just shown that the unit of velocity varies directly as the unit of length when the unit of time is kept constant. There- fore, when all are allowed to vary together, the unit of velocity will vary directly as the unit of length, and in- versely as the unit of time. (See Todhunter's Algebra, Art. 425.) 9. As above stated the mathematical expression for any physical quantity always consists of two factors, one MEASUEE OF VELOCITY. 5 being the unit of the same kind as the thing considered, the other representing the number of such units in the quantity considered, and constituting the numerical mea- sure of such quantity. The complete representation of a physical quantity in mathematical language must there- fore consist of two symbols, representing these two factors respectively. The unit is sometimes represented by a capital letter placed in square brackets ; e.g. the unit of time thus [T]. Now the equations used in almost all mathematical investigations are equations between the numerical measures of quantities, and not between the quantities themselves ; the symbol representing the unit is therefore omitted, since the unit itself does not enter into the equations, and in consequence, the habit of re- presenting only the numerical measures of quantities has become so general, that even when the quantities them- selves are considered, but one symbol is generally used, the corresponding unit being understood. Now any quantity being represented by the product of the unit of the same kind, and the number of such units contained in the quantity considered, it is obvious that if the unit change, the quantity measured remaining the same, the number of units contained in it will be changed in the inverse ratio of the unit. Hence the numerical measure of any given quantity varies inversely as the unit in terms of which it is measured. For example, a stick 72 whose length is 72 inches will measure , or 6, feet, and 72 36 , or 2, yards. 10. Applying the principles of the preceding article to the measurement of velocity, we see that the numerical measure of any velocity will vary inversely as the unit of velocity ; and it has been shown that the unit of velocity varies directly as the unit of length and inversely as the unit of time. Hence the numerical measure of a velocity varies inversely as the unit of length and directly as the unit of time. Thus, a velocity of 10,560 feet per minute is equivalent to a velocity of 2 miles per minute, or of 120 miles per hour, or of 176 feet per second. Other examples 6 APPLICATION OF ALGEBRAICAL SIGNS. of the change of units will be found at the end of this chapter. The British standard unit of length is the imperial yard, which is denned to be the distance between the centres of the lines engraved on two gold plugs in a bronze bar kept in the Exchequer chambers, and known as the im- perial standard yard, the temperature of the bar being 62 Fahrenheit. The unit generally adopted by engineers is one-third of this distance, and is called a foot. As before stated, the unit of time generally adopted is the second of mean solar time. 11. If distances measured in one direction along a line be considered positive, it is usual to consider distances measured in the opposite direction as negative. Thus if distances measured from A towards B be reckoned posi- tive, those measured from B towards A or from A towards C will be negative. The same convention is extended to velocities. If a point move along a line in the direction in which distances are reckoned positive, its velocity is considered positive, but if it move in the opposite direc- tion its velocity is considered negative. Thus, if a point move from A to B, it increases its distance from A, measured in the positive direction, and its velocity is accordingly positive ; but if it move from B towards A it diminishes its distance from A, and its velocity is considered negative. Again, if the point move from A towards G, though it increases its distance from A con- sidered numerically, yet such distance being negative is decreased algebraically, and the velocity of the point is accordingly reckoned negative. In a similar way it will be seen that the velocity of a .point moving from C to- wards A will be positive. This convention being adopted, the velocity of a point which continues to move in the same direction will not change sign when the point passes through A. This is, of course, as it should be, since there is no more reason why the velocity of a point should change sign when the point passes through A than at any other point of its path. UNIT OF ACCELERATION. 7 12. Acceleration is the rate of change of velocity. It is said to be uniform when equal increments of velo- city are generated in equal intervals of time. If this be not the case the acceleration is variable. It is measured, when uniform, by the velocity generated in a unit of time ; when variable, it is measured, at any instant, " by the velocity which would be generated in a unit of time, were the acceleration to remain constant during that unit, and the same as at the proposed in- stant." All that has been said respecting variable velo- city applies to the measurement of variable accelerations. In order that the above may furnish a proper measure of acceleration, it will be seen that the unit of acceleration must be that of a point whose velocity is increased by the unit of velocity in the unit of time. If the unit of length vary, the unit of time remaining the same, it has been shown that the unit of velocity will vary in the same ratio ; hence, the unit of acceleration will also vary in the same ratio, and therefore when the unit of time remains constant, the unit of acceleration varies directly as the unit of length. 13. Next suppose the unit of length to remain constant, but the unit of time to vary. Then we have seen that the unit of velocity varies inversely as the unit of time. Now if the acceleration were always allowed the same time for the generation of the unit of velocity, the unit of acceleration would then vary directly as the unit of velocity, that is, inversely as the unit of time. But the time allowed for the generation of the unit of velocity does not remain constant ; it is in fact the unit of time, and therefore varies as that unit varies. Now if the time be diminished in which any given velocity is generated the acceleration must be proportionately increased ; and if the time be increased, the acceleration must be pro- portionately diminished. Hence, if the unit of velocity could be kept constant, the unit of acceleration would vary inversely as the unit of time, simply because the time during which the unit of velocity must be generated is changed. But it has been shown that if this latter were kept constant, the unit of acceleration would vary 8 MEASURE OF ACCELERATION. inversely as the unit of time, solely on account of the change in the unit of velocity. Hence, taking both reasons into account, when the unit of length remains constant, the unit of acceleration must vary inversely as the square of the unit of time. Also it has been shown that the unit of acceleration varies directly as the unit of length when the unit of time remains constant. There- fore, when ah 1 three are allowed to vary together, the unit of acceleration must vary directly as the unit of length, and inversely as the square of the unit of time. 14. The numerical measure of any acceleration varies inversely as the unit of acceleration. Therefore the numerical measure of a given acceleration varies inversely as the unit of length and directly as the square of the unit of time. The above reasoning will be rendered much clearer by the consideration of an example. Suppose a certain acceleration to be represented by 32 when a second and a foot are the units of time and length rexpecticely ; what will be the measure of the same accelera- tion when a minute and a yard are units ? With the acceleration in 1" there is generated a velocity per 1" of 32 feet ; .-. 1" 60" ,,32x60 feet; .-. 60" 60" 32 x 60 2 feet. But 32 x 60 2 feet are equivalent to 38,400 yards. Hence with the given acceleration in one minute there is gene- rated a velocity of 38,400 yards per minute, and therefore when a minute and a yard are the units of time and length respectively the acceleration will be represented numerically by 38,400. 15. An acceleration is reckoned positive when the velo- city of the moving point tends to increase algebraically ; if the velocity tend to diminish algebraically the acclera- tion is considered negative. Thus an acceleration is con- sidered positive if, with it, a positive velocity increase numerically or a negative velocity decrease numerically ; while an acceleration with which a positive velocity de- creases or a negative velocity increases numerically is COMPOSITION OF VELOCITIES. 9 considered negative. An acceleration with which the numerical measure of a velocity tends to decrease is fre- quently called a retardation. 16. A velocity is completely known if we know its magnitude and direction. Now a straight line can be drawn in any direction and of any length ; if then a straight line be drawn in the direction in which a point is moving and of such length as to contain as many units of length as there are units of velocity in the velocity of the point, such a straight line will represent in every respect this velocity. Similarly an acceleration is completely known if we know its magnitude and the direction of the velocity generated. An acceleration may therefore be represented in every respect by a straight line drawn in the direction of the velocity generated, and containing as many units of length as there are units of acceleration in the accelera- tion in question. Moreover, since an acceleration is measured by the number of units of velocity generated in the unit of time, the straight line which represents an acceleration in magnitude and direction may also com- pletely represent the velocity , generated in the unit of time to which the acceleration corresponds. 17. Suppose a point to be moving with two indepen- dent velocities in any directions ; then at any instant the point must be moving in some definite direction and with some definite velocity ; this velocity must therefore be equivalent to the two independent velocities, and is called their resultant ; the independent velocities themselves, considered with reference to their resultant, are called components. The same reasoning must apply to any number of independent velocities with which a point may be moving, and which must therefore be equivalent to a single resultant. Let a point be supposed to move with two independent velocities, and let it be required to find the actual or resultant velocity of the point. For ex- ample, the point may move with a given velocity along a straight tube, while the tube, always remaining parallel to its original direction, slides with a given velocity and in a given direction along a fixed plane. The problem 10 PARALLELOGRAM OF VELOCITIES. will then be to determine the velocity of the point and the direction of its motion relative to the fixed plane. The two independent velocities in this case are the velo- city of the point along the tube, and the velocity of the tube relative to the fixed plane. 18. If the two velocities be in the same straight line it is obvious that the resultant velocity is the algebraical sum. of the two ; velocities in one direction being con- sidered positive and those in the opposite direction ne- gative ; and similarly the resultant of any number of independent velocities in the same straight line is the algebraical sum of the component velocities. If the independent velocities be not in the same straight line their resultant must be found by help of the following proposition, known as the " parallelogram of velocities." PROP. If a point be moving with two independent velo- cities represented in magnitude and direction by two straight lines drawn from a point, the resultant velocity will be re- presented in magnitude and direction by the diagonal, drawn from that point, of the parallelogram constructed upon these two straight lines as adjacent sides. Let ABj AC represent in magnitude and direction the velocities, which we denote by w, v respectively. Then AS, AC may denote the spaces passed over in the unit of time by points moving with the velocities u, v respectively. Complete the parallelogram ABDC, and draw the diagonal AD. Then AD shall represent in magnitude and direc- PARALLELOGRAM OF VELOCITIES. 11 tion the resultant of the velocities represented by AB and AC respectively. Let the moving point be denoted by P, and suppose P to move along a straight tube OK with uniform velocity u, while the end of the tube moves uniformly along the straight line AC, with velocity v, the tube remaining always parallel to AB. Then, supposing the point P to start from A when the tube is in the position AB, if at the end of any time r we take, along AC, AO equal to vr, and draw OK parallel to AB, OK will be the position of the tube ; and taking OP equal to tir, P will be the posi- tion of the point. Draw PN parallel to AC; then .'. AN : AO :: ur : vr :: u : v :: AB : AC. Therefore the parallelograms ON, CB are similar. But similar parallelograms which have a common angle are about the same diagonal ; therefore P lies on the diagonal ^4 D ; and since T may be any interval we please, the point P will always lie on AD or AD produced ; in other words, AD is the path of P, and therefore always represents its resultant velocity in direction. Also the resultant velo- city of P is constant in magnitude : for AP, the space passed over during the interval r, is always proportional to AN or UT, that is to T, since u is constant. The space passed over in any time is therefore proportional to that time, and the velocity of the moving point is therefore constant in magnitude. Again, at the end of the unit of time after leaving A the end of the tube will have reached C, and the point P will consequently be at D. Hence in one unit of time P will have moved from A to D, and we have just shown that its path is the diagonal AD and its velocity uniform. Hence the straight line AD represents in magnitude and direction the resultant velocity of the point P, that is, the resultant of the velocities represented by AB and AC respectively. Therefore, if. etc. Q.E.D. 12 RELATIVE VELOCITY. 19. If a particle be moving with more than two inde- pendent velocities, we can find their resultant by finding first the resultant of any two, then compounding that with a third, and so on. Also, since velocities, like forces, are subject to the parallelogram law of composition and resolution, the propositions which are true for a system of forces acting at a point on account of the forces being subject to this law, are also true for a number of indepen- dent velocities with which a point may be moving. Thus, if the straight lines representing the velocities are equal, and parallel to, and in the same sense as, the sides of a closed polygon taken in order, the point is at rest ; if they form all but one of the sides taken in order of such a polygon, their resultant is represented by the remaining side taken in the reverse order. Similarly we have the triangle of velocities, the parallelepiped of velocities, and so on. 20. If the velocities of two moving points, A and B, be given relative to certain points which we consider fixed, we can, by help of the " parallelogram of velocities," determine the velocity of A relative to B. Suppose, for example, that B is a point fixed on some surface which moves in any direction without rotation, and with known velocity, so that every point of the sur- face has the same velocity as J5, and that A is a point which moves on that surface, the velocity of A, at any instant, relative to fixed objects around being known ; then the velocity of A relative to the surface, that is, rela- tive to B, can be found. For instance, B may be a point fixed on the deck of a ship which is moving uniformly 011 a still sea, and A some point moving about on deck ; the velocities of A and B relative to the water being given, the velocity of A relative to the ship's deck can be found. If the points A and B be moving in the same straight line and in the same direction with uniform velocities u and v respectively, it is obvious that the distance between A and B is increased or diminished during each unit of time by a space numerically represented by u v, accord- ing as A is in front of, or behind, B ; u v is therefore the velocity of A relative to B. If u v be negative, it shows EELATIVE VELOCITY. 13 that the velocity of A relative to 1> is in the direction opposite to that in which the points are moving. If A and B be moving in opposite directions with velo- cities represented numerically by u and v respectively, it may be shown that the velocity of A relative to B is nu- merically represented by u + v ; but if u and v represent not only the numerical but the algebraical values of these velocities, then u and v will be of opposite signs, and the velocity of A relative to B will be represented by u v, as in the case in which A and B are moving in the same direction. 21. In each of the above cases the velocity of A rela- tive to B may be found by the following process. Let a velocity equal and opposite to that of B be supposed given to both A and B. Then B will be brought to rest, and the same velocity being impressed on both A and 7J, their relative velocity will be unaffected. For example, if a man be walking on the deck of a ship, his velocity relative to the deck, or any point upon it, is altogether indepen- dent of the ship's motion, and will remain the same if the ship be brought to rest. The velocity of A will then be u v ; and since B will have been brought to rest, this will be the velocity of A relative to B. Hence if two points A and B be moving in the same straight line with velocities represented in magnitude and direction by u and v respectively, the velocity of A relative to B will be represented in magnitude and direction by u v. If A and B be moving in parallel straight lines with velocities represented respectively by u and v, we may show, by precisely the same method as that adopted above, that the velocity of A relative to B is u v, as before. 22. Suppose A and B to move with velocities repre- 14 RELATIVE VELOCITY. seiited by u and v respectively, but not in the same or parallel straight lines. Let the velocity of A be repre- sented in magnitude and direction by OH, and that of B by OK. Then the velocity of A relative to B will be represented by KH. For the velocity of A may be considered as the re- sultant of two velocities ; viz. the velocity of B, and the velocity of A relative to B. Hence the velocity repre- sented by OH is the resultant of two independent velocities, one of which, that of B : is represented by OK, and the other is the velocity of A relative to B. But the velocity represented by OH is, by the " parallelogram of velocities," the resultant of velocities represented by OK and KH. Therefore the velocity of A relative to B is represented in magnitude and direction by KH. If this velocity be denoted by w, we have, since KH*=OH 2 +OK 2 -20H. OKcosKOH, w 2 = u 2 + v 2 2uv cos HOK. 23. The same result may be obtained in a different way. As before, let OH, OK represent the velocities of A and B respectively. Suppose a velocity equal and opposite to that of B to be impressed upon both A and B. Let OL represent in magnitude and direction this velocity. Then OL is equal to KO, and in the same straight line with it. Also by this means B will be brought to rest while A is made to move with two inde- pendent velocities ; viz. its original velocity represented by OH and the velocity we have supposed impressed PARALLELOGRAM OF ACCELERATIONS. 15 upon it, which is represented by OL. Complete the parallelogram OH PL ; then, by the " parallelogram of velocities," the diagonal OP represents in magnitude and direction the resultant velocity of A. But since B is now at rest, the resultant velocity of A is the same as its velocity relative to B. Also the velocity of A relative to B will have been unchanged by impressing the same velocity on both A and B. Hence OP represents in magnitude and direction the velocity of A relative to B. If, then, we wish to find the velocity of a moving point A relative to another moving point B, we may impress on both A and B a velocity equal and opposite to that of B, and the resultant velocity of A will then be the velocity required. That this process leads to the same result as the method of the preceding article is at once obvious, for KOPH is a parallelogram, and therefore OP is equal and parallel to KH. If the velocities of A and B be not uniform, the velocity of A relative to B at any instant may be found as above, OH and OK representing in this case the velocities of A and B respectively at the instant in question. The velocity of B relative to A is of course equal and opposite to that of A relative to B. 24. Accelerations, also, like forces and velocities, may be resolved and compounded according to the parallelo- gram law. This we proceed to prove. PROP. If a point be moving with two independent ac- celerations represented in magnitude and direction by two straight lines drawn from a point, the resultant accelera- tion icill be represented in magnitude and direction by the diagonal, drawn from that point, of the parallelogram constructed on the two straight lines, representing the acce- lerations, as adjacent sides. For simplicity of expression, suppose a second to be the unit of time. Let AB represent the initial velocity of the point ; EC, BD the accelerations. Then BC, BD repre- sent the velocities generated in one second corresponding to the respective accelerations taken singly. Now since 16 PAEALLELOGBAM OF ACCELEEATIONS. the accelerations are independent, their combined effect produced at any instant is the sum of the effects corre- sponding to each considered singly at that instant, and hence the final change of velocity produced in one second under the two accelerations together is the same as if each had existed separately during one second, since the effect of an acceleration is independent of the velocity of the moving point. Suppose then the acceleration repre- sented by 13 C to exist by itself for one second ; the velocity generated in that interval is represented by BC, and compounding this with the original velocity, AB, of the moving point, it follows from the parallelogram of velocities that the resultant velocity is represented by AC. Now suppose the acceleration represented by BD to exist singly for a second, the point moving, initially, with the velocity represented by AC. Draw CE equal and parallel to BD ; then CE represents the velocity generated in one second corresponding to the acceleration represented by BD. Combining this with the velocity which the point already possesses, viz. that represented by AC, the final velocity is, by the parallelogram of velo- cities, represented by AE. But the effect when the two accelerations exist together for one second is the same as the whole effect produced when each exists separately for one second. Hence the effect when the two accelera- tions exist together for a second is to change the velocity of the moving point from that represented by AB to that represented by AE. The velocity generated when the two accelerations exist together for a second is therefore EELATIVB ACCELEKATION. 17 represented by BE, since BE represents that velocity which, when compounded with the velocity represented by AB, produces a resultant represented by AE. But the whole velocity generated in a second is the measure of the resultant acceleration ; hence BE represents the acceleration which is the resultant of the accelerations represented by BC and BD. But BE is the diagonal of the parallelogram constructed upon the straight lines 5(7, BD as adjacent sides. Therefore, if, etc. Q.E.D. Hence, accelerations, like forces and velocities, are sub- ject to the parallelogram law of composition and resolution, and any propositions true of forces in consequence of their being subject to this law must also be true of accelerations. We have therefore the triangle, polygon, etc., of accele- rations, and any number of accelerations may be com- pounded in the same way as a system of forces acting at a point. 25. By help of the preceding proposition, if the acce- lerations of moving points A and B be given, we can, as in the corresponding case of velocities, find the acce- leration of A relative to B. Let OH, OK represent in magnitude and direction the accelerations of A and B respectively. Suppose an acce- leration equal and opposite to that of B impressed upon both A and B. This cannot affect the relative motions of A and J5, and therefore does not affect their relative G. D. c 18 RELATIVE ACCELERATION. accelerations. (For we may suppose A and B to be points moving about inside a closed surface ; then if the same motion in space be impressed upon the surface and every- thing within it, this will obviously not affect the motions of A and B relative to the surface or each other.) Let OL represent the acceleration equal and opposite to that of B. Then OL is equal to KO and in the same straight line with it. Now an acceleration equal and opposite to that of B having been impressed upon it, it will be moving with no acceleration, that is, with uniform velocity. Suppose a velocity equal and opposite to that of B impressed on both A and B ; this will not affect their relative velocity, but B will thereby be brought to rest. Complete the parallelogram OHFL, and draw the diagonal OP. Then the acceleration of A is the resultant of two independent accelerations represented respectively by OH and OL, and is therefore, by the parallelogram of accelerations, represented in magnitude and direction by the diagonal OP of the parallelogram constructed upon OH and OL ; and, since B is now at rest and not possessed of any acceleration, this is the acceleration of A relative to B. Since KOPH is a parallelogram, OP is equal and parallel to KH; hence KH represents in magnitude and direction the acceleration of A relative to B. From the above investigation we see that in order to find the acceleration of a moving point A, relative to another moving point 5, we have only to impress on both A and B an acceleration equal and opposite to that of B. The resultant acceleration of A is then the original ac- celeration of A relative to B. Or, more briefly, if the accelerations of A and B be represented in magnitude and direction by two straight lines OLT, OK respectively drawn from a point 0, the acceleration of A relative to B will be represented in magnitude and direction by the straight line KH. 26. If the accelerations of A and B be not uniform, the acceleration of A relative to B may be found at any instant by the above process, OH, OK representing the accelerations of A and B respectively at that instant. ANGULAR VELOCITY. 19 The acceleration of B relative to A is obviously equal and opposite to that of A relative to B. If the accelerations of A and B be in the same straight line and be represented numerically by a and /3 respec- tively, the acceleration of A relative to B will be repre- sented by a/3, according as the accelerations of A and B are in the same or in opposite directions. If the accelerations be represented algebraically by a and /3, the acceleration of A relative to B will be represented in magnitude and direction by a /?. 27. If a point move with uniform velocity v, the space passed over by it in t units of time is equal to vt units of length. For the distance passed over in each unit of time is v units of length, and therefore the distance passed over in t units of time is ft units of length. If a point move with a certain acceleration always in the direction of motion and represented by /", the velocity generated in t units of time and corresponding to this acceleration is represented by ft. For the velocity gene- rated in each unit of time is numerically equal to the acceleration, and is therefore /' units of velocity ; therefore the velocity generated in t units of time is ft units of velocity. If the direction of the acceleration be always that of the motion of the point, and its magnitude be constant and represented by /", and if the point be originally moving with u units of velocity, its velocity after t units of time will be represented by u + ft. If a point be moving with an initial velocity in a certain direction, and then continue moving for a given time with an acceleration in some other constant direction, since with a given acceleration the velocity generated at any instant is independent of the velocity or direction of motion of the moving point, it follows that the final velo- city of the point may be determined by finding the velo- city generated in the direction of the acceleration, and compounding this with the original velocity of the point according to the parallelogram law. 28. If P represent a point which moves in any manner 20 ANGULAR VELOCITY. relative to"0, and OX be a line through O and fixed in direction, the rate of change of the angle POX is the angular velocity of P about 0. If P be moving directly towards or from its angular velocity about is zero. If P be moving at right angles to OP with velocity v, and P l be its position after a very short .time r, then PP l = vr and the angle POP l described in time T is re- pp Vr presented by ^ = - . Hence the angular velocity of P n* about is represented by ., p . 29. If P be moving with velocity v in a direction making an angle 6 with OP, and P 2 be its position after JC a very short time T, then PP 2 is represented by vr. Draw P.,N perpendicular to OP. Then P 2 A T =PP 2 sin = vr sin 0. Also the angle P0P 2 = _^_ = ^I-^ - and the angular i -, ^ , -. -, v sin 6 v sin 6 velocity about O is represented by ^-^- or n Ti > since \J I 2 LXJ by taking r sufficiently small OP a may be made to differ from OP by a quantity as small as we please. KINETICS. 21 30. It the point P is moving along any line XP (which may be straight or curved) the rate of increase of the area POX is called the areal velocity of P about O. Referring to the figure of the last article, the area of the triangle POP*, is represented by ^P 2 N. OP=\ . VT sin 6 . OP=\ VT sin d.r if OP be represented by r. Hence the areal velocity will be represented by ^ vr sin 6. If to represent the angular velocity, then o> - or v sin d = a>r. Hence the areal velocity is represented by !r. 31. The rate of change of angular velocity is called angular acceleration. It is measured when uniform and when variable according to the same principles as linear acceleration. Rate of change of areal velocity is called areal accele- ration. KINETICS. 32. Hitherto we have been considering simply motion without any reference to the agencies producing it, or the properties of the thing moved. This portion of the sub- ject, or the geometry of motion, is frequently called Kine- matics. We must now consider motion with reference to the agencies producing it, and the things in which it is produced. The term Dynamics is sometimes confined to this section of our subject, but it is more frequently known as Kinetics, while Dynamics is (improperly) under- stood to embrace both this and Kinematics. Matter, like space and time, must be considered as one of the inevitable primary conceptions of the mind, of which no satisfactory definition can be given. Many of its properties are known to us with more or less of scientific exactitude by our every-day experience, and it is to this we must refer for a distinct conception of matter. 33. The most characteristic manner in which matter affects our senses is through the effort required to produce in it sudden changes of motion. If a small mirror be held 22 MASS AND FORCE. in the hand a reflected sunbeam may be made to dance about in any arbitrary manner without any sensible effort on the part of the operator, for the small exertion of which he is conscious is the same whether the sun be shining on the mirror or not. But it is far from easy to move a half hundred-weight quickly aside and then sharply to bring it to rest, even though it be suspended by a very long string which supports its weight and allows of its being moved almost in a horizontal plane. We say that the half hundred-weight is matter, while we call the sunbeam immaterial. The character of matter, in virtue of which an effort is required to rapidly change its motion, is sometimes called inertia, but as this term only conveys the same idea as the word mass, when properly understood, it will not be frequently employed in subsequent pages. 34. A body is a quantity of matter limited in every direction. A particle is a portion of matter whose dimensions in every direction are indefinitely small, and which may therefore be treated as a physical point. We however adopt the convention of conceiving particles to contain a finite quantity of matter, though it is contrary to ex- perience that an indefinitely small body should contain a finite quantity of anything. The conception of a par- ticle simply enables us to treat a small body as though it were indefinitely small, and thus enables us to neglect motions of rotation which, if considered, would remove our investigation to the domain of " Rigid Dynamics." Force is that which, produces or tends to produce motion in matter, or modifies or tends to modify existing motion. The mass of a body or particle is the quantity of matter which it contains. 35. The masses of two or more particles are said to be equal when the same force acting similarly upon them for the same time generates in them the same velocity. If two bodies of equal mass be connected so as to form one, the mass of the body so formed is double that of DENSITY. 23 each of its constituents. Similarly, if three equal masses be connected together, we get a body of triple mass, and so on. We thus arrive at a system of measurement appli- cable to masses. The British standard unit of mass is the imperial standard pound Avoirdupois, a mass of platinum kept at the Exchequer chambers. All masses may be expressed in terms of this unit, or of other units deduced therefrom. The French imperial standard of mass is Borda's platinum kilogramme, which was origi- nally constructed so as to contain the same amount of matter as a cubic decimetre of distilled water at the temperature corresponding to its maximum density. The pound and the kilogramme are standard of mass, not of force. After stating the laws of motion we shall see how unequal masses may be compared by observing the effects of forces upon them. We are not at present in a position to explain any system of measurement applicable to forces. Newton's second law of motion will however provide us with the means of comparing forces, and consequently of measur- ing them in terms of some unit. 36. The density of a body, when uniform, is the ratio of the mass of any volume of the body to that of an equal volume of some standard substance. When variable, an approximation to the density at any point is obtained by determining the number of units of mass in a very small volume including the point and dividing by the measure of the volume. If the volume could be indefinitely diminished the result would give accurately the density at the point in question. Hence, when the density of a body is variable its measure at any point is the ultimate ratio of the number of units of mass in any volume containing that point to the number of units of volume in the same when such volume is indefinitely diminished. The standard substance is generally so chosen that the unit of volume of the substance contains the unit of mass of matter. The unity of density is consequently the density of a uniform substance, of which the matter 24 KINETIC ENEEGY. contained in the unit of volume is the unit of mass. Then, the density of a body, whose density is uniform, will be measured by the number of units of mass con- tained in the unit of volume, or, which is the same thing, the ratio of the number of units of mass contained in any portion of the body to the number expressing the volume of that portion. If the density of a substance be uniform, and numerically represented by p, the mass of V units of volume of the wtb- stance will be Vp units of mass. For each unit of volume contains p units of mass, and therefore V units of volume will contain Vp units of mass. 37. The momentum of a particle is the product of its mass and its velocity. The unit of momentum is conse- quently the momentum of the unit of mass moving with the unit of velocity, and the momentum of any moving mass is measured in terms of this unit. The phrase " quantity of motion " was used by Newton in place of the more modern term " momentum." 38. The vis viva or kinetic energy of a moving particle may be denned as one half the product of its momentum and its velocity. The vis viva used to be denned as the product of the mass and the square of the velocity of the moving par- ticle, and the kinetic energy was called the semi-vis viva ; but the above definition is more convenient, and has moreover the advantage arising from the fact that both the momentum and the velocity of a particle admit of a physical interpretation, while no meaning has been assigned to the square of a velocity. 39. All the theorems hitherto given have been deduced from abstract reasoning, and it is impossible for us to con- ceive of any order of things in which these theorems should not be true ; but in order to determine the mutual relations between force and matter, or, in other words, the effect of forces upon matter, we must have re- course to experiment. The conclusions to which such experiments lead us are embodied in three statements, generally known as the laws of motion, and first given LAWS OF MOTION. 25 by Newton. As enunciated by him these laws are as follows : LAW I. Every body will continue in it* state of rest or of uniform motion in a straight line, except in so far as it is compelled by impressed force to change that state. LAW II. Change of motion is proportional to the im- pressed force, and takes place in the direction in ichich that force acts. LAW III. Action and reaction are always equal and opposite. 40. The evidence upon which these laws are accepted may be stated as follows : In the first place, our daily observations of phsenomena around us, and the results of rough experiments, lead us to infer the probable truth of the principles enunciated in the statement of the laws ; it is then found that the more nearly we make the conditions, which obtain in our expe- riments, approximate to the ideal conditions of the case to which the laws are immediately applicable, the more nearly do the results of our experiments coincide with the principles laid down in these laws ; and lastly, the results of long and complicated calculations based on the assumption of their truth are exactly in accordance with natural phsenomena. This last evidence is that on which we chiefly rely, and it amounts almost to an absolute proof of the points at issue. Our acceptance of all other physical laws rests on precisely the same kind of evidence. As an example we may refer to the moon, whose motion is calculated on the assumption of the truth of the laws stated above, and of the law of gravitation ; and, notwithstanding the extreme complexity of the calculation, we are enabled to determine the moon's position at any instant for years in advance with such precision as to be within the limits of error of the most exact instruments. It was also by the assumption of the truth of these same laws that Prof. Adams and M. Le Verrier were enabled to calculate the position and orbit of the planet Neptune before it had been seen. 26 THEORETICAL MEASUREMENT OF TIME. 41. We shall now consider each of these laws sepa- rately, and trace them into some of their consequences. LAW I. Every body will continue in it* state of rest or of uniform motion in a (straight line, except in so far as it is compelled by impressed force to change that state. This law supplies us in the first instance with the definition of force given above. For if a body do not continue in its state of rest or of uniform motion in a straight line it must be under the action of force ; so that force is that which tends to change a body's state of rest or of uniform motion in a straight line. Secondly the law indicates a mode of measuring time. The velocity of a body is uniform when the body passes over equal distances in equal intervals of time. Suppose there are two bodies A and B and that no force acts upon either of them which has a tendency to change its motion relative to the other. Then by the first law of motion B must move uniformly in a straight line relative to A. Hence the intervals of time during which B moves over equal distances relative to A must be equal. Therefore when applied to this system of only two bodies, the first law of motion simply defines equal intervals of time as those during which B moves over equal distances relative to A, but it states no law of nature. Now suppose a third body C introduced and consider its motion relative to A. If C move in a straight line relative to A and pass over equal distances in the intervals during which B passes over equal distances, then B being our time-keeper, it follows that C moves uniformly relative to A and is therefore, by Law I., under the action of no force which has a tendency to change its motion relative to A. If, however, C do not fulfil this condition, some force must act upon it in such a way as to change its motion relative to A. The statement that if both B and C be under the action of no forces tending to change their motion relative to A, then C will move over equal distances in the same intervals during which B moves over equal distances, is one which considered a priori might or might not have been true, and it is in this statement that the law con- sists. FIRST LAW OF MOTION. 27 42. If we were furnished only with the system con- sisting of A, />', and C\ we might have no better reason for supposing B to be under the action of no force, than for supposing C to be free from the action of force. In this case the choice of B as our time-keeper would be a purely arbitrary choice. But suppose there were a large number of bodies B, C, l)...all moving relative to A, and suppose that of this multitude a certain number Z, M, etc., all moved over equal distances relative to A in the same intervals during which K moved over equal distances. Then all these bodies A", L, 3f, etc., agree in indicating the same intervals of time as equal intervals, that is, they all provide us with the same measure of time, and if there be no apparent reason why all these bodies should be similarly acted on by forces, we have very good reason to believe that each of them is under the action of no forces and to accept their joint testimony as the basis of our measurement of time ; while if the rest of the bodies B : C, D, etc., neither agree with A", L, M, etc., nor yet among themselves, we have good reason to believe that these are acted upon by forces, and to reject them all as means of measuring time. It is from the contemplation of such a system of bodies that our highest conception of the measurement of time is derived, and though we sometimes speak of equal intervals of time as those during which the earth turns through equal angles, we really ultimately refer to the joint testimony of all the heavenly bodies, after duly allowing for all the forces which we know to act upon them, in determining our measurement of time, and thus we may ask whether the length of the day (or the rate of rotation of the earth) is the same now as formerly, a question which would be absurd if our funda- mental notions of the measurement of time were based upon the earth's rotation simply. (See Maxwell's Matter and Motion.} 43. The first law of motion attributes to matter the property known as inertia, by virtue of which a finite force acting during a finite time is required, to produce a finite change in the velocity of a finite quantity of mat- ter. In other words, it states that any particle of matter 28 SECOND LAW OF MOTION. lias no power in itself of changing, of its own accord, any velocity with which it may be moving. Now that a body, if at rest, will continue so, if no force act upon it to disturb it, every one will at once admit as in accordance with every-day observation ; but that a body in motion will continue to move uniformly in a straight line is not quite so obvious, because we never have an opportunity of observing the motion of a body under the action of no forces. If a stone be projected along a horizontal plane it will at length come to rest, but if the stone and plane be made smoother the stone will continue longer in motion, and this leads us to believe that if all opposing forces were removed, the stone would never come to rest. If, however, we reflect on the facts that we have no no- tion whatever of absolute motion ; that all motion which comes under our notice is essentially relative ; that a particle appears at rest when it is moving in the same direction and with the same velocity as ourselves ; and that we have no means of ascertaining whether a particle is absolutely at rest or not, we see that there is no dis- tinction in kind between uniform motion in a straight line and absolute rest, the latter being in fact only a particular case of the former. The assertion, therefore, that a particle under the action of no forces will remain at rest is entirely without meaning, since we never know whether a particle is absolutely at rest or not, unless it be also true that a particle under the action of no forces will continue to move uniformly in a straight line if it be once in a state of motion, so that this latter part of the law is a necessary consequence of the former. 44. LAW II. Change of motion is proportional to the impressed force, and takes place in the direction in which that force acts. The phrase "change of motion" here is equivalent to change of quantity of motion or of momentum. If the force be finite it will require a finite time to produce a sensible change of motion, and the change of momentum produced by it will depend upon the time during which it acts. The change of motion contemplated must then be understood to be the change of momentum produced SECOND LAW OF MOTION. 29 per unit of time, or the rate of change of momentum.* If the force be variable, the rate of change of momentum is measured at any instant by the momentum which would be generated in the unit of time if the force remained constant during that unit and the same as at the proposed instant, and this the law asserts to be proportional to the intensity of the force at that instant. Now the momen- tum of a moving particle is the product of the mass into the velocity ; if, then, the mass remain constant the change of momentum is measured by the product of the mass into the change of velocity, and the rate of change of momentum by the product of the mass and the rate of change of velocity. But rate of change of velocity is acceleration ; therefore this law states that any force P acting on a particle of mass m is proportional to the pro- duct of the mass m on which it acts, and the acceleration /' produced therein by the force ; /. P GO mf. Hence we may put P equal to 7cmf] where k is some con- stant. 45. If the unit of force be taken as that force which produces the unit of acceleration in the unit of mass, or, which acting on the unit of mass for the unit of time generates therein the unit of velocity, then, if we put m equal to unity, that is, take the unit of mass, and /' equal to unity, that is, introduce the condition that the accelera- tion produced therein shall be the unit of acceleration, we must have the force producing the acceleration equal to the unit of force, or P equal to unity. Hence fc must also be equal to unity, and we have the equation P=mf. The meaning of this equation is as follows : The num- ber of unit* of force in any force in equal to the product of the number of twits of mans in any particle on which it may ct, and the number of units of acceleration produced in that maxs by the force in question. * Newton made the second law of motion refer to the whole c/ianye of momentum produced, which is the equivalent of the im- 30 MEASURE OP FORCE. It should be carefully noted that the unit of force im- plied in the above equation is that force ichich acting on unit maw for unit time produce* unit velocity. This unit of force is implied in all dynamical equations in which no numerical constant is introduced as a factor to change the unit. 46. Suppose that a second and a foot are the units of time and length respectively, a pound being the unit of mass, and that we require to know the corresponding unit of force, in order that the above equation may be true. We may argue thus : The unit of force in this case is that force which acting on the mass of a pound for one second generates in it a velocity of one foot per second ; now we know from the results of experiments, some of which will be described hereafter, that a force equal to the weight of a pound in London, at the sea level, if acting on the mass of a pound generates in it in one second, if free to move, a velocity of nearly 32 - 2 feet per second; and hence the unit of force is ^^ of the weight of a pound, or rather less than fja'a the weight of half an ounce. This is the British absolute dynamical unit of force, and is called a poundal. In order that the equation P=mf may be universally true when a pound, a second, and a foot are the units of mass, time, and length respectively, all forces must be expressed in terms of this unit. 47. If a centimetre be taken as the unit of length, and a gramme as the unit of mass, a second being the unit of time, the unit of force is that which acting on a gramme for a second produces a velocity of a centimetre per second, and is called a dyne. The weight of a gramme at the sea level in Paris is equal to about 981 dynes. 48. Again, from the equation Pmf we see that a force is measured dynamically by the momentum which it generates in the unit of time. We have thus a method of comparing the magnitude of one force with that of an- other, and are enabled to express any force in terms of the i>ulxe of the force; but the interpretation given in the text is more convenient. IMPULSE OF A FORCE. 31 absolute unit above denned, or of any other unit we may adopt. In fact, the second law of motion provides us with an absolute measure of force. Suppose P units of force to act uniformly for t seconds on m units of mass, producing /' units of acceleration. p Then /'= . Also if v units of velocity be produced in m p the end we have v = ft= -t. or Pt = mv. Hence the pro- m duct of the force into the time during which it acts is numerically equal to the momentum produced by it. Hence the second law of motion enables us to assign a physical meaning to momentum. It is the effect or pro- duct of a force acting for an interval of time. A given force acting for a given time will always generate the same amount of momentum, whatever be the mass upon which it acts, and this momentum is measured by the algebraical product of the measure of the force and of the time during which it acts. 49. If in the interval during which a force acts its magnitude vary, the acceleration produced by it will vary proportionally. If we split up the time of action of the force into intervals so short that we may consider the force as uniform during each interval, the momentum generated during any interval T during which the force is equal to P units will be equal to PT, and so for every interval. Therefore the whole momentum generated may be found by multiplying the number of seconds in each interval by the measure of the force, supposed constant during that interval, and adding the results. If the num- ber of intervals be increased and their length diminished indefinitely, we arrive at the case of a continuously vary- ing force. Those acquainted with the Integral Calculus will see that the above may be shortly expressed by saying that the whole momentum generated by a force is equivalent to the " time-integral" of the force itself. DEF. The tvhole momentum generated by a force is called the impuhe of the force. 50. If we know the impulse of a force and the time 32 IMPULSE OF A FORCE. during which it acts, we have only to divide the impulse by the time in order to obtain the measure of the force, supposing it uniform, or the time-average of the force, supposing it variable. Sometimes a finite momentum is generated in a time so short we are unable to measure it. In this case the force producing it must be very great, but we cannot tell how great because we do not know the time during which it acts. We know the whole momentum generated, that is, the impulse of the force, but not the force itself. In other cases we may be able to find the momentum produced and the whole time during which the force acts, but may be unable to trace the variation of the force. In this case we can find the impulse of the force and its (time) average value, but not the magnitude of the force at any particular instant. 51. Until recently it was customary to find the follow- ing definitions in works on Dynamics : A finite force is one which requires a finite time to generate a finite momentum. An impulse is a force which generates a finite mo- mentum in an indefinitely short time. An impulse is measured by the whole momentum generated by it. Now we have no experience of a finite momentum being generated in an indefinitely short time, and an in- finite force would be required to produce such an effect. Moreover, if an impulse is measured by the whole mo- mentum generated by it, it is something quite different from force, being of one dimension higher in time. According then to this system there were two classes of forces ; viz. finite forces and impulses, which could not be compared with one another, being, in fact, of different dimensions. 52. The only mode of measuring force now recognised is that indicated by the second law of motion. If the whole momentum generated by a force, or the impulse of the force, be given, then if the force take a long time to generate this momentum the force itself must be cor- respondingly small ; if it take a very short time the force IMPULSE OF A FORCE. 33 must be correspondingly great. If we cannot measure the time we cannot determine the force, and have then to content ourselves with knowing the momentum gene- rated by it, or its impulse, a term invented simply to hide our ignorance of the force itself. A force considered only with reference to the whole momentum generated by it during its action is called an impulsive force. Generally the term impulsive force is applied only to forces which act for a time so short that no sensible changes take place in the configuration of the system during their action, and hence in consider- ing their effect we need take no account of forces which require a much longer interval to produce any appreci- able effect. This is, in fact, the only practical point in distinguish- ing between impulsive forces and other forces. In the case of impulsive forces it is generally understood that their time of action is so short that no change takes place in the position of the body acted upon, and no effects are produced by the other forces acting on the body while the impulsive forces last, so that, in considering the effects of these forces, we do not complicate the problem by the introduction of other forces which are much less intense but more continuous. Examples illustrating this distinction will be met with in some of the problems re- lating to the motion of bodies connected by a string and in the chapter on Collision. 53. Suppose a force to generate a finite momentum in ToWth of a second. The force may at first increase, then reach a maximum and subsequently diminish, vanishing at the end of the above-mentioned interval. Except by very refined methods we should be quite unable to mea- sure, even approximately, the time during which the force acted, and could therefore form no idea even of the average magnitude of the force, much less could we detect its variation, and we should have to speak simply of its impulse. But suppose our faculties or instruments so much improved that we could appreciate Tono^iroth of a second ; then we might not only measure the time during which the force acted, but determine its mean a. D. D 34 PHYSICAL INDEPENDENCE OF FORCES. value for each, separate millionth of a second during its action. The force would, in fact, be as completely subject to our measurement as a force which acts for a quarter of an hour would be to an observer who could measure and appreciate no interval of time less than a second. Measurements of this character have been actually carried out by Sir Frederick Abel and Captain Noble in their determination of the velocity of a shot at different positions in a rifled gun, and their deduction of the pres- sure which must be exerted upon it in the several posi- tions by the products of combustion of the powder. An important practical result of these experiments was the determination of the form of the rifled grooves that the twisting couple upon the shot might be uniform, and thus the wear of the groove uniform notwith- standing the variation of the pressure of the powder gases. If the grooves were regular helices, it is clear that the velocity of rotation of the shot would be pro- portional to the linear velocity, and hence the twisting couple would be always proportional to the pressure of the powder gas and the grooves would wear most in the part of the barrel where the pressure was greatest. By making the pitch of the screw at all points proportional to the pressure of the powder gas the twisting couple is kept constant, notwithstanding the great variation of pressure in the barrel. Of course other very important objects were the determination of the requisite strength of the gun at different portions of its length, and the adaptation of the powder to the character of the projectile. 54. Suppose several forces to act at once on a particle either at rest or in motion ; then, the second law of motion being true for every one of these forces, and the effect of a force being completely determined by that law, it follows that each must have the same effect, in so far as the change of motion produced by it is concerned, as if it were the only force in action ; we may therefore infer that When any number of forces act simultaneously on a body, tchrther at rest or in motion, each produces the same change hi the body's motion as if it alone acted on the body at rest. SECOND LAW OF MOTION. 35 This statement expresses the principle of the "physical independence of forces" From it, taken in conjunction with Newton's second law of motion, it follows that if any number of forces act on a particle, initially either at rest or in motion, the equation P=mf will be true for each, f being the acceleration produced by the force P, and this acceleration will be in the direction in which P acts. From this result, coupled with the " parallelogram of accelerations " proved above, the " parallelogram of forces " immediately follows, and we have at once the whole subject of Statics. Precisely the same remarks apply to impulses, which may therefore be resolved and compounded in the same way as forces. 55. In all the preceding cases, change of velocity must be estimated in accordance with the parallelogram law. Thus, if a straight line AB represent in magnitude and direction the initial velocity of the particle, and A C its final velocity, B C will represent the whole change of velocity produced. 56. When it is stated in the second law of motion that " change of motion is proportional to the impressed force," it is not implied that the whole change of motion should be that of a single particle, or even of a single body, in the usual sense of the term ; nor is it necessary that all the changes of motion in the different parts of a complicated system, which act on each other by means of their connections, should be at any instant in the same direction. The change of motion of each particle of the system will be in the direction of the resultant of all the forces acting upon it ; but some of these forces are intro- duced by the connections of the system, and the direction of the resultant force on the particle depends on the nature of these connections. Since, however, action and reaction between the particles of the system must be equal and opposite it follows that the resultant momen- tum generated in the system, as determined by com- pounding the momenta of all the particles according 36 WEIGHT. to the parallelogram law, is proportional to the resultant of all the forces acting on the system and is the same as this resultant force would produce in the same time on a single particle. This is practically D'Alembert's Principle. When a system is such that the velocity of one particle being given in magnitude and direction that of all the others can be found, the fact that the resultant momen- tum of the system corresponds to the effect of the re- sultant of the forces acting upon it will enable us to determine completely the motion of the system when known forces have acted upon it for a given time. We shall meet with examples of this hereafter. 57. As an application of the second law of motion we may take the following. It has been established as an experimental fact that if two or more bodies of different materials, as, for example, a sovereign and a feather, be allowed to fall in vacuo simultaneously from rest, if let fall from any, the same, height, they will reach the ground together. Hence their velocities at every instant during their fall must be the same. Now since the velocities of all the bodies at any subsequent instant are the same, it follows that the accelerations under which they are mov- ing are the same for all. The force acting on each is, therefore, by the second law of motion, proportional to its mass. But this force is that with which it tends to fall towards the earth ; in other words, its weight. The weight of each body is therefore proportional to its mass, and in- dependent of the Wind of material of which it is formed. It follows then, from this experiment, that the earth attracts all kinds of matter alike. 58. The acceleration of a body falling freely in vacuo is found to vary slightly with the latitude, and also with the elevation above the sea-level. This acceleration is generally denoted by gr, and when we say that at any place g is equal to 32, we mean that the velocity generated per second in a body falling freely under the action of gravity at that place is a velocity of 32 feet per second The value of g at the sea-level in the latitude of Edinburgh WEIGHT. 37 is 32'2 very nearly. It may be mentioned here that a body is said to be moving freely when it is acted upon by no forces except those under consideration. If in the equation P=mf we put for P the weight W of the body, we know that the acceleration produced is g ; hence for /"we must write g, and we get the equation W mg, the unit of weight, or of force, being in this case the absolute dynamical unit of force. 59. The iceight of a body is the force with which it tends to move towards the earth, and is equal and opposite to the force which must be exerted in order to support it. It is equal to the attraction of the earth for the body diminished (according to the parallelogram law) by the force which is necessary to cause the body to participate in the diurnal rotation of the earth, and which produces its whole effect in causing acceleration towards the earth's polar axis when the body is at rest relative to the earth's surface. 60. We have seen that the weight of a body is propor- tional to its mass ; therefore bodies whose weights at any given place are equal must contain equal quantities of matter, and we thus obtain a means of determining the equality of two masses by the use of the common balance. Hence, also, we have a practical method of comparing the quantities of matter in two or more different bodies by determining how many masses, each equal to the standard, are required to balance each of the bodies whose masses are to be compared. Sub-multiples of the standard may be obtained at first by weighing out a quantity of matter equal to the standard unit of mass and dividing it by trial into the required number of equal parts, the equality of the parts being tested by the balance. In this way a system of " weights " for the measurement of mass may theoretically be obtained. It should be borne in mind that the ultimate object of weighing things is not generally to ascertain their weight, that is, the force with which the earth attracts them, but to determine the quantity of matter contained in them, 38 BALANCES. and this is rendered possible only by the fact that the earth attracts all kinds of matter alike, a fact which we have learned by observing that all bodies fall with the same acceleration in vacuo. The same conclusion also follows from the result of certain experiments on pendu- lums, which show that, other things being the same, the time of oscillation of a pendulum is independent of the material of which it is composed. 61. The characteristic difference between the results obtained by a spring balance and by a pair of scales is this. A pair of scales is used to determine at once how many units of mass are contained in the body weighed. This is done by determining how many units of mass are attracted by the earth with the same force as the body in question, and since, as above stated, all kinds of matter are attracted alike by the earth, the result is the number of units of mass contained in the body ; and this test is entirely independent of the absolute intensity of gravita- tion, that is, of the value of g, at the place. The spring balance, on the other hand, simply measures the force with which the earth attracts the body weighed ; the apparent weight of a body will therefore depend on the intensity of gravity at the place. Now the acceleration produced by gravity in a body falling freely is less at the equator than at the poles, and increases continuously with the latitude ; it also diminishes as the altitude above the sea-level is increased. The apparent weight of a loody will consequently be less in low than in high latitudes, and a spring balance will be disadvantageous to a mer- chant buying goods in England and selling them at Cape Coast Castle ; while if a pair of scales or a steelyard be used in its place, the apparent weight of the merchandise will be independent of the latitude. 62. DEF. A force is said to do work when it moves its point of application. An agent is said to do work when it overcomes resis- tance. Since action and reaction are exactly equal and oppo- site it matters not whether we consider the work done by an agent as proportional to the force exerted, and the dis- WORK. 39 tance through which the point of application is moved in the direction of the force, or as proportional to the resis- tance overcome and the distance through which it is overcome. The work done by a force whose magnitude and direc- tion remain constant, is proportional to the product of the intensity of the force into the distance through which its point of application has been moved in the direction of the force. From this it will be seen that, if the point of application move always in a direction perpendicular to that of the force, the latter does no work. Thus no work is done by gravity in the case of a particle moving on a horizontal plane, and when a particle moves on any smooth surface no work is done by the force which the surface exerts upon it. If, on the other hand, a heavy body be lifted from the ground, the agent raising it does work upon it, and the work done is proportional to the product of the weight of the body and the vertical height through which it is raised. In this case the body is moved in the direc- tion opposite to that in which its weight acts, and the work done by the earth's attraction is accordingly nega- tive. When the work done by a force is negative, that is, when its point of application moves in the direction opposite to that in which the force acts, this is frequently expressed by saying that work is done against the force. In the above case work is done by the agent lifting the heavy body and against the earth's attraction. 63. If the amount of work which is done by the unit of force, when its point of application moves through the unit of length in the direction of the force, be taken as the unit of work, then the measure of the work done by a force whose magnitude and direction remain constant will be the product of the numbers representing respec- tively the force and the distance traversed by its point of application in the direction of the force. Choosing it so that this condition may be satisfied, the British absolute unit of work is that done by the absolute dynamical unit of force, or poundal, when its point of application moves through one foot in the direction of the force, and is called 40 WORK. a foot-poundal. It is equal to the work done in lifting rather less than half an ounce one foot high. The unit of work generally adopted by engineers is the foot-pound, that is, the work done against gravity by an agent in raising the mass of a pound through the ver- tical height of one foot. Now the mass of a pound being invariable, its weight varies with the locality on account of the variation of g. The foot-pound is therefore not an invariable standard but depends on the locality, and is consequently unsuited for a scientific unit of work. The foot-pound contains g absolute units of work because the weight of a pound is equivalent to g absolute units of force. The unit of work belonging to the centimetre-gramme- second system, or the C.G.S. unit of work, is that done by a dyne in working through a centimetre and is called an erg. It is rather more than the work done in lifting a milligramme against gravity through one centimetre, for since g is equal to about 981 centimetre-second units of acceleration it follows that the weight of a gramme is about 981 dynes and that the work done in lifting a gramme through one centimetre is 981 ergs. 64. "When either the magnitude or direction of a force varies, or if both of them vary, the work done by the force during any finite displacement cannot be defined as above. In this case the work done during any indefinitely small displacement may be found by supposing the magnitude and direction of the force constant during that displace- ment, and estimating the work done in accordance with the above definition : taking the sum of all such elements of work done during the consecutive small displacements, which together make up the finite displacement, we obtain the whole work done by the force during such finite displacement. The effect or product of a force, when its point of appli- cation moves over any distance in the direction of the force, is a certain amount of work, and this work is measured by the algebraical product of the measure of the force and of the distance moved over by its point of application in its direction. HORSE-POWER. 41 65. DEF. The rate at which an agent works is measured when uniform by the amount of work done by it in the unit of time : when variable it is measured at any instant by the amount which would be done by it in the unit of time, if the rate remained uniform during that unit and the same as at the proposed instant. The rate at which work is done by a force is the product of the force and the velocity of its point of application in the direction of the force. DEF. The power of an agent is proportional to the rate at which it can work. An agent capable of performing 33,000 foot-pounds of work per minute is said to be of one Horse Power. Thus, when we say that the actual horse- power of an engine is ten, we mean that the engine is able to perform 330,000 foot-pounds of work per minute. The nominal horse-power of a steam-engine depends only on the number and measurements of its cylinders, and the nature of the engine, and not upon the actual rate at which it can work, which varies with the pressure of steam in the boiler and is limited by the strength of the latter. The letters H. p. are often used as abbreviations of the words horse-power. It will be seen that the horse-power, like the foot-pound, is not an absolute unit, but depends on the intensity of the earth's attraction at the place. All such units are sometimes classed together under the name of gravitation units. The British absolute unit of power is that of an agent which can perform one foot-poundal per second. The C.G.S. unit of power is that of an agent capable of doing one erg per second. As this is an inconveniently small unit another unit based on the same system is adopted by electricians for the rate of doing work, and is called a Watt. The watt is 10,000,000 ergs per second. It should be noticed that the watt, like the horse-power, is not a quantity of work, but a rate of doing work, and is of the nature of work divided by time. The horse-power is equivalent to 745'8 watts. 66. DEF. The amount of work which a system is cap- 42 ENERGY. able of doing in passing from its present condition to some standard condition is called its energy. A system may possess energy in virtue of its configura- tion, or the relative positions of its parts. Thus a distorted spring can do work in returning to its natural form ; the system consisting of the earth and a raised weight can do work by their mutual approach. The energy which a system possesses in virtue of its configuration is called potential energy. 67. If we catch a cricket ball when the ball strikes the hand it exerts a pressure upon it, and unless the hand be made to move in the direction of the ball's motion it may inflict a serious injury. If the hand be withdrawn some- what as the ball strikes it, the ball exerts pressure upon the hand while it moves in the direction of the pressure. It therefore does work, and we see that work may be done when two bodies collide, in virtue of their relative motion. In order that work may be done we must have at least two bodies moving relatively to one another, for if the cricket ball be not acted upon by any other body it will go on moving indefinitely and have no opportunity of doing work. Similarly, in order that a system may have potential energy we must have at least two bodies or portions of a body capable of changing their relative positions. The energy which a system possesses in virtue of the relative motions of its parts is called kinetic energy. 68. If a moving body such as a cannon shot strike the earth and come to rest relative to it, it will not sensibly affect the earth's motion, in consequence of the enormous mass of the earth as compared with the shot. In this case we may assume that the shot loses the whole of the velocity which it had relative to the earth, and it will be shown hereafter (Chapter II.) that in such a case the number of units of work done is the same as the number of units of kinetic energy possessed by the shot as defined in Art. (38). Whenever we speak of the kinetic energy of a single moving particle, we mean the kinetic energy possessed by the particle and the earth in virtue of their THIRD LAW OF MOTION. 43 relative motion ; similarly when we speak of the potential energy of a raised weight, we mean the potential energy possessed by the weight and the earth in virtue of their relative positions. 69. LAW III. Action and reaction are always equal and opposite. If a body A press against another body B with which it is in contact, B will exert an equal pressure on A, but in the opposite direction. But the application of this law is not confined to the mutual actions between surfaces in contact, the statement being true for all kinds of me- chanical action whatsoever. Thus, while in accordance with the law of gravitation the ' sun attracts the earth with a certain force, the earth also attracts the sun Avith an equal force, and these two attractions are, of course, in opposite directions. Again, the earth attracts a falling rain-drop with a certain force, while the rain- drop attracts the earth with an equal force. The result is that while the rain-drop moves towards the earth on account of its attraction, the earth also moves towards the rain-drop under the influence of the attraction of the latter, but the mass of the earth being enormously greater than that of the rain-drop while the forces on the two arising from their mutual attractions are equal, the motion produced thereby in the earth is all but incom- parably less than that produced in the rain-drop, and is consequently quite insensible. The third law of motion is also applicable to electrical attractions and repulsions, to the actions of magnets and of conductors conveying electric currents on themselves and each other, and, in fact, to all cases of mechanical action. 70. The first law of motion states the property of force and gives us, so to speak, a qualitative test of its presence, for from it we infer that when a body changes its state of rest or of uniform motion in a straight line it is under the action of force. The second law of motion raises force to the dignity of a mathematical quantity and explains how it is to be measured. Combining this law with a purely Jcinematical 44 THIRB LAW OF MOTION. theorem, viz. the parallelogram of accelerations, we learn how to find the resultant of any number of forces, and the subject of statics follows. The third law of motion is a brief summary of a number of phenomena having a general resemblance but differing in their details, and is intended rather to be a conventional mode of expressing in as few words as possible a series of experimental results, than to connote the several physical phenomena which it includes. 71. If one body attract or repel another, the second attracts or repels the first with an equal and opposite force. This is true not only of the attractions of the heavenly bodies but of molecular forces between the particles of matter constituting any solid, liquid or gas. (Were this not the case we might have a body A attracting another body B with a force greater than that with which B attracts A. Then considering A and B together as form- ing one system there would be a resultant force upon the system acting from B towards A and making its centre of inertia move with an acceleration from B to J., while no force is applied by external agency, and this is contrary to experience.) If a body A press against another body B both bodies being at rest, then every one will at once admit that B presses A with a force equal and opposite to that which A exerts upon B. Thus if a pound weight rest on a horizontal table we know that the table exerts a vertical pressure on the pound equal to the weight of a pound, because it supports it, and no one will doubt that the pressure of the weight on the table is equal to the weight of a pound, (in fact, some persons will consider the latter statement more obvious than the former,) so that in this case action and reaction are equal and opposite. 72. Suppose the finger pressed against a piece of soft putty or other material so as to penetrate it; the question may be asked " Is the pressure of the putty on the finger in this case equal to the pressure of the finger on the putty ? and if so why does the finger penetrate the putty?" Consider a very thin section of the finger THIED LAW OF MOTION. 45 which includes all the portion in contact with the putty ; let m denote the mass of the section and f its acceleration towards the interior of the putty. Then the resultant force on the section is mf towards the putty, and the pressure exerted on the section by the rest of the finger must be greater than the pressure, of the putty upon it by the quantity mf. But by taking the section of the finger sufficiently thin we can make m as small as we please. Hence the pressure exerted by the putty upon an Indefinitely thin section of the finger in contact with it is indefinitely nearly equal to the pressure exerted by the rest of the finger upon that section, and this latter pres- sure differs by the indefinitely small quantity, mf, from the pressure exerted by the finger on the putty. In this case, then, the action and reaction between the finger and the putty are equal and opposite. If the pressure of the hand on the knuckle end of the finger be greater than that which the putty can exert upon the other end, there will be a resultant force upon the finger towards the putty, and this force will produce acceleration in the finger which will therefore penetrate the putty with an accelerated motion, and the pressure of the finger upon the putty will be equal only to the pressure exerted by the putty on the finger, and not equal to that exerted by the hand on the knuckle end of the finger, the difference being that force required to produce acceleration in the finger, so that action and re- action between the end of the finger and the putty are equal and opposite. If the pressure exerted by the hand on the finger be equal to that exerted by the putty upon it the finger will remain at rest or penetrate the putty with uniform velocity. 73. Suppose two particles whose masses are respectively m and m' to be connected, and the particle of mass in to be acted on by a force P in the direction of the line join- ing the two. The acceleration of two particles will be the same as that of one particle of mass m + m , and will p therefore be denoted by .. Now the force required m + m to produce this acceleration in the particle of mass in' is 46 THIRD LAW OF MOTION. Pm - and this is therefore the force with which the first m + m particle acts on the second and is in the same direction as P. Also the force required to produce the acceleration P Pm - ; in the first particle is - and this therefore re- ni + m m + m presents the r exultant force upon it. But one of the forces applied to the particle is P, and the other is the reaction of the second particle; this latter is therefore in the Pm direction opposite to that of P and equal to P -- , that Pm' is. to -,, and is therefore equal and opposite to the m + m action of the first particle upon the second. Hence in this case action and reaction between the particles are equal and opposite. We see then that in all cases in which force is exerted between portion* of matter, whether at rest or in motion, action and reaction are equal and opposite, and this is as true for the molecular forces which act between the ulti- mate molecules of matter as for the forces of gravitation between the heavenly bodies. 74. If m denote the mass of a particle, and f its accele- ration, the force mf required to produce this acceleration is called the effective force on the particle, and is identical with the resultant of all the forces acting on it. Imagine any connected system, and let m denote the mass of one of its particles and f its acceleration. Then the effective force on the particle will be denoted by mf, and this must be equivalent to the resultant of all the forces acting on the particle, and will therefore if reversed maintain equilibrium with them ; and this is true for all the particles in the system. Therefore the reversed effec- tive forces of all the particles will balance all the other forces acting throughout the system. But the forces ap- plied to any particle consist in the most general case of two classes ; viz. those impressed upon it by external agency, and the forces exerted by the other parts of the system which we may call Internal forces. But to each force be- THIED LAW OF MOTION. 47 tween the parts of the system there is (by the third law of motion) an equal and opposite reaction, so that the inter- nal forces taken throughout the .system are in equilibrium amongst themselves. Hence the effective forces of all the particles when reversed will maintain equilibrium with the forces impressed from without. This is D'Alembert's principle and it follows at once from Newton's laws of motion. All that has been said above is true however great the forces may be, or however small the time during which they act, and will therefore apply to forces which act only during an interval too short for measurement. 75. It must be remembered that the action and reac- tion contemplated in the third law of motion are action and reaction bet ween portions of matter. There is nothing in nature corresponding to a " force of resistance against acceleration " which is supposed by many writers to be exerted by matter, and it cannot be too strongly urged that reversed effective forces have no real existence. But Newton's third law of motion admits of a wider interpre- tation than this. In the scholium to this law Newton says : If the action of an agent be measured by its force and velocity conjointly ; and if similarly the reaction of the re- sistance be measured by the velocities and amounts of its several constituents conjointly, whether these arise from fric- tion, cohesion, toeight, or acceleration ; action and reaction in all combinations of machines will be equal and opposite. Now it has been shown that the product of a force into the velocity of its point of application in the direction of the force is the rate at which it works. Newton then in this scholium measures the action of an agent by the rate at which it works, and similarly he measures the reaction of the resistances by the rate at which -work is done against them. The work done against the resistance arising from acceleration mentioned in the scholium is the work done by the effective forces of the system on account of their producing acceleration and is expended in generating kinetic energy in the system. Hence the measure of the reaction arising from acceleration is the 48 CONSERVATION OF ENERGY. rate at which kinetic energy is being generated in the system, and the scholium when interpreted into modern phraseology will stand thus : If the action of an agent be measured by the rate at ichich it works, and similarly the reaction of the resistances arising from friction, cohesion, and weight by the rate at which work is done against them, and if we include amongst the measures of these reactions the rate at which kinetic energy is being generated in the system ; action and reaction in all me- chanical combinations will be equal and opposite. 76. The statement thus interpreted is nothing more nor less than the enunciation of the great principle of the conservation of energy. If an agent work upon a system and there be no opposing forces such as friction, etc., then the rate of change of the kinetic energy of the system is precisely equivalent to the rate at which the agent works, and therefore the whole change of kinetic energy pro- duced in the system in any time is equivalent to the work done on the system by the agent. If, however, the action of the agent be opposed by forces of the nature of friction, etc., the rate of change of kinetic energy in the system is less than the rate at which the agent works by the rate at which work is being done against these opposing forces, and the whole change of the kinetic energy of the system produced in any time is less than the work done by the agent in that time by the work done against friction, etc., and converted into heat or other forms of energy. Together with fric- tion must be classed all forces whose action does not remain constant in magnitude and direction whether the system be at rest or moving in any way whatever. 77. "When the agent works against forces of the same nature as weight, with which we include all those which are independent of the time, and of the velocity of the system, depending only on its position and configuration, and which are sometimes called conservative forces, then the rate of change of kinetic energy in the system is less than the rate at which the agent works by the rate at which a work is done against these other forces, and THIRD LAW OF MOTION. 49 the work done against these forces in any time becomes potential energy in the system, and can be converted into kinetic energy by leaving the system free to return to its original position and configuration. If friction, or other forces of like nature, act on a system, then, when the motion of the system is reversed, these forces are also reversed, and the work done against them is not converted into kinetic energy on leaving the system free to return to its original configuration, having been at first converted into heat, sound, the energy of electric currents, or other forms of energy. 78. We see then that Newton's third law of motion consists of two distinctly different principles. The first states the equality of the action and reaction between portions of matter (and is equally true of electrified and magnetized matter), in which case both action and re- action are of the nature of forces such as those considered in statics, or as a particular case of these, of great forces acting for a short time, and the effects of which we find convenient to treat as impulses. The second consists of the great principle of conservation of energy, and asserts the equality of action and reaction when these terms do not imply statical forces but rates of doing work ; that being called action by which work is done and that against which work is done being treated as a reaction, the rate of increase of kinetic energy being included amongst the latter in the statement that action and reaction are equal and opposite. The endeavour to make the law apply in its first sense to the action of a force on a particle, free to move, was the origin of the introduction of a force of resistance to acceleration equivalent to the effective force reversed, to which there is nothing in nature at all corresponding. 79. Many modern writers, following out the -sugges- tions of the late Dr. Whewell, have enunciated the prin- ciple of " the physical independence of forces " as their second law of motion, and Newton's second law they have called the third law of motion. The third law, as given by Newton, they have then either assumed as an axiom or treated as a fourth law. The prevailing tendency amongst G. D. E 50 THTED LAW OF MOTION. mathematicians has recently been to return to the laws as enunciated by Newton, and in this form they have been given above. 80. If a particle of mass m be acted upon by a force P which would cause it to move, were it free to do so, with an acceleration /", and be prevented from moving by a string or some other means of constraint, then the string or other constraint must exert on the particle a force equal to P, but in the opposite direction ; and, since action and reaction are equal and opposite, it follows that the particle will exert a force upon the string or other constraint equal to, and in the same direction as, the force P. Thus if a heavy particle be suspended in equilibrium by a vertical string, it exerts a downward force on the string equal to its own weight. Similarly a heavy body at rest on a horizontal table exerts a pressure on the table vertically downwards, and equal to the weight of the body. 81. If the particle whose motion is constrained by a string, or other means of constraint, be moving with a velocity uniform in magnitude and direction, since the motion of the particle is not accelerated there can be no resultant force acting upon it, and the conditions are pre- cisely the same as if the particle were at rest. If the particle acted on by the force P, instead of mov- ing with an acceleration /", be constrained by some other means to move with an acceleration f in the direction in which P acts, we can find immediately the force exerted by the constraint. For since the force P would produce in a particle of mass m an acceleration denoted by /", we must have P equal to mf. Also, the particle moves in the direction in which P acts with an acceleration /i ; the resultant force on it must therefore be in the direction of P, and numerically equal to mf^ The force exerted by the constraint must therefore be in the direction in which P acts, and algebraically represented by mf t mf, or by m(f l f). If /*! be less than /*, this shows that the force exerted by the constraint is in the direction opposite to that in which P acts, and numerically represented by m (/-A). EXAMPLES ON LAWS OF MOTION. 51 If the actual acceleration of the particle be in the direction opposite to that in which P acts, it will be of negative sign. Suppose it to be f v Then the force exerted by the constraint must be in the direction oppo- site to that of 7 J , and numerically equal to m (f + / 1 ), that is to mf\ + P. If the actual acceleration of the particle be not in the straight line in which P acts, then we may find the resultant force which must act on the particle in order to produce this acceleration, and the force which is exerted by the constraint is that which must be compounded with P to produce this resultant. This can be found at once by help of the parallelogram of forces. 82. As an example of the preceding article, suppose a heavy particle to rest on a horizontal plane, while the plane moves vertically downwards with an acceleration f ; what will be the pressure of the particle on the plane, the acceleration produced by gravity in a particle falling freely being denoted by g ? The pressure of the particle on the plane must, by the third law of motion, be equal and opposite to that of the plane on the particle, and the only forces acting on the particle are the earth's attraction and the pressure of the plane. Now the particle moves downwards with an ac- celeration /"; the resultant force upon it must therefore act downwards and be numerically equal to mf. But the attraction of the earth upon the particle acts downwards and is numerically equal to mg. Hence if /"be less than g, the pressure of the plane on the particle acts upwards and is equal to mg - mf, that is, to m (g f)', while the pressure of the particle on the plane, which is equal and opposite to this, acts vertically downwards and is numeri- cally represented by the same expression, viz. m (gf). If f be greater than , and the weight R placed upon it, and that at a particular instant the system is left free to move. Then since the weight Q is equal to P the weight of Q and R together is greater than that of P; they will therefore descend, and P will be raised. The time required for the top of Q to reach the ring D is then accurately measured, and as the top of Q passes through this ring the weight R is removed. Then, since the weights of P and Q are equal, there is no force to change the motion of the system, which will therefore continue uniform until the weight Q strikes the platform E. The time elapsing between the instants when the weight R is removed from Q and when Q strikes the platform E is accurately observed, and the distance between the ring D and the top of Q when resting on the platform is measured by the graduations on the bar SF. Hence the velocity of Q between D and E can be at once found. We shall now describe in detail a few experiments with Attwood's machine which have an important bearing on some of the foregoing articles. EXPEEIMENTS WITH ATTWOOD'S MACHINE. 57 87. EXP. I. Let the equal weights P and Q be formed of the same material, M being the mass of each, and let m be the mass of R. Let h be the distance between the ring D and the top of Q when resting on the platform ; t the time elapsing between the commencement of the motion and the instant when the weight R is removed from Q, and t' the interval between this instant and that at which Q strikes the platform E. Then if the same weights be used, and the distance between D and E kept constant, while the depth of the ring T) below the point at which Q is liberated is varied so as to vary t, it is found that t' varies inversely as t. Now the velocity of P or of Q after the latter has passed through D is constant, and therefore equal to -,. Hence v the velocity varies as t ; that is, the velocity generated in the moving system by the weight of R is proportional to the time during which this weight acts upon the system. But the velocity generated by a force is pro- portional to the time during which that force acts only when the acceleration produced by it, and therefore the force itself, is constant, the mass moved being supposed to remain unaltered. Therefore the force acting upon the system, that is, the weight of a mass m (since m is the mass of R), is constant and independent of the velocity with which the mass is moving. This being found by experiment to be true for any body used in place of R, it follows that the earth's attraction on any mass of matter, that is, its weight, is independent of the velocity with which it is moving. We may therefore introduce a symbol ic to represent throughout the motion the weight of the mass m, that is, of R. 88. EXP. II. If we neglect the masses of the string and of the pulley 6 Y , the whole mass moved is 2M + m, and the force producing the motion is w. Now suppose the weights P and Q to be changed, P remaining always equal to Q, while R is unaltered. Then 2M is changed, and the ratio in which it is changed may be supposed known, for P, Q, and R may be formed by joining to- gether equal weights of the same homogeneous matter, 58 EXPEBIMENTS WITH ATTWOOD'S MACHINE. whose masses are consequently equal. By adding to or removing from P and Q the same number of such small equal bodies, the sum of the masses of P, Q, and R, that is, the whole mass moved, may be changed in various known ratios. Then it is found from the experiment that the velocity of Q between D and E varies as ,=-,-, . Also, as was found in the preceding experiment, when 23/+ m remains constant, the velocity of Q after passing D varies as t. Hence the velocity generated in the unit of time by the constant force w varies as =-=-= - ; that is, ' in the inverse ratio of the whole mass moved. The result of this experiment agrees with that deduced in Article 85, in which the second law of motion was taken as the basis of the investigation. 89. EXP. III. Suppose, as in the preceding experi- ment, that P, Q, and E are all made up of a number of equal small masses of the same material, and let any equal numbers of these small weights be removed from P and Q and attached to E. Then w : the force producing motion, is changed in a new ratio ; for the weights of each of the small bodies making up Ti being the same, the weight of R must be proportional to the number of these small weights contained in it. Also, the whole mass moved is the same as before, and it is found that -,, that is, the velocity of the system after Q has passed through Z), varies directly as the product wt. Therefore the velocity generated in the unit of time varies as ic. But the mass remaining always the same the momentum varies as the velocity ; therefore, the momentum generated in the unit of time varies as w, the force producing the motion. This result is in accordance with that deduced in Article 85 from the second law of motion, and may be taken as direct evidence in favour of the truth of that law. It also shows that the momentum generated by a force in the unit of time is a proper measure of that force, the unit of force in such case being that force which generates the unit of momentum in the unit of time. EXPERIMENTS WITH ATTWOOD's MACHINE. . 59 90. EXP. IV. Let the bodies P and Q be exchanged for bodies of other, the same, material, whose weights, as determined by a balance, are the same as those of P and Q (e.g. brass weights exchanged for iron, platinum, glass, china, or other material), while R remains unchanged, so that the force producing motion remains unaltered. Then experiment shows that the velocity generated in the unit of time is independent of the material of which P and Q are composed. But by the second law of motion the momentum generated in the unit of time is always the same. Hence, since both the velocity and the momentum generated in the unit of time are independent of the material of which P and Q are made, it follows that the whole mass moved, and, therefore, the masses of P and Q are independent of the kind of that material. But P and Q were taken so that their weights should remain unchanged ; hence, if the weights of a number of bodies of different material be equal, their masses will also be equal : another proof of the statement that the earth exerts the same attraction on all kinds of matter alike. 91. The first experiment has shown us that the at- traction of the earth on a moving body is independent of the velocity with which it is moving. The second, that the momentum generated in the unit of time by a con- stant force is constant and independent of the mass in which it is generated. The third, that the momentum generated in the unit of time in the same mass is pro- portional to the force producing motion. Combining the results of the second and third experiments, we see that the momentum generated in any mass in the unit of time is proportional to the force producing motion, and hence we obtain a dynamical measure of force. From the result of the fourth experiment we infer that the earth exerts the same attraction on all kinds of matter, or that the weight of a body depends only on the quantity of matter it contains, and is independent of the Icindj or quality, of that matter. 92. In the preceding experiments suppose -, equal to V c. Then v is the velocity of Q when its top passes 60 MEASURE OF g. through the ring I) that is, the velocity generated in a mass 2M+ m by the weight of the mass m acting upon it for a time t (see Art. 85) ; and the experiment shows that this velocity is proportional to #=7- t. We have there- 2J/+ m fore, m m g being a constant throughout the series of experiments. Therefore the velocity generated in the unit of time is m 9 and since the whole mass moved is '2M+ m, it follows that the momentum generated in the unit of time is mg. But if the unit of force be that force which acting on the unit of mass for the unit of time generates in it the unit of velocity, it has been shown that the measure of any force will be the number of units of momentum generated by it in a given time. Hence the weight of the mass m is represented numerically by mg, or w = mg. The value of g is found to be always the same at the same place, but to vary with the latitude of the place of observation, being greater near the poles than at the equator, and increasing continuously with the latitude. It is also less at great heights above the sea-level than at that level. In the latitude of Edinburgh at the sea-level, when a second and a foot are taken as the units of time and space, the value of g is found to be 32'2 very nearly. The value of g at the sea-level in latitude X is approxi- mately represented by the expression G(l- 0-0025659 cos 2X), where G obviously represents the acceleration of gravity at the sea-level in latitude 45, and is equal to 32' 1703 foot-second, or to 980*533 centimetre-second units. 93. We have seen that the momentum generated in a moving system in the unit of time, by the weight of a ATTWOOD'S MACHINE. 61 mass nij is represented by mg. Now, in the case of a particle falling freely, the force producing motion is its weight while the mass moved is simply its own mass. Hence the velocity generated in the unit of time will be g, which is therefore numerically equal to the accelera- tion produced by gravity in a body falling freely ; and when we say that at a particular place g is equal to 32*2, we imply that at that place the attraction of the earth is such as to generate in a body falling freely a velocity of 32*2 feet per second during each second of its fall. Such motion is called uniformly accelerated motion. This result might of course have been obtained directly from observations on bodies falling freely, but since it is very difficult to measure great velocities, especially when they are continually varying, and since the resistance of the air becomes considerable when the velocity of the moving body is great, Attwood's machine is convenient, inasmuch as the velocity to be actually observed is uniform, and, by sufficiently diminishing the ratio of m to M, can be made as small as we please. 94. In the above explanation of the use of Attwood's machine we have for simplicity neglected the motion of the pulley C. The mass of this pulley is, however, generally comparable with that of P or Q, and the points on the circumference of C have the same velocity as P or Q, while the points within the circumference are moving with less velocity. Were the whole mass of the pulley collected at the circumference we should only have to add this mass to that of the other moving bodies to obtain the whole mass moved ; but since that is not the case a complete correction may be made by adding a term M', less than the number of units of mass in the mass of C, to the quantity 23/+w, and thus employing %M+ M + m as the expression for the whole mass moved. To determine theoretically the value of M' we require some knowledge of Rigid Dynamics ; it is sufficient here to state that it admits of exact determination. The motion of the friction rollers is so slow and their mo- mentum consequently so small that they may be alto- gether neglected. 62 EVIDENCE IN FAVOUE OF THE LAWS OF MOTION. The error arising from the friction of the pulley can be best compensated by making the two weights, which we have supposed equal, slightly unequal, giving such pre- ponderance to the weight which is to descend during the experiment that when the weight m is removed, the system, if just started, may continue to move with perfect uniformity. If the suspending cord be silk its weight may be made so small in comparison with the weight of the suspended masses that the error introduced by it is almost negligible. We may however get rid of the error altogether by hanging a piece of the same cord from the bottoms of the two weights, forming a sort of tail-rope. The weight of cord on each side of the pulley will then always be the same and it will only be necessary to include the whole mass of the cord with the mass moved. 95. It is almost unnecessary to remark that the experi- ments detailed above have never been carried out pre- cisely in the form there given, nor would the results of such experiments, were they attempted, accurately corre- spond with those stated, the principal cause of the discrepancies being the mass of the pulley C, for the experiments described were supposed to be conducted with an apparatus in which the mass of the pulley was insensible. They however serve to illustrate the method which might be pursued by a person ignorant of dynamics, and investigating the laws of motion and of gravitation by means of Attwood's machine, the results at whi ch he would arrive agreeing sufficiently nearly with those stated above to suggest the probable truth of those laws which have been deduced from what would be the result of experiments performed under the ideal conditions which we have supposed to obtain. The evidence on which these laws are accepted must then be looked for from the agreement with observed phenomena of calculations based on the assumption of their truth. The experiments described also illustrate the methods in which such experiments may be varied so that the " question asked " in each may be in the simplest form possible, and the answer obtained may be simply the answer to that question. FUNDAMENTAL UNITS. 63 and not involve the answers to several others from which it could be disentangled only by repeatedly varying the conditions. 96. It may be well here briefly to review the several units which have been already defined, and to show how each is connected with the three fundamental units of time, length, and mass. These three units are chosen arbitrarily, and upon them the magnitudes of all others employed in dynamical science depend. The unit of time universally adopted throughout the world for scientific purposes is the second of mean solar time, and is therefore ultimately derived from observa- tions of the earth's rotation. All measurements of velo- cities amount, therefore, simply to a comparison of the motion of the body considered with that of the earth about its axis, time being employed merely as the con- necting link. The unit of length adopted in Britain by engineers is the foot. This is the third part of the distance between the centres of two lines engraved on two gold plugs, sunk in a bar of bronze, which is now kept at the Exchequer Chambers, and known as the Imperial Standard Yard, the temperature of the bar at the time of observation being 62 Fahrenheit. The unit of mass adopted in British measurements is the Imperial pound, that is, the quantity of matter con- tained in a certain mass of Platinum, kept at present in the Exchequer Chambers, and known as the Imperial Standard Pound Avoirdupois. 97. We pass on now to the consideration of the units derived from the three fundamental units of time, length, and mass. The unit of velocity is the velocity of a point which passes over the unit of length in the unit of time. If a second and a foot are the units of time and length, the unit of velocity is the velocity of a point which passes over one foot in a second. Suppose that when the unit of time is T seconds and the 64 MEASURE OF ACCELERATION. unit of length a- feet, the velocity of a certain point in denoted by u ; what icill be the measure of this velocity when the unit of time is t seconds and the unit of length s feet ? The number of feet in u units of length, each contain- ing a feet, is ua. Let v be the measure required. Then in T seconds the point passes over no- feet ; .'. in 1 second U feet ; T in t seconds tur- feet. T But Now fa- -3 feet are equivalent to f- . - of the new units H ' ,. F 66 EXAMPLES ON CHANGE OF UNITS. 99. As examples we may take the following : Ex. 1. The acceleration produced by gravity in a particle falling freely being denoted by 32, when a second and a foot are the units of time and length, what tcill be the measure of this acceleration when a day and the length of the earths radius are units, the latter being supposed equal to 4000 miles ? Gravity in 1" generates a velocity per l"of 32 feet ; /. 1 day I",,32x24x60 2 ft.; /. 1 day 1 day of 32 x 24 3 x 60 4 ft. Now 32 x 24* x 60 4 feet are equivalent to %&** 5280 x 4000 times the earth's radius. The required measure is there- , 32x24 2 x60 4 fore - or 100. Ex. 2. An acceleration which, when a second and a foot are units, is represented by 32*2, is represented by 9660, a yard being the unit of length. Find the unit of time, Let t seconds be the unit of time. Then with the acceleration in t" there is generated a velocity per t" of 9660 yds. V i" 966 11 11 )J 11 11 ^ 11 ' 11 v , v , 9660 * *- 11 A-> JJ A-> that is, of 9660 2 x3 feet. But with the acceleration in one second there is gene- rated a velocity of 32'2 feet per second ; 9660 x 3 o and the unit of time is thirty seconds, or half a minute. 101. The unit of momentum is the momentum pos- sessed by the unit of mass when moving with the unit of velocity. UNITS OF MOMENTUM AND FOECE. 67 The unit of momentum, therefore, varies directly as the unit of mass and directly as the unit of velocity : but this latter varies directly as the unit of length and inversely as the unit of time. Therefore the unit ot momentum varies directly as the unit of mass, directly as the unit of length, and inversely as the unit of time. Hence, since the numerical measure of any quantity varies inversely as the unit in terms of which it is measured, it follows that the numerical measure of a given momentum varies inversely as the unit of mass, inversely as the unit of length, and directly as the unit of time. If the measure of a momentum referred to any known set of units be given, its measure in terms of any other system of units can be found by the same method as that adopted above in the case of acceleration, provided that each of the second system of units is known in terms of the corresponding unit in the first set. 102. The unit of force is that force which generates the unit of momentum in the unit of time : or, that force which acting upon the unit of mass for the unit of time generates in it the unit of velocity. If a second, a foot, and a pound be taken as the units of time, length, and mass respectively, the unit of force is that force which acting on the mass of a pound for a second generates in it a velocity of one foot per second. This force we have shown to be rather less than the weight of half an ounce. A certain force is represented by P when r seconds, a- feet, and /j, pounds are the units of time, length, and mass respec- tively : what will be the measure of this force when t seconds, s feet, and m pounds are the respective units ? Since the force is represented by P in the first system of units, and a force is measured by the momentum which it will generate in the unit of time, it follows that the force acting on a mass of /u, pounds for r seconds will generate in it a velocity of P units of length, that is, of Pa feet, per r seconds. 68 MEASURE OF FORCE. .'. the force acting on //, Ibs. for T" generates a velocity per T" of Per feet ; .'. the force acting on 1 Ib. for T" generates a velocity per T" of Pay/, feet ; /. the force acting on 1 Ib. for 1" generates a velocity per T" of Poyi- feet ; :. the force acting on 1 Ib. for 1" generates a velocity per 1" of Po~p~2 feet ; .'. the force acting on 1 Ib. for 1" generates a velocity per t" of Po"/^ feet ; /.the force acting on 1 Ib. for t" generates a velocity t 2 per t" of Per/* 2 feet ; :. the force acting on m Ibs. for t" generates a velocity m T* But Po-^ feet are equivalent to P* 7 ^ -- of the TO ' -r s ' m ' T* new units of length. Hence the force acting on the new unit of mass for the new unit of time will generate in it a .g velocity of P- . ^ . a new units of length per the new s m r 2 unit of time. The measure of the force, expressed in 559 > 676 > ver y nearly. 110. Ex. 2. If an agent working at the rate of one horse-power perform the unit of work in the unit of time, and the acceleration produced by gravity in a body falling freely be the unit of acceleration, a pound being the unit of mass, find the units of time and length, it being given that g is equal to 32 when a second and a foot are units. Let t seconds be the unit of time. The agent performs 33,000 foot-pounds of work per minute, or 550 foot-pounds per second. Therefore in the unit of time, that is, in t seconds, it performs 550 foot-pounds. But it performs the unit of work in the unit of time. Therefore 550 foot-pounds is the unit of work. Now the unit of force is that force which acting on the unit of mass produces in it the unit of acceleration. EXAMPLES OF THE CHANGE OF UNITS. 75 Therefore the unit of force is that force which, acting on the mass of a pound, produces in it the same accelera- tion as is produced in it by gravity. The unit of force is therefore equal to the weight of a pound. Again, the unit of work is the work done by the unit of force when its point of application moves through a distance equal to the unit of length, and in the direction in which the force acts, and the unit of work has been shown to be 550 foot-pounds. Therefore the unit of length is 550 feet. Again, the unit of acceleration is the acceleration of a point in which in the unit of time there is generated a velo- city with which if a point move it will pass over the unit of length in the unit of time. /. with the unit of acceleration in t" there is generated a velocity per t" of 550 feet. *- 1 11 *- 11 11 11 11 n *- jj . ?> But with the unit of acceleration in one second there is generated a velocity of 32 feet per second. Therefore 550 oo ~r . t- 650 _17 B_ 32 The unit of time is therefore 17 T \- seconds. Also the unit of length was shown to be 550 feet. Therefore the unit of length is equal to 550x17^ feet; = 9453ifeet, 111. The unit of density is the density of a uniform substance, the unit of volume of which contains the unit of mass. It will be seen that the unit of density, and therefore 11 *- T) 11 11 11 11 11 ,, 76 UNIT OF DENSITY. the measure of the density of any substance, is indepen- dent of the unit of time, and depends only on the units of length and mass. Also the unit of density must vary directly as the unit of mass if the unit of volume remain constant ; for if the unit of mass be changed, the mass of the unit of volume of the substance whose density is unity will be changed in the same ratio. Again, if the unit of volume be changed, the unit of mass remaining the same, the volume which must contain the unit of mass of the substance whose density is unity will be changed in the same ratio : therefore, the standard substance must be changed so that the mass of a given volume of it may be changed in the inverse ratio of the unit of volume. Therefore, if the unit of mass remain constant, the density of the substance whose density is unity, that is, the unity of density, will vary inversely as the unit of volume. If the unit of volume be the volume of a cube whose edge is the unit of length, it follows that the unit of density varies in- versely as the cube of the unit of length. Also it has been shown that the unit of density varies directly as the unit of mass when the unit of volume remains constant. Hence, when all are allowed to vary together, the unit of density will vary directly as the unit of mass, and inversely as the cube of the unit of length. Since the numerical measure of any quantity varies inversely as the unit in terms of which it is measured, it follows that the numerical measure of the density of any substance varies inversely as the unit of mass, and directly as the cube of the unit of length. 112. Suppose the density of a substance to be repre- sented by p when /j, pounds is the unit of mass, and a- feet the unit of length : tvhat will be the measure of the same density when in pounds is the unit of mass and s feet the unit of length ? The unit of volume of the substance contains p units of mass when ^ pounds and a- feet are units of mass and length. EXAMPLES OF THE CHANGE OF UNITS. 77 Therefore jt 0'3 But pa '-^pounds are equivalent to p *- .'- units of mass o- 3 r m a* each containing m pounds. Therefore the new unit of volume of the substance if oO contains p ..;, new units of mass. m a 6 The measure of the density in terms of the new system of units is therefore p ^ .'--. m cr 3 113. Ex. If the unit of force be the tceight of one ounce, and the mass of a cubic foot of the substance whose density is unity be 162 pounds, the unit of time being one second, ichat is the unit of length, g being equal to 32 ichen a second and a foot are units ? Let the unit of length be s feet. Then the unit ot volume is s 3 cubic feet. Since a cubic foot of the standard substance contains 162 pounds of matter, the unit of volume of this sub- stance must contain 162s 3 pounds. But the mass of the unit of volume of the standard substance is the unit of mass. Therefore the unit of mass is 162s 3 pounds. Now the unit of force is that forjce which acting on the unit of mass for the unit of time generates a velocity of the unit of length per the unit of time. Therefore, the weight of oz. Ibs. feet. 1 generates in 1" in a mass of 162s 3 a velocity per 1" of s lib. 162s 3 sx!6 1 1 Ib. s x 16 x 162s 3 . But the weight of one pound acting on the mass of one 78 APPENDIX ON THE DIMENSIONS OF UNITS. pound for a second generates a velocity of 32 feet per second. Therefore 162.s- 4 x 16 = 32 : The unit of length is therefore one-third of a foot, that is, 4 inches. APPENDIX ON THE DIMENSIONS OF UNITS. 114. A LINE, whether straight or curved, being " length without breadth," possesses only one degree of extension in space. In the neighbourhood of any point, not being a singular point on a curve, the line extends in only one direction and its opposite, and is the simplest form of extension of which we can conceive. A line is therefore said to be of one dimension in space, a point being that which has no dimensions, but position only. The magnitude of a line is completely known when we know its length, and is expressed in terms of the unit of length. Thus, if we say that the length of a straight line is /, we mean that it contains I units of length, and if [L] denote the unit of length, the complete expression for the length of the line is I [L]. Now if the unit of length [L] vary while the line considered remains invari- able, the expression I [L] must remain constant while [L] varies, and therefore Z must vary inversely as [L] ; in other words, the numerical measure of a given line varies inversely as the unit of length in terms of which the line is measured, and therefore inversely as the number ex- pressing the unit of length in terms of some invariable unit. 115. A superficies, whether plane or curved, possesses " only length and breadth," and is therefore extended in two directions. A superficies is consequently said to be of tiro dimensions in space, and its magnitude is completely known when we know its area. All areas are measured in terms of the unit of area which is the unit of two dimensions in space. Thus when we say that the area of a superficies is , \ve mean that the figure con- UNIT OF ABEA. 79 tains a units of area, and if [A] denote the unit of area, the complete expression for the magnitude of the super- ficies is a [A], and, as in the case of a line, we see that if [ A] varies while the figure considered remains unchanged, a must vary inversely as [A], or the numerical measure of an area varies inversely as the unit of area. Now, invariably in mathematical investigations, and generally in practical measurements, the unit of are^a adopted is the area of a square the length of whose side is the unit of length: and this may be represented by [L] 2 . It must be borne in mind that the symbol [L] 3 does not represent an Algebraical product, since [L] is not a number, but it represents a geometrical quantity of a totally different nature from that expressed by [L], which is defined as the area of a square whose side is equal to the quantity [L]. 116. It may be proved geometrically that the number of units of area in any rectangle, when the unit of area is defined as above, is equal to the product of the numbers of units of length in two adjacent sides, and therefore the number of units of area in any square is equal to the algebraical square of the number of units of length in its side. The areas of different squares are therefore to each other in the duplicate ratio of the number of units of length in their respective sides, and therefore different units of area are to one another in the ratio of the squares of the numbers expressing the measures of the correspond- ing units of length in terms of one and the same unit. The words placed in italics are very important; from them it follows, since the numerical measure of an area varies inversely as the unit of area, that the numerical measure of an area varies inversely as the square of the number expressing the unit of length in terms of some in- variable unit. For the sake of brevity it is frequently stated that the numerical measure of an area varies in- versely as the square of the unit of length, but the two latter terms of the proportion must be reduced to nu- merical measures before they can be Algebraically com- pared. With this understanding we shall in future generally use the abbreviated form of expression. 80 UNIT OF VOLUME AND MASS. 117. A solid is extended in three directions in space, since it possesses length, breadth, and thickness, and is said to be of three dimensions in space. Its magnitude is known when we know its volume, and is expressed in terms of the unit of volume. Thus when we say that the volume of a solid is c, we mean that it contains c units of volume, and denoting the unit of volume by [C] the com- plete expression for this is c [(7], and as in the other cases considered, the numerical measure of a given volume varies inversely as the unit of volume. 118. The unit of volume generally adopted is the volume of a cube, whose edge is equal to the unit of length, and this may be represented by [Z,] 3 , this symbol representing a physical quantity of a totally different nature from those considered in the preceding articles, and not capable of being measured in terms of them. By reasoning similar to that of the preceding articles it may be shown that, when the unit of length varies, the volume of this cube varies directly as the cube of the number expressing the unit of length in terms of some constant unit, and we therefore infer that the numerical measure of a volume varies inversely as the cube of the number ex- pressing the unit of length in terms of a constant unit, or, as it is generally expressed, inversely as the cube of the unit of length. 119. We have thus seen that the numerical measures of quantities of one, two, and three dimensions in space vary inversely as the first, second, and third powers of the unit of length respectively. We shall not proceed to consider higher dimensions in space because most persons are unable to form a distinct conception of space of more than three dimensions, but shall now turn our attention to physical quantities of a different nature. 120. The magnitude or duration of an interval of time is measured in terms of the unit of time, and if we say that its measure is t we mean that it consists of t units of time, and the complete expression for the interval is t[T}, where [T] denotes the unit of time. The numerical measure of a given interval will therefore vary inversely as the unit of time in terms of which it is measured. By UNIT OF VELOCITY. 81 analogy an interval of time may be said to be of one dimension in time. At present we know of nothing of more than one dimension in time, though we shall pre- sently meet with quantities of negative dimensions in time. 121. The quantity of matter in a body, that is, its mass, is measured in terms of the unit of mass, and if its numerical measure be m, the complete expression for the mass of the body is m [I/], where [M] represents the unit of mass. The numerical measure of a quantity of matter varies inversely as the unit of mass, while mass, like time, being a primary conception, its unit is completely arbi- trary. A quantity of matter may be said to be of one dimension in mass. 122. Having thus considered the primary units we proceed to the consideration of complex or derived units ; that is, of units derived from the three fundamental units of time, length, and mass. The unit of velocity is the velocity of a point which passes over the unit of space in the unit of time. If v denote the measure of the velocity of a point and [ V] the unit of velocity, the complete expression for the velocity will be v [V], where v represents a number only, and [V] a physical quantity. Now the space passed over in the time t by a point moving with uniform velocity v is equal to vt units of length, and is therefore completely repre- sented by vt [L\, or by I [L], where I is a number equal to vt. But the complete representation of the product of the velocity v [V] in the time t [ T] is v [V] t [T]. Hence we have the equation vt[V](T}=vt[L], or [V][I]=[L]. Therefore W-ffi > = [L](Tl[ l (n), or the unit of velocity is of the nature of the line of unit length divided by the unit of time. This is expressed by saying that the unit of velocity is of one dimension in G. D. G 82 UNIT OF ACCELERATION. length, and of minus one dimension in time, as shown by equation (n). The unit of velocity therefore varies directly as the unit of length, and inversely as the unit of time, and since the numerical measure of a given velocity varies inversely as the unit of velocity, the numerical measure of a velocity varies inversely as the unit of length, and directly as the unit of time. 123. The unit of acceleration is that of a point whose velocity is changed by the unit of velocity in the unit of time. If [F] denote the unit of acceleration, the complete representation of an acceleration whose measure is f is f[F], and /"varies inversely as [F] if the acceleration remain constant. Also the velocity generated in t units of time is ft units of velocity, and the complete representa- tion of the product when a point moves with an accelera- tion f [F] for a time t [T] is the product f [F] t [T]. Hence we have the equation ft[F](T]=ft(V]; therefore [F] = U, or an acceleration may be said to be of one dimension in velocity and of minus one dimension in time. But by the preceding article therefore [F] = [L][T]">. The unit of acceleration is therefore of one dimension in length, and of minus two dimensions in time, and varies directly as the unit of length,. and inversely as the square of the unit of time. Hence, since the measure of any ac- celeration varies inversely as the unit of acceleration, it follows that the numerical measure of a given accelera- tion varies inversely as the unit of length and directly as the square of the unit of time. 124. "VVe may remark that whenever in the definition of a quantity the word per is introduced, it implies that the quantity is of minus one dimension in the element immediately succeeding the per, and if the word per is UNITS OF MOMENTUM AND FOECE. 83 introduced twice in the definition, each time it is intro- duced it implies minus one dimension in the quantity immediately succeeding. Thus velocity is space passed over per unit of time, and is of minus one dimension in time, while the unit of acceleration is that which gene- rates per unit of time a velocity of unit length per unit of time, and is of minus two dimensions in time. Similarly, if the price of oranges be so much per dozen oranges, the price is of minus one dimension in oranges, and the pro- duct of the price into a number of oranges is an amount of money, snowing that the price is also of one dimension in money, and the measure of the price will vary directly as the number of oranges per which it is quoted. 125. The unit of momentum is the momentum possessed by the unit of mass moving with the unit of velocity. If [K] represent the unit of momentum, a momentum whose measure is Jc will be represented by fr [7v], and if this momentum remain constant Jc must vary inversely as [A"]. Also the momentum of a mass ni moving with velocity v is mv units of momentum, that is mv [K]. But the momentum of a particle is possessed by it in virtue of the two factors, its mass and velocity, and is completely represented by their product, that is, by m [M] v [ V] ; hence mv [K] = mv [M][V], therefore [K] = [M] [V] = [M][L](T\-. The unit of momentum is therefore of one dimension in mass, one in length, and minus one in time, and the measure of a given momentum will therefore vary in- versely as the unit of mass, inversely as the unit of length, and directly as the unit of time. 126. The unit of force is that force which generates the unit of momentum in the unit of time, and the pro- duct of the numerical measure of a force into the number of units of time during which it acts is the number of units of momentum generated thereby. If p be the numerical measure of a force, and [P] represent the unit of force, the complete representation of the force is p [P], and 84 UNIT OF WOEK. while the force remains constant p varies inversely as [P]. But the effect of a force p acting for a time t is to generate pt units of momentum, and the complete representation of the product of a force p [P] during a time t [T] is p [P] t [T]. Hence pt [P][T\=pt[K\. Therefore [P] [T\ = [K\. = [M][L][T}-\ and [P] = [M] [L] [T]- 2 . Hence the unit of force is of one dimension in mass, one in length, and minus two in time, and the measure of a given force will therefore vary inversely as the unit of mass, inversely as the unit of length, and directly as the square of the unit of time. These conclusions might have been deduced from the consideration that force is that which produces accelera- tion in mass, and is measured by the product of the mass and of the acceleration produced therein. 127. The unit of work [ W] is that which is performed by the unit of force when its point of application moves over the unit of length in the direction of the force. The work done by any force measured in terms of this unit of work is equal to the product of the measure of the force and the number of units of length passed over by its point of application in the direction of the force. Hence if the point of application of a force p [P] move over a space I [L] in the direction of the force, the work done is completely represented by jpZ [W]. But the product of a force p [P] working through a distance I [L] is completely represented by p [P] I [L]. Hence pl[W]=pl[P}[L], or The unit of work: is therefore of one dimension in mass, two in length, and minus two in time, and the numerical measure of a given amount of work will therefore vary inversely as the unit of mass, inversely as the square of the unit of length, and directly as the square of the unit of time. UNITS OF ENERGY AND DENSITY. 85 128. The unit of kinetic energy [E] is twice the energy possessed by a particle moving with unit momentum and unit velocity, or, in other words, by a particle of unit mass moving with the unit of velocity. The kinetic energy of any moving particle is measured in terms of this unit by one-half the product of its momentum and velocity. Hence if its momentum be 7c [K] and its velocity v [F], its energy is -kv [E]. But one-half the L. product of the momentum and velocity is represented completely by Hence } lev [E] = ^1cv [K] [ F], 2i 2i therefore [E] = [K] [ V] = [M] [ F] The unit of energy is therefore of one dimension in mass, two in space, and minus two in time. It is there- fore of precisely the same nature as the unit of work, and we shall see (Art. 146) that when work is done on a free particle, an equivalent amount of energy is always given to it ; this could not be the case were work and energy of different natures, that is, of different dimensions in the three fundamental mechanical units. The numerical measure of the kinetic energy of a moving particle will, of course, vary inversely as the unit of mass, inversely as the square of the unit of length, and directly as the square of the unit of time. 129. The unit of density [D] is the density of a body the unit of volume of which contains the unit of mass. The density of any homogeneous body is measured in terms of this unit by the number of units of mass per unit of volume, and we therefore infer that while the unit of density is of one dimension in mass it is of minus one dimension in volume, that is, of minus three dimen- sions in length, since volume is of three dimensions in space. The same conclusion might have been derived from the consideration that the mass of any homogeneous 86 UNIT OF IMPULSE. body is the product of its volume and density, or, if v' [L\* be its volume, d [D] its density, its mass is v'd [ J/], whence, as in other cases, we get . The numerical measure of the density of a given body will therefore vary inversely as the unit of mass, and directly as the cube of the unit of length. 130. If the body considered be of .the nature of a string, then its linear density, that is, its mass per unit of length, is of one dimension in mass, and of minus one in length, as shown by the per, and its measure will vary directly as the unit of length, and inversely as the unit of mass. If the body be a thin lamina, its surface density, that is, its mass per unit of area, is of one dimension in mass and of minus one in area, that is, of minus two in space, and the numerical measure of its density will therefore vary inversely as the unit of mass, and directly as the square of the unit of length. 131. The unit of impulse is the impulse of a force which generates the unit of momentum. An impulse is measured by the number of units of momentum gene- rated. Hence, if [/] denote the unit of impulse, and i the measure of a given impulse, the momentum correspond- ing to the impulse i [I] is i units of momentum, and this is denoted by i [A'], where [K] represents the unit of momentum, and we have therefore or = [M] [L] [T]- 1 . (Art. 125.) The unit of impulse is therefore of one dimension in mass, one in length, and minus one in time, and the numerical measure of an impulse varies inversely as the unit of mass, inversely as the unit of length, and directly as the unit of time. ASTRONOMICAL UNIT OF MASS. 87 Impulse is of the same dimensions and therefore of the same nature as momentum, just as kinetic energy is ot the same dimensions and nature as work. 132. As another example, we will consider the dimen- sions of the Astronomical unit of mass in terms of the fundamental units. The Astronomical unit of mass is defined as that quantity of matter which acting upon an equal quantity of matter at unit distance (each mass being supposed condensed at a point), attracts it with the unit of force. The law of gravitation is, that the attrac- tion between two material particles varies directly as the product of their masses, and inversely as the square ot the distance between them. Hence, if ?%, m 2 denote the masses of two particles expressed in terms of the Astro- nomical unit, and r the measure of the distance between them, the number of units of force with which each attracts the other is y??1? " 3 . Hence, if \N] denote the unit *M of mass, we have m 1 [A 7 ] m 2 [N] __ m^m^ .--, ~~ where [P] represents the unit of force. But the unit of force being that force which acting on the unit of mass produces in it the unit of acceleration, is of one dimension in mass, one in length, and minus two in time, and is therefore represented by [N] [L] [T]~ z (Art. 126), or therefore r [L\ therefore rrr ' LAI therefore [N] =[L] 3 [T]- 2 . The Astronomical unit of mass is therefore of three dimensions in space and of minus two in time. The unit of density is of one dimension in mass and of minus three in length. Hence the Astronomical unit 88 METRIC UNIT OF WORK. of density is of minus two dimensions in time only. The Astronomical unit of density therefore depends solely on the unit of time and is independent of the unit of length. 133. Each of the above units might, of course, be ex- pressed in terms of any of the others instead of the funda- mental units ; for example, the unit of mass might be said to be of one dimension in force, minus one in length, and of two in time ; the unit of acceleration, of one dimension in work or energy, minus one in length, and minus one in mass ; the unit of time, of one dimension in velocity, and of minus one in acceleration, and so on ; but it is very seldom indeed that such expressions are of service. It is highly important to remember the dimensions of all the units in terms of the fundamental units of mass, length, and time, since then we can immediately find the measure of any quantity in terms of any given system of units. 134. For example, suppose it, required to find how many units of work of the metric system are equivalent to a foot- pound, the unit of mass being the gramme, that is, 15'432 grains, and the unit of length the centimetre, that is, '03281 feet the value of g in British measure being 32'1912. The unit of work is of one dimension in mass, two in length, and minus two in time. Hence the measure of a quantity of work varies inversely as the unit of mass, inversely as the square of the unit of length, and directly as the square of the unit of time. In the present exam- ple the unit of time remains unaltered. The foot-pound is equal to 32'1912 British absolute units of work. Hence the measure of a foot-pound, referred to the above metric units, is 321912 x ^ = 13,564,400 nearly. 135. The units recommended for general use by the Committee of the British Association are those belonging to the centimetre-gramme-second system, in which the unit of length is the centimetre or '03281 feet, and the unit of mass the gramme, or 15'432 grains, the unit of time being one second. BEITISH ASSOCIATION UNITS. 89 The C.G.S. unit of force is that force which acting on a gramme of matter for a second generates in it a velocity of one centimetre per second, and is called a dyne. The C.G.S. unit of work is the work done by the dyne in working through a centimetre, and is called an erg, or ergon. The unit of energy being mechanically equivalent to the unit of work the same name is applied to it. The C.G.S. unit of power is the power of doing work at the rate of one erg per second. A rate of doing work equal to 10 7 ergs per second is called a icatt. For multiplication or division by a million the prefixes mega-, or megal-, and micro-, may be employed ; thus, a megadyne is a million of dynes and a microdyne the millionth of a dyne. The prefixes kilo-, hecto-, deca-, deci-, centi-, mini-, may also be employed in their usual senses, that is to say, the prefixes kilo-, hecto-, deca-, deci-, centi-, milli-, imply respectively multiplication by 1000, 100, 10, so that a kilo-gramme is a thousand grammes while a cen- tigramme is the one-hundredth of a gramme. For the expression of high decimal multiples and sub- multiples the exponent of the power of ten which serves as multiplier is denoted, when positive, by an appended cardinal number, and when negative, by a prefixed ordinal number. Thus 10 9 grammes constitute a gramme-nine, while -- - of a gramme constitutes a ninth-gramme. A megalerg is equivalent to an erg-six, while a microdyne is a sixth- dyne. The weight ot a gramme is about 980 dynes or rather less than a kilodyne. The weight of a kilogramme is rather less than a mega- dyne. The dyne is about 1'02 of the weight of a milligramme 90 BRITISH ASSOCIATION UNITS. at any point of the earth's surface ; and the megadyne about T02 of the weight of a kilogramme. The IcHogrammetre^ or work done against gravity in lifting a kilogramme one metre in height, is rather less than the ergon-eight, being about 98,000,000 ergs. The gramme-centimetre is rather less than the Ttilerg, be- ing about 980 ergs. The value of g in the above statements is taken as 980 C.G.S. units of acceleration. The weight of a pound is about 445,048 dynes. A foot-pound is equivalent to about 13,564.400 erg*, or rather more than 13 megalergs. One Jiorse-poicer is about 7*46 erg-nines per second, or 746 watts. Nearly the whole of this article has been taken almost verbatim from the report of the British Association for the Advancement of Science for 1873. 136. The dimensions of the principal mechanical units have already been given, but for the sake of convenience they are tabulated below. TABLE SHOWING THE DIMENSIONS OF UNITS. Length = [L], Mass - [If], Time = [T], Volume "M ) T] 3 Density . . = V] \T ]-3 Velocity "* J rri i j ^ T\ ? -1 Acceleration ... ... \T] T] -2 Momentum or Impulse = a m- 1 , Force ... ......= M] [T \T]~* Work or Energy . . = [M] T L-*- j ? TO' 8 , Power or Kate of Doing Work . . . = M L TO' 3 - 137. Having determined the dimensions of all the quantities with which we have to deal in terms of the fundamental units, it is easy to translate any set of measurements from one system to another. The process is in fact little else than a process of construing with the assistance of the table of dimensions as a vocabulary. TRANSMISSION OP POWER BY A BELT. 91 When it is required to find the fundamental units on which any system of measurements is founded, it is fre- quently advisable to assume symbols for the ratios of the units to the foot, pound and second, and then to translate all the measurements into the foot-pound-second system. Illustrations of the process will be found in the examples of Arts. 138-142. It should be noticed that the numerical measure of the kinetic energy of a particle is one-half the product of the momentum and velocity. Hence the kinetic energy of unit mass moving with unit velocity is only half a unit, and the unit of kinetic energy is that possessed by two units of mass moving with unit velocity or by unit mass moving with A /2 units of velocity. The unit of work, \vhich corresponds to the unit of kinetic energy is that done by the unit of force in working through unit dis- tance. Now acting on the unit of mass the unit of force will produce unit of acceleration, and will therefore in unit time generate unit velocity. The energy of the unit of mass moving with this velocity is only half a unit of energy, so that, acting on a unit of mass originally at rest, the unit of force in unit time produces only half a unit of energy, and in the next chapter we shall see that during this time the body is moved through only half the unit of length, so that the force does only half a unit of work. 137 a. Power is rate of doing work. Now the work done by an agent is the product of the force exerted into the distance through which its point of application is moved in its own direction. Hence the rate at which an agent works is the product of the force exerted and the velocity of the point of application in the direction of the force. Thus the tractive force of a locomotive expressed in pounds' weight, multiplied by the speed at which it is running expressed in feet per second, gives its rate of working in foot-pounds per second, and dividing by 550 we obtain its rate of working in horxe-power. Similarly, if a rope is employed for traction, the tension of the rope multiplied by its velocity gives the power transmitted. 137 @. When a belt or strap passing over a pulley or 92 TRANSMISSION OF POWER BY A BELT. drum is employed to drive a shaft the power transmitted may be determined in the same manner, but in this case the belt pulls both at the top and bottom of the pulley, and it is the difference of the tensions in the advancing and receding portions of the belt which we have to con- sider. For suppose the belt is equally tight throughout. Then the pulley is solicited in opposite directions by the two portions of the belt acting upon its circumference with equal forces, and if the pulley meet with no resis- tance and be already in motion, it will continue to move, but no power will be transmitted. If the motion of the pulley were resisted, its speed could only be maintained by the receding portion of the belt exerting a greater pull upon it than the approaching portion. Suppose that 7\ is the tension in the part of the belt which is leaving the pulley, T 2 the tension in that which is approaching the pulley, and suppose the velocity of the belt to be v feet per second. Then in one second the work done by the receding part of the belt is vT^ but of this vT 2 units are employed in overcoming the pull of the approaching por- tion or slack part of the belt ; hence the work done per second on the pulley is v (T t T 2 ) which is therefore the measure of the power transmitted. By expressing v in feet per second, r l\ and T 2 in pounds' weight, and dividing the expression v (T^ T 2 } by B50, we obtain the horse- power transmitted. The maximum ratio of 7\ to T 2 is determined by the condition that the belt shall not slip, and as machinery is generally subject to vibration a certain margin of safety is required in determining this ratio. It will depend on the nature of the pulley and of the belt and on the angle of contact. If the "driver" and "follower" are of the same diameter the angle of contact will be about 180. If the driver and follower are nearly in the same horizontal plane the angle of contact may be made somewhat greater than 180 by making the lower portion of the belt the tight portion and the upper the slack portion, especially if the pulleys are at a considerable distance apart, for the " sag " of the slack portion, being greater than that of the tight portion, increases the angle of con- tact, Under such circumstances with wrought or cast EXAMPLES OF CHANGE OF UNITS. 93 iron pulleys and clean leather belts r l\ may with safety be taken to be about three times T 2 . 1377. When energy is transmitted along a shaft it is the couple exerted and the angular velocity of the shaft that determines the power transmitted. For suppose the moment of the couple exerted on the shaft to be G foot- pounds and that the shaft turns through the angle a in fy one second. Then we may suppose the couple due to pounds' weight acting at the extremity of an arm of- length r feet. In turning through an angle a the ex- tremity of the arm would move through ar feet and the work done by the force would be ar foot-pounds, that is Ga foot-pounds, a being, of course, expressed in radians (circular measure). Hence if a shaft be exposed to a couple of G foot-pounds and turn through the angle a in a second, the power transmitted is Ga foot-pounds per second. If the shaft make n revolutions per second a = 2w7r and the power transmitted is 2mr6r foot-pounds per second. The horse-power transmitted is obtained by dividing this expression by 550. 137S. For example : A dynamo absorbs 25 H.P. and is driven by a belt running at 100 feet per second, the tension in the tight portion being 2 times that in the slack portion of the belt. Find the tensions. 3 Hence the difference of the tensions is ~ T%. The power absorbed is 550 x 25 foot-pounds per second and the velo- city of the belt is 100 feet per second. Hence x 100 - 550 x 25. and 2 T 1 =|2 7 ,=229i. The tension of the tight portion is therefore equal to the weight of 229-J- pounds ; that of the slack portion to the weight of 91-| pounds. 94 EXAMPLES OF CHANGE OF UNITS. 138. The following are examples of units : Ex. I. If 10,000 foot-pounds be the unit of work, the weight ofb cwt. the unit of force and 1 cwt. the unit of maw, find the units of length and time. Let L feet be the unit of length and T seconds the unit of time. The dimensions of force are *- r/ !L an( ^ those of work [M] [L]* a T*O = - - - \TV* 112 L Hence the unit of force will be equal to -j^ foot- 112 L 2 pound-second units and the unit of work to ^ foot- pound-second units. But the weight of 5 cwt. is 560 x 32 poundals and 10,000 foot-pounds is 320,000 foot-poundals. = 660x32, and -f- = 320,000, 139. Ex. IT. If the unit of velocity be that of a point which passes over 8 feet in 3 seconds and the unit of accelera- tion be that in virtue of which a velocity of 60 miles an hour is generated in 55 seconds, find the units of space and time. Let L feet be the unit of length, and T seconds the unit of time. The unit of velocity will be ^ feet per second, and the unit of acceleration will be y foot-second units. Q Now a velocity of 8 feet in 3 seconds is Q feet per second ; 60 miles an hour is 88 feet per second, and the EXAMPLES ON THE CHANGE OF UNITS. 95 acceleration in virtue of which this velocity is produced L.XK A S8 S f 1 * in oo seconds is r ~ or r toot-second units. 5o 5 L = 8 ,L_ = S .'. rr\ o and y 1;i pj, 5 40 /. T=c> and L ^. 140. Ex. III. 7w the first of a system of units the measure of a certain acceleration is 500 times its measure in the second system, but the measure of a certain velocity is only BO times its measure in the second system. Find the ratios of the units of length and time. Let L i feet and 2\ seconds be the units in the first system ; L 2 feet and T 2 seconds the units in the second system. Suppose that the acceleration is f foot-second units and the velocity v feet per second. In the first system the acceleration will be represented y 2 rp by f - r -- and the velocity by v T -. In the second system J ^i -^i T 2 T. they will be represented by f Y~ and v ^ respectively. rji 2 1 2 T T. and v r = 50 v j^. ^1 7j 2 Hence T^ = 10 T 2 and Z, = g L 2 , J- 1 + r . -. -/-'i -L or y-=10 and ^ 5- 141. Ex. IV. 7/'^e density of wrought iron be denoted by 12, and the work done in lifting 1 cwt. vertically through 20 feet by 14, a pound being the unit of mass, find the units of length and time, it being given that 1 cub. foot of wrought iron contains 480 Ibs. Let L feet be the unit of length and T seconds the unit of time. The unity of density is therefore that of a sub- 96 EXAMPLES ON THE CHANGE OF UNITS. stance of which U cub. ft. contain 1 Ib. and the unit of L* work is rpt foot-poundals. The density of wrought iron will therefore be repre- sented by 480L 3 and the work done in lifting 1 cwt. through 20 feet by 112 * 112 x 20 x 32T 72 and f-n - = 14, * 142. Ex. V. In a certain system of absolute units the acceleration of gravity is denoted by 10, the kinetic energy of a train of 200 tons travelling at 60 miles an hour by 40, and its momentum by 2. Find the units of length, mass, and time. Let L feet be the unit of length, M pounds the unit of mass, and T seconds the unit of time. The acceleration of gravity is 32 foot-second units, and ^2 will be denoted by 32 -j^- . 200x2240x88% 1 he energy 01 the train is n ~ ioot-poundals, A -n i, j j ^ 200 x 2240 x 88 2 T 2 and will be denoted by ^ Its momentum is 200x2240x88 foot-second-pound T units, and will be denoted by 200 x 2240 x 88 . ^2 '. 32 T = 10 (1), 200 x 2240 x 88 2 T 2 2 in ^ QG 'ML 200x2240x88.-^ = 2 (3). EXAMINATION ON CHAPTER I. 97 From (2) and (3) 44 = 20 (4), /TTJ But 32 4- = 10... ...(1), 16' and M= 200 x 2240 x 44 v- jj = 200 x 2240 x 20 by (4) = 8,960,000. EXAMINATION ON CHAPTER I. NOTE. In all the examples, except where otherwise stated, the numerical value of o referred to a foot and a second as units of length and time is taken to be 32. 1. How must a physical quantity be measured? Of what does the complete representation of any physical quantity consist ? 2. Define the velocity of a point. If a second be the unit of time and an acre be repre- sented by 10, what will be the measure of a velocity of 45 miles an hour ? 3. If a train move from rest with uniform acceleration, and in five minutes attain a velocity of 60 miles per hour, find the measure of its acceleration when a second is the unit of time and a foot the unit of length. 4. What must we know about an acceleration in order that it may be completely defined ? Show that an acce- leration can at any time be represented by a straight line. B. A ship is sailing due North with a velocity of 10 knots an hour, while another is steaming South- West at the rate of 15 knots an hour. Find the velocity of the second ship relative to the first. G. D. H 98 EXAMINATION ON CHAPTER I. 6. A ship whose head points N.N.E. is steaming at the rate of 16 knots an hour in a current which flows E.S.E. at the rate of 4 knots an hour, find the velocity of the ship relative to the sea-bottom. 7. Explain what is meant by the resultant of two independent accelerations, and what by the acceleration of one point relative to another. 8. Define the density of a body. If one pound be the unit of mass and a yard the unit of length, find the measure of the density of water, it being given that a cubic foot of water contains 1,000 ozs. 9. State Newton's second law of motion, and explain briefly the nature of the evidence on which our accep- tance of this and other plrysical laws is based. 10. What do you understand by the physical inde- pendence of forces ? 11. What is the dynamical unit of force? If the unit of mass be a ton, the unit of length a yard, and the unit of time a minute, compare the unit of force with the weight of one pound, taking g equal to 32. when a foot and second are units. 12. During what time must a constant force equal to the weight of one ton act upon a train of 100 tons to generate in it a velocity of 40 miles per hour ? 13. Upon what experimental evidence do we base the assertion that the attraction of the earth upon any body is proportional to its mass and independent of the nature of the material of which it is formed ? 14. Supposing the attraction of gravitation at the equator to be '995 of its value in London, if a person sell goods in London by a spring balance accurately graduated at the equator, how much per cent, on the selling price does he gain in excess of his fair profit ? 15. If the cage of a lift be descending with an accelera- tion represented by T^J is cos $ equal to the coefficient of friction. 15. If the mass of the Scotch express be 150 tons, and the resistances to its motion arising from the air, friction, etc., amount to 16 Ibs. weight per ton, when the train is going at the rate of 60 miles an hour on a level plain, find the horse-power of the engine which can just keep it going at that rate. 16. If, in the preceding example, the driving-wheels of the engine be 8 feet 2 inches in diameter, and during each revolution of the wheels two pistons make each a complete stroke (to and fro), the diameter of each piston being 18 inches, and the length of the stroke 28 inches, find the mean effective pressure of the steam on each square inch of the pistons necessary to drive the train at 60 miles an hour, the slip of the driving wheels on the metals and the friction of the working parts of the engine being neglected. Note. The steam must do as much work per revolu- tion of the driving wheel on the pistons as is required to drive the train. 17. Find the actual horse-power of an engine which can just propel an ironclad ship at the rate of 16 knots an hour ; the resistance to the ship's motion when steam- ing at that rate being equal to the weight of 50 tons, and a knot being taken equal to 6078 feet. 18. If a point situated at the orthocentre of a triangle have three component velocities, represented in magni- tude and direction by its distances from the angular points of the triangle, show that its resultant velocity will tend to the centre of the circle circumscribing the triangle, and will be represented by twice the distance of the point from the centre. 19. A train travels at the rate of 45 miles an hour ; rain is falling vertically, but owing to the motion of the train the drops appear, as they fall past the windows, to EXAMPLES. 103 make an angle tan" 1 1'B with the vertical. Find the velo- city of the raindrops. 20. If in Attwood's machine the string can bear a tension equal to only one-fourth the sum of the weights, show that the least acceleration possible is -SL. v 2 21. A shot of mass m is fired from a gun of mass M with a velocity u relative to the gun : show that, if the mass of the powder be neglected, the velocity of the shot is , and that of the gun. -_ relative to the m + M m~+ M ground. 22. A smooth wedge, whose angle is a, has one face in contact with a horizontal plane. Find the acceleration with which it must be made to move that a heavy particle may be in relative equilibrium on its inclined surface. 23. If a be the distance between two moving points at any time, V their relative velocity, and u, v the resolved parts of V in, and perpendicular to, the direction of a, show that their distance when they are nearest to each /~fU other is -- , and that the time of arriving at this nearest y distance is ==; y* 24. If the acceleration caused by gravity be the unit of acceleration, and the velocity of a mile in B minutes the unit of velocity, find the unit of length. 25. If the units of length and time be a yard and a minute respectively, and the unit of force the weight of 32 Ibs., find the unit of mass. 26. If the unit of time be 5 minutes, and the unit of length 5 yards, find the value of g. 27. If the acceleration of a falling body be the unit of acceleration, and a velocity of 3 miles an hour the unit of velocity, find the units of space and time. 28. If the unit of velocity be the velocity of a point which passes over a feet in t seconds, and the unit of 104 EXAMPLES. acceleration that of a point which acquires in T seconds a velocity of b feet per T seconds, find the units of length and time. 29. Two nations estimate the acceleration of gravity by numbers in the ratio of 300 to 1, but the velocity of the earth in space by numbers in the ratio of B to 1. Find the ratios of their units of time and length. 30. If the area of a ten-acre field be represented by 100, and the acceleration of a heavy falling particle by 58, find the unit of time. 31. If the unit of force be equal to the weight of 5 Ibs. and the unit of acceleration, when referred to a foot and a second, be denoted by 3, find the unit of mass. 32. A constant force acts upon a particle during 3 seconds from rest, and then ceases ; in the next 3 seconds it is found that the particle describes 180 feet : find the velocity of the particle at the end of the second second of its motion, and the numerical value of its acceleration (1) when a second, (2) when a minute, is taken as unit of time, the unit of length being 1 foot. 33. If the unit of velocity be a velocity of a feet per t seconds, and if the weight of 1 pound be the unit of force, and a pound the unit of mass, find the units of length and time. 34. If /"j, f 2 be the measures of an acceleration when m + n seconds and m n seconds are the respective units of time, and a feet and b feet the respective units of length, show that the measure becomes - (*/fiti ^f-J-rf when 2n seconds are taken as unit of time and c feet as unit of length. 35. The measures of an acceleration and a velocity, when referred to (a + b) feet (m + n) seconds, and (a b) feet (m n} seconds respectively, are in the inverse ratio of their measures when referred to (a b) feet (m n) seconds, and (a + b) feet (m + n) seconds. Their measures, EXAMPLES. 105 when referred to a feet m seconds, and 6 feet n seconds, are as ma : nb ; show that =-v? 36. If f be the measure of an acceleration when a feet and t seconds are units of space and time, and f its measure when a feet and t' seconds are units, and if the acceleration be measured by f+f when c feet and r seconds are units, show that __ I c a a" 37. A shot of 700 Ibs. is moving with a velocity of 1200 feet per second ; find the numerical measure of its kinetic energy when a ton is the unit of mass, a yard the unit of length, and a minute the unit of time. 38. If the unit of length be a yard, the unit of accele- ration that produced by gravity in a body falling freely, and the unit of density that of water, find the number of units of work required to raise a ton of ore from the bottom of a mine 600 fathoms deep, assuming that a cubic foot of water contains 1000 ozs. 39. Supposing the earth to be a sphere of 4000 miles radius and mean density 5 - 5 times that of water ; if its mass be represented by a billion, the density of water by 10, and the work done in lifting a ton a yard high be the unit of work, find the units of mass, length, and time, a cubic foot of water containing 1000 ozs. 40. A circle revolves with uniform velocity in its own plane about its centre. The centre moves with varying velocity along a straight line. Find the velocity parallel to this line, at any instant, of a point on the circumfer- ence, and deduce the acceleration of the centre necessary for this point to be always moving at right angles to the line. 41. The number expressing the weight of a cubic foot of water (in absolute units) is y^-th of that expressing its volume, -^th of that expressing its mass, and T uoth of the 106 EXAMPLES. number expressing the work done in lifting it one foot. Find the units of length, mass, and time. 42. The resistance to the motion of a train is equal to the weight of 12 Ibs. per ton of its mass. An engine in drawing a train of 300 tons along a level line at 50 miles per hour consumes 20 Ibs. of coal per minute. If the heat required to raise the temperature of one pound of water through 1 C. be capable of doing 1390 foot-pounds of work, and if the combustion of 1 Ib. of coal be capable of raising 80 Ibs. of water from the freezing to the boiling point, find how much of the whole heat generated is use- fully employed by the engine. 43. Given that a quadrant of the earth's circumference is 10 centimetres, and that the mean density of the earth is 5'67, prove that the unit of force will be the attraction of two spheres each of 3928 grammes, whose centres are a centimetre apart ; the acceleration of gravity at the earth's surface being 981, when a centimetre, second, and gramme are the units of length, time, and mass. (N.B. Two spheres attract as if they were particles of the same mass situated at their centres.) 44. In a certain system of absolute units the accele- ration produced by gravity in a body falling freely is represented by 3, the kinetic energy of a 600 Ib. shot, moving with a velocity of 1600 feet per second, is denoted by 100 and its momentum by 10 ; find the units of length, mass, and time, assuming that g is equal to 32 in the foot-pound-second system. 45. It is required to transmit 50 H.P. by a single leather belt, running at 100 feet per second, the tension of the tight portion being three times that of the slack. Find the width of belt necessary if the working tension be limited to 75 Ibs. weight for a belt one inch wide. CHAPTER II. ON UNIFORM, AND UNIFORMLY ACCELERATED, MOTION. 143. IF a point move with uniform velocity w, the space passed over in t units of time will be denoted by vt. For the space passed over in each unit of time is v units of length, and therefore the space passed over in t units of time will be vt units of length. If a material particle be under the action of no external force, it will remain at rest or move uniformly in a straight line. Hence, if such a particle be moving at any instant with velocity #, it will retain that velocity, and the space passed over in t units of time will be vt. 144. If a point move with a constant acceleration f in the direction of motion, the velocity generated in each unit of time is f units of velocity, and therefore the velocity generated in t units of time will be ft units of velocity. Hence, if the initial velocity of the point be it, the velocity, v, at the end of t units of time will be u + ft, and if the point be initially at rest, its velocity at the end of t units of time will be ft, or v = ft. If the acceleration be in the direction opposite to that of the initial velocity, we must give the negative sign to /', and the above formula will still be true. If a material particle of mass m be acted on by a con- stant force Pin the direction of its motion, the momentum generated by the force at any instant will be in that direction, and the rate at which the momentum is gene- rated will be constant, being proportional to the force. Hence the acceleration of the particle will always be in the direction of its motion and be constant. If f repre- sents this acceleration, we have mf= P, the unit of force 107 108 UNIFORMLY ACCELERATED MOTION. being properly chosen (see Art. 45). Hence if u be the initial velocity of the particle, the velocity at the end of p t units of time will be u + ft, or u H t , and its momentum iiL will be mu + Pt. 145. To find the space passed over in t units of time by a point moving from rest with uniform acceleration f we may proceed as follows. Divide the time t into n equal intervals, each equal to r. Then the velocities of the point at the beginning of the 1st, 2nd, 3rd,...M th intervals will be denoted by 0, /V, 2/r. ..(w l)/r respec- tively, since its acceleration is uniform. Also the velo- cities of the point at the end of the 1st, 2nd, 3rd...?& tu intervals will be /V, 2/V, 3/V,...w/r respectively. Hence, if the point moved during each interval with the velocity which it had at the beginning of the interval, the space passed over in the t units of time would be . T + /T . T + 2/V . T + . . . + (n - 1) fr . T _n(n-V) f o / ' Again, if the point moved during each interval, with the velocity which it had at the end of the interval, the space passed over in the t units of time would be /T . T + 2/V . T + 3/r . + . . . + nfr . T Now the velocity of the point at any instant of a given interval is intermediate between its velocities at the be- ginning and end of that interval. Hence the space passed over during any interval is intermediate between that which would be passed over by the point if its velocity KINETIC ENERGY EQUIVALENT TO WORK DONE. 109 remained constant during the interval, and the same as at the beginning of the interval, and that which would be passed over by the point if its velocity were the same throughout the interval as at the end of the interval. Hence the space actually passed over by the point during the t units of time will be intermediate between that which it would pass over under the first of the above hypotheses and that which it would pass over if the second hypothesis were realised, that is, the space actually passed over by the point is intermediate between and this is true however great n may be. Now as wo diminish each interval and consequently increase w, each of the above hypotheses more nearly corresponds with what actually takes place ; and if we make each interval indefinitely small, and therefore n indefinitely great, - vanishes, and the above expressions for the space passed over ultimately coincide with each other, and therefore with the expression for the space actually passed over by the point, which is intermediate between the two. Hence the space passed over in t units of time by a point moving from rest with an acceleration f in the direction of motion is ~ ft~ units of length. a 146. From this we see that if a material particle, of mass m, move from rest under the action of a constant force P, the space passed over in t units of time will be p y -t 2 . The work done upon the particle will therefore (Art. 62) be P . f- . t 2 . Its velocity will be t and its 2w m momentum Pt. Its kinetic energy will therefore (Art. 1 P P 2 38) be _ Pt . t, that is .= t 2 . and is therefore numerically 2 m ' zm equal to the work done upon the particle by the force. If at the end of the t units of time we suppose the 110 KINETIC ENERGY AND MOMENTUM. direction of the force reversed so as to retard the motion, the intensity of the force remaining the same as before, the v.elocity of the particle will be destroyed in the same time as that in which it was generated, and the velocity at any instant during the retardation will be the same as at the corresponding instant when the acceleration was positive. Hence the space passed over by the particle, while the velocity is being destroyed by the force P, is the same as that passed over during the production of that velocity. Hence the work done against the force by the particle while being brought to rest is the same as that done by the force upon the particle when the velocity of the latter was being increased, and is therefore nume- rically equal to the kinetic energy of the particle. The kinetic energy of a particle is therefore numerically equal to the amount of work which it is capable of doing in being brought to rest, and we thus see a reason for the term "energy" being applied to one half the product of the momentum and velocity of a moving particle. 147. The effect of a force when its point of application moves in its direction is to do work, and we now see that if it act upon a particle its effect is to generate an amount of kinetic energy numerically equal to the work done. Now these effects must be one and the same, and therefore kinetic energy is mechanically equivalent to work. The effect of a given force, when its point of applica- tion moves over a given space in the direction of the force, is always to generate the same amount of kinetic energy represented by the Arithmetical product of these two factors ; hence while the effect of a force acting for an interval of time is to generate an amount of momentum measured by the Algebraical product of these two factors, the effect of a force when its point of application is moved in its direction is to generate an amount of kinetic energy measured by the Arithmetical product of these factors. The distinction between these two products of a force is very important. If a given force act for a given time upon a particle, the distance through which it will move it will vary inversely as the mass, and therefore the kinetic energy generated will vary inversely as the mass UNIFOEMLY ACCELERATED MOTION. Ill while the momentum generated is invariable, but if the distance through which the particle is moved, instead of the time, remain the same, the kinetic energy generated will remain invariable, but the momentum produced will be proportional to the square root of the mass. 148. If a point move under the influence of a constant acceleration f in the direction of motion, but start with an initial velocity M, a process precisely similar to that adopted in Art. 145 will enable us to find the space passed over in any given time. For, as before, dividing the time t into n equal intervals each equal to r, the velocity of the point at the beginning of the respective intervals will be u, u + /V, u + 2/V, . . . u + n 1 /V, and its velocity at the end of these intervals will be u + /V, u + 2/V, u + 3/V, ... u + nfr respectively. Hence the space actually passed over in t units of time will be intermediate between u.r + (u + /V) T + (u + 2/V) r + . . . + (u + n - 1 . /V) T and (u +fr}r + (u + 2/V) T + (u + 3/V) T + . . . + (u + nfr) r ; that is, between ut + i ft* (l - *\ and ut + i ft* ( 1 + -V 2 \ n) 2 \ n) and, increasing n indefinitely, each of these becomes ulti- mately equal to ut + ^ ft 3 , which is therefore the space a passed over by the point in t units of time. 149. The space passed over by a point moving under the influence of a constant acceleration in the direction of motion may also be found in the following way. Let time be represented by lengths measured along the line AB, and let AB represent t units of time and contain t units of length. Let the velocity of the point at any time be represented by a straight line drawn perpen- UNIFORMLY ACCELEEATED MOTION. dicular to AB from that point in AE which corresponds to the time in question, the line containing as many units .4. J< L .M B of length as the point possesses units of velocity. Let the time t be divided into n equal intervals, and let the straight line AB be divided into corresponding portions at the points K, L, etc. Through A, K, L, ... B, let lines AC, KN, LP, ... BD, be drawn perpendicular to AB, and representing the velocity of the moving point at the corresponding times. Draw the straight line CM ... E parallel to AB. Then, if u be the initial velocity of the point, and fits constant acceleration, AC u, KN=u + '- = u+f. AK. LP= u + ' - = u + f . AL. and so on. There- n n fore JcNf.Ckj lP=f.Cl, and so for the other points. Hence (7, N, P, ... D all lie in a straight line. Draw the straight line CD passing through each of the points N, P, etc. Complete the inner and outer series of parallelo- grams as in the figure. Then, if the moving point were to move during each interval with the velocity which it has at the beginning of the interval, the space passed over during any interval represented by LM will, since LP represents the velocity during that interval, contain as many units of length as the parallelogram PM contains units of area ; and, this being true for each of the other intervals, it follows that the number of units of length which would be passed over by the point in the time represented by AB, that is, in t units of time, if it moved during each interval with the velocity which it actually has at the beginning of the interval, would be equal to UNIFORMLY ACCELERATED MOTION. 113 the number of units of area contained in the sum of the inscribed parallelograms. Similarly the space passed over by the point in the same time, if it moved during each interval with the velocity which it actually has at the end of the interval, would be represented by the sum of the areas of the outer series of parallelograms. And this is true, however great may be the number of intervals into which the time t is divided. But the actual space passed over by the moving point must be intermediate between these two ; and when the number of intervals into which the time is divided is indefinitely increased, the sums of the areas of the inner and outer series of parallelograms ultimately coincide with the area of the figure CABD. Hence the number of units of length passed over by the point in the time represented by AB is equal to the number of units of area in CABD. But the figure CABD is made up of the rectangle CABE and the triangle CED, Therefore its area is equal to CA.AB+l CE.ED. Now CA represents the initial velocity and therefore contains u units of length, DB represents the velocity at the end of time t and therefore contains u +ft units of length ; hence DE contains ft units of length, and AB or CE contains t units of length. Hence the area CABD contains ut + ~. ft. t, or ut + ^ff 2 units of area. Therefore _ _ the space passed over in t units of time by a point starting with initial velocity w, and moving with a constant ac- celeration, /', in the direction of motion, contains ut + /%- a units of length. If the point start from rest, u = 0, and the figure will contain no rectangle corresponding to CABE. The space passed over in the time represented by AB will then be represented by the area of the triangle CED, and will therefore be ft* units of length. Since DE contains ft units of length, and CE contains t G. D. I 114 SPACE DESCRIBED DURING A GIVEN INTERVAL. units, the ratio -^-^ is numerically equal to /", and therefore JiiL> the acceleration is represented geometrically in the figure by the tangent of the angle DCE. (Compare Newton, Lemma x.) 150. If a material particle be under the action of a con- stant force in the direction of its motion, its acceleration will be constant, and the above investigation determines the space passed over by the particle in any given time, P substituting for /' the expression - , when m is the mass of the particle and P the force acting upon it. As an example we may take the following. A particle is allowed to fall from rest under the action of gravity only ; find the space moved through by the particle in 4 seconds, supposing g = 32 token a foot and a second are units. Here the only force acting on the particle is its weight, which is constant and equal to mg. Hence it will move with uniform acceleration g. The space passed over in -1 seconds will therefore be ^g . 4 a ft. = Sg ft. = 256 ft. ft 151. If we wish to find the space passed over in any particular second, the th for example, by a point moving with uniform acceleration, we may find the space passed over in n seconds and subtract from it that passed over in n 1 seconds ; or we may find the velocity at the begin- ning of the n th second, and then find the space passed over in 1 second by a point starting with this initial velocity and moving under the given acceleration. For example, let it be required to find the space passed over by a particle falling freely, during the 7 th second of its fall, g being supposed equal to 32. The space passed over in 7 seconds by a particle falling from rest is =g . 7 2 ft. = 16 x 7 2 ft. That passed over in 6 a seconds is 16 x G z ft. The difference, or 16 x 13 ft., is the space passed over during the 7 th second. EQUATIONS FOR UNIFORMLY ACCELERATED MOTION. 115 Or we may proceed thus. The velocity at the end of the 6 th second is 32 x 6 ft. Hence the particle at the be- ginning of the 7 th second has an initial velocity of 32 x 6 ft. per second, and will therefore during that second pass over a distance equal to j(32 x 6) + O a ft., or 32 x 6| ft., putting t= 1 in the formula ut + ft*. 152. If a point move from rest with a constant accelera- tion f in the direction of motion, and if v represent its velocity at the end of t units of time, and .v the space passed over by the point during that time, then we have v=ft (1), .. l/v* (<2\ O ~l ( \ijt From (1) and (2) we get 9 Cl J? / O \ 2 = 2/ (3). These three equations are very important, and should be remembered. If it be a material particle of mass m that is in motion, we obtain by multiplying each side of equation (3) by m and dividing by 2. mv 2 = mf . -s 1 . Now mf is the force which must act on the particle of mass m to produce the acceleration f ; and since .v is the space moved through by the particle in the direction of the force, mf . s is the work done upon the particle by the force, and we thus see that the kinetic energy of the par- ticle is equivalent to the whole amount of work that has been done upon it since the commencement of the motion. 153. If a point start with velocity u. and move with a constant acceleration /' in the direction of motion, and if v 116 EQUATIONS FOR UNIFORMLY ACCELERATED MOTION. represent its velocity at the end of t units of time, and * the space passed over by it during that time, we have = + /* ............... (1), & ............ (2). Squaring (1) and multiplying (2) by 2/", we see that v 2 = ti 2 + 2f ............ (3). If in this case it be a material particle of mass m that is in motion, then, multiplying each side of equation (3) by m, dividing by 2 and transposing, we get mv 2 = mu 2 mf . s. 2 i Here as before mf . s represents the work done upon the particle by the force producing acceleration, and the ex- pression on the left-hand side of the equation represents the increase of the kinetic energy of the particle. Hence the increase of the kinetic energy of the particle during any time is equivalent to the amount of work done upon it during that time by the force producing acceleration. If the direction of the acceleration of the moving point be opposite to that of its initial velocity, we have only to write /' for /' in the above expressions, and the equations will still be true. 154. It must be borne in mind that in all the above investigations the units of force, work, etc., are the abso- lute units belonging to the system of fundamental units employed. Thus if we adopt the foot-pound-second sys- tem the unit of force is the poundal which is of the 9 weight of a pound ; the unit of work is a foot-poundal or of a foot-pound, and the unit of energy is equivalent to & the unit of work and is therefore also a foot-poundal. EXAMPLES. 117 Hence since the kinetic energy of a particle of mass m moving with velocity v is - mv 2 foot-poundals it is equal a to = mv 2 foot-pounds. %9 155. For example, the number of foot-pounds of work which can be done by a 600 Ib. shot moving with a velocity of 1200 feet per second in coming to rest is 600 x 1200* = 600x1200* 2g 64 foot-pounds. Again, if in the course of one minute a steam gun pro- ject one hundred 4 Ib. shots with a velocity of 1200 feet per second the work it does in one minute is 100 4 x ; = 9000000 %g foot-pounds, and the horse-power at which the gun works is 33000 ~ 11' 156. As illustrations of this portion of the subject we may take the following Examples. Ex. 1. A heavy particle is projected vertically upwards with velocity u : supposing its weight to be the only force acting upon it, it is required to completely determine the motion. The particle will in this case move with a constant acceleration g downwards. We must therefore write g for f in the equations of the preceding Article. Its velocity, v, at the end of the time t will therefore be given by The space * passed over in t units of time will be given by 1 ?// /"I/"* I 11 t II v rtW'' \ ** I 2 \ > 118 EXAMPLES. If we take / equal to , we see by (i) that the velocity tj of the particle is zero, that is, the particle at this instant is at rest. If t be taken greater than this the velocity becomes negative, showing that the particle is descending. Again, from equation (ii) we see that the space passed over ?/ 1 7/ during the time - is - ; and since at this time the par- g <*9 ticle begins to descend, this is the greatest height to which it will rise ; or, if h denote this greatest height, we have Let w be the weight of the particle, m its mass ; then, if we multiply each side of this last equation by ic, that is mg, we get wh fjnu 2 , 2 which shows that the particle attains its greatest height and comes to rest when the work which it has done against its weight is numerically equal to the kinetic energy pos- sessed by the particle at the beginning of the motion. 7/ If we make t numerically greater than , the value of .S 9 Sin a " Comparing these two equations we see that v 2 = 2g sin a . s, or - mv 2 = w sin a . s. Now ., of each particle at the end of time t is given by M m v = at y ' EXAMPLES. 125 and the space s moved over by each particle during the time t is equal to The kinetic energy of the system is the sum of the kinetic energies of the two particles, that is Mv 2 + - mv* N + w A a at ] y J 2 Also the work done by gravity upon the particle A is ., 2J/+w * and the work done against gravity in raising the particle B is 1 M m 9.. 2 man. or w.^-r^ - flrr. y Hence the whole work done by gravity upon the system is m that is and is therefore equal to the kinetic energy of the system. 163. In the preceding problem suppose the string to ex- tend below the particle B, and when the system has been in motion for t seconds let a third particle C, of mass m', initially at rest, be attached to the end of the string below the particle whose mass is m, the string remaining stretched throughout. Determine the motion and the impulse of the tensions of the portions of the string. 126 MOTION OF BODIES CONNECTED BY A STRING. The common velocity of the particles A and B at the end of t seconds is, by the preceding investigation. - qt. Let this be denoted by v. M+ m y Now when C is attached, the 'string between B and C becomes suddenly tight, and C moves off with a jerk. Let u be the velocity with which C starts off, then the im- pulse of the tension acting upon C must be equal to m'u, which must therefore be the measure of the impulse of the tension of the string between B and C. Now imme- diately after C has been attached, the three particles A, #, and C must be moving with the same velocity, viz. u, since the string remains stretched (and we suppose the string inelastic). The velocity of A is therefore changed from v to u, u being, of course, less than v. The impulse of the jerk acting upon A must therefore be measured by M (v u\ which is consequently the effect of the impulsive tension of the string between A and B ; and this must be the same throughout. Hence an impulse represented by M (v u\ due to the impulsive tension of the string above B, is exerted upon the particle B vertically up- wards. But an impulse in'u, due to the impulsive tension of the string between B and 6 Y , is exerted upon B ver- tically downwards. The resultant impulse upon B is MOTION OF BODIES CONNECTED BY A STRING. 127 therefore m'u - M (v - u) vertically downwards. But the velocity of .Bis changed thereby from v to ?<, and B moves upwards throughout. Therefore m'u M (v 11) = m (v - u}, or (M+ m + m') u = (M+ m) v ; M+m ' u = TF v M+m + in M-m^ f " Now the impulse of the tension of the string between A and B is M (v u\ and substituting for v and , we get for this impulse Mm' (M-m) ' ^ ' Also the impulse of the tension of the string between B and C is m'u, that is _ M+m + m' ' After the impulse the system is moving with velocity , and it may be shown, as in the previous example, that the acceleration of A downwards and of B and C up- wards is now Mm m' M+m + m' ^' The velocity at the end of t' seconds after C was attached will therefore be M m m' ,, U + -^ - , Of. M+m+m * and the space moved over during these t' seconds by each particle will be ,/ 1 Mm m' ,,., ut + _- .at 3 . 2 M+m + m b If m + m' be greater than M the motion will be retar- ded after C is attached, and finally be in the opposite direction. 164. In the previous example the velocity, u, of the 128 MOTION OF BODIES CONNECTED BY A STRING. system immediately after C was attached might have been determined from the consideration, that since no impulsive force external to the system of these particles acts upon it in the direction of the motion,* the momentum of the whole system immediately before and immediately after C is attached to the string must be the same, since the weights of the particles, which are finite forces, cannot generate a finite momentum in an indefinitely shorb time. But the momentum of C before being attached to the string was zero, since it was at rest. Hence we must have (M + m) v = (M+ m + m) u, whence u , v , v. M+m + m as in the previous solution. 165. Before quitting this part of our subject we will consider one other example. The two particles A and B, whose masses are M and m respectively, being connected as in the precious example.*, after the system has been in motion for t second,*, a third particle 0, of mass m', originally at rest, is attached to the string above A ; it is required to find the subsequent motion, the string being inelastic. As before, if v be the velocity of the system at the end of t seconds from the commencement of the motion, we shall have -. .... M+m Now when the particle C is attached above A, there will be an impulsive tension of the string between A and 6 T , and C will move off with an initial velocity, which we will denote by u. Also, since the string between A and C is inelastic, it will remain stretched, and A and C will proceed with a common velocity, viz. u. The im- pulse of the jerk which must act upon C to make it start * The pressure of the smooth peg is always at right angles to the motion of the part of the string on which it acts and therefore can- not change the velocity. MOTION OF BODIES CONNECTED BY A STRING. 129 off with the velocity u is mu, which is therefore the effect of the impulsive tension of the string between A and C. Hence the particle A receives an impulse upwards, de- C4- noted by m'u. But the velocity of A is changed by this impulse from v to u. Therefore m'u = M v u or Mv M+m' M (M- m) .(B). Now since the velocity of A is diminished, and there is no impulsive force acting upon B to diminish its velocity, immediately after the impulse B will be moving faster than C and A. Hence the string between C and B will become slack, and B will be moving as a particle acted upon by its own weight only, and projected vertically upwards with an initial velocity v. The velocity of B at the end of t' seconds after C is attached will therefore be v gt', and the height through which C will have risen in that time will be denoted by G. D. 130 MOTION OF BODIES CONNECTED BY A STRING. While the string above C is slack, A and C will be fall- ing freely, having started with an initial velocity u. Hence the common velocity of A and C at the end of the time t' will be u + gt' ; and the space through which they will have fallen during this time will be Now the string will become tight when C has descended through a space equal to that through which B has ascended in the same time. Hence, if the string become tight at the end of t' seconds after C was attached, we must have or gt' = v MI ,, v u whence t and substituting for v and u their values from (A.) and (B), we get ., = M-m . M(M-m) ~ M+m (M+m')(M+m) m (M- m) . ,px (M+m')(M+m) ' and the space through which the system will have moved since the attachment of C will be M m . m' (M ni) . 1 f m' (M m) ") 2 , 2 M+m 9 ' (M+m'}(M+m) 2 9 i (M+ m')(M+ m) j _(,! ^ ) m'(M-m)* I 2 M+ m'j (M+ m')(M+ mf y MOTION OF BODIES CONNECTED BY A STRING. 131 Also the velocity of C and A when the string becomes tight will be u + at'= U (M-m) m' (M-m) (M+ m'\M+ m) * (M+ m')(M+ m} y M- m the same as the velocity of the system before C was attached. Also the upward velocity of B when the string becomes tight is ,/ Mm . TO' Mm n , AN ^ /r *,-, --- at - ---- , . ^ r ---- at. -by (A) and (C)} , y y M M-m ~ M+m ' M+n? ' Now when the string is tight all the parts of the system must be moving with a common velocity. Hence when the string becomes tight there will be an impulsive tension. To determine the common velocity immediately after the tightening of the string, we may adopt the method of the last article. The momentum of the whole system, reckoned as though all the bodies were moving in the same direction, since the smooth peg simply serves to change the direction of the string, must be the same before and after the tightening of the string. Hence, if v' be the common velocity immediately after the string becomes tight, we must have (M + m + m') v' = (3f + m') *r. gt + -. . %=2gt nt (M+mf + Mm M-m M+m ' M+m y ' m, f , (M+mf + Mm M-m Therefore v = -}. --- ;-W; r TV. -r? -- fl- (M+m)(M+m + m] M+m The effect of the impulsive tension of the string may be determined from the consideration that by it the velocity of the particle J5, whose mass is m, is changed from 132 TIME OF DESCEIBING A GIVEN PATH. M M- m , M+ m' M+ m After the string has become tight, since it is inelastic, it will remain tight, and the system starting with the velo- city v will move with a constant acceleration equal to M+m' m M+m' + m g ' whence the velocity at any subsequent time, and the space passed over during any time, can be immediately found. 166. "We will leave to the reader to investigate the motion of the system when, after the two weights A and B have been in motion for t seconds, the weight (7, origi- nally at rest, is suddenly attached to the string between B and the peg, and at a considerable distance above B. The reason for introducing the last condition will be readily seen. The whole problem is very similar to that last investigated, and will well repay the student for the time he will require to examine it. 167. If a point be moving with a known acceleration in the direction of its motion, and its initial velocity be given, the time in which it will describe any portion of its path can be immediately found from the equation which is a quadratic equation for finding t. If u and /' be of the same sign, one of the roots of this equation is negative. The positive root is of course the one required. The negative root gives the time before the earliest time considered in the question, at which, if the point had been at a distance s in the positive direc- tion from the point from which we have supposed it to start, and moving with a proper velocity in a direction opposite to its acceleration, it would have passed through the point from which we have supposed it to start, and have returned to it with a velocity w, at the instant at which we have supposed it to start. TIME OF DESCENT DOWN CHOEDS OF A VERTICAL CIRCLE . 133 If, u and f be of opposite signs both the values of t are frequently admissible, as for example, in the case of a par- ticle projected vertically upwards, when the smaller value of t gives the time in which it will reach a point verti- cally above the point of projection and at a distance S from it during its ascent, and the larger value of t gives the time at which it will reach the same point in its descent. 168. As an example of the preceding article we will consider the following problem. A heavy particle falls from rest at the highest point of a vertical circle of diameter a down a smooth chord of the circle. Find the time of descent. Let m be the mass of the particle ; then mg will repre- sent its weight. Let 6 be the inclination of the chord to the vertical, s its length. Then s a cos 6. Now the resultant force on the particle acts down the chord and is equal to mg cos 6. The acceleration pro- duced by this force in the particle whose mass is m is g cos 6. Hence the particle moves down the chord with uniform acceleration g cos 0, and if t be the time of descent, we shall have t 2 . or 8 = ;grcos 6 . 2 a cos -g cos . & 134 STRAIGHT LINES OP QUICKEST DESCENT. Therefore and is independent of the inclination of the chord. Hence the time of descent down all chords of a vertical circle from the highest point is the same, and equal to the time of descent down the diameter. Similarly the time of descent from any point of a ver- tical circle along the chord to the lowest point is the same as the time down the vertical diameter. Since all the sections of a sphere by vertical planes through its highest point are equal circles, it follows that the times of descent of a particle down all chords from the highest point of a sphere are the same. 169. If it be required to find the straight line of quick- est descent to any plane curve AB from a point P in its own plane, we have only to describe a vertical circle having P for its highest point and touching the curve. Then the chord PQ drawn from P to the point of contact of the curve and circle is the line of quickest descent required. For if PQ be not the line of quickest descent, let some other straight line as PK be that of quickest descent. Let P/T cut the circle in L. Then the time down PK is greater than that down PL. But the time down PL is equal to the time down PQ ; therefore the time down PA" is greater than that down PQ, which is contrary to the hypothesis STRAIGHT LINES OF QUICKEST DESCENT. 135 that PA" is the straight line of quickest descent from P to the curve. If it be required to find the straight line of quickest descent from any point P to a given surface, it is only necessary to describe a sphere having P for its highest point and touching the surface. The chord drawn from P to the point of contact will then be the straight line of quickest descent from P to the surface. 170. If a particle be in motion under the action of a force always in the direction of its motion, since the work done by a force is measured by the product of the force and the distance moved over by its point of application in the direction of the force, it follows that the work done on the particle during any time is measured by the pro- duct of the force and the space passed over by the particle during that time. Hence the measure of the rate at which the force does work on the particle at any instant is the product of the force and the velocity of the particle at that instant, and is therefore proportional to the velocity. Hence if a particle move from rest with a constant acce- leration in the direction of its motion, the rate at which work is done upon it is proportional to the time during which it has been moving. "We have said that the work done during any time r by a uniform force acting upon a particle in the direction of its motion is measured by the product of the force and the space passed over by the particle during the time T. Now if v' be the mean velocity of the particle during the time T, v its velocity at the beginning, and v + u its velocity at the end of the time, we have $ = v + M, and the space a passed over during the time T will be represented by V'T. Hence the work done by the force during the time T may be represented by the product of the force, the time T, and the mean velocity v' of the particle. But the product of the force into the time is the measure of the momentum generated during that time. Hence the work done by the force during any time is measured by the product of the mean velocity of the particle and the momentum gene- rated during that time. 136 KINETIC ENERGY EQUIVALENT TO WORK DONE. Now the kinetic energy of a moving particle has been defined as one-half the product of its velocity and its momentum. It is therefore represented by - mv . #, or a 2 mv 2 , if v be the velocity and m the mass of the particle. Now suppose that, during the time T, v is changed tov + u. (By taking T small enough u can be made as small as we please.) Then the kinetic energy becomes - m (v + ii)~. 23 The increment of the kinetic energy is therefore muv + ^ mu 2 . Hence we have for the increment of the kinetic energy the expression mu (^ + o u\. Now mu is the momentum generated during the time T and v + ~ u is the 2. mean velocity of the particle during that time, since the velocity changes uniformly. Hence the increment of the kinetic energy of the particle during any time is measured by the product of the momentum generated during that time and the mean velocity of the particle during the interval. The increment of the kinetic energy of the par- ticle is therefore equivalent to the work done upon it by the force producing acceleration. From this investigation we see a reason for the definition we have given of kinetic energy. 171. If the force acting on the particle in the -direction of its motion be not uniform, then if we make r very small we may consider the force uniform during the interval T, and the above investigation will hold. Hence in this case the increment of the kinetic energy produced in any very small time is equivalent to the work done during that time, and this being always true, it follows that the in- crement during any finite time, being the sum of the increments during the intervals into which that time may be divided, is equivalent to the work done on the particle during that time. Hence if a particle move from rest under the action of KINETIC ENERGY EQUIVALENT TO WORK DONE. 137 a force in the direction of its motion, whether its magni- tude be constant or variable, the whole kinetic energy of the particle at any time will be equivalent to the whole amount of work done upon it by the force. If the force be not in the direction of motion it may be resolved into two, one in that direction and one perpen- dicular to it. Now the latter component does no work, since its point of application moves always in a direction perpendicular to that of the force, and it may be shown that it does not increase the velocity of the particle, but simply tends to change the direction of its motion, and therefore does not alter its kinetic energy. Hence the former component is the only one which we need consider, and it follows that the whole change of the kinetic energy of the particle during any time is equivalent to the work done upon it during that time. This result is a case of the Principle of the Conservation of Energy. 172. From the preceding articles it follows that if the resultant force acting on a particle at every point of its path be known, as well as the velocity at any given point, the velocity at any other point can be found if we can find the work done by the forces in passing from the one point to the other. Some examples illustrating this result will be found in Chapter V. Since the rate at which work is done upon a particle by a force acting in the direction of its motion is measured by the product of the force and the velocity of the particle, it follows that if this rate be uniform, the force must vary inversely as the velocity of the particle. Hence if a par- ticle start from rest, it is impossible for an agent to do work upon it at a finite rate at the commencement of the motion, since in order to do this, it would be necessary to exert an infinite force. For example, it is impossible for an engine in starting a train to work up to its full horse-power. 173. We may illustrate this subject by the following examples. 138 EXAMPLES ON HORSE-POWER. Ex. 1. A train whose mass is 100 tons (including the engine) is drawn by an engine of 150 horse-poicer. The resistance to motion on a level line dueto friction being equi- valent to a force of 14 Ibs? weight for every ton in motion, and the resistance of the air being neglected, find the maxi- mum speed which the engine is capable of sustaining on a level line. The resistance to the motion of the train due to friction, etc., is equal to the weight of 1400 Ibs. Hence if v be the velocity of the train in feet per second, the rate at which its engine works is 1400t> foot-pounds per second. But it is capable of doing 150 x 550 foot-pounds per second. Hence we must have, if v be the velocity required, 1400^ = 150x550, _ 15 x 55 . or the velocity of the train is -^ feet per second, that is 40^ miles per hour. 174. Ex. 2. If the train described in the preceding ex- ample be moving at a particular instant with a velocity of 15 miles per hour, and the steam turned off, how far trill it run before coming to rest ? The kinetic energy of the train is 100x_2240 x 22 2 foot-pounds, and the resistance to the motion is equal to the weight of 1400 pounds. The train will come to rest when the work done against the resistance is equal to the original kinetic energy. Hence if s feet be the distance which the train runs 1400s = 20 100 x 2240 x 22 2 2^x1400 = 1210, and the train will come to rest after running 1210 feet. EXAMPLES ON HOESE-POWER. 139 175. Ex. 3. Find the horse-power of an engine required to drag a train of 100 ton* up an incline of 1 in 50 icith a velocity of 30 -miles per hour, the friction being equal to the weight of 1400 M*, \ The line is inclined to the horizon at an angle whose sine is . Therefore the resolved part of the weight of the train down the incline is 2 tons' weight. Hence the whole force tending to stop the motion of the train up the plane is equal to the weight of 2 tons and 1400 Ibs., that is of 5880 Ibs. Now a velocity of 30 miles per hour is 44 feet per second. Hence the engine must do 5880 x 44 foot-pounds of work per second, and the horse-power at ftQQf) v 44 which it works must therefore be ' X ZZ = 470|. The ooU engine must therefore be of not less than 470 2 horse- power. Since the pressure of the train upon the metals and bearings of the wheels when on an incline whose inclina- tion is a, is less than the corresponding pressures on a level line in the ratio of cos a to 1, the friction will be less in the same ratio. Hence if the resistance to the motion of the train, due to friction when on an incline of 1 in 50 be 1400 Ibs. weight, the corresponding resistance when on a level line will be 1400x50 70000 - 1 v/2499 pounds' weight. 176. Referring to the example in Art. 156, we see that if a particle be projected vertically upwards with velocity U 2 u it will rise to a height -- , and then come to rest. If the *9 mass of the particle be denoted by m its weight will be u z represented by nig, and in falling from the height _- it f& might be made to lift a weight, less than itself, but differ- ing by as small a quantity as we please, through the same 140 CONSERVATION OP ENERGY. ^ % height ; that is, it may be made to do mg x o , or - , ff units of work. Now the kinetic energy of the particle when projected is represented by , and we see that a this amount of kinetic energy can be converted into the capacity or poicer of doing the same number of units of work due to the separation of the body from the earth, i.e. into the same number of units of potential energy. If the particle be allowed to fall freely from the height 7/2 Q-, its kinetic energy on reaching the point of projection , N ~sr ........................ ( )( G. D. M 162 VELOCITY DUE TO FALLING FROM DIRECTRIX. and BK u cos a . t u? cos a sin a But since the curve is a parabola, if I be the latus rectum we must have j _ 2u~ cos 2 a , . >. ~7~ The range is equal to twice BK, since the parabola is symmetrical about AK. Hence if r be the range, Zu~ cos a sin a T = 9 u 2 sin 2a .(0). g In order that the range may be a maximum we must have sin 2a = l, and therefore a = 45. Hence for a given velocity of projection an elevation of 45 gives the greatest horizontal range. 189. The directrix being at a height above A equal to one-fourth of the latus rectum, its height above BC is .............................. ' Now the height of Q above BC at the end of the time t is u sin a . t - ^gt 2 . Hence the distance of Q below the direc- a trix is gt* .............. (F). Also the velocity of the particle is the resultant of its hori- zontal velocity, u cos a, and its vertical velocity, u sin a gt, and these are at right angles. Hence if v denote the velo- city of the projectile at the end of the time , v = {u 2 cos 2 a + (u sin a gtfY* + # 2 2 }* ............ (G). EXAMPLES OF PROJECTILES. 163 Now the distance, S, of Q below the directrix is given by u 2 1 ^ = o ~ Msino .t + ~ ?ir 2 2 COS 2 a ,, a . Hence BM= u cos a.t = - . (tan a tan 0). and BP= cos 2< 2 cos 2 a a cos (tana tan 0), which gives the range on the inclined plane. GREATEST RANGE ON AN INCLINED PLANE. 167 This expression for the range we may put into another form, thus : u 2 BP= ~ n (2 sin a cos a - 2 cos 2 a tan 0} gcosd n 2 -= 5-^ (sin 2a cos 0-2 cos 2 a sin 6} g cos 2 6 = u . (sin 2a cos 6 cos 2a sin sin 0) G. D. 178 PROBLEMS ON PROJECTILES. Hence we obtain for the height AP in feet the ex- pression 1 20 2 50 ay/2 Q. These two expressions, (1) and (2), for the height of AP must of course be equivalent. Hence, equating them, we have (a a \ v/3-1. *\60* 6.25V v/3 _ y/3-1 3.50 2 -73- g = 250X3- -v/3) Substituting this value of a in either of the above ex- pressions for the height AP, we obtain for this height /r> /o\ 9 /rj .(d v/o)/ -- (p S' = (2- v/3). (3- v/3) 2000 -(9-5^3) ; g and if we take g = 32. we obtain for the height of the tower 26-542968.'.. feet, This last example is useful as indicating how such questions may be solved, without any reference to the parabola, by simply considering the horizontal and vertical motions independently, and finding the time of flight, 205. Ex. 3. A shot is fired at an elevation of 30 so as to strike an object at a distance of 2,500 feet, and on an PROBLEMS ON PROJECTILES. 179 ascent of 1 in 40. Find the velocity of projection, neglecting the resistance of the air. This is of course equivalent to finding the velocity of projection in order that the range on a plane through the point of projection inclined to the horizon at the angle sin" 1 -^ may be 2500 feet, the direction of the projection being inclined at an angle of 30 to the horizon. "We may, therefore, at once use the formula of Art. 193 for the range on an inclined plane, or we may proceed thus : u \/'3 Let u be the velocity of projection. Then is the ft horizontal velocity of the shot. Now the horizontal distance between the point of projection and the object hit is 2500^* ~ ty feet, Hence the time of flight is 4U 2500- rx '- j^ seconds 40 u a -', and its acceleration g cot a. sin a CHAPTER IV. ON COLLISION. 206. IF a particle of mass m be moving with a velocity v, and be retarded by a constant force which brings it to rest in time t, then the measure of this force we have seen to be . In fact the primary notion of momentum V is the effect, or product, of a force acting during a finite time upon matter free to move, and it follows from the second law of motion that that which is produced by a given force acting for a given time on any quantity of matter free to move, is always the same amount of momentum, and this is proportional to the algebraical product of the force and the time during which it acts. Now, suppose the time t during which the particle is brought to rest to be made very small. Then the force required to bring it to rest is very large, and if we suppose t so small that we are unable to measure it, then the force becomes very great, but we are unable to obtain its measure. In this case, then, we are compelled to adopt some other mode of considering the question. Since we are unable to measure the time during which the velocity of the particle is being destroyed, we leave the element of time out of consideration altogether ; the force we call an impulsive force, and we measure its impulse by the whole momentum which it destroys. Hence it will be seen that the nature of an impulse is totally different from that of a force, and the two things cannot be compared, for an impulse is the same as the ultimate effect of a finite force acting for a finite time. Indeed, we speak of the impulse instead of the force itself 188 DIRECT AND OBLIQUE IMPACT. 189 simply on account of our inability to measure very short intervals of time, and to observe what takes place during them, there being no case in nature in which a finite change of motion is produced in an indefinitely short time ; for, whenever a finite velocity is generated or destroyed in nature, a finite time is occupied in the process, though we are frequently unable to measure it even approximately. For example, if two balls strike one another, each of them will be more or less compressed or indented, and they will remain in contact for a finite time ; though the harder the balls, other things being the same, the less will they be compressed, and the shorter will be the time during which they will remain in contact, We may here notice that some bodies, when they become indented by impact, retain the indentation, while others more or less completely resume their original form. The first class are generally called inelastic, and the second class are called elastic bodies. 207. DEF. Two bodies are said to impinge directly upon one another when the surface of either at the point of contact is perpendicular to the direction of their relative velocity. When this condition is not fulfilled, the impact is said to be oblique. From the definition of direct impact it follows that when two bodies impinge directly upon one another, the mutual action between them is entirely in the same straight line as the velocity of one relative to the other. Newton found that if he allowed two bodies to impinge directly upon one another, their relative velocity after impact bore a constant ratio to that before impact, so long as the materials of which the bodies were composed were unchanged, but was in the opposite direction. The numerical value of this ratio is called the coefficient of mutual elasticity of the two substances, and is generally denoted by e. (See Art. 251.) If the incidence be oblique but the surfaces of the bodies smooth, then the whole action between the two is in the direction of the common normal to the surfaces at 190 COEFFICIENT OF ELASTICITY. the point of incidence. In this case the component of the relative velocity of the bodies in the direction of the common normal is changed by impact in the ratio of 1 to e, and reversed in direction, while the component of the relative velocity in a direction perpendicular to this remains unchanged. If the surfaces of the bodies be rough, there is also a tangential action between them, and the change in the relative velocity is not wholly in the direction of the common normal to the surface at the point of contact. When e is equal to 1, or the velocity of one body relative to the other is the same after impact as before, but in the opposite direction, the bodies are said to be perfectly elastic. If e be equal to 0, or the bodies after impact go on moving together, they are said to be inelastic. No bodies occurring in nature are either perfectly elastic or inelastic. Glass and some crystals are amongst the most elastic bodies known, while clay, putty, etc., have very little elasticity. 208. If the two bodies are of the same material, the numerical value of the ratio of their relative velocities after and before impact is called the coefficient of elasticity of the particular material of which they are composed. It should be observed that the coefficient of elasticity is a very different quantity from the elasticity or modulus of elasticity of a substance, as those terms are employed in connection with the strength of materials. The elasticity or modulus of elasticity is properly defined as the limiting ratio of - . when each is indefinitely diminished, and strain for every kind of strain which may be produced this ratio gives a corresponding modulus of elasticity. The coeffi- cient of elasticity, as defined above, is quite independent of any modulus of this sort. For example, the coefficient of elasticity, as measured by ratio of velocity after and before impact, may be much greater for india-rubber than for wrought iron, but the ratio - .- is enormously greater strain in the case of wrought iron than in the case of india- IMPACT ON A MOVING PLANE. 191 rubber for every kind of strain except cubic compression. For a further discussion of the coefficient of elasticity see Arts. 235-241. "We may remark that the impact of two spheres is direct when the line joining their centres at the moment of impact is in the direction of their relative motion. We shall in this chapter consider the impact of particles and spheres against one another and against planes only, and, except when the contrary is stated, we shall suppose their surfaces smooth. 209. The simplest case of impact is that of a particle impinging directly upon a fixed inelastic plane. In this case, since the coefficient of mutual elasticity is zero.^ the particle will come to rest. Now, if m be the mass of the particle and v its velocity, its momentum will be denoted by mv, and as this is entirely destroyed by the impact, the measure of the impulse which the plane exerts on the particle is mv ; but since action and reaction are equal and opposite, this is also the measure of the impulse of the pressure exerted by the particle on the plane. 210. Next, suppose a particle of mass m, and moving with velocity v, to impinge directly on a fixed plane, the coefficient of elasticity being e. In this case the particle will rebound from the plane with velocity ev. The impulse exerted by the plane on the particle must there- fore destroy the velocity v with which it is moving, and generate a velocity ev in the opposite direction. Hence the whole change of the particle's velocity is numerically equal to v + ev, and the whole change in its momentum to mv (1 + e). Hence the measure of the impulse between the particle and plane is mv (1 + e). If the plane be perfectly elastic we have e equal to 1, and the impulse is measured by 2mv. 211. Next, suppose the plane on which the particle impinges to be moving with a velocity V in the same direction as the particle before impact. Then the velocity of the particle relative to the plane is v V before impact, and after impact it is e (v V), but in the opposite direc- tion. If we adopt the usual convention with respect to 192 OBLIQUE IMPACT ON A FIXED PLANE. sign, we may denote the velocity of the particle relative to the plane after impact by e (v V). The whole change of the velocity of the particle is (v F)(l + e), and the change of its momentum is m (v F)(l + e), without regard to sign. This last expression is the measure of the whole impulse between the particle and the plane. Since the velocity of the particle before impact was #, and the change of velocity produced by the impact is (v F)(l + e) in a direction opposite to that of v, it fol- lows that the velocity of the particle after impact is v (v F)(l + e), and is in the same direction as before, or in the opposite direction, according as the sign of this expression is positive or negative. If e be zero, or the plane inelastic, the velocity after impact is F, as of course it should be. If e be unity or the elasticity perfect, the velocity after impact is 2F , and this is in the same direction as before, or in the oppo- site direction, according as F is greater or less than - . a The above investigations are true for spherical balls as well as particles, provided that their centres of gravity coincide with their centres of figure, and that they have no motion of rotation unless their surfaces be perfectly smooth. If the spheres be rough, and they have a motion of rotation, or if their centres of gravity do not coincide with their centres of figure, the problem becomes much more complicated, and requires the principles of Rigid Dynamics, to which subject all problems on the motion of spheres or of any rigid bodies of finite dimensions properly belong. 212. Suppose a particle of mass m, moving with velocity v along the line QP, to impinge obliquely at P, upon the smooth plane AB, the coefficient of elasticity between the particle and plane being e. It is required to find the motion of the particle immediately after impact, and the impulse on the plane. Let PT be the direction of motion after impact, PN the normal at P to the plane. Let the angle QPN be denoted by a, and the angle TPN by 6. Then the velocity of the particle IMPACT ON A ROUGH PLANE. 193 before impact may be resolved into two components ; viz., v sin a along the plane, and v cos a perpendicular to the plane. Since the plane is smooth, it is only the latter component which is altered by the impact, and this is replaced by a velocity ev cos a in the opposite direction. The velocity after impact is therefore the resultant of the two velocities, v sin a along the plane and ev cos a per- pendicular to the plane, and is therefore numerically represented by v v / sin 2 a + e 2 cos 2 a ; and if PT be the direction of motion after impact, cot TPN= ev cos a v sin a ' or cot e cot a. If the elasticity be perfect, or e equal to 1, we have 6 equal to a, or the angle of reflection equal to the angle of incidence. It remains to determine the impulse of the pressure between the particle and the plane. The component of the velocity of the particle perpendicular to the plane before impact is v cos a, and after impact it is ev cos a in the opposite direction. The whole change of velo- city produced in the particle by the impulse is therefore v cos a (1 + e), and the change of momentum is measured by mv cos a (1 + e\ which is therefore the measure of the impulse. 213. If a particle impinge obliquely on a rough plane G. D. o 194 DIRECT IMPACT OF TWO BALLS. whose coefficient of friction is /A, then, besides the impul- sive pressure, there will be an impulsive friction called into play, which will be at each instant proportional to the pressure, and such that its effect is to diminish the velocity of the particle parallel to the plane. Referring to the figure of the preceding Article, since the velocity of the particle perpendicular to the plane is reversed in direction and diminished in the ratio of 1 to e by the impact, it follows, as before, that the measure of the impulse is mv cos a (1 + e). Now, the impulse of the friction called into play is p times this, that is [j, . mv cos a (1 + e), and diminishes the particle's velocity parallel to the plane. Hence the velocity, parallel to the plane, of the particle o , ,1 N rru f -n alter impact is v sin a pv cos a (1 + e). Therefore, it, as m the preceding example, 6 denote the angle the direction of motion after reflection makes with the normal, we have ev cos a cot = v sin a pv cos a (1 + e) ecos a sin a p, cos a (1 + e) 1 which gives the direction of motion after impact, and the velocity can be found as in the previous case. (See also Art. 222, p. 203.) If the surface upon which the particle impinges be curved, the effect of the impact is the same as if the surface were replaced by its tangent plane at the point at which the particle strikes it. Thus in the preceding Article, if PN be the normal at P to the surface AB, the whole of the reasoning will be equally true whether the surface AB be plane or curved. 214. Hitherto we have considered the obstacle against which the moving particle impinges to be either fixed, or made to move in such a manner that its velocity is un- affected by the collision. We proceed now to the con- sideration of the collision of two particles, or two spheres, each free to move. It will be seen that in order that two spheres may impinge directly upon one another, the line DIRECT IMPACT OP TWO BALLS. 195 joining their centres at the time of impact must be the line of motion of either relative to the other. Suppose two particles, or spheres, whose masses are re- spectively M and /, to be moving in the same direction with velocities V and v respectively, of which V is the greater, and to impinge directly upon each other ; it is required to find the motion of each after impact, the co- efficient of elasticity being e. Let F 1? i\ be their respective velocities after impact. Then whatever be the impulsive action between them at the moment of impact, the impulse on the first must be equal and opposite to that exerted on the second. Hence the momentum generated in the first must be numerically equal but opposite in direction to that generated in the second. Hence, whatever momentum reckoned in the posi- tive direction may be lost by the first, the same amount must be gained by the second, and vice versd; consequently the algebraical sum of the momenta of the two balls must be the same after impact as before. Therefore MV + mv t = MV+ mv .................. (I.). Again, the relative velocity after impact is to that before impact as e to 1, and the velocity of the first relative to the second before impact is F v, and after impact it is V^ v^. Therefore Vi-Vi=-e(V-v) .................. (II.). These two equations, (I.) and (II.) determine Fj and v^. From (II.) we have F 1 = t5 1 -e(F-) ............... (HI). Substituting in (I.) we get (M+ mX = MV+ mv + eM( V- v) ; _ MV + mv + eM( V- v} n v , . Vi -- - - - - =r-= - ............... ( -L V . ) M+m Hence from (III.) MV+ mv - em( V- v) v _ l ~ 196 IMPULSE BETWEEN TWO BALLS. If the balls be inelastic, e is zero, and from equation (II.) we have V 1 equal to v^ or the}' proceed with a common velocity ; in other words, they do not separate. Substi- tuting in equation (I.) we have, in this particular case, v , .. ...(VI). M+m This ot course follows immediately from the general expressions given above for F x and i? l5 by putting e equal to in them. 215. If the balls before impact are moving in opposite directions, we have merely to give opposite signs to V and v. It remains to determine the impulse of the pressure between the two balls. "We have determined the velocity of each after impact on the assumption that the changes of their respective momenta are equal and opposite. To find the impulse, we need only consider the change of momentum of one of the balls. The velocity of the first before impact is F, and after impact its velocity is F,, hence the change of its momen- tum is M(V- Fj), that is '-tQl ) M+m Hence if I denote the impulse between the balls, we have r 71-r/T/ Ml'+mv-em (V-v)\ I Mi V ^ -> ...(VII.). ~M+m> We may notice that if the masses of the balls are equal and the elasticity perfect, or e equal to 1, it follows from equations (IV.) and (V.) that v l V and V 1 = v, or the balls exchange their velocities. If in the equations we make m infinite, we obtain the same result as in Art. 211 ; in fact, we revert to the case of impact against a moving obstacle whose velocity is unchanged by the collision. OBLIQUE IMPACT OF TWO SPHEEES. 197 216. We proceed now to consider the case of the impact of two smooth spheres not moving in the same or in opposite directions. We shall investigate only the case in which the centres of the two spheres are moving in the same plane. The solution of the general case is precisely similar, but the geometry is more difficult. Let the two spheres be called A and B respectively, M denoting the mass of A, and m that of B. Let KG be the straight line drawn through 0, 0' the centres of the spheres at the moment of impact. Let DO -H\ be the direction in which the sphere A is moving before impact, and V its velocity, EO' the direction of motion of B before impact, and v its velocity. Let F,, v l be their respective velocities after impact, and OH, O'L the direc- tions in which they are respectively moving. Let the angles DOK, EOK, HOG, LO'G be denoted by a, ft, 6, respectively. Let e be the coefficient of mutual elasticity. The velocity of A before impact may be resolved into two components, V cos a along 00' and V sin a perpen- dicular to 0(7, while that of B may be resolved into v cos ft and v sin ft in the same directions respectively. Now, since the spheres are smooth, the whole action between them is in the line 00. Hence the components of their velocities perpendicular to this line remain un- affected. We have therefore V sin a = FL sin 0\ ,-, , V sin ft = v< sin d> j > v* 198 OBLIQUE IMPACT OF TWO SPHERES. Since action and reaction are equal and opposite, the impulse upon A is equal and opposite to that upon B ; hence the change of A's momentum must be equal and opposite to that of the momentum of B, and the change in each takes place wholly in the direction of the impulse ; that is, along the line 00'. Hence, equating the momen- tum lost by A to that gained by B, we have M(V cos a FL cos 0) = m (i\ cos v cos /3), or M V 1 cos + mv^ cos $ = JfFcos a + mv cos /3 (H.). Also the velocity, in the direction of the impulse, of either sphere relative to the other after impact is to that before impact as e to 1. Hence V 1 cos 6 v 1 cos = e(Fcos a v cos /3) (III.). The four equations (I.), (II.), and (IH.) completely deter- mine F 17 1? 6 and in terms of the given quantities. From equations (II.) and (HI.) we obtain, precisely as in the preceding Article, v /, MV cos a + mv cos ft-em( V cos a -v cos /8) r 1 COS = v sin /3, hence the components of the velocities of A and B along and perpendicular to 00' are known, and the values of Fj and ^ can be at once written down. Again, from (I.) and (IV.), tan = Vain a (M+m) _ (VI>) VM cos a + mv cos @ em (F cos a v cos fi) tan = ^ sin ^ (M+m) ^^ MV cos a + wit- cos ft + eM (I cos a - u cos yS) Hence the direction of motion of each sphere, after impact is found, ACTION BETWEEN ELASTIC BALLS. 199 To find the impulse of the pressure between the balls, we observe that the momentum of A in the direction 00' before the impact is MV cos a, and this is changed by the pressure into MV^ cos 6. The measure of the impulse is therefore M (Fcos a - V l cos ff) ; or, if this impulse be denoted by /, we have T MJV _ -^^"cos a + mv cos/3 - em(Fcos a v cos/3)) \ M+m ) ( Vcos a ~ v cos /3) (1 + e) . . . . .(VHI.). M+m 217. In all the cases we have investigated the expres- sion for the impulse involves the factor 1 + e, and is there- fore greater in the ratio of 1 + e to 1 than it would have been had e been zero, but all other circumstances the same. Thus if, in any case of collision, I' measure the impulse when the coefficient of elasticity is zero, and I be the measure of the impulse when, all other things being the same, the coefficient of elasticity is e, we have Now, if we examine somewhat more closely into what takes place when two bodies strike one another, we find that, in the first place, each of them becomes compressed or indented ; but if they are elastic, they subsequently recover more or less completely their original form. If they are inelastic, they remain indented, and move on together with a common velocity. Now, up to the instant of greatest compression the action is the same whether the balls are elastic or inelastic, and therefore, at this instant, even though the balls be elastic, they will be moving with a common velocity, and the change of momentum produced in either ball up to this instant will be the same as though they were inelastic. This change of momentum is sometimes improperly called the " force of compression" (See Art. 239.) We have denoted it by /'. Now, in the case of elastic balls, after the compression, or indentation, has attained its maximum, the balls begin 200 FORCES OF COMPRESSION AND OF RESTITUTION. to recover their form. The parts which have been com- pressed consequently swell out against one another, and the force which they exert on one another serves to separate the balls. The impulse of this force is some- times improperly called the " force of restitution" The time taken by the balls to recover their form, and there- fore the time during which this force acts, is so short, that we are unable to measure it, and we are consequently compelled, in respect of the force of restitution, like that of compression, to consider the whole momentum gene- rated ; in other words, to consider the impulse. Let this impulse be denoted by /". Then the whole change of momentum produced in either ball during the impact is that due to the force of compression, together with that due to the force of restitution, and is therefore numerically equal to /' + /". Hence /= r + r. But /= T (1 + e) ; therefore . T' = eT, or the "force of restitution " is equal to e times the " force of compression." If e be equal to unity, or the elasticity be perfect, the " forces of restitution and compression " are equal. 218. In some treatises the coefficient of elasticity is denned as the ratio of the " force of restitution " to that of compression, and it is stated as the result of experi- ment that this is constant for the same materials. It should, however, be borne in mind that the element observed in experiments on this subject is not the forces which act during the collision, but the velocities of the balls before and after impact, and the measures of the forces of compression and restitution are subsequently deduced from the results of these observations by the Second Law of Motion. It would therefore seem that the method adopted in the preceding Articles is the more natural way of treating the subject. The coefficient of elasticity is sometimes called " the coefficient of restitution" 219. If when the two bodies impinge upon one another EXAMPLES ON COLLISION. 201 they be acted upon by some " finite " force, as, for exam- ple, gravity, then, since the time during which they remain in contact is so short that we cannot measure it, the eifect produced by the finite force in that time will also be immeasurably small. "We may therefore neglect it altogether while considering what happens during the collision. The subsequent motion of the bodies will, however, of course depend upon the forces which act upon them. 220. As illustrations of the preceding Articles, we will consider a few examples. Ex. 1. A ball of 8 pounds, moving with a velocity of 12 feet per second, strikes directly a ball of 12 pounds, moving with a velocity of 8 feet per second in the opposite direction, the coefficient of elasticity being ^. Find the i velocity of each after impact, and the impulse of the pressure bettceen the two. Let V denote the velocity of the first ball in feet per second, v that of the second, after impact. We will con- sider velocity positive when in the direction in which the first ball is moving before impact. Now, since the im- pulses on the two balls are equal and opposite, the momentum gained by the one is equal to that lost by the other ; hence the sum of the momenta of the two is the same after impact as before. Now, one pound being taken as the unit of mass, and a velocity of a foot per second as the unit of velocity, the momentum of the first ball before impact will be represented by 8 x 12, and that of the second by - 12 x 8. Therefore 8. F+12y = (8xl2)-(12x8) = ............ (I.). Again, the velocity before impact of the first ball rela- tive to the second is 12 + 8 feet per second, while after impact it is V v. Therefore, since the coefficient of elasticity is ^, we have F-t>=-i(12 (II.), 202 EXAMPLES ON COLLISION. The equations (I.) and (II.) determine V and v. From (I.) v=-*V. 3 Therefore, from (II.), JJF=-1Q, or V= -6. Hence v = 4, and thus the velocities after impact are found. In this particular case we see that the velocity of each is re- versed in direction and its measure reduced by one half. To find the impulse, we observe that the momentum of the first ball before impact was represented by 8 x 12, while its momentum after impact is 8 x ( 6). Hence the change of momentum produced by the impact is represented by 8 x 18 or 144 ; that is, the impulse is equivalent to a velocity of 144 feet per second generated in one pound of matter. The same results might of course be obtained from the formulae of Art. 214. 221. Ex. 2. A ball falls from rest at a height of 20 feet above a fixed horizontal plane. Find the height to which 3 it will rebound, the coefficient of elasticity being ^ and the value of g being 32 foot-second units. Let the velocity of the ball, when it strikes the plane, be denoted by v. Then, since it has fallen from rest through 20 feet under an acceleration denoted by 32, the equation v 2 = 2gs becomes in this case tf 2 = 64x20 = 1280. EXAMPLES ON COLLISION. 203 If the upward velocity of the ball after impact be v feet per second, we have v' = ev, and therefore = 720. Also, if h be the height to which it rises in the rebound, v' 2 h = , , 720 Or h = -^-j- : or the ball rises to a height of 11^ feet. 222. Ex. 3. A particle is projected from a point dis- tant 20 feet from a rough vertical wall, with a velocity of 60 feet per second, and at an elevation of 60, its plane of motion being perpendicular to that of the wall. Find its velocity and the direction of its motion immediately after striking the wall, the coefficient of elasticity being - , and the coefficient of friction ^. The initial velocity of the particle may be resolved into a velocity of 30 feet per second in a horizontal direc- A/3 tion, and 60 ~-or 30 v/3feet per second vertically upwards. Now its horizontal velocity remains constant until it strikes the wall. Hence the time elapsing before it strikes the wall, that is, while it moves over a horizontal distance of 20 feet, is two-thirds of a second. Its vertical velocity 2 when it strikes the wall is 30 -/3 - ^ g feet per second. 3 Now, since the coefficient of elasticity is ^, its horizontal velocity after impact is 18 feet per second away from the wall. Hence the whole change of its horizontal velocity is represented by 48 feet per second, and if m denote the mass of the particle, the impulse of the pressure on the 204 EXAMPLES ON COLLISION. wall will be represented by ra.48. Now, when the particle strikes the waD, the component of its velocity parallel to the wall is 30 Hence the vertical component of the V COS C7 ball's velocity when it reaches B is * a v sin 0. But v cos v since the coefficient of elasticity is , its vertical velocity on leaving B is numerically equal to one-half of this. Hence we must have / ____ t? sin 6\ = v sin 0, 2\v cos ) or 2r, 3v sin 9 ; v cos 6 :. v* sin 20 = ?# ........................ (II.). o This equation, together with equation (I.), determines v and 6. From (I.) and (II.) we have :. sin 20= 2 cos 2 0; ;. sin 6 = cos 6 ; :. = 45 Hence sin 20 = 1, 2 end we have v 2 = EXAMPLES ON COLLISION. 207 If we take g equal to 32, we get 8 or the velocity of projection must be -.- - feet per second, v o and the elevation of projection we have seen to be 45. 224. Ex. B. An engine whose mass is 40 tons, and 20 coal-trucks, each of 15 tons, are at rest on a horizontal line, there being an interval of one foot between the engine and the first truck, and between each truck and the next succeeding. The engine starts off and strikes the first truck, which then strikes the second, and so on down the train, the trucks being each inelastic. Supposing the engine to be constantly impelled by a force equal to the weight of one ton, find the velocity with which the last truck starts and the whole time occupied in starting the train, neglecting friction, and taking g equal to 32. Let v be the velocity of the engine when it strikes the first truck. Then, since the mass of the engine is 40 tons, and it is acted upon by a force equal to the weight of 1 ton, it moves over the one foot between its initial position and the first truck with a constant acceleration denoted by JL Hence we have by Art. 1B2 Now, when it strikes the first truck, the two will pro- ceed with a common velocity. Let v t denote this velocity. Then, since the momentum of the engine and truck im- mediately after impact is equal to that of the engine before impact, we have (40 + 15X = 400', or v * l 55 2 ' 208 EXAMPLES ON COLLISION. "We have now a mass of 55 tons impelled by a constant force equal to the weight of one ton, and therefore moving with constant acceleration /=,. Hence if v/ denote its velo- 55 city when it strikes the second truck, since the space over which it has passed under this acceleration is one foot, we have Therefore <8 = 2g.4g_+ 2 g . 66. Let t> 2 denote the common velocity immediately after striking the second truck ; then, since the whole momentum is unaltered, (66 + 16)%= 55V. Therefore _/55\ 2 20. (40 + 55) '(TO)- = 20. 55 2 40 + 55 70 2 We have now a mass of 70 tons moving under a con- stant force equal to the weight of one ton. Hence if v% denote its velocity when the third truck is struck, 70' 70 2 When the third truck is struck, the mass in motion is changed from 70 tons to 85 tons. Hence if v 3 denote the common velocity immediately after striking, 70 , S = QK*>8- EXAMPLES ON COLLISION. 209 , 2 /70\ s g 40 + 55 + 70 t:i ~ ~ \85/ ' 3 ' 70 2 = 2 40 + 55 + 70 g ' 85 2 Proceeding in this way, we see that the common velo- city, ; 19 , immediately after starting the 19th truck, is given by the equation 40 + 55 + 70+. ..+310 and if v 19 ' be the velocity of the rest of the train at the instant when the last truck is struck, ti* & _ /\* * - - / 19 ^19 -325, 40 + 55 + 70+. ..+310 + 325 or = 2f. - But if v 20 denote the common velocity immediately after starting the last truck, . 40 + 55 + 70+. ..+325 f\Y* ^ " " * * "* = 64. 325 2 40 + 55 + 70+. ..+325 340 2 " 3650 ' 340* 3650 340 2 ' _ 8. ^3650 Hence the last truck starts with a velocity of 1'421 feet per second, very nearly. G. D. p 210 EXAMPLES ON COLLISION. Since the force producing motion is always equal to the weight of a ton, if we take a ton for the unit of mass, the momentum generated in t seconds will be repre- sented by gt. Now since the momentum of the train is unaltered by the successive impacts, the momentum of the train when the last truck starts must be that produced by a force equal to the weight of a ton in the time t , if t be the time which has elapsed since the engine started. Hence since the mass of the train is 340 tons, 340^80 9 or t = __ 8^3650 32 = 15-103..., or the time required to start the train is rather more than 15' 1 seconds. 225. Suppose n equal particles, each of mass TW, and moving with a velocity v, to impinge directly upon a fixed inelastic plane surface during t seconds. The surface will in the course of the t seconds receive n impulses, each re- presented by mv, and the whole momentum destroyed by the surface in the t seconds is represented by nmv. Now suppose the intervals between successive impacts to be equal to one another, and the velocity of each particle remaining the same, suppose n to increase while m dimi- nishes in the inverse ratio, so that mn remains constant and equal to M say. Then the sum of the impulses upon the surface during t seconds is the same as before, and would generate a momentum represented by Mv, and this is true, however great n may be. But if n become indefinitely great, we can no longer distinguish any inter- val between successive impulsive pressures ; in fact the action upon the surface becomes a continuous and uni- form pressure, and the momentum destroyed in t seconds by the reaction of the surface, which is always equal to this pressure, is Mv. Hence the momentum which would be destroyed in one second by this pressure is equal to CONTINUOUS IMPACT. 211 , which is therefore the measure of the pressure. The reaction of the surface is, in fact, measured at any time by the rate at which momentum is being destroyed by it. 226. The pressure of the wind, or of a jet of water upon any object which it strikes, may be taken as an illustration of continuous pressure produced by a quick succession of impacts. The air, or water, which comes in contact with the obstacle in the course of a second consists of a very great number of particles, each of which strikes the obstacle with a certain velocity, and the im- pacts are in such quick succession that they link them- selves together into a continuous pressure. 227. In the case of a jet of water striking a wall, if the anea of the section of the jet remains always the same, the amount of matter which strikes the wall in a second will be proportional to the velocity ; and since the momentum of each particle of water is proportional to its velocity, it follows that the change of momentum produced by the reaction of the wall in each second is proportional to the square of the velocity of the water, supposing this con- stant, and hence the pressure on the wall is proportional to the square of the velocity of the jet. 228. For example, suppose a jet of water, the area of whose transverse section is one square inch, to impinge directly upon a wall with a velocity of 128 feet per second, the coefficient of elasticity being ^. We proceed to find the pressure on the wall. Since the area of the section of the jet is one square inch and the velocity of the water 128 feet per second, the volume of water which strikes the wall in one second is 128 x 12 cubic inches, and since one cubic foot of water 128 x 12 contains 1000 ozs., the mass of this is . 1000 ozs. or 500 - pounds. Now the velocity of this before impact is y 128 feet per second towards the wall, and after impact it 212 CONTINUOUS IMPACT. -i 90 is - feet per second away from the wall, since the co- efficient of elasticity is ^. Hence the change of velocity produced by the impact is 136 feet per second. The change of momentum produced by the reaction of the 1 QA P\OO wall in one second is therefore - - units of momen- y turn. But the weight of one pound generates in the mass of one pound in one second a velocity of 32 feet per second, or the weight of one pound generates in one second 32 units of momentum. The reaction of the wall is therefore equivalent to the weight of - "-^on pounds, y x o& and since action and reaction are equal and opposite, it follows that the pressure of the jet upon the wall is equal to the weight of - pounds, that is, of 236^ Ibs. y x o& This result affords an explanation of the great mechanical effect produced by the jet from a fire-engine. The result that the pressure varies as the square of the velocity is the basis of the ordinary theory of fluid resistances. 229. A perfectly flexible uniform chain, the mass of each unit of length of which is m, hangs vertically from its upper extremity with the lower end just in contact with an inelastic table. If the chain be allowed to fall, find the pressure upon the table at any instant during the motion. If a series of particles are arranged in a vertical line, and then all allowed to fall freely at the same instant, their velocities at any subsequent time will be the same, and the distance through which they will have fallen will be the same for each. Hence they will always remain in a vertical straight line at the same distance apart as when they started, although they are perfectly free. Conse- quently if they be so connected together that they are unable to alter their distances from each other, there will be no stress on the connections and the motion will be unaffected thereby. Hence, in the case of the falling chain, each particle of the chain will fall as if it were FALLING CHAIN. 213 free. Therefore at the end of t seconds, after the motion has commenced, the velocity of each particle of the chain will be gt, and the space through which the upper part will have fallen will be ^ gt 2 . There will therefore be a a length of chain measured by ^ gt 2 coiled up upon the a table. Now if the velocity of the chain were to remain constant during one second, and the same as at the time tj the length of chain which would be brought to rest upon the table during that second would be gt, and its mass mgt. Also, its velocity being gt, the momentum which would be destroyed during the second would be mg~t 2 . Hence mg~t 2 is the expression for the rate at which momentum is being destroyed at the end of t seconds after the commencement of the motion. But the rate of change of momentum is the measure of the force pro- ducing that change. Hence at the end of the time t the reaction of the table required to destroy the momentum of the falling chain is numerically equal to mg 2 t 2 , and therefore the pressure exerted by the falling portion of the chain in coming to rest upon the table is denoted by mg 2 t 2 . But at the end of the time t there is a length of chain, denoted by ^ gt 2 , coiled up on the table, and the a weight of this is 3 Hence the whole pressure upon the table is ^ m . g 2 t 2 , that a is, three times the weight of the portion of chain coiled up. If the lower end of the chain had been at a height h above the table, the length of chain coiled up at the end of t seconds would have been ^gt 2 -h, provided ^ gt 2 were greater than h, and the whole pressure upon the table would have been 3 Q mg 2 t 2 mgh. 214 FALLING CHAIN. 230. A perfectly flexible uniform chain, the mass of each unit of length of ichich is m, is coiled up in the hand, and one end is attached to a fixed point. Suddenly the hand is removed ; it is required to find the force upon the fixed point at any instant before the ichole of the chain has come to rest. Each particle of the chain which remains in the coil will, at any subsequent time, be falling freely, but succes- sive portions of the coil will be brought to rest, and hang vertically from the fixed point. The velocity of every particle in the coil at the end of the time t after the com- mencement of the motion will be gt, and the space through which the coil will have fallen will be denoted by ^ gt 2 . 2i Hence the weight of chain hanging from the fixed point will be ^ mg 2 t 2 . Also, as in the preceding example, the 4 rate of destruction of momentum will be mg 2 t 2 . Hence 3 the whole force upon the fixed point will be ^mg 2 t 2 , or a three times the weight of chain hanging vertically from the point. The tension at any point of the chain distant h below the highest point will be less than the tension at the highest point by the weight of the length h of the chain, 3 and will therefore be denoted by - mg*& mgh. a 231. We will give two other examples of falling chains. A flexible chain is suspended from a fixed point, and hangs vertically with its lower end just touching an inelastic horizontal table ; it is then allowed to fall. Supposing the density at any point of the chain to be proportional to the distance from the lower end, find the pressure on the table at any subsequent time. As in the preceding example, the chain will fall freely, and its velocity at the end of t seconds from the com- mencement of the motion will be gt, while the length of chain coiled upon the table will be ^ gt 2 . FALLING CHAIN OF VAEIABLE DENSITY. 215 Let the density of the chain at a point distant . = v sin /3 j" Also a _ M Fcos a + mv cos /3 em(Vcos a v cos , ^-^ .MFcos a + mv cos 8 + eM(Vcos a vcos ^ ' v< cos 6 = - , f , M+m Now the kinetic energy of the two balls before impact is ), that is a o { MV 2 (cos 3 a + sin 2 a) + mv 2 (cos 2 /3 + sin 2 /9) } . Let us denote this quantity by E. Then if E^ denote the kinetic energy after the impact #1=5 { M F^cos^ + sin 2 0) + mv*(ca&4> + sin 3 ^) } . Now from equations (I.) 2 (MVf sin 2 + raVsin 2 <) - g (MV*sw* a + mv 2 sin 2 yS) . . . (III.) KINETIC ENEEGY AFTER IMPACT. 221 and, precisely as in the preceding article, we may show from equations (II.) that 2 ( 1 (1 e~)Mm( Fcosa cos/3) 2 x-ry v since we have only to write Fcos a, v cos /3, Fjcostf, v^ cos for F, v, F x and ^ respectively in the equations of that article. Hence, adding equations (III.) and (IV.), we get o" - ' s As before, we see that if the elasticity be perfect the kinetic energy is the same after impact as before, but if it be imperfect there is a loss of energy by the impact. 235. Suppose we have a cylinder of some compres- sible material, say one foot in length, and suppose that when a pressure equal to the weight of 100 Ibs. is applied at the ends the length is diminished by '01 inch. Then, provided we keep within the limits of elasticity, a pressure equal to the weight of 200 Ibs. would shorten the cylinder by nearly *02 inch, and so on in proportion. The weight of one pound would dimmish the length of the cylinder by '0001 inch, or by T- onnn Aof the original length. If the cross section of the cylinder be equal to the unit of area the elasticity of the material as measured by the quotient - r- becomes 120000, the weight of a strain pound being taken as unit of force. 236. Now suppose that the cylinder has been shortened by the application of pressure at the ends, and let the pressure now be gradually diminished. Suppose that as soon as the pressure is diminished ever so little the length of the cylinder begins to increase, and increases in such a way, as the pressure continues to be diminished, that the length under any given pressure is exactly the same as 222 KINETIC ENEEGY AFTER IMPACT. under the same pressure when the cylinder was being com- pressed, and that this continues until the cylinder has regained its original length. In this case just as much work will be done by the elastic force during the restitu- tion of form of the cylinder as was done upon the cylinder during the compression. Such a cylinder might be alter- nately compressed and allowed to expand without any loss of mechanical energy, and without any heat being produced. Its elasticity or power of regaining its original form may be considered as perfect. 237. On the other hand, it may happen that when the body has been compressed, though a very great pressure may have been required to produce the deformation, yet the whole of this pressure may be removed without the body showing any tendency to return to its original form. In such a body the elasticity, or power of restitution of form, may be considered as zero, and nearly the whole of the work done upon the body during its compression has its equivalent generally in heat produced within the body. 238. Between these two extremes we may have any number of intermediate links. Suppose, for example, that in the case assumed above the cylinder has been loaded with 1000 Ibs. and therefore shortened by *1 inch. Now suppose that the pressure is diminished, and that the cylinder retains its length of 11 '9 inches until the pressure has been reduced to 500 Ibs.' weight, and that it only begins to return when the pressure is reduced below this. Suppose also that throughout the restoration of form the pressure corresponding to any length of the cylinder is exactly half what it was during the compression, so that when the length of the cylinder is 11/95 in. the pressure is equal to the weight of 250 Ibs., and when the length is 11'99 in. the pressure is equal to the weight of 50 Ibs., and so on. Then exactly one-half of the amount of work done in compressing the cylinder will be restored by the elastic forces during the restoration of form, the other half being mostly converted into heat within the imperfectly elastic material. Generally, if the pressure at any stage during the restoration of form be n times that at the correspond- KINETIC ENERGY AFTER IMPACT. 223 ing stage of the compression, the work done by the elastic forces during the restoration will be n times that done against them during the compression, while the energy permanently transformed into heat or otherwise wasted (as far as mechanics is concerned) will be 1 n times the work done in compressing the body. 239. Now suppose that two bodies are compressed by impinging against each other, and suppose, for the sake of simplicity, that they are both of the same material, whose elasticity is such that the pressure exerted at any stage during the restoration of form is n times that re- quired to produce further compression at the same stage, and that the restoration of form is complete. Now let E denote the number of units of work done in compressing the bodies up to the condition of maximum compression, that is, up to the instant when the relative motion of the centres of gravity 0f the bodies is zero. Then up to this instant E units of kinetic energy will have been lost by the moving bodies, having been expended in doing work in producing compression. During the restoration of form nE units of work will be done by the elastic forces and expended in producing motion in the bodies, so that nE units of kinetic energy will be returned to the system ; the remaining (1 n}E units of energy being converted into heat or otherwise disposed of. Comparing these ex- pressions with those obtained in Art. 233, for the kinetic energy lost by impact, we see that n must be equal to e 2 . Hence e, the coefficient of elasticity, must be equal to the square root of the ratio of the pressures between the balls at the same stage of the restitution and compression re- spectively. Hence e is equal to the square root of the ratio of the true force of restitution to the true force of com- pression in any, the same, configuration. To use the terms " force of restitution " and " force of compression " in the sense explained in Art. 217 is consequently not only to use words in their wrong senses, but to convey a false impression of what the relation between the forces of restitution and compression really is. 240. The coefficient of elasticity, e, is the ratio be- tween the impulses of the forces of restitution and com- 224 ENERGY DISSIPATED BY IMPULSIVE ACTIONS. pression. and differs from the ratio of the forces, since the time occupied by the compression is only e times the time occupied by the restitution. The basis of this last state- ment we will now consider. From the instant of greatest compression up to that of complete restoration of form the centres of gravity of the bodies move relatively to one another over exactly the same distance as from the commencement to the end of the compression. Also at any stage in the restoration the pressure between the balls is n times as great as at the same stage (i.e., the same configuration) during the compression. Hence the accele- ration is n times as great as the retardation during the compression. Now the time taken to travel over any distance from rest with uniform acceleration is inversely proportional to the square root of the acceleration (for s^^ft 2 ), and it may be easily shown geometrically that if two points travel over the same distance from rest with accelerations which vary in any manner, but which bear a constant ratio, n, to one another, when the points are in the same positions, then the times taken by two points to travel the same distance will be to one another in the ratio of 1 to */n. Also the velocities at the end of the times will be in the ratio of ^/n to 1, for though the first acceleration is n times that of the second, the time during which the velocity of the first point is being increased is j- times that during which the velocity of the second is being increased, so that the final velocities generated are in the ratio of , given by the equation 1 + e e being the coefficient of elasticity of the balls. 36. A heavy ball is thrown horizontally from A so as to hit a point B after one rebound from a horizontal plane C. Supposing e to be the coefficient of elasticity and the height of B from the plane to be e 2 times that of A, the height of A being such that a body would drop from it to the plane in 1 second, show that the point C where the ball must hit the plane divides the horizontal distance between A and B into two parts which are as 1 : e. 37. A ball is projected from a point A at an elevation of 45 against a vertical wall BC, and in a vertical plane perpendicular to the wall ; after impact at C it strikes the ground between A and B, and arrives at A after n rebounds. Find the ratio of BC to AB in terms of the coefficient of elasticity of the ball. 38. A parabola is placed with its axis vertical and vertex downwards. A perfectly elastic ball dropped ver- tically strikes the parabola with the velocity acquired in falling freely from rest through a space equal to one-fourth of the latus rectum ; find where it must strike the parabola that after reflection it may pass through the vertex. 39. A particle is dropped from a point in a fixed cir- cular hoop whose plane is vertical, the elasticity being perfect. Find the condition that after two rebounds it may rise vertically and determine in what ways this may happen. 40. A circular arc has its plane vertical. A perfectly elastic ball is projected from the arc along a horizontal diameter, and after one rebound at the arc returns to the point of projection. Show that the latera recta of the two parabolas described are as 4 to 1, and determine the velocity of projection. 236 EXAMPLES. 41. A number of balls whose elasticity is ^ ( x/2 - 1) are let fall on an inclined plane, and each strikes it the second time twice as far down as it did the first time. Show that the points from which they fall lie in a plane perpendicular to the inclined plane, and intersecting it in a horizontal straight line. 42. A smooth sphere stands on a horizontal plane to which it is fixed, and from its highest point a perfectly elastic ball is projected in a direction inclined 45 to the vertical. Find the velocity of projection in order that the ball may strike the sphere once only, at an angular dis- tance of 45 from the vertex, and prove that in that case the ball will strike the plane at a distance from the point of contact of the sphere equal to its diameter. 43. An elastic ball is projected from a point in a smooth inclined plane in the vertical plane containing the line of greatest slope on the plane. Find the condition that after three reflections it may return to the point of projection. 44. Each of two planes is inclined 45 3 to the horizon, and they intersect in a horizontal straight line ; from any point in one of them it is possible to project a perfectly elastic ball in a plane perpendicular to the intersection of the planes, so as to return to the point of projection if the velocity of projection be not less than that acquired in sliding from the point of projection to the intersection of the planes. 45. Two vertical walls are inclined to one another at an acute angle a. A perfectly elastic ball projected hori- zontally from a point distant c from the ground and b from the intersection of the walls, comes to the ground, after striking both of them, at the same point as if it had fallen from rest. Find the direction of projection, and show that the space through which the ball would rise if projected vertically with the velocity of projection is equal to (6 sin are the inclinations to the horizon of the directions of motion of the particles at any, the same, moment pre- vious to their impact. 65. An elastic sphere is at rest on a plane. The plane and sphere are simultaneously hit by another smooth sphere, whose coefficients of elasticity with the first sphere and plane are the same and equal to e. Determine e when the directions of motion of the second sphere, after and before impact, are equally inclined to the plane. 66. A very small elastic ball is projected with a given velocity from one extremity of a diameter of a horizontal circular hoop, which rests on a smooth horizontal table, and after reflection at the curve passes through the other extremity of the diameter. Find the coefficient of elas- ticity in order that the whole time occupied in the motion may be n times that of describing the diameter with the initial velocity ; and the greatest and least values n can have. 67. A ball of given mass lies touching a smooth wall. Another moving at right angles to the wall impinges on it obliquely. The balls being inelastic, find their velocities immediately after impact. 68. An imperfectly elastic ball is projected from a given point in a horizontal plane against a smooth vertical wall EXAMPLES. 241 in a direction making a given angle with the vertical : find where it strikes the horizontal plane, and prove that the locus of these points for different vertical planes of projec- tion is an ellipse. 69. A heavy chain hangs vertically from its upper end with its lower end just in contact with a smooth plane in- clined at an angle a to the horizon : if the string be allowed to fall, find the pressure on the plane at any time. 70. Two smooth equal balls are placed in contact on a smooth table ; a third equal ball strikes them simultane- ously and remains at rest after the impact ; show that the coefficient of restitution is -. o 71. A uniform flexible chain of indefinite length, the mass of the unit of length of which is m, lies coiled on the ground while another portion of the same chain forms a coil on a platform at a height h above the ground, the in- termediate portion passing round the barrel of a windlass placed above the second coil. An engine which can do H units of work per the unit of time is employed to wind up the chain from the ground and to let it fall into the upper coil. Show that the velocity of the chain can never exceed the value of v determined from the equation mghv + mv 3 = H. 72. A, B, C are three perfectly elastic balls of equal mass lying on a horizontal plane. If A and B are con- nected by a tight inelastic string, and C is projected so as to strike A directly with velocity F, prove that C will rebound with velocity y COS 2 9 3 + sin 2 ~6>' where 6 is equal to the angle BAG and is less than a right angle. 73. A bullet is fired in the direction towards a second equal bullet, which is let fall at the same instant. Prove that the two bullets will meet, and that if they coalesce the latus rectum of their joint path will be one quarter of the latus rectum of the original path of the first bullet. G, D. E 242 EXAMPLES. 74. A bucket and a counterpoise, connected by an in- elastic string passing over a pulley, just balance one an- other, and an elastic ball is dropped into the centre of the bucket from a distance h above it ; find the time that elapses before the ball ceases to rebound, and prove that the whole descent of the bucket during this interval is A /i i j 7a f> _,. -; - r , where m. M are the masses of the ball and 2M +m(l- ey bucket and e is the coefficient of restitution. 75. Two equal particles A, B, of imperfect elasticity e, move with equal uniform velocity in the same straight line. B impinges perpendicularly on a wall. Show that there will always be two impacts between A and B, and two between B and the wall, and that if there is a third collision between the balls, 76. A particle is projected from a point in a smooth plane inclined at an angle a to the horizon, in a vertical plane which cuts the inclined plane in a horizontal line, and at an angle 6 to the horizon. Prove that after n rebounds the space traversed in the direction of the line of greatest slope on the inclined plane is a eCl-e"- 1 ) a sin a tan . ^ -- ' 1-e where a is the horizontal space described, and e the co- efficient of restitution. 77. A ball having descended to the lowest point of a circle through an arc whose chord is C drives an equal ball up an arc whose chord is c : show that the common elasticity (e) of the two balls is given by the relation 1: e::C:2c-C. 78. An elastic ball being projected at any elevation is continually reflected from a horizontal plane, and the sum of the areas of all the parabolas described : area of the first parabola :; 8 : 7. Find the elasticity of the ball. CHAPTER V. MISCELLANEOUS. 245. A particle falls down a smooth curved tube ; it is required to find its velocity at any point of the tube. Since the tube is smooth, the only force which it exerts upon the particle is at right angles to the direction of the particle's motion. Hence no work is done upon the par- ticle by the action of the tube, and the only force which does work upon the particle is its weight, so that the kinetic energy generated must be the equivalent of the work so done, and therefore depend simply on the vertical height through which the particle has fallen (Art. 171). Its velocity is therefore the same as if it had fallen ver- tically through the same difference of level. Hence, if a particle starting with velocity u move along a smooth tube (or other surface) through a vertical distance h, its velocity at the end of the distance will be i/u z + %gh. The same will be true if the particle be constrained by an inextensible string so that it moves always at right angles to the string. If the particle start from rest at A its velocity at any point, whose vertical depth below A is ft, will be 246. Similarly if a particle be projected up a smooth tube with velocity u, its velocity after rising through a vertical height h will be -v/w* 2gh. If h be the greatest height to which it will rise, we must have u'* 2gh, or h = . Hence the particle will rise up the tube to the same vertical height to which it would rise if it were free and projected vertically upwards with velocity u. 243 244 PARTICLE FALLING DOWN A BENT TUBE. 247. If the tube be bent, as in the subjoined figure, and the particle fall down one arm from a height h above the lowest point, it will rise up the other arm to the same height, for its velocity at the lowest point will be Vtyh, and this will just carry it to a height h up the other arm. The particle will afterwards descend again, and will con- tinue to oscillate to a height h on each side of the lowest point of the tube. 248. If a particle move subject to any constraints whatever, such that the force exerted upon the particle by the constraint is always perpendicular to the direction in which it is moving, no work is done upon or against the particle by the means of constraint. Hence the change in the kinetic energy of the particle produced during its motion from one point to another must be equivalent to the work done upon it by the forces to which it is subject, the action of the constraints being left out of consideration. Hence, if a particle be moving under the action of gravity but constrained by any smooth surfaces, inextensible strings, or system of Motionless link-work, the change in its kinetic energy will depend only on the vertical distance through which it has risen or fallen. We have already, under the head of projectiles, con- sidered a case in which the force acting upon a moveable particle is inclined to the direction of motion, and we found that in this case the kinetic energy of the moving particle depended only on its vertical distance below the directrix of the parabola which it described. The motion of the earth about the sun affords another example of the PARTICLE FALLING DOWN A SMOOTH TUBE. 245 motion of a body under the action of a conservative force but subject to no constraints, and the velocity of the earth in its orbit depends only on its distance from the sun. 249. As an illustration of the preceding articles we will take the following example. A number of heavy particles slide from rest at the vertex down a smooth tube in the form of a parabola whose axis is vertical, and are allowed to quit the tube at different points. Find the locus of the foci of the trajectories subsequently described by them. Let S be the focus of the parabolic tube, ZX its directrix, AL the tangent at the vertex. Suppose a particle to quit the tube at any point P. Draw PKH parallel to the axis. Let PT be the tangent at P to the curve of the tube. Then, since the particle starts from rest at .4, its velocity at P is that due to falling freely from the point K. AK must therefore be the directrix of the para- bola subsequently described by the particle (see Art. 187), and since PT is the direction of the particle's motion at P, it is a tangent to this parabola. Now, since PK is the perpendicular on the directrix and PT the tangent at P, if F be the focus of the parabola, the angle FPT is equal to TPK. Hence F lies in SP. Also FP is equal to PK 246 NEWTON'S EXPERIMENT ON IMPACT. and SP to PH. Hence 8F is equal to HK, that is, to SA. The locus of the foci is therefore a circle whose centre is S and which passes through A. Since SF is equal to SA, S is a point on the trajectory described by the particle which leaves the tube at P. Hence S is a point on the parabolas described by each of the particles after leaving the tube. These parabolas are in fact the same as would be described by a series of particles projected in different directions from $, each with the velocity which it would acquire in falling freely from A to S, and the curve of the tube is the envelope of all these parabolas. (See Art. 198.) 250. In Art. 245, we have supposed the particle to be constrained to move in a smooth tube, but the proof there given will be equally true (Art. 248) if the constraint be produced by any other means, provided it exert no force upon the particle in the direction of its motion or in the opposite direction. For example, if a particle slide down a smooth surface of any form whatever, or if it be fastened to one end of a string of constant length (and whose mass may be neglected), the other end of the string being attached to a fixed point. In all these cases the change of the velocity of the particle will depend only upon the vertical height through which it has fallen. 251. We are now in a position to understand the method by which Newton arrived at the law of impact, enunciated in Art. 207, and the determination of the coefficient of elasticity for different substances from the results of experiment. A and B are two spherical balls suspended by strings from fixed points so that the centres of both are free to move in the same vertical plane. Let this plane be that of the paper. Now, if the diameters of the balls are small compared with the length of the strings, we may, without introducing any considerable error, suppose them to move as particles situated at their centres of gravity. The length of each string and its point of suspension are care- fully adjusted, so that when at rest the balls may be just in contact and the line joining their centres may be NEWTON'S EXPEEIMENT ON IMPACT. 247 horizontal and in the vertical plane in which they move. Let a, b denote their centres in this position. Let the centre of the ball B be raised to ', its strings being kept tight, and then let B be allowed to fall from rest. Let the vertical height of B' above the line ab be h feet. Then the velocity of the centre of the ball when it strikes A being denoted by n we have and this is the velocity of B relative to A just before impact. Also, since at the instant of impact B is at its lowest position, its centre will be moving horizontally, that is, along ba. After impact, the centre of the ball A will move along the arc of a vertical circle. Let A' be the extreme point which it reaches, and let the vertical height of A above the line ab be denoted by k. Then, since the ball comes to rest at A, the velocity v with which it left a must be given by the equation Again, after impact, the ball B will, on leaving A, either 248 COEFFICIENT OF ELASTICITY. return towards H or continue moving along its circle in the same direction as before, but with a diminished velocity ; or it may, as a particular case, come to rest at once. Suppose it to return towards R and its centre to rise to a vertical height h' above the line ab. Let v' be its velocity immediately after impact with A. Then, since the greatest height to which it rises above ab is /*', we must have or v'* Hence the velocity of the centre of B relative to that of A immediately after impact, that is, (v + #'), is numeri- cally equal to - v%g ( V~k + vh'). Now Newton found by measuring the heights 7z, Ik and h' that so long as the materials of which the balls were composed were the same the ratio / + vh' Vh was always constant, whatever were the relative dimen- sions of the balls or the height h to which B was raised. But this ratio, with a negative sign prefixed, is the ratio of the velocity of B relative to A after and before impact. Hence this latter ratio is constant. "We have called the ratio '. the coefficient of elasticity, and have V h denoted it by e. It is always less than 1. 252. We have said that after impact upon A the ball B may return towards B, may come to rest, or may go on in the same direction as before, but with diminished velocity. Its behaviour in this respect will be determined by the value of e, and the ratio of the masses of the balls. If it come at once to rest, h' is zero and e becomes ~- t If V/i it proceed in the same direction as before, but with velocity ?", then its velocity relative to A immediately after impact is (v /') ; and if h" be the height to which it rises, we have for the ratio of the relative velocities, after and THE CYCLOID. 249 , and this expres- The coefficient of _ < before impact, the expression ^ ' - \/ fi sion will be found to be constant. elasticity will in this case be _*_ If the balls be inelastic they will proceed after impact with the same velocity, and will therefore rise to the same /-i /~h" height. Hence 7c = h" and v /=-- becomes zero, as of v h course it should. No known bodies are, however, per- fectly inelastic. 253. "We propose now to investigate the motion of a particle constrained to move under the action of gravity upon a smooth cycloid whose axis is vertical and vertex downwards. Before doing this we must examine some of the properties of the cycloid. The proofs given in the following articles are due to Dr. "W. H. Besant, of St. John's College. DEF. A cycloid is the curve generated by a point in the circumference of a circle, while the circle roll* (without dipping) along a straight line. Suppose PQK to be the circle, Q the point fixed in its circumference, and the circle to roll along the under side of the straight line AB, starting from the position in which Q is in contact with the line at B. Then Q will generate the cycloid BOA. If be the position of Q when the diameter through Q is perpendicular to AB, then is called the vertex of the cycloid, and OZ>, the diameter of the generating circle which passes through 0, 250 TIME DOWN THE AKC OF A CYCLOID. is called its axis. The points A and B are the cusps of the curve, and the line AB its base. If the diameter of the generating circle be denoted by , then OD is equal to a, and AB to the circumference of the circle, that is, to Tra. The length a of the diameter of the generating circle is called the parameter of the cycloid. 254. Let Q be any point on the cycloid, and P the point at which the generating circle touches the base when K the tracing point is at Q. Let Q' be the position of the tracing point when the circle has turned through the in- definitely small angle from this position. Then as the circle begins to roll, it turns about the point P as an instan- taneous centre of rotation, and (f> being the angle through which it turns, Now if a particle starting from rest at A fall down the arc of the cycloid, which we suppose smooth, its velocity at Q will be given by the equation .PN. If we suppose this velocity to remain constant while the particle moves over the indefinitely small arc QQ', then the time taken to pass from Q to Q' will be denoted by -QQ-. Call this time T, then QQ' PQ THE ISOCHRONISM OF THE CYCLOID. By similar triangles PKi PQ.r.Pd'.PN, 251 or and this is true for each indefinitely small arc into which AO may be divided. Hence the time in which the particle will descend from A to is TT /y ^-, since the circle turns i/ through two right angles while the tracing point passes from A to 0. 255. A particle starts from rest at any point in the arc of a smooth cycloid whose axis is vertical and vertex down- wards ; to find the time of descent to the vertex. Let T be the point from which the particle starts. Through T draw TC parallel to AD, and let a second cycloid OA be drawn, having its vertex at and OC for axis. Let OD be denoted by a, OC by a. Let Q be any point on the first cycloid between T and 0, Q' a contiguous point. Draw Qq, Q'q parallel to AD, meeting the second cycloid in q, q' respectively, and let Qq meet PK in N. Draw Q'L perpendicular to Qq. 252 THE ISOCHEONISM OF THE CYCLOID. QQ' is ultimately perpendicular to jHQ, that is, QQ' coin- cides with QK when Q' is indefinitely near to Q. Hence QQ' : QL :: KQ : KN, Q'L KN' But KN: KQ :: KQ : PK; ' KN KN' QL KN KN' Similarly qq' _ I a' QL~ V KN' Now if a particle slide from rest at A' down the arc of the second cycloid, its velocity at q is the same as the velocity at Q of the particle which slides from T, since the vertical height through which each will have fallen is the same. Hence the time taken by the second particle to slide down qq' is to that taken by the first to slide down QQ' as arc qq' is to arc QQ', that is, as v'' to */a : and this is true for each pair of corresponding elements into which the arcs A'O, TO can be divided. Therefore the time taken by the first particle to slide down TO is to that taken by the second particle to slide down A'O as Ja to vV. But the time taken by the second particle to slide down A'O is, by the preceding article, wy . Therefore the time taken by the first to slide from T to is equal to ir\/ -. or the time from Tto is the same 2# as from A to 0. Hence the time taken by a particle to fall from rest at any point of the cycloid to the vertex is the same. This property is called the " isQchromsm of the cycloid" LENGTH OF THE AEG OF A CYCLOID. 253 The particle after passing will then ascend the cycloid to the same height as the point T from which it has fallen. If we denote by T' the point at which it comes to rest, the time from T to T is or V|. The particle will then return from T through to I 7 , and the whole time occupied in a complete oscillation or " swing-swang," that is, in passing from T to T and back again, is and is constant, however great or small the arc of vibration may be, provided the particle do not leave the curve. 256. Let Q be an}'- point on the cycloid, and PK the corresponding vertical diameter of the circle. Suppose the r> JL circle to roll through the indefinitely small angle <, and Q thereby to move to Q. Let PK be the diameter which then becomes vertical. Then the angle PCP is equal to 6. Join QIC. Draw KT parallel to DB and let PK meet KT in H. Draw HF perpendicular to QK. Then QF is ultimately equal to QA", and KF= QK- QK'. 254 THE CYCLOIDAL PENDULUM. Now when It is from experiments of this description that the most exact values of g have been determined. 2B9. As an illustration of the preceding articles we will determine the length of a simple pendulum which will perform a semi-oscillation in one second in London, the value of g being supposed equal to 32' 19. A pendulum which performs a semi-oscillation in one second is called a seconds' pendulum. By ki beats " of a pendulum are always meant semi-oscillations. If I be the length of the seconds' pendulum in feet, we have 1-3. _ 3-1416 2 -3-262... G. D, 258 NORMAL ACCELERATION. or the length of the seconds' pendulum is about 39*144 inches. 260. The reason of the isochronism of the pendulum is simply that the acceleration with which it moves is always proportional to its distance (measured along the path of the bob) from its position of rest. If this condition be fulfilled, it matters not whether the body be moving in a straight or curved line, it will still be isochronous in its vibrations, and the solution of the whole problem will be exactly similar to that of the pendulum. Hence, if a particle of mass M be free to move in any path under the action of a force along its path towards some point and always equal to fj,Md, where d represents the distance of the particle from the point measured along the path, the particle will perform isochronous vibrations about 0, whose o_ period will be - , no matter what may be the amplitude _ v> of the oscillation. Similarly, if a rigid body be capable of turning about a fixed axis, and it can be shown that its angular accele- ration towards its position of rest is always proportional to the angle through which it has been deflected, the body will oscillate about its position of rest according to the same law as the pendulum, its time of oscillation being independent of the amplitude. If the angular acceleration towards the position of rest be /i#, where 6 represents the angle through which the body has been deflected from o_ rest, the period of a complete vibration will be - . v /* Any sounding body which is emitting a pure tone exe- cutes its vibrations in accordance with the same law as the pendulum, the acceleration of each point being always proportional to its distance from its position of rest, and such oscillations are, therefore, frequently called harmonic vibrations. 261. Suppose a particle to be describing the curve APB, and let v be its velocity at P ; then, if PT be the tangent at P, the direction of its motion at P is along PT. Let Q be a point on the curve very near to P, v' the NORMAL ACCELERATION. 259 velocity of the particle at Q, QT' the tangent at Q, and let the normals at P and Q intersect in 0. Then, when Q is indefinitely near to P, is the centre of curvature of the curve APB at P. Let the angle POQ be denoted by 0, then T'KT is equal to 6. Now the velocity of the particle when at P is entirely along PT; its velocity parallel to OP is therefore zero. The velocity of the particle when at Q is v' along Qy, and its velocity parallel to PO is therefore v' sin 0. Now the time occupied by the particle in moving from P to Q ~T~)S~\ TJ/^1 lies between and " p, and if PQ be indefinitely small, v v ' we may take v equal v', and the time from P to Q becomes PO PO . Hence during the time ~, a velocity represented by v sin 6 is generated in the particle in a direction parallel to PO. The measure of the acceleration which will T)/~\ generate this velocity in the time is v sin 6 . ^-, or PQ XT Now v sn sn PQ* Hence the particle must, while passing from P to Q, be 260 NORMAL FORCE. moving with an acceleration Avhose measure is the limit of the expression v^y-p-. -'-.-, when is indefinitely dimi- JrLl u nished. But the limit of , when 6 is indefinitely small, a PQ is unity, and the limit of - is P0, that is, the radius of a curvature at P. Let this be denoted by p. Then the acceleration of the particle at P in the direction PO is measured by . Hence, if the mass of the particle be denoted by m, it must be acted upon by a force in the direction PO represented by . P If the velocity remain constant, the particle has no acceleration in the direction of motion. Hence the resul- tant force upon it is a force - acting inwards along the P normal at P. If the curve described be a circle of radius r, then p is equal to r, and the particle is always acted upon by a force TYtl] towards the centre of the circle. r In the case of motion in a circle, if o> denote the angular velocity about the centre, v = tar and the force towards the centre becomes mo> 2 r. If the particle make n complete revolutions per second, and the force towards the centre becomes This method of determining the acceleration of a point along the normal to the curve in which it is moving is due to Dr. Besant. 262. In the case of a particle being prevented from leaving the circle by a string attached to the centre, this force is supplied by the tension of the string. Hence the string must exert a force upon the particle represented by CONICAL PENDULUM. 261 ?2 m , and since action and reaction are equal and opposite, it follows that the particle exerts a force upon the string acting from the centre of the circle, and also represented v* by m . This action of the particle upon the string or v other means of constraint is frequently called centrifugal force. It should always be borne in mind that the force acting upon the particle is towards the centre of the circle, but that the action of the particle upon its means of constraint is in the opposite direction, and is properly termed centrifugal force. 263. As an example of the preceding article we may take the following. Suppose a particle P, of mass m, to be attached to one end of a string of length Z, the other end of which is fixed at A. The particle is made to describe a horizontal circle with uniform velocity, such that it makes n complete revolutions per second. It is required to find the inclination, 0, of the string to the vertical, and the tension of the string. Let be the centre of the circle described by the particle. The velocity of the particle is ^irn . OP. The acceleration towards is (27rn) 2 . OP, and the resultant force which must act on the particle to produce this acce- leration is m . (27rw) 2 , OP, 262 CONICAL PENDULUM. This is, therefore, equal to the horizontal component of the tension in AP. The vertical component must balance the weight of the particle, and therefore be equal to mg. But AO is to OP as the vertical component of the tension in AP is to the horizontal component. .-. AO : OP :: mg : m(2irw) 8 OP; An & or ^ " - a - l - / /f\ ~Ts> U1 ~"o (27T) 8 6> 2 The vertical height therefore of the point of suspension, A, above the particle is equal to -^, and depends only on the value of " g " and on the angular velocity of the particle. This result is of great importance. The tension, T, in AP may be determined at once, for its vertical component is equal to mg, the weight of the particle, .-. T: mg :: AP : AO ff . Or we may proceed thus : The tension in AP is to its horizontal component as AP to OP. :. TimtfOP:: AP: OP- :. T=ma>*AP To determine the angle 6 we have A0 ~ T mg -JL EXAMPLES OF CENTBIFUGAL FOECE. 263 or . 4W 2 7T 2 Z For example, suppose I equal to 2 feet and m to be 20 pounds, and that the system makes 10 revolutions per second, then taking g equal to 32, we have T=20.20 2 7T 2 .2 = 160007r 2 , or the tension of the string is 160007T 2 dynamical units of force, that is, equal to the weight of BOOir 2 pounds. If the string in this example be replaced by a rigid rod, which can turn about A in a ball and socket joint, we obtain the instrument known as a conical pendulum. 263a. If a number of particles be suspended by strings or thin rods of unequal lengths from the same point A, and all be made to execute the same number of revolutions per minute, all the particles will revolve in the same horizontal plane at the depth -j below A, and if the particles are equal the tensions of the several rods or strings will be proportional to their respective lengths. If the length of one of the pendulums is less than ^ 2 that pendulum will hang verti- cally and simply rotate about its own axis. 263/3. The pendulum bob, for in the case of a spherical bob the centre of the ball will behave in the same way as the particle considered above, may be supported on a bar bent to the form of any curve instead of being suspended by a string or rod. If the bar be smooth, its pressure on the ball must be along its normal, and we have therefore simply to replace the string AP, of Art. 264, by the normal to the curve of the bar. If the bar be bent into such a curve that the subnormal, which corresponds to AO, is constant and equal to -^, the bob will be in relative equilibrium upon the bar in every position. If the angular velocity be ever so little less than the bob will sink to the lowest position oh the bar. If it be ever so little greater than o> the bob will 264 PARABOLIC GOVERNOR. rise to the highest possible position, and such an instrument will be infinitely sensitive in recording any change of angular velocity from the standard velocity to. The curve whose subnormal is constant is a parabola, and the bar must therefore be bent into the form of a parabola, with its axis vertical and coinciding with the axis of rotation, and its 20 latus rectum must be -. ar If the balls be made to rotate by being connected with the crank shaft of a steam-engine, and be connected by suit- able mechanism with the steam supply, so that the steam is gradually cut off as the balls fly out, we have the so-called " parabolic governor," the principal fault of which is that it is too sensitive and is liable to cause, by the completeness with which it cuts off the steam when the speed is only slightly increased, greater fluctuations of speed than those it is designed to prevent. An approximation to a parabolic governor is frequently obtained by suspending the balls by rods from two points at the extremities of a horizontal cross arm, so that the points of suspension are at some distance from the axis of rotation. The rods are provided with slots, and are made to cross one another and the axis. The centre of suspension of each ball being on the opposite side of the axis from the ball itself, may be regarded as the centre of curvature of a parabolic arc over which the ball may be sup- posed to move for a small portion of its path, and for the corresponding velocity of rotation the governor is very sen- sitive. It will be seen that as the balls fly out the point at which the rods cross the axis moves upwards, as well as the balls themselves, and for a certain position of the balls the upward velocity of this point of intersection is equal to the upward velocity of the balls themselves, so that the line cor- responding to AO is unchanged, and the governor therefore behaves like a parabolic governor for a small range. The student should make drawings of these different forms of governors for himself, to help him the better to understand this article. 264. The earth's equatorial radius being taken as 4000 miles, it is required to find the force necessary to prevent a EXAMPLES OF CENTRIFUGAL FORCE. 265 particle of mass m at the equator from leaving its surface on account of the diurnal rotation. The time in which the earth makes a complete rotation O?* about its axis is a sidereal day, that is, about ^-^ mean solar ODD days, or nearly 86,164 seconds. The velocity of a point on ., , ,, ,,.-,, f 27r . 4000 x 5280 the equator due to the rotation is therefore - ^ feet per second. Hence the force which must act towards the centre of the earth to prevent a particle of mass m from leaving the surface must be 47r 2 .4000 2 .5280 2 ,, ^=r dynamical units 01 lorce 86164*. 4000. 5280 = 11203m dynamical units of force, very nearly. Hence the resultant force upon a particle of mass m at the equator must act towards the centre of the earth and be equal to '11203m units of force in order that it may be at rest on the surface. Now suppose the force with which the earth attracts the particle to be denoted by mf, and the pressure of the particle on the ground to be mg. Then the pressure of the ground on the particle is also mg, and the resultant force upon it is mf mg towards the earth's centre ; :. m(/-gr) = -11203m, or the apparent weight of the particle viz., mg is less than the force with which the earth attracts it by '11203w. If we suppose the value of g at the equator to be 32, we see that the weight of a body in its neighbourhood is diminished by about 4 -^ of the whole weight on account of the earth's rotation, assuming the earth's radius to be 4,000 miles. The effect of the earth's rotation upon bodies at the equator is actually to diminish their apparent weight by about ^03 of the whole. This force produces its whole effect 266 TENSION IN A EUNNING BELT. in changing the direction of the body's motion, that is, in producing in it acceleration towards the earth's centre. The value of g at the equator is rather less than 32, but the earth's equatorial radius is only about 3962 miles instead of 4000 miles, as we have taken it. 265. Suppose a string, the mass of a unit of length of which is m, to form a circle of radius r, and to revolve in its own plane with uniform angular velocity w. If we consider a very small length .s of the string, the mass of the element will be ms, and the force which must act upon it towards the centre of the circle will be msuPr. The string must therefore be acted upon by a normal force on each element towards the centre at the rate of ma)*r units of force per unit of length, and this force must be supplied by the tension of the string. The tension, T, may be determined from the following consideration. Suppose the radius of the circle to diminish by a very small quantity h. The circumference will then be re- duced by 27r7i, and the work done by the tension will be 2TrhT. But if the tension be employed to balance a normal pressure of ma> 2 r units of force per unit of length, the work done against this pressure will be %irr . muPr . h. Hence or T= mvPn 2 ' mv 2 , if v denote the linear velocity of the string. This result is of very great importance. It shows that in the case of a string revolving in a circle the tension depends on the linear density and the square of the velocity, and is in- dependent of the size of the circle. This tension deter- mines a limit to the peripheral speed of fly-wheels and the wheels of locomotives and railway carriages. If the tension in iron or steel be restricted to 4,000 Ibs. per square inch, the velocity cannot exceed 197 feet per second, and no advantage is gained by increasing the diameter of the wheels. Since the tension is unaffected by the radius of the circle, it follows that if a rope or belt be guided to move in any curve of varying curvature, the least tension con- VELOCITY OF A WAVE IN A STRETCHED STRING. 267 sistent with the motion will still be given by the expres- sion mv*. In the case of a belt running over pulleys, a constant tension mv 2 will be required throughout the belt to keep it in its curved path without producing any pressure on the pulleys, and without transmitting any power. The requisite tension to produce the necessary grip of the pulleys and to transmit the power must there- fore be added to this, and the result is to limit the speed of a belt of given strength and weight, and therefore the power which can be transmitted by it. For example, if the working tension of a leather belt be restricted to 400 Ibs. on the square inch (about one eighth of its breaking load), the density of the leather being 64 Ibs. per cubic foot, the maximum speed at which the belt can be run without transmitting any power is 170 feet per second, and the maximum power is transmitted at a speed of about 98 feet per second when the power transmitted amounts to about 32 H.P. per square inch of section of the belt. Another very important consequence follows from the same expression for the tension in a running belt or string. If the belt is running over any guides it will hang between the guides in certain curves, and as the belt runs, these curves will remain as stationary waves upon the belt which will run past them. Now the passage of the material of the belt through the stationary wave is precisely the same action as the transmission of a wave at the same speed along a stationary belt. Hence we con- clude that in a string whose linear density is m, while its tension is T, the velocity of transmission of a wave will be given by the equation T=mv 2 , /T or v /w _. . m This determines the velocity of transmission of a wave in a stretched string such as the strings of a musical instrument, and from it follows at once the number of vibrations which such a string of known length can execute per second, or the pitch of the fundamental note of the string. 268 HAEMONIC MOTION. 265 that HARMONIC MOTION. 269 is, to \/fA ON, and is proportional to sin PON. The acce- leration is proportional to ON, that is, to cos PON, and the displacement from is also proportional to cos PON. Rectilinear harmonic motion may be regarded as the projection of uniform circular motion on a plane perpen- dicular to the plane of the circle. The above investigation of harmonic motion affords a proof, quite independent of the properties of the cycloid, of the theorem that if a point move with an acceleration towards a fixed point and always proportional to its distance from that point, the time of oscillation is inde- pendent of the amplitude. 266. When two or more bodies are so connected that if the motion of one of them be given that of each of the others is known, we can, by help of the equations ex- pressing the geometrical connections of the system, find the motion of each part and the forces between the parts when the external forces acting on the system are known. We have seen examples of this in the cases of weights connected by a string over a pulley. The general method of solution of problems of this class is to take the accelera- tion of one of the parts as the unknown quantity, then by help of the geometrical equations the accelerations of all the other parts can be expressed in terms of this. The accelerations of all the parts being thus expressed in terms of one unknown quantity, the resultant force upon each can also be so expressed ; hence the reactions between the parts can be expressed in terms of this one unknown ; and finally, the resultant force on the first part of the system being expressed in terms of the external forces upon it, and the reactions of the other parts of the system, it can be expressed in terms of the unknown acceleration. But it is this force which produces that acceleration in the first part of the system, and this furnishes us with another expression for the same force. Equating these two expres- sions we have an equation to determine the unknown acceleration. This process will be best understood by an example. 267. Ex. A smooth wedge A whose angle is a and mass 270 MOTION OF A SYSTEM OF BODIES. M rests on a smooth plane inclined at the same angle a to the horizon, so that one face of the wedge is horizontal. On the upper surface of the wedge is placed a weight B of mass m. Find the motion of the system and the pressures between the parts. The upper surface of the wedge being smooth, all the forces upon the weight B are vertical, and therefore this weight will descend in a vertical straight line. Also, since the weight remains on the top of the wedge, which is horizontal, the vertical motion of the wedge must be the same as that of the particle, and therefore its vertical acceleration must be the same. Also the motion of the wedge is always along the inclined plane, and therefore its acceleration must be in that direction. Let / denote the acceleration of jB, f that of the wedge A, then, since their vertical accelerations are the same, f sin a = f; therefore f = . ' . sin a Therefore the resultant force on the wedge must be .- sin a along the plane. Let P denote the pressure of the weight B on the wedge. Then resolving along the plane, sin a P= M f Ma sin a a But the resultant force upon B is mg P, acting verti- cally downwards, and this must therefore be the force required to produce an acceleration f in the mass m. Hence mq . i- + Mq = mf. sin a a or f(M+m}gsin 2 a : M+msm*a MOTION OF A CONNECTED SYSTEM. 271 this determines the acceleration of the weight B. The acceleration of the wedge along the plane is (M + m)g sin a M + m sin 2 a and the pressure, P, between the weight B and the wedge mMg cos 3 a is M+ wsin a a ularly to the that the pressure between the wedge and plane is Resolving perpendicularly to the inclined plane we see at the pressure between the that is, M g C os a * 268. "When the connections between the parts of a system are such that the motion of all can be expressed in terms of that of one of them, and there are no sudden changes of velocity in any part of the system, we may fre- quently determine the motion from the consideration that the kinetic energy of the system is equivalent to the work done upon it by external forces. We may illustrate this by the following example. 269. Ex. A weight of 64 Ibs. is supported in equilibrium by a weight of 4 Ibs. in a system of pulleys in which each string is vertical. If a half-pound weight be added to the 4 Ib. weight, determine the motion of the system,, neglecting the friction and inertia of the pulleys, strings, etc. Since a weight of 4 Ibs. supports in equilibrium a weight of 64 Ibs. it follows from the principle of vertical velocities that if the former fall through a very small space the latter will rise through T V of that space, and since the strings are vertical the system is always similar through- out the motion. Hence the geometrical connections must be such that in each displacement throughout the motion the 4 Ib. weight will move through 16 times the space moved through by the 64 Ib. weight, and its velocity and acceleration will therefore be 16 times that of the larger weight. Hence, since the geometrical connections are un- disturbed by adding the half-pound weight, the same will 272 MOTION OF A CONNECTED SYSTEM. be true in the motion we are considering. Let f be the f acceleration of the 4i- Ibs. Then /= is the acceleration of ID the 64 Ib. weight, and f will remain constant throughout the motion, since the conditions are always the same. Hence the kinetic energy of the system at the end of time t from the commencement of the motion will be 1 1 f 2 / 2 1Q ~ . 4 J . f 2 t* + i . 64 . LL , or ^ f z t\ units of kinetic energy. Also, the space described by the 4^ Ib. weight will be - ft 2 units, and that through which the 64 Ib. weight has a 1 f ascended will be ^ . 4^P- Hence the work done on the 't J.O system by gravity will be - g ^ ft 2 units of work. _ & Therefore \ gfl* = ^ /*** ; .-. f= ^ g, 2 and the 4^ Ibs. will descend with uniform acceleration g. xy 270. Or we may proceed thus: The tension of the string supporting the " weight " will always be 16 times that of the string supporting the " power," whether the system be at rest or in motion, since the weights of the strings and pulleys are neglected. Let T denote the latter tension in absolute units, then 1QT will represent the former. The acceleration of the " power " will therefore T 16T be Q--TT downwards, and that of the "weight" -^- ff upwards. But the acceleration of the " power " is always 16 times that of the weight ; T and the acceleration of the 4 Ib. weight is g- 41 , or ^2 19 9, as before. INITIAL ACTIONS. 273 271. We purpose now to give a few examples of a class of problems not unfrequently proposed, namely the following : Suppose a system of particles at rest and in equilibrium under given constraints, and let one of these constraints be suddenly removed. It is required to find the change instantaneously produced in the action of the other con- straints. The general method to be adopted in order to determine the initial actions of the remaining constraints is to find the direction and acceleration with which each particle begins to move. If we multiply the expression for this acceleration by the mass of the particle, the product is the measure of the resultant force upon the particle, and this resultant force being determined in magnitude and direc- tion, we have sufficient equations for determining all the forces in the system at the commencement of the motion. 272. Ex. 1. A particle of mass m is suspended from two points in the same horizontal line by two strings of equal lengths. One of the strings is suddenly cut. It is required to find the initial change of tension of the other string. Let P be the particle in its position of rest, A, B the points of suspension. Let AB = 2a, and let the length of each string be /. Let the angle EPA be equal to 2a. If G. D. T 274 INITIAL TENSIONS. T be the tension of each string when there is equilibrium, we have by resolving vertically /y>j rt f-w~i llvlJ 2 cos a' Jn _ n z where cos a = _ _ . I Now suppose the string AP suddenly cut. Then BP remaining of invariable length, the particle will begin to move at right angles to BP. Hence the resultant force upon it must be in this direction. Therefore, if T be the tension of the string BP, immediately after cutting AP, we have T mg cos a, and the initial change of tension is 2 cos a If a be less than 45 the tension of BP is suddenly in- creased, and if a be greater than 45, or I less than -/2. a, the tension of BP is suddenly diminished, by cutting AP. The resultant force upon the particle immediately after cutting the string AP is mg sin a, acting in a direction perpendicular to BP. Hence the initial acceleration of the particle is in this direction and is numerically equal to g sin a. 273. Ex. 2. A string having it* ends fastened to two fixed points A and B in the same horizontal straight line has four equal particles, each of mass m, attached to it at equal intervals. If while the system is at rest the string he cut in the middle, it is required to find the instantaneous change of tension of the other portions of the string. When at rest the portion QR of the string will be hori- zontal, and the system will be symmetrical about the vertical line through the middle point of the string. Let a, denote the inclination of ES to the horizon, a 2 that of BS. Then, if !T represent the tension of QR when the INITIAL TENSIONS. 275 system is in equilibrium, T^ that of RS, and T 2 that of BS, we have T 2 cos a. 2 = T cos a x = T (1). Also T, sin a. = raff, or J 1 , = -^ ^- , sin oj' and Tg sin a 2 7\ sin a 1 = mg, T 2 =^-> sin a 3 If the length of the string be known as well as the distance AB, we have sufficient equations for determining a v and a<>. Let TI represent the tension of RS, and T 2 ' that of SB, immediately after cutting the string between Q and R. Then since SB remains of invariable length, the direction in which S moves is always perpendicular to $5, and hence at the beginning of the motion its acceleration along SB must be zero. Therefore T 2 ' T cos (a 2 a x ) = mg sin a 2 ; therefore T 2 = mg sin a 2 + J 1 / cos (a 3 aj. Also the resultant force upon >S perpendicular to BS is Twgr cos a 2 TI sin (a 2 aj, and its acceleration in this direction is therefore T' g cos a 2 -i- sin (a 3 a^, 276 INITIAL TENSIONS. and the component of this in the direction ES is ( T' ~) -j g cos a 2 - 1 - sin (a 2 - a { } > sm (a 2 a x ). L 111 J Now since SB is of invariable length the velocity of R in the direction RS must always be the same as that of S in that direction. Hence the acceleration of R in the direction RS must be equal to that of S in the same direc- tion. The resultant force upon R in the direction RS is T^ mg sin a n and its acceleration in this direction is therefore T' i q sin a, . m Hence we have rp' rp, i- a sm a< q cos a sm (. ,) ! sin 2 (a, a.) ; w w therefore m , _ sin a x + cos a 2 sin (a 2 a t ) l + sin^a,-^) sin a 2 cos (a 2 a,) s:: T^O 1 + sin a (a 2 flj) ' But jy = wgr sin a 2 + T cos (a 2 - a : ) ; therefore ^ , 2 sin a 9 71 = mq . 1 + sm^ (a, - aj Hence we have found the tensions of the parts of the string immediately before and immediately after the sec- tion. The differences between the corresponding tensions will of course be the instantaneous changes required. By proceeding precisely in the same way we might find the tension of the different portions of the string, immediately after cutting it at any point, whatever be the number of weights suspended from it. PRESSURE EXERTED BY A STEAM HAMMER. 277 274. We will conclude this chapter with the following example, illustrating the application of the theory of energy to problems connected with uniformly accelerated motion. Ex. A Nasmyth hammer, moving in a vertical direction, is driven by steam pressure on a circular piston 40 inches in diameter. The mass of the hammer and piston together in 25 tons, and the pressure of steam on the upper side of the piston is equal to the weight of 50 Ibs. per square inch more than on the lower side. Supposing the hammer after falling through 59 inches to strike a mass of iron, compress it verti- cally through one inch, and then come to rest, find the mean pressure exerted by the hammer upon the iron. The whole force acting upon the hammer before striking the iron is equal to the weight of .20 2 .50 and the whole mass moved being 25 tons, it will fall with uniform acceleration represented by 14J The velocity, v, which the hammer will acquire in fall- ing through 59 inches, is given by the equation .8 = 9 /I 57T\ 59 9 \ + 14/12' and this velocity is destroyed by a constant force, while the hammer moves over one inch. Hence if f denote the uniform acceleration with which the hammer moves throughout that inch, we have = 2dl + n . 12 , or f = 278 PEESSUEE EXEETED BY A STEAM HAMMEE. Therefore the resultant force on the hammer, while com- pressing the iron, is equal to the weight of 25 x 59l + tons. But the downward force upon it due to its weight, and the pressure of the steam, is equal to the weight of 261 + tcms. Hence the vertical pressure of the iron on the hammer, and therefore of the hammer on the iron, is equal to the weight of 60 x 25 (1 + -^j tons, or about 3183 tons. 275. We might have obtained the result of the preced- ing article from the consideration that if a mass move under uniform acceleration, the change of its kinetic energy in any time is always numerically equal to the work done upon it during the interval. If the acceleration under which the body is moving be suddenly changed during the motion, this principle is true for each portion of the motion, and therefore throughout the whole. Now the hammer starts from rest, and finally comes to rest. Hence the whole work done upon it must be zero. But it falls altogether through 5 feet under a constant force equal to the weight of 25f 1 + ^j tons. Hence the work done upon it by this downward force is 125 ( 1 + ^J foot- tons. Therefore the work which the hammer must do upon the iron is 125 M + ^ J foot-tons. But it compresses the iron through of a foot. Hence the mean pressure La which it exerts upon the iron must be equal to the weight of 1600 l + tons. EXAMINATION ON CHAPTER V. 279 EXAMINATION ON CHAPTER V. 1. A particle is projected from the vertex of a smooth, parabolic tube, whose axis is vertical, and latus rectum equal to 4, along the tube with a velocity represented by \/%ag. Find the velocity of the particle at any point in the tube in terms of the focal distance of the point. 2. Assuming that on ".descending a mine, g varies directly as the distance from the earth's centre, find the number of beats lost in a day by a pendulum which beats seconds at the sea-level, when carried down a mine to a depth of 400 fathoms, supposing the earth a sphere of 4000 miles radius. 3. Show that the time of oscillation of a particle under the action of gravity about the lowest point of a vertical circle of radius 2, is greater than the time of oscillation on a cycloid the diameter of whose generating circle is a, if the arc of oscillation in the circle be of finite length. 4. A particle slides down the surface of a right circular cylinder whose axis is horizontal from rest on the highest generating line. Find the pressure on the cylinder in any subsequent position of the particle, and the point where the particle will leave the surface. 5. Supposing the mass of the bob of a conical pendulum to be 20 Ibs., and the length of the string to be 3 feet, find the inclination of the string to the vertical when the bob is making 3 revolutions per second, and its tension. 6. A heavy particle is attached to a string 5 feet long, and swung round in a vertical circle. Find its velocity at the highest point in order that the string may just remain tight. 7. Explain the action of the conical pendulum as a regulator or " governor" for a steam engine. 8. In the case proposed in question 1, find the pressure of the particle upon the tube at any point of its path. 280 EXAMPLES ON CHAPTER V. 9. A train goes round a curve whose radius (i.e. the radius of the curve lying midway between the two metals) is 150 yards, at the rate of 50 miles per hour. Find the height to which one of the metals must be raised above the other in order that the whole pressure of each carriage on the metals may be perpendicular to the floor of the carriage, the breadth of the gauge being 4 ft. 8^ ins. 10. A heavy particle is suspended from the angular points of an equilateral triangle whose plane is horizontal, by means of three strings each equal in length to one side of the triangle. If one of the strings be cut, find the initial change of tension of the other two. 11. A uniform endless string, of length 2?r, is rotating in its own plane with uniform angular velocity to, under the action of no external forces. Find the tension of the string, the mass of each unit of length being m. 12. The mass of a smooth wedge whose angle is 30 is 10 Ibs., and it rests on a smooth plane inclined 30 C to the horizon, so that the upper surface of the wedge is hori- zontal. A weight of 2 Ibs. is placed on the top of the wedge. Find its acceleration and the pressure of the weight on the wedge. EXAMPLES ON CHAPTER V. 1. The value of g at Greenwich being 32-1912 and at Trinidad 32-0913, find how many beats a Greenwich seconds' pendulum would lose in a day at Trinidad. 2. Show that the acceleration of a particle oscillating in a smooth cycloidal tube whose axis is vertical, is at any point proportional to its distance from the vertex measured along the curve. 3. Find the inclination to the vertical of a conical pen- dulum 20 inches long, and making 200 revolutions per minute. 4. A body whose mass is 10 Ibs. is suspended by a string from a point in the roof of a railway carriage, which is describing a curve of 509 feet radius at the rate of 45 EXAMPLES. 281 miles an hour. Find the inclination of the string to the vertical when it is in relative equilibrium, and the tension of the string. 5. Find the difference in the pressures exerted on the metals by a train of 200 tons when going due East, and when going due West, along a horizontal rail at 60 miles an hour in latitude 60. 6. A pendulum which at A beats seconds, gains 2 beats an hour at B. Compare the weights of the same substance at the different places. 7. Two very small imperfectly elastic balls are let fall simultaneously from different points, their centres moving on the same cycloid whose axis is vertical and vertex downwards. Show that all their impacts will take place at the vertex, and find the ultimate range of vibration when the impacts have ceased. 8. The length of a pendulum which vibrates 30 times in a minute in 156'8 inches. Find the space through which a particle will fall from rest in one second under the action of gravity. 9. A free body falls from rest through nearly 301|- yards in one- eighth of a minute in the latitude of Green- wich. How far would a body fall from rest in a quarter of a minute at a place where the length of the seconds' pendulum is '999 of its length at Greenwich ? 10. The attraction of a planet of mass m on a given body at a point distant r from its centre, r being greater than the radius of the planet, varies as -, t . The mass of the earth is 49 times that of a certain planet, while its radius is 4 times that of the planet. Prove that a seconds' pendulum carried to the planet would oscillate in about T seconds. 4 11. If a simple pendulum 39i inches long oscillate in one second, what is the length of a pendulum which makes 3540 beats in an hour ? 12. The horizontal attraction of a mountain on a par- 282 EXAMPLES. ticle at a certain place is such as would produce in it an acceleration denoted by - g. Show that a seconds' pen- Its 21600 dulum at that place will gain ^ beats in a day, very nearly. 13. Show that a pendulum one mile long would oscil- 121 1 late in about . = y22 minutes. 14. A seconds' pendulum is carried to the top of a mountain 3000 feet high ; assuming that the force of gravity varies inversely as the square of the distance from the earth's centre, and that the earth's radius is 4000 miles, find the number of oscillations lost in a day, neglecting the attraction of the mountain. IB. A railway train is moving uniformly along a curve at the rate of 60 miles per hour, and in one of the carriages a pendulum which would ordinarily beat seconds, is observed to oscillate 121 times in two minutes. Show that the radius of the curve is very nearly a quarter of a mile. Supposing a stone dropped from the window of one of the carnages, find how much farther from the centre of the curve is the point at which it strikes the ground than the point vertically beneath that from which it falls, the height of the latter point above the ground being 6 feet. 16. A particle is projected horizontally with a given velocity from the highest point of a smooth sphere. Find the point where it leaves the sphere. 17. Find the greatest velocity with which a particle may be projected horizontally from the highest point of a sphere, so as to begin to move on the surface of the sphere. 18. A smooth straight tube is made to describe a right circular cone whose axis is vertical and semivertical angle equal to a, with uniform velocity, the vertical plane through the tube turning about the axis of the cone with uniform velocity to. Find where a particle will be in relative equi- librium in the tube. EXAMPLES. 283 19. Show that if a heavy particle fall from a cusp down the arc of a smooth cycloid whose axis is vertical and vertex downwards, its pressure on the curve at its lowest point will be equal to twice its weight. 20. Supposing the earth's orbit about the sun to be a circle of 93,000,000 miles radius, and the earth to describe this orbit with uniform velocity in 365 days, express the force exerted by the sun on a pound of matter at the earth's surface in British absolute units, neglecting the magnitude of the earth in comparison with the sun's distance. 21. If different points be describing different circles uniformly with accelerations proportional to their radii, their periodic times will be the same. 22. A heavy particle is placed very near the highest point of a smooth vertical circle ; show that the latus rectum of the parabola which it describes after leaving the circle is to the radius of the circle as 16 : 27. If, retaining the same highest point, the circle vary in size, show that the locus of the focus of the parabolic path of a particle so flying off is a straight line. 23. A lamina in the form of a regular hexagon of side a is placed flat on a smooth horizontal plane and fastened to the plane. A string of length equal to the perimeter of the polygon is wound round it, one end being attached to an angular point, and the other end carrying a particle of mass m. If the. particle be projected horizontally at right angles to the string with velocity w, find the time after which the string will be wound up again, and its greatest and least tensions. 24. A skater, whose weight is 12 stone, cuts on the outside edge a circle of 3 yards radius, with uniformly de- creasing velocity, just coming to rest after completing the circle in 6 seconds. Find the direction and magnitude, when he is half-way round, of his pressure on the ice. 25. A particle suspended from a point by a string of length a is projected from its lowest position with velocity 7<5 Vga ; show that it will pass through the point of 284 EXAMPLES. suspension, and that the direction of its motion at that point will make an angle cos" 1 - with the horizon. o 26. If a wheel of radius a roll on the lower side of a horizontal plane so that its centre moves in a straight line with uniform velocity v/gra, any point on its circumference will move in the same manner as a heavy particle starting from the cusp of a smooth cycloidal arc whose axis is vertical, and sliding down it. 27. A particle starts from the extremity of a smooth cycloidal arc whose axis is vertical ; show that when it has fallen through half the distance measured along the arc to the vertex, it will have accomplished f of its vertical descent, and two-thirds of the time of descent will have elapsed. 28. Three equal smooth spheres are placed in contact on a horizontal plane, and are connected where they touch. A fourth equal smooth sphere is placed so as to be sup- ported by the other three. If the connections between the lower spheres be simultaneously broken, show that the pressure between each and the upper sphere is instan- taneously diminished by one-seventh. 29. A heavy uniform string rests on a smooth hori- zontal table with one end pinned to the table and - of its n length hanging over the edge of the table ; if the pin be removed the resultant pressure on the table will be instan- taneously diminished by of the weight of the string. IV 30. A smooth wedge, whose vertical angle is 30 and mass 10 Ibs., is placed on a smooth plane inclined 45 to the horizon, the edge of the wedge being horizontal and directed upwards. On the top of the wedge (which is inclined at an angle of 15 to the horizon) is placed a smooth weight of 5 Ibs. Determine the motion, and the pressures between the weight and the wedge, and between the wedge and the plane. 31. In a system of pulleys in which all the strings are EXAMPLES. 285 vertical, a weight of one pound can support a weight of 32 Ibs. If a weight of one ounce be added to the one- pound weight, determine the motion of the system and the space through which the 32 Ib. weight will be raised in one minute, neglecting the friction and inertia of the pulleys, and the rigidity and inertia of the ropes. 32. A small smooth ball is running horizontally round the inside of a hemispherical bowl of given size. Show how to determine its height above the bottom of the bowl by observation of the time taken in making each circuit. 33. A circular elastic band is placed round a wheel, the circumference of which is twice the natural length of the band ; if the wheel be made to revolve with constant angular velocity, find the pressure of the band on the wheel. 34. A heavy particle is placed very near the vertex of a smooth cycloid having its axis vertical and vertex up- wards ; find where the particle runs off the curve and prove that it falls on the base of the cycloid at the distance i !T + y/3 J a from the centre of the base, a being the radius of the generating circle. 35. Pendulums which beat seconds correctly in London (0 = 32 '19) and Edinburgh (- 47TV 2 This is therefore the measure of the pressure exerted upon each square foot of the plane by particles which impinge upon it at angles of incidence between a and a - 6. Hence the whole pressure upon a square foot of the plane is the sum of all such quantities for values of a lying between zero and ^. But when their breadth is indefi- 49 nitely small, the sum of the volumes of all the cylin- drical shells similar to the shell PM is equal to the volume of the hemisphere, since these shells make up the hemi- sphere. Hence the whole pressure n of the gas on each square foot of the plane is given by n = I 2nmv . volume of hemisphere 2 3 3 = 2nm = - nmv* o _pv 2 PRESSURE OF A SIMPLE GAS. 293 282. Now m being the mass of one of the particles, and v its velocity, ^ mv 2 is its kinetic energy, and therefore 19 nmv 2 , or pv 2 , is the aggregate kinetic energy of all the particles in a cubic foot of the gas. Hence we see that if the velocities of all the particles were the same, the pres- sure of the gas upon a unit of area of a plane exposed to its action would bo numerically equal to two-thirds the kinetic energy of a unit of volume of the gas. Now the particles of any actual gas are not only moving in different directions, but they are also moving with different velocities. Suppose a cubic foot of the gas to contain n^ particles each moving with a velocity equal to w 17 n 2 each moving with a velocity v 2 , and so on, where n i + /n 2 +etc. n. Thus, the mass of each particle being m, and the kinetic energy of a cubic foot of the gas being denoted by E, we have E=~{n 1 v ] 2 + n 2 v we may wr it e The quantity v 2 is the mean of the squares of the velo- cities of the particles : v is sometimes called " the velocity of mean square." The pressure produced by the gas upon any area will be the sum of the pressures produced by the systems of particles moving with the velocities 1? v 2 , &c., respectively. Now the pressure exerted on a square foot by the particles moving with velocity v, we have seen to be - n^mvf, 6 where n t is the number of such particles in each cubic foot of gas, a,nd similar expressions hold for the pressures exerted on a square foot by each of the other systems of particles. Hence the whole pressure, P, exerted on a square foot by the gas is given by 294 PRESSURE OF A SIMPLE GAS. P= K (njmvf + n^mv/ + . . .&c.) o = - nmv 2 o and this is true however many quantities similar to v l we may consider. Hence the pressure exerted by any gas on each unit of area of a plane is numerically equal to two-thirds the kinetic energy of a unit of volume of the gas. It will be observed that the energy possessed, as well as the pressure exerted, by any gas is the same as if each particle were moving with the " velocity of mean square." 283. In the preceding article we have found the pres- sure exerted by a gas upon each unit of area of a plane. If a curved surface be exposed to the action of a gas, the pressure upon any small portion of the surface will be the same as if it were plane, and hence if S denote the area of the whole surface, the whole pressure upon it will be 1 Q pv*S. The whole pressure which a gas exerts upon any unit of area of a surface exposed to its action is called the pressure of the gas. The pressure of a gas may, under certain circumstances, vary from point to point. In such case the pressure at any point is measured by the pres- sure which would be exerted on the unit of area if the pressure of the gas were uniform over that area, and the same as at the proposed point. 284. If the densities of two gases be different, but their pressures the same, we have pv 2 the same for each. Hence, the pressures being the same for each, v 2 is in- versely proportional to the density of the gas. Also the kinetic energy possessed by a given volume of all gases at the same pressure is the same, for the pressure is numeri- ABSOLUTE TEMPERATURE. 295 cally equal to two-thirds of the kinetic energy possessed by a unit of volume of the gas. If v remain constant for any particular gas, then P, its pressure, varies directly as p. Now if V be the volume of a given mass M of gas, M=pV. Hence pec , and therefore P oc -, or PV is constant. Again, the equation may be put in the form or PV=\Mv\ o Therefore the product of the pressure and the volume of a given mass of gas varies directly as its kinetic energy. But (see Besant's Hydrostatics, p. 73) the product of the pressure and the volume of a given mass of gas increases uniformly with the temperature. Hence the kinetic energy of a given mass of gas increases uniformly with its temperature. But the kinetic energy of a mass M varies as Mv 2 . Therefore v~ increases uniformly with the temperature. Hence the value of v for any particular kind of gas depends only on its temperature. Let T de- note this temperature ; then if the zero of temperature be so chosen that T and v vanish together, T will always be proportional to v 2 , and if V be the volume occupied by a constant mass M of the gas, we have PV oc T. The tem- perature is then called the absolute temperature of the gas. But if E denote the kinetic energy of a mass M of the gas, PV oc E. Therefore E oc T 1 , or the kinetic energy of a given mass of gas varies as its absolute temperature (i.e. The specific heat is constant). Again, since for a given mass of gas PV oc T, if P be kept constant, V oc T ; the principle of the air thermo- meter. 29G VELOCIT? OF MEAN SQUARE. 285. Before proceeding further with this subject we will find the value of v for some of the principal simple gases at a given temperature. We commence with Hydrogen. The mass of a cubic foot of hydrogen, which, at the temperature of melting ice, exerts a pressure equal to the weight of 2116-4 pounds upon each square foot of surface exposed to its action, in a place where g = 32'2, is known from experiment to be '005592 pounds. The pressure upon a square foot of the surface is 2116'4 x 32 - 2 absolute units of force. Hence P=2116'4 x 32-2. Also the mass of a cubic foot of the gas being '005592 pounds, we have p = '005592. Hence, since v 2 is determined from the equation , 2116-4 x 32-2 "= 3 ' -005592 = 36593916; therefore v 6097 ; or the velocity of mean square for particles of hydrogen at this temperature is a velocity of 6097 feet per second. The velocity of some of the particles may be consider- ably greater than this, and that of others less, but the velocity of the majority of the particles will be not very widely different from this quantity. The pressure of any gas we have seen to be proportional to the kinetic energy of the unit of volume of the gas. Hence, if the pressures of two different gases be the same, the value of v 2 for each will vary inversely as its density. Now the density of oxygen is found by experiment to be always 16 times that of hydrogen at the same tempera- ture and pressure. Hence the value of v for oxygen is one-fourth that for hydrogen : that is, the velocity of mean square for oxygen at the temperature of melting ice is 1524-25 feet per second. The density of nitrogen is 14 times that of hydrogen at the same temperature and pressure. Hence, at the RELATION BETWEEN VELOCITY AND TEMPERATURE. 297 temperature of melting ice, the velocity of mean square , ., - 6097, , for nitrogen is , feet per second. 286. We have seen that if P represent the pressure and V the volume of a given mass M of gas, then PV in- creases uniformly with the temperature. Now it is found from experiment that the increase of PV for an increase of temperature of one degree centigrade is ^=^ of its value at the temperature of melting ice, and is the same for all gases. Hence the zero of absolute temperature is the same for all gases, and is 273 centigrade below the tem- perature of melting ice. From this it follows that if T c represent the tempera- ture of a gas in degrees centigrade reckoned from the zero of the centigrade scale, that is, the temperature of melting ice, the absolute temperature T of the gas will be equal to 273 + T c . We have seen that the value of v 2 for any particular gas is proportional to the absolute temperature. Hence if the value of v for any gas at the temperature of melting ice be known, its value at any temperature T e can be found by multiplying this value of v by v/273 + T c . 287. Suppose a mass M of gas to be contained in a cylinder of transverse section A, closed by a piston which is made to move with a small uniform velocity u away from the gas. Suppose the piston to be at the distance x from the bottom of the cylinder. When the piston is at a distance x from the bottom of the cylinder, the volume occupied by the gas is Ax, and M. its density p is therefore equal to -j- . J\.x Suppose a cubic foot of the gas to contain n^ particles moving with velocity 1? in directions making an angle a 1 with the axis of the cylinder. Then of these particles ^n l j are moving towards the piston. The velocity of each of 298 WORK DONE DURING EXPANSION. these relative to the piston is ^cosc^ u. Hence the number of these particles which strike each square foot of the area of the piston in a second is f ^n l (v 1 cos a t M). a Now by the impact the velocity of each relative to the piston is reversed. The change of the velocity of each is therefore 2 (v l cos a^ ii); and if m be the mass of each the pressure exerted on each unit of area of the piston by particles moving with velocity v i in directions making an angle a 1 with the axis of the cylinder will be n v m (v v cos a x uf. Now during any small time r the piston will move over a space ur in the direction of this pressure, and u and r being both small we may suppose the pres- sure to remain uniform during this displacement. Hence the work done during the time r upon the piston by the pressure exerted by the particular set of particles under our consideration is A . njn (v 1 cos a^ iCf . ur. Again, the component of the velocity of each particle perpendicular to the axis of the cylinder is unaltered by impact on the piston while the component parallel to this axis is changed from v 1 cos 04 to v l cos a x 2ti. Hence the kinetic energy lost by each particle on account of the col- lision is m cos ctj) 2 (v l cos a t 2) 2 j = Qmu (v v cos a^ Also, the rate at which these particles strike the piston is ^(v 1 cos a t M) per second on each unit of area ; and there- tO fore the number which strike the piston in the small time T is 7 ^(v l cos a x - u} AT ; and since the kinetic energy lost a by each particle is 2mw (v t cos a x ), the kinetic energy- lost by the system of particles we are considering is numerically equal to n^m (v 1 cos a^ u) 2 Aur, that is, to the work done upon the piston by this particular set of particles. Similarly, if we consider the set of particles moving with any other velocity, and whose directions of motion EXPANSION ACCOMPANIED BY COOLING. 299 make any other angle with the axis of the cylinder, we obtain the same result ; and, this being true for each set of particles, is true for the whole gas. Hence the loss of kinetic energy of the gas within the cylinder during any small time r is equal to the work done upon the piston by the pressure. Also, this being true for each small interval of time T, it follows that the whole work done upon the piston during any finite time is numerically equal to the whole loss of kinetic energy sustained by the gas. If the gas were contained within any other form of envelope, and this were allowed to expand in any way, it might be shown that the whole work done upon the envelope by the pressure of the gas is numerically equal to the kinetic energy lost by the gas. 288. Since the temperature of a given mass of gas is proportional to its kinetic energy, it follows that if a gas be allowed to expand and do work upon the vessel which contains it, the temperature of the gas will be diminished, and the fall of temperature will be proportional to the work done by the gas. If in the case considered in the preceding article u be very small compared with v, the pressure upon the piston during any very small interval of time T may be con- sidered uniform, and the same as if the piston were at rest. The pressure upon the piston will therefore be - pv 2 A, and the work done upon it in time T will be ^pv 2 o o AUT ; this will therefore be the measure of the kinetic energy lost by the gas. But the whole kinetic energy of the gas is - Mv 2 or - pv 2 Ax ; and, since the absolute tem- & perature of a gas is proportional to the kinetic energy of a given mass, it follows that if T were the absolute tem- perature when its volume was F, and T the loss of temperature in expanding to the volume F+ V where V is very small compared with F, T' : T :: m : x :: V : F, or V T' . = T, where V and T are each indefinitely small. 300 EXPANSION INTO VACUUM. In exactly the same way it may be shown that if a given mass of gas be compressed its temperature will be raised ; and if the change of volume be very small, the increase of temperature will be approximately proportional to the decrease of volume. Also, whatever be the change of volume of the gas, it may be shown, as in the case of expansion, that the increase of temperature of a given mass of the gas is pro- portional to the work done upon it by the agent com- pressing it, and that the increase of its kinetic energy is numerically equal to this amount of work. 289. If we make the piston in the case investigated in Art. 287 move through a given space with a velocity greater than that of any of the particles of the gas, and then suddenly come to rest, it is obvious that none of the particles of gas will impinge upon it during its motion, since they will be unable to overtake it, and when they do impinge upon it the piston will be at rest : and in this case the numerical measure of the velocity of each particle will be unaltered by impact, and therefore the kinetic energy of the gas will be unchanged. Hence also its temperature will be unchanged. Of course we cannot experimentally make a piston move with a velocity greater than that of any of the particles of a gas, but if two chambers be separated by a diaphragm, one of them containing gas and the other a vacuum, and if the diaphragm be suddenly removed, the effect will be the same as if it were moved with infinite velocity to the extremity of the vacuum chamber. For the diaphragm substitute a tap or valve closing a pipe which connects the two otherwise closed chambers, and we have an experimental realization of the hypothesis ; and we infer that if by such a contrivance a gas be allowed to expand into vacuum, its temperature will be unaltered by the expansion. This result was obtained experimentally by Dr. Joule. 290. We have seen (Art. 285) that the velocity of mean square of a gas at a given pressure is inversely proportional to the square root of its density. Suppose two gases at the DIFFUSION OF GASES. 301 same temperature and pressure to be separated by a porous diaphragm. Then at first the number of molecules of each gas which will pass through the diaphragm into the other will be proportional to the velocity of mean square of the particles, and we have an explanation of Graham's law of diffusion, viz., that gases diffuse into one another at rates inversely proportional to the square roots of their densities. Again, some of the gas in the first compartment having passed into the second and existing there as gas will return through the partition into the first. The amount of the first gas passing through from the first compart- ment to the second in the unit of time will depend on the density of the gas in the first compartment. The amount of the first gas passing from the second compartment into the first will depend on the density of this gas in the second compartment. When the density of the first gas in the second compartment is the same as in the first the amount passing per second from the first to the second will be equal to that passing in the other direction, and then it will appear as if there were no diffusion going on at all. The same will be true for the second gas, so that ultimately the two gases will be uniformly mixed in both compartments. Except that the collisions impede the process of diffusion " different gases act to one another as vacua." It is from the rate of diffusion of gases that we estimate the number of collisions or " encounters " which the molecules suffer per second, and hence deduce the number of molecules present in a unit of volume. 291. It may be shown that if two different sets of molecules are in communication the average kinetic energy of each will be finally the same. Hence the mass of a molecule of a gas must be inversely proportional to the mean square of the velocity at a given temperature, and we have seen that this is inversely proportional to the density of the gas. Hence the mass of each molecule is proportional to the density of the gas, and the number of molecules per unit volume is the same for all gases at the same temperature and pressure. This is in accord- ance with the fact that the chemical combining weight 302 CONCLUSION. of a gas is proportional to its density, and with Dalton's Atomic Theory. 292. The mode in which we have treated this subject in the preceding articles is not that which we should have adopted had our object been to develop its relations to the dynamical theory of heat. "We have indeed only introduced the notion of temperature because it enables us to define in a few words the condition of the gas we are considering. The definition given of temperature in treatises on heat is of course different from that which we have given, though the connection between the two is intimate. The subject of gases has been introduced here simply because it affords an example of some of the methods adopted in treating problems on elementary dynamics, the mode of investigation being merely an extension of that followed in the chain problems of Chapter IV. The student who desires a further acquaint- ance with the subject is referred to the articles upon it in the last chapter of Professor Clerk Maxwell's Theory of Heat. 293. The following method of determining the average value of cos 2 a is due to Dr. J. A. Fleming, of St. John's College. Referring to the figure of Art. 281 and supposing it to revolve about the line AB, let a- denote the area of the zone PQ and a the angle AOP, PQ being indefinitely small. Then, denoting the radius of the sphere by v : the area of the annulus MN is a cos a, while PN v cos a. Hence the volume of the cylindrical shell generated by PM is O stant. But the sum of all such zones as cr makes up the sur- face of the hemisphere, or 2?ry 2 . Hence, 2 (er) = %Trv 2 and EXAMPLES. 303 jvpMa ~3' or the average value of cos 2 a for uniform distribution in space is ^. MISCELLANEOUS EXAMPLES. 1. The mass of each of two hammers is 30 tons, and it is moved through 3 ft. 6 in. by a constant force equal to half the weight of the hammer, the two moving in oppo- site directions towards one another. An inelastic mass of iron placed between the two is thus compressed so that its thickness is diminished by one inch. Supposing the pres- sure exerted upon the iron by each hammer to be constant throughout the compression, find its measure in pounds' weight, and the time during which the pressure acts. Show that the diminution of thickness of the iron pro- duced by the blow of the two hammers is twice that which either hammer would produce if the mass of iron were placed against a fixed anvil. 2. Two particles A and -6, of masses 8m and m respec- tively, lie together at a point on a smooth horizontal plane, connected by a string of insensible mass which lies loose on the plane : B is projected at an elevation of 30 with a velocity equal to g ; if the string become tight the instant before B reaches the plane again, and break when it has produced half the impulse it would have produced if it had not broken, and if the particle rebound at an elevation of 30, show that the coefficient of elasticity between it and the plane is -. 3. A number of heavy particles are projected from the same point, (1) with the same vertical velocity, (2) with the same horizontal velocity. Show that in each case the locus of the foci of their paths is a parabola with its focus 304 EXAMPLES. at the point of projection and axis vertical, but in (1) the vertex is upwards, and in (2) downwards. 4. Prove that the angular velocity of a projectile about the focus of its path varies inversely as its distance from the focus. 5. Three equal particles are projected from the angular points of a triangle along the sides taken in order with velocities proportional to the sides along which they move. Prove that their centre of gravity remains at rest. Hence show that if P, Q, R, be points in the sides BC\ CA. and AJB respectively of the triangle ABC, such that PP C'Q A P rP = ~AO i = 7? 7?' t ^ ien ^ e centre ^ g rav ity of the triangle PQR coincides with that of ABC. 6. A parabola is placed with its axis vertical and vertex upwards. Prove that the square of the time of quickest descent from a given point in the axis along a chord to the curve varies as the sum of the latus rectum and the horizontal chord through that point. 7. A solid smooth cylinder of radius r lies on a smooth horizontal plane, to which it is fastened, and an inelastic sphere of radius 2r moves along the plane in a direction at right angles to the axis of the cylinder. Find the con- dition that it may pass over the cylinder. If the sphere be elastic and the modulus of elasticity be greater than - , prove that it cannot in any case pass over the cylinder, and if e be less than ^, find the con- o dition that the sphere may, after its first ascent, fall on the top of the cyclinder. 8. A particle is oscillating on the arc of a smooth cycloid whose axis is vertical. Show that the sum of its kinetic energies at any two points where the directions of motion are at right angles to each other is constant. 9. A parabolic tube is placed with its axis vertical and vertex downwards, and a particle of elasticity e starting EXAMPLES. 305 from the vertex with a given velocity . Show that a heavy particle will rest in the tube in any position if a> = /\/ . OJM 16. Two particles are projected from the same point at the same time with different velocities and in different directions; find the curve described by their centre of gravity. 17. Two given weights, whose masses are M and m respectively, are connected by an inextensible string which passes over a smooth fixed pulley. The system being initially at rest, determine the weight which let fall at the beginning of the motion from a point vertically above the ascending weight so as to impinge upon it will instantaneously reduce the system to rest. 18. A ball is projected from the middle point of one side of a billiard table so as to strike in succession one of the sides adjacent to it, the side opposite to it, and a ball placed in the centre of the table. Show that if a and b be the lengths of the sides of the table, and e the coefficient of elasticity of the ball and cushion, the inclination of the direction of projection to the side a of the table from which it is projected must be n->a i + e EXAMPLES. 307 10. A string charged with n + m + 1 equal weights fixed at equal intervals along it, and which would just rest on a smooth inclined plane with m of the weights hanging over the top, is placed on the plane with the (m + l) th weight just over the top ; show that if a be the distance between each two adjacent weights, the velocity which the string will have acquired, at the instant the last weight slips off' the plane, will be soo u/;> ) Teo t>t)u> ^ t ' i ' T"3o- 44. 600 feet. 1200 Ibs. 7| seconds. CHAPTEE II. EXAMINATION. 1. 88ft. 2. 256 feet. 4 5 \ sees. 3. 2560. 4. 8 A /J, ft. per second. V do 5. 5 seconds. 6. Sv'JTft. per second, ^b seconds. 7. 1'424... seconds. 28'08 ft. per second, nearly. 9. -9463... sec. 10. 1 A / ?-.- sec. 2 V cos 2 15 11. 160w foot-second units. 12. 8. ANSWEBS. 315 EXAMPLES. 2. '-seconds. 2Qr ^ 1+B./3 Pressure hetween weight and wedge = 40cos 1 ^~ LOA/2 Ihs' y weight, or, g poundals. Pressure between wedge and plane = 45 ,72+20^3 cos 15 y Ibs.' weight,- or, ^ g poundals. Acceleration of weight =g . 3\/3 o 31. Acceleration of weight ~o^- do Space = 102f feet. THE END. Butler & Tanner, Ttie Selwood Piintiog Works, Frume, and London. November 1889. A CLASSIFIED LIST OF EDUCATIONAL WORKS PUBLISHED BY GEORGE BELL & SONS. Cambridge Calendar. Published Annually (August). 6s. 6<7. Student's Guide to the University of Cambridge. f>s. Qd. Oxford : Its Life and Schools. Is. 6d. The Selioolmastcr's Calendar. Published Annually (Decemler). Is. BIBLIOTHECA CLASSICA. A Series of Greek and Latin Authors, with English Notes, edited by eminent Scholars. 8vo. %* The Worts with an asterisk (*) prefixed can only le had in Hit Sets of :!(> F