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Les diagrammes suivants illustrent la mdthode. by errata ned to lent une pelure, fapon d 1 2 3 1 2 3 4 5 6 32X N 1 M CA V A 315". 530 NATURAL PHILOSOPHY, ONTAIMMi THE ELEMENTS OF MECHANICS HYDROSTATICS, AND OPTICS, .SCHOOLS, COLLEGEvS, CANDIDATES 'FOR MATRICULATION AT THE UNIVERSITY OF LONDON. ^\)N -_.._.. \ uv KICHxVKD WUKMELL, M.A., ii.Sc, Author of Mechanics, Et^drustatU-s, 4'c- ^.>w \- \x SIXTH EDITION. A. & W. MACK IN LAY, PUCLISilEKS, HALIFAX, N.S. .An>.( c('i,x\x. [The riijht of Translation is rnrjcrved.] f I'iJ... •4i 'I PREFACE. The present work is intended chietly for students preparing for Matriculation at tlie University of Lon- don. It is introductory to the Autlior's works for the Examinations suhsequent to jVIatriculation, and follows the same form and order ; so that students in pursuing the subjects further will be able easily to connect what they have to learn with what they have already learnt. At the same time the book by no means represents the narrowest view wliich may be taken of the Curriculum of the London University, but will, it is hoped, be found useful in schools generally, as containing a systematic explanation of the more elementary prin- ciples cf this branch of Natural Philosophy, and being calculated to lay a sound foundation for a more advanced studv ol llic siiliicil. « ml ^1 II CONTENTS. \ le uni be a in- mg MECHANICS. Introduction J; orcc ••• •■• ««« ••• olAllGo ... ... ... ■) Magnitude of a forco Direction of a force ... ... .« Grapliic representation offerees Trausmissibility of forces Resultant of forces in the same straight line Composition and resolution of forces Forces in the same plane meeting at a point Deductions from the parallelograms of forces Conditions of equilibrium Polygon of forces ... ... ., Triauf'le of forces ... ... Parallel forces ... ... ... •« Composition of parallel forces... ... Composition of unlike parallel forces Centre of parallel forces Moments ... ... ... ,„ ( 1 . 3 . 4 n I rr i 19 21 23 24. 2(5 2S 28 31 33 S')^'l^(c Tiii CONTENTS. PAGE Centre of gravity ... ... .». 3G To find the C. G. of the surface of a triD^'^lo ... 39 To find the C. G. of three equal particles placed at tho angles A B C of a triangle ... 40 Properties of the centre of gravity ... ... ... 43 To find geometrically tho C. G. of anj rectilineal figure 45 The mechanical powers ... 50 Machines 50 Tlio Lever 52 Balances 5i Wheel and axle 5S Toothed wheels 5 'J The pulley 05 Several movable pulleys with scparrio strings 07 The inclined plane 73 The wedge 76 Tho screw 77 Condition of equilibrium when a weight is supported by a screw 73 Compound machines • • • 79 VVi'Il^ .•• ••• ••• ••• ••• • • • 83 L/i^AMICS ••• ••• ••• ••• a * • 88 Motion ... .•• ••• . *,. • • • 83 Direction of velocity • • • 89 Graphic roproseutation of motiou • « • 90 Motion in a straight line \7ith uniform acceleration • • • 90 Forces producing motion ... ... ... . . • 97 Falliug bodies ••• ... ... •«. • •• 99 Jjiass ,•• ..« ••• «•• »« . • • 107 Attwood's Machine .,. ... ... ••• . • « HI Momeutum... ... ... ••« n%. . . . IIG Newton's Third Law of Motion .. k 120 PAGE 30 , 39 I . 40 . 43 40 ,. 50 .. 50 .. 52 .. 51 .. 53 05 07 73 70 77 . 73 ,. 79 .. 83 .. 88 .. 8S .. 89 ... 90 ... 90 ... 97 ... 99 ... 107 ... Ul ... 110 ... 120 JV CO-NTENTS. IZ HYDROSTATICS. PAGE Introduction !•« • • • 1 The Transmission of pressure • •• • •• • • • 3 The Bramah press ... f • t • • ■ Pressure of a fluid arising from its weight • «• • •• « • • 12 The pressure on the sides of a vessel «•• • • • IS The pressure on a solid immersed in water • •• • • • 2i Specific gravity • •* • • • 21 To find the specific gravity • • • • •• « • « 29 Of a solid heavier than water • •• • • • 31 Of a solid lij^hter than v/ater... • • • • • • • • • 3;? Of a liquid • •• • • • S3 Of a gas .. ... ... • • • • • • • ■ • 34 Of a solid body soluble in water ... • •* • > • 35 Density •.. ... • • « • • • • • • 37 Problems on the density of liquids • •• • • • 39 Pressure of the air • •• • • • • • • 42 Magdeburg hemispheres ... • « • ... 45 The barometer • • • • • • « • . 40 Marriotte's law • • • • •• 47 Relation of pressure and temperature • •• • •• 49 The weather-glass... • •• • • • 50 The com non pump ,., • • • ^•» • •• 51 The lifting pump • •• • • • 52 The Archimedian scretv • • • • • • •• 53 The forcing pump • •• • •• 53 The fire-engine ... ,., • •• • •• • • • 54 The siphon «• • • • • 55 The air pump ... ... • • ' • t t • • • 57 The siphon gaugo «•« • • « 59 The diving-bell • • • * • • « • a GO CONTL.NTS. orxics. PAGE 1 • «• e • « ■ • • •• • • ■ • • • • • •• • a Definitions... ... ... ... Propagation of liglit ... ... Velocity of light Light incident on a surface ... Law of Reflection MiiTora ... ,,. ,,, xHIia'qCo ••• ••• ••• ••• ••• To explain the foriniition of images with a plane mirror To draw tiio paths of the rays Continued reflections ... ... .< Parallel mirrors Inclined Mirrors ... ... ... ... 10 Spherical mirrors ... ... ... ... ... 15 Principal focus of concave mirrDrs... Ileal and virtual foci... Images with concave mirrors Convex mirrors The size of images with spherical mirrors ... Refraction Laws of refraction ... ... ... Experimental proof ... ... ... Total reflection ., ... ... ... Effects of refraction... Prisma ... ... ... ... Lenses ... ... ... ••* Converging leusos ... ... Diverging lenses ... ... ••• 15 18 19 21 23 25 27 30 31 34 3G 37 39 46 PAQE MECHANICS. rror o *j 3 4 6 6 7 8 9 9 10 15 15 18 19 21 23 25 27 30 31 34 3G 37 39 46 -♦♦- L-INTRODUCTION. 1. A BODY is at rest when it constantly keeps the same position in space, and it is in motion when it occupies successively dilTerent positions. 2. To illustrate these conditions, let us place a stone or other body on a horizontal table. If left to itself, the stone will remain for an indefinite length of time in the same position. Imagine the table to be suddenly removed ; the stone will no longer remain at rest, but will fall to the earth. Why does it take that particular direction ? A physicist would tell us that there is a constant attraction between the earth and the stone, and that the movement of the stone is due to this cause. Whenever a body at rest is made to move, there always exists a cause of the motion, the action of which is due to the presence, near or remote, of another body ; in other words, a body without life cannot move of itself. 3. If the stone be projected along a level road, the speed with which it leaves the hand will not be main- tained, but will be gradually diminished, until finally il 2 INTRODUCTION, ihQ stone will stop in its course. If, instead of the road, the frozen surface of a lake be chosen, the same stone thrown with the same force will travel much further on the ice than on the road. And if, instead of the irregular stone, we roll a smooth ball of ivory on the ice, the distance traversed will be greater still. It is evident, therefore, that the stone is gradually stopped }>y the resistances it encounters. Similarly, whenever a body ceases to move, it does so because its motion is destroyed by the resistances it meets with. The more we diminish these resistances, the longer and the further will the body move ; and, consequently, if we imagine that they are all suppressed, we shall be led to the conclusion that the body under these circumstances would continue to move for an indefinite length of time ; in other words, that a body cannot of itself alter its speed. 4. Neither can it change the direction of its motion. If no obstacle be encountered in its course, the ivory ball thrown on the ice will turn neither to the right nor to the left. It is true that a stone thrown into the air returns to the ground, but this is because its weight tends constantly to bring it to the earth. Conceive the weight and the resistance of the air removed, and the stone will continue to move in a straight line with uni- form speed. This fundamental principle in Mechanics is termed the First Law of Motion, and may bo enunciated thus : — When a body is not acted on by any external agent, if it he in a state of rest it will remain so, and if it he in motion it will continue to move in a straight line with uniform speed. FORCE. 8 Force. 5. Wli'?n wc raise a burden, set in motion a body at rest, or arrest a body ah'eady in motion, we are conscious of exerting a certain eflbrt, and tbe term force, which we give tliis effort, is naturally applied to similar causes. A force is therefore any cause tending io change a hodfs state of rest or motion. For a long time the forces employed by man were only those furnished by human and animal labour; but, in proportion as progress in science has been made, we have not only utilised these natural mechanical forces, but have subjected to our use others which were before unknown ; as, for example, forces arising from the motion of the air, a running stream, the changing of water into vapour by heat, chemical action, electricity, &c, 6. Forces from different sources, however, may bo compared with one anotluir, and are capable of nume- rical valuation. The science of Mechanics does not consider the nature of the forces, but treats only of the properties common to all; and a force will be regarded as fully determined when we know the point at wMcli it is applied, its direction, and its intensity. 7. Forces do not always have the effect of producing or modifying the motion of a body ; other forces acting at the same time may counteract them. A weight held in the hand is not the less heavy because it does not fall ; it is prevented frora falling by a force exactly STATICS. equal to it, but opposite in direction. If the weight be placed on the table, it is still prevented from falling by a force exerted by the table. This force is termed a reaction, or a resistance. Whenever a body acted on by a force is at rest, we may at once conclude that there exists a resistance which counteracts the force. When two or more forces neutralise one another, so as to keep the body or the particle on which they act at rest, they are said to be in equilibrium. 8. The science of Mechanics may bo conveniently divided into two parts. Statics treats of forces in Equilibrium. Dynamics treats of forces which produce motion. J! IL— STATICS. 9. Three things have to be considered in a force. 1st. Its intensity or magnitude. 2nd. The point of application. 3rd. The direction. The I^c gnitude of a Force. 10. Two forces are equal when, if applied at the same point in opposite directions, they will be in equilibrium. One force is twice another when the first will counter- act two forces, each equal to the second, applied to the same point, but in the opposite direction ; so also one force is thrice another when three of the latter TnB MAONITUDE OP A FORCE. w'll be required to keep the former in equilibrium, and so on. 11. Tlie magnitude of a force is estimated by stating how many times a given unit will be equal to it. The unit usually employed is a pound. Although forces are obtained from different sources, they may all be compared ^nth the pound weight. We will describe an instrument wiiich may be used to compare forces of different kinds with tlie unit of weight. 12. Figure 1 represents a spring balance for this purpose, con- sisting of an elastic band of steel, BOA, to the ends of which metallic graduated arcs are attached. The outer arc, da, is fixed to the lower arm, passes through an aperture in the upper, and terminates in a ring, F. The inner arc is attached to th.e upper arm, passes freely through the lower, and terminates in a hook, o. Place the ring on a firm support, and hang a weight of one pound upon the hook. The two branches of the steel band will approach each other until the elasticity of the bar counter- balances the weight. Mark on the arc, a d, the position of the bar. Suspend in a similar manner Tveights of 1 lb., 2 lbs., 3 lbs., Fig. 1. I&c, and make the corrcsprnuling marks until the *n 6 DYNAMETEUS graduation of the arc is completed. Any force wliicli when applied to the instrument produces a certain deflection is equal to the weight which would produce the same. A spring balance used to compare forces differing in kind may be termed a dynameter. Figures 2 and 3 represent other forms of the instru- ment. If the ends of the bar, a o b in Fi g. 1 were Fig. 2. Fig. 3. united to a similar bar, by joints at a and b, the strength of the instrument would be greatly increased, and by means of such an arrangement we should be able to compare even forces of great magnitude with the imit of weight. If we wish, for example, to mea- sure the force with which a horse draws a carriage along a paved road, it will only be necessary to inter- pose the dynameter between the horse and carriage, attaching one to the ring, and the other to the hooh. ce wliich I certain 1 produce JTering in tie instru- r, 1 were ,nd B, tlie increased, should be le with the to mea- la carriage (y to inter- carriage, lie hooK. HErr.ESENTATION OF FORCES. The Direction of a Force. 13. When a body is suspended by a threaa (Fig. 4), the thread takes a determined direc- tion, termed the vertical ; if it be suddenly cut, the body will fall in the direction of the vertical. This is therefore termed the direction of the weight. The direction of a force applied to a point is the straight line along ivhich the point would be displaced if it were free to move. Graphic Kepresentation of Forces. Fig. 4. 11. In the three properties we have described, forces ; may be represented by straight lines. For example, ■ the direction of the line may represent the direction of the force, the extremity of the line the point of appli- cation of the force, and if we agree to represent a unit of weight by a unit of length, as for instance, by taking a line of three inches to represent a force of tlirce lbs., then the magnitude of the line will repre- sent the magnitude of the force. The Transmissibility of Forces. 15. If a weight be attached by a cord to the spring balance in Fig. 1, the effect will be the same at what- ever point in the cord the weight be tied. Similarly, a force may be applied to a body directly » or by the interposition of a rigid rod, and, supposing the rod to be supported indcuendently, the result will bo the B w^ I fi 8 FOrX'ES IN THE SAME STRAIGHT LINE. same. The general principle here illustrated may be stated thus. A force may he applied at any point in the line of its direction, provided this p)oint he connected with the first point of application hy a rigid and inex- tensihle straight line. Let A B (Fig. 5) be a string, at the extremities of which equal forces are acting in opposite directions, it -B F Fig. 5. is clear that the string will be in equilibrium. Take any other point, c, in the string, and remove the force, F, from a to c, still the force will be in equili- brium. Hence we may consiujr the force applied at one end to be transmitted throughout the string, and we may suppose two opposite forces at any point, each equal to f. Either of these forces is termed the tension of the string. Suppose the string to pass round a smooth peg, ring or surface, in this case also the tension of the string is the same at every 2'>oint, Hesultant of Forces in the same Straight Line, 16. If we hang two weights of 1 lb. each to the dynameter, the indication of the instrument will be precisely the same as if a single 2 lb. weight were attached to it. Three weights of 1 lb. each produce the same effect as a single 3 lb. weight. Similarly, forces of 3 lbs. and 5 lbs. may be replaced by a single force of 8 lbs. And, generally, it is evident that when two or more forces act upon a point in the same straight Urn E. :ed may be ny point in >e connected I and inex' remities of rections, it FORCES IN TDE SAME STRAIGHT LINE. 9 in the same dircctiouy their effect ivill be equivalent to that of a single force equal to their sum. Conversely, if for a single force two others, the sum of which is equal to the frst, be substituted, the effect ivill be the same. Let ns now apply at the point o two unequal forces in opposite directions, the one, f', acting horizontally from left to right, and the other, f, horizontally from right to left (Fig. 6). If f' be a force of 2 lbs. and F of 3 lbs., we may substitute three single forces of Lum. Tahe remove the be in equili- j applied at string, and any point, termed the pass round :ase also the nt. F' — > ght Line, ach to the cut ^"ill be } reiglit were produce the larly, forces single force at xi'hen two straight lins ■\. .1 Fig. 6. 1 lb. each for the latter, and two of them will then be in equilibrium with the former. There remains a free force of 1 lb. acting from right to left. The body under the influence of the two forces is therefore in the same condition as if it were acted on by a force, F — f', in the direction of the greater. "Whenever two unequal forces act on a particle in the same straight line, but in opposite directions, the greater may bo divided into two parts, one of which is equal to the less, and will therefore neutralise it, and an effec- tive force will remain in the direction of the greater, equal to the difference of the forces. The single force ivhich tvill produce the same effect as several forces acting together, is termed the residiant of the forces ; hence the general proposition explained above may be enunciated thus. Wlien any number of forces act upon the same point, in the same straight line, their resultant may be found ■i® ■V \ f I ! iL 10 COMPOSITION AND RESOLUTION. by taking the difference between the sum of all the force." acting in one direction, and the sum of all the forces acting in the other, the direction of the resul- tant being that of the greater sum. Let us agree to write a + sign before all the forces acting in one direction, and a — sign before all those acting in the opposite direction, and then we may state the rule thus : — The resultant of a number of forces acting in the same straight line is equal to the algebrai- cal sum of the forces. ] 7. The forces which are thus combined are termed the Components^ and the process of finding the resul- tant is termed the Composition of Forces, The con- verse operation, namely, that of finding Component forces which shall have a given resultant^ is termed the Eesolution of Forces, The resultant is the single force which will produce the same effect as the com- ponents ; hence, if to a system of forces a new force be added, exactly equal to the resultant, and opposite in direction, this force will keep the system in equi- librium (Fig. 7). Thus the problem to find the resultant also gives the force required to keep the system in equili- brium. Fig. 7. - i%^x QUESTIONS ON THE REPRESENTATION OF FORCES. 11 Exercises. 1. If a force ■whicli will support a "weight of 18 lbs. be repre- sented by a line 3 ft. long, how long must be the line which will represent a weight of 1 i cwts. ? — Ans. 28 ft. 2. In the above case, what weight wiU a line of 17 in. repre- sent ?— ins. 8.^ lbs. 3. The force required to balance two others acting together in the same straight lino is five times one-fourth of larger force, what multiple is it of the smaller ? — Ans. 5. 4. Wlien a ton is represented by a line 5 ft. 4 in. long, what force will a Hue 7 in. represent ? — Ans. 2 cwt. 21 lbs. 5. The weight of a cubic foot of a substance is 6 cwt., and is represented by ii line a foot long, what will be the length of the line required to represent 720 cubic inches of the same substance ? —Ans. 5 inches. 6. "Wlien two forces act together they have a resultant of 12 lbs., and when they act iu opposite directions their resultant is 2 lbs. ; find the forces. — Ans. 5 lbs. and 7 lbs. 7. A string supports a weight of 4 lbs. at its extremity, another weight of 5 lbs. above the first, and a third of 7 lbs. above the second : find the tensions of the three parts of the string. If the tension of tho middle part be represented by a line 11 J in. long, what will be the length of the lines required to represent the others ?—Ans. Tensions, 4 lbs., 9 lbs., 16 lbs. ; lines, 5 in. ond 20 in. i Fig. 8. 12 III.— THE COMPC^ITION AND RESOLUTION OP rOUCES. Forces in tho same Plane meeting: at a Point. 18. Suppose two forces not in tlie same straight line act on a point m (Fig. 8). If left to the action of these forces only, the point will begin to move in a certain direction. There mnst, therefore, be a single force v»'hicli will produce the same effect, and therefore, also, a force which will counteract this effect. We have to consider how we can find the mag- nitude and direction of this force. Over two pulleys, m and n (Fig. 9), capable of moving without friction, pass a fine thread o^ silk. To the mid- dle point B tie a very light strip of wood e d. Take a number of small equal weights, and attach two of the 1 to each end of the thread, and three to the extremity e of the strip e d. The point b will descend to a definite position, and will then remain at rest. Let us examine the position of equilibrium. First, wc shall find that the rod e d will bisect the angle M B N J and this will always be the case whatever tho Fig. 9. PARALLELOGRAM OF FORCES. 13 ON OP Point. •aiglit line 111 of these begin to a. There »gle force ue effect, rrhich will e have to the mag- s,M andN if moving ass a fine the mid- ery light Take all equal h two of of the to the itrip E D. scend to and will First, he angle lever the weights, provided that those at m and n are equal. A^irain, take a certain length — an inch, for example — to rcnresent the weight 7?, and measure off on b n and R M as many inches as there are weights at each end of the thread and through the points A and thus dctermiued, draw a d and c d parallel to b c and A b respectively. Tliese lines will meet tlic rod in the i^amo point d, and it will be found that the line b d will contain as many inches as there are weights attached to E. Instead of equal weights sus- pend 3 /? at one end of the tlireacl, 2 ;; at the other, and 4 p to the rod (Fig. 10). The form of the figure will be clianged, but if we measure on n M three units of length, and (haw a D parallel to b c, also measure on b c two units, and draw c d parallel to b a, still we shall find that these straight lines intersect at a point d on the rod, and that the line b d contains as many nnits of length as there are weights at ^.. This experiment illustrates the fact, that, in order to find the resultant of two forces acting on a point, we must take on their directions lengths proportional to the forces, and complete on them a parallelogram. The diagonal of the parallelogram through the point of application will represent the resultant of the forces in magnitude and direction. 19. We may here remark that the thread must bo Fig. 10. iiv I 5 14 TAUALLELOGRAM OF FORCES. tied to the rod, in order that there may be equilibrium with unequal weights at m and n. If the weight at u be attached by a ring, as in F* \ 11, the tension in Fig. 11. the two parts of the string must be equal. The parallelogram in this case is equilateral. The proposition above explained is termed the Parallelogram of Forces. 20. 1/ two forces acting on a particle he represented in magnitude and direction hy two adjacent sides of a parallelogram^ the resultant of these forces will he repre- sented in magnitude and direction hy that diagonal of the parallelogram which passes through the px^rticle. From this important law we may make several deductions. (i.) If two Cvqual forces f, f', act on a point m, the resultant bisects the angle between them (Fig. 12). (ii.) Three equal forces making with each other angles of 120° (Fig. 13) will be in equilibrium, for the resultant of any two will be equal and opposite to the third. (iii.) We may always replace a single force by two others which v»ill have the same effect as the single force, and which will act on the same point in any given directions. (iv.) When the two forces act at right angles, the resultant will be found by adding the squares of the ^ J.*; nnn root 1 ric;-l thei ■ ''J? 1. resp 1 ■1 vt ^ % '%:> ''( 2. 5311 1 ■V; '1 'i f ...■? pre wh( ' i for( thr J BC ''It % ; rop \ rati i ,1 ■ PARALLELOGRAM OF FORCES. 15 tension in numbers, measuring the forces, nnd taking the sf|uaro root of tlic sum. Thus, if P and q be the forces acting on the point at right angles to one another, and R their resultant, then r'' = p'^ + Q^ Examples, 1. FiuJ the resultant of two forces of 8 lbs. and 15 lbs. respectively acting ou a particle at riglit angles to one another. p2 = 82 = 61 Q2 = 152 ^ 225 2S9 17 lbs. 2. If the resultant of two forces acting at right angles be 53 lbs. and one of the forces be 45 lbs., find the other force. E2 = 532 = 2809 r2 = 453 ^ 2025 Subtract .*. Q^ 784 .-. Q = 28 21. Remark, that the unit of length taken to re- present a unit of force may be anything we please, but when a certain line has been taken to represent a given force, the unit is fixed, and must be maintained throughout the problem. Example 3. — A boat is held at rest in a stream by cords, A 0, B (Fig. 14), attached to stakes in the opposite banks, and a I rope, C 0, fastened to the boat C. Show how to find the ratio of the tensions in the cords. 16 TARALLELOGRAM OF FORCES. Tako any Icngtli A' to represent tho tension in A. Tlirougb A' draw a lino parallel to O B, and produce C to meet it. From tlie point of intersection draw a parallel to A, meeting B in B'. The ratio of O A' to O B' will bo the ratio of the tensions. The parallelogram might have been constructed on any two of tho lines. If tho absolute value of one of tho tensions is known, (A'0\ is known, and the tension in the other cords may now bo found. R' M Fig. 12. i R Fig. 13. Fig. 15. Fig. 14. 4. Show how to find the magnitudes of the forces when tho magnitude and direction of tho resultant are given, and the directions of tho forces. Let M represent the point on which tho forces act (Fig. 15), draw the lines M A, M B, having respectively the directions of the two forces, and the line M C, having the direction of tho PARALLELOGRAM OF F0RCE3. 17 lOA. Througli meet it. From meeting O B in of the tensions. i on any two of isions is known. ken A'0\ 50 y now bo fouucl. (= luUant, Take Bomo unit of length to represent a unit of light, and on tho scale chosen cut off M C to represent the ignitude of tho resultant R. Through C draw C A parallel to 13, and C B parallel to LI A, then M A, M B, will represent tho rocs in magnitude, for tho forces represented by these lines will l tho parallelogram of forces have for resultant the given force R. 5. Two forces which act at right angles on a particle havo 10 ratio of 9 to 40, and their resultant is 123 lbs. ; find tho bgnitudes of the forces. Since tho forces are as 9 to 40, one is nine times some unit of ^Lich tho other is 40 times. LetP = 9a .'. Q = 40a P2 = 81 a' Q2 = IGOO a- R2= 1081 a= (11a)' but R == 123 lbs. .-. 41a = 123 lbs. a = 3 lbs. •. P = 27 lbs. andQ= 120 lbs. L-ces when tho ven, and the act (Fig. 15), directions of 3ction of the Exercises. 1. Define the term force, and show that all forces may be bompared with iveiglit. 2. In what respects may a force bo represented by a straight 10? 8. What is the Principle of tho Transmissibility of Forco ? 4. Defipe the terra Rcsidta7it, and show how to find the tosultan*: of two or more forces acting in tho same straight line. 5. Enunciate the Parallelogram of Forces. 6. Show that if the directions of two forces acting on a point iclude an angle, their resultant is less than if they act in the [ame straight lino and in the same direction. 7. "What must bo the directions of two forces that their 3sultant may bo the least possible ? 8. Three ropes, P A, Q A, R A, are knotted together at the |oint A J P A is attached to a tree, Q A and R A are pulled by 18 TARALLELOGRAM OF FORCES. two men ; liaving given the angle Q A H, and tlie force exerted by each man, show how to find the pressure on the tree. 9. Show how to find the resultant of two forces Avhen the directions of the two forces, the direction of the resultant, and the magiiitude of one force are given. 10. Show how to find the direction of the resultant of two forces when the magnitudes of one force and the resultant as well as the directions of the two forces are given. 11. Show how to- find the magnitude of two forces, having given their directions and the direction and magnitude of their resultant. 12. Two forces, P and Q, act on a point at right angles ; P = IGlbs., Q =- 03 lbs.; find U.—Ans. 65 lbs. 13. If P = 13 and Q = Si, the angle being right, find R.— Alls. 85. 14. The angle between P and Q is right, P = 112 cwt. and R = 113 ; find Q.—Ans. 15 cwt. 15. The ratio of P to Q is 33 : 56, and the resultant is 130 lbs. ; find P and Q, the angle PQ being right.— .4ns. 66 lbs. and 112 lbs. 16. The rosultant being 85 lbs. and Q 77 lbs., find P when L PQ is right.— ^ns. 36 lbs. 17. P, the greater of two forces, is 3 '2 lbs. less than R, which is 304 lbs. ; find Q, the less, PQ being a right angle. — Ans. 13-6 lbs. 18. The smaller of two forces which act at right angles is 7*2 lbs., and the sum of the resultant and the larger force is 259*2 lbs. ; find the resultant and the larger force. — .4ns. 129*7 and 129-5. 19. For a given vertical force a horizontal and an oblique force have to be substituted; the horizontal force equals the given force j find the magnitude and direction of the oblique force. 20. Two forces which act at right angles are to one another as 16 to 03, and the resultant is 13 lbs. ; find the forces. — Ans. 3-2 and 12-0. ES. PARALLELOGRAM OF FORCES. 19 id the force exerted n tlie tree. forces Avlien the the resultant, and G resultant of two he resultant as well two forces, havincr magnitude of their t at right angles ; bs. ig right, find R. — P = 112 cwt. and 1 the resultant is Lglit. — Ans. 66 lbs. 33., find P when ;ss than R, which Ight angle. — Ans. ■j right angles is le larger force is )rce. — Ans. 129*7 and an oblique force equals the of the oblique ,0 one another as le forces. — Ans. [y.-APPLICATIONS OF THE PARALLELOGRAIM OF rOIlCES. 2. By means of the Parallelogram of Forces we y always find tlie resultant of two forces acting on a nt when we know the magnitudes of the forces and angle between them, but the magnitude of the ultant can be calculated numerically without the aid trigonometry only in the case of a few angles, namely, when the given angle is one of the angles of t^ following triangles : — (i.) A right-angled triangle in which the acute iaglcs are respectively 30° and 60°. Such a triangle 18 half an equilateral triangle, and it is easily proved ^at The shortest side = half the longest. The third side = ^^^ x the shortest, (ii.) An isosceles right-angled triangle (in which of irse the angles ire 45°, 45°, and 90°.) The hypothenuse = either side X n/2. 23. To find the diagonal of the parallelogram when two sides and included angle are given. B /■ -y M^ A D Fig. 16. jLot A M B c be the parallelogram, and let c d be the I'pendicular on m a ^rom the point c. 20 PAIIALLELOGUAM OF FOUCES. First, let the anprle at m be acute, as in Fior. 16, then MC^ M D^ + D C« or since M D = M A + A D MC- = (MA + AD)2 + DC' = M A2 + A C2 + 2 M A-A D. obtuse, as in Fig. 17, Secondly, let the angle at m then M C^ = M D^ + D C^ = (M A — A D)2 + D C^ = M A2 + A C^ — 2 M A-A D. Now it has been shown above that a d can be easily found from a c when the angle at m is either G0% 30°, 45^ (180°— 60°), (1S0°— 30°), or (180°— 45°). Tbus.if CAD = GO°, DA= - A C, and CD = '^ AC=-8C6 AC. 2 2 /3 1 * if CAD=30°, DA= •— AC, and CD= - AC. AC V2AC if OAD=45°, DA=DC=-r;= — ^ =-707 AC. 24. To find the resultant of two forces p and q, when they act at one of the simple angles. Let M B represent Q, .•. A C also represents Q, and let A M represent P. Then, if ^ B M A = 60°, ll' = P^ + Q2 + P Q. ifZ.BMA = 30°, 112 ^ pa ^. q2 +^3.pQ. if ^ B M A =^ 45°, R2 = P2 + Q2 -i-^2-P Q. if ^ B M A = 120°, r.2 = p2 + Q2 _ p Q. if I. B M A = 150°, r.2 = p2 + Q2 — ^3-P Q. if:.BMA = 135°, K2 = P2 + Q2_^2-PQ. Example. Find the resultant of two forces of 12 lbs. and 8 lbs. respectively, acting at an angle of 60*. U!f— CE3, rARALLELOGRAM OF FORCES. 21 e, as in Fig. 16, + 2 MA- AD. se, as in Fig. 17, y AAD. A D can be easily s either GO", 30^ 0°— 45°). 2 h -, AC. =•707 AC. forces p and q, (es. its Q, and let A M PQ. V3PQ. V2PQ. PQ. V3PQ. -n/2P Q. forces of 12 lbs. of 60^ in Fig. IG let m b = 12, ma == 8, and z. t!M a = ^ then AC = 12, and since ^ dac = 60°, a d =::: [.-. DC = (^y/o. id M D = M A + a D = 8 + 6 = 14 \.'. M c- =^i D^ + n c^ = 14= + (Cv/3)2 = 196 + 108 .-. R =1^/(304) = 17-43558. we may apply directly the formula— M C^ = M A^ + A 0= + 2 M A • A D .-. R- = 8' + 12^ + 2 X 8 X G = 304. Conditions of Equilibrium. 25. Any relations which must hold amongst the forces when there is equilibrium, and such that there is always equilibrium when the relations are fulfilled, are termed the Conditions of Equilibrium. 20. The condition of equilibrium of two forces acting on a point is tliat they be equal and opposite. 27. When three forces act upon a point, the condition of equilibrium is that any one of them shall be equal *nd opposite to the resultant of the other two. 28. When the three forces act at points in a solid body, and are not parallel, one condition of equilibrium us that their directions shall meet in the same point. 29. It is evident that we may add to or subtract from a system of forces any set of forces which are in Equilibrium amongst themselves. For example, if wo wish to find the resultant of three forces of 30 lbs., 50 lbs., and CO lbs., acting at angles of 120°, since tlirce forces of SO lbs. acting at 120° will be in equili- brium, we may subtract 30 lbs. from each of the forces, and find the resultant of the remaining 20 lbs. and 80 lbs. acting at an angle of 120°. ,:J,^MW» I> w 22 CONDITIONS OF EQUILIDRIUM. Exercises. 1. Find the resultant of two equal forces of 15 lbs. acting on a point at an angle of 60°. 2. Find the resultant of equal forces of 20 lbs. acting on a point at an angle of 30°. 3. Find the resultant of equal forces of 40 Idlogrammcs, tlio angle between them being 45°. 4. Find the resultant of equal forces of P lbs., the angle between them being 1.20°. — Ans. P lbs. 5. Find the resultant of equal forces of 100 lbs., the angle between them bjing 135°. — Ans. 7G"5. 6. Tkvo forces of 10 lbs. and 42 lbs. act upon a point at an angle of 120°; find their resultant. — Ans. 38 lbs. 7. Forces oi 8 and 12 lbs. act at an angle of C0°; find the resultant. — Ans. 17'43. 8. Forces of 9 and 11 lbs. act at an angle of 120°; find their resultant. — Ans. 10*14. 9. The resultant of two forces, P and Q, is perpendicular to F ; show that it is less than Q. 10. Two fox'ces, each equal to 50 lbs., act at an angle of 150°; find their resultant. — Ans. 25-8819. 11. At what angle must two equal forces act to produce the same effect as one of them ? 12. Two rafters, making an angle of 60°, suj^port a chandelier weighing 90 lbs. ; what will be the pressure along each rafter? —Ans. 30V3 lbs., or 51-96 lbs. 13. Fur a given vertical force two other forces are to be sulj- etituted, one horizontal and the other making an angle of 45° with the vertical ; find the magnitude of the forces. 14. The resultant of two equal forces acting upon a point at iin angle of 90° is 10 lbs. ; find the value of each component. — Ans. 7-07107 lbs. 15. The resultant of two equal forces acting upon a point at an angle of 135° is 10 lbs. ; find the value of each component.— Ans. 13-0359 lbs. 16. The resultant of two forces acting upon a point at an angle of 150° is 95 lbs.; one of the components is 55 lbs. j find the other component. — Ans. 80,^3 lbs., or 138'56. #11 ■M' iii :m m d] lUM. THE POLYGON OF FORCES. 23 )f 15 lbs. acting on a ■ 20 lbs. acting on a to kilogrammes, tlio if P lbs., the angle 100 lbs., tlie angle upon a point at an Ubs. glo of C0°; fina the 3 of 120° J find tbcir , is porpendicnlar to at an angle of 150°; 3 act to produce the support a chandelier along each rafter ? orces are to be suh- ng an angle of 45^^ brces. ng upon a point at each component. — ing upon a point at jacli component. — point at an angle 55 lbs. J find the The Polygon of Forces. 30. Wlien any number of forces f, p', f", f'", act upon a point M, wc may find tlio resultant of the whole by combining tv;o, f and f', by the Parallelogram of Forces (Fig. 18), then their resultant r' and the third force f", and r-o on. When this construction is made, it will be found that by neglecting the successive diagonals it amounts to the following rule. 31. From any point draw lines parallel and propor- tional to the forces; from ^IliG extremity of one of these lines draw a line )arallGl and equal to the lext ; from the end of this [drav/ a lino parallel and equal to the third, and so on. [Tho line joining the open part of the polygon thus formed will be parallel and equal to the line reprcsent- [ing the resultant. (Fig. 18.) 32. If the polygon be a closed polygon, the forces [haye no resultant, that is to say, there is equilibrium. JFor example, let ma, m b, m c, m d, represent forces [acting on the point m. From A draw a b', parallel and equal to m b ; from b' draw b' c', parallel and equal to M c ; and from c' draw o' d', parallel and equal to m d ; [then M d' represents the resultant in magnitude and direction. If the polygon be closed, so that d' coincides [•with M, then the forces are in equilibrium. c Fig. IS. tSj^^""'^^^^' ! i! 1^ *"J 21: THE rOLYGON OF FORCES. B ^>C A particular case of this proposition is termed the triangle of forces. 33. When three forces acthg on a particle can he reprcse7iied in marj- nitude and direction ly the three sides of a triangle taken in order j they will he in eqiiilihrium. oJr. The converse of this propo- sition is also true. When three forces acting on a 'particle are in cqicili' hrium, the sides of my triannle which are parallel to the lines of action of the forces are also proportional to the forces. 35. Let R s T be three forces acting on a point (Fig. 19), and let the sides of the triangle a b o be respectively parallel to the forces, thus A B parallel to n BO ,, to s CA ^, to T Then the first of the above propositions states that if ABiBCiOAasn: s : t, then the forces arc in equilibrium ; and the second states that if the forces are known to be in equilibrium, and the triangle A b c bo constructed >Yitli sides parallel to the forces, then AB : BC : CA as II : s : T. 36. It must be remembered that these propositions apply only to forces acting on the same point. \^ A Fig. 20. o; P fir EXEIICISES ON THE POLYGON OF FORCES. 25 is termed tlie •es activg on a 3711 ed in onag- hj tlie three I in order J they f tills propo- en three forces xre in eqicili- triannle ivhich es of action of )ortional to the ig on a point angle a b o be states that if forces arc in the forces arc triangle a b c forces, then propositions \int. Exercises. 1. Show Low to find tlie resultant of a Dumber of forces acting on a point. 2. Tiiree equal forces, P, Q, and R, act on a point, the angles A A P Q and Q R are right angles ; find their resultant. 3. Three forces, P, Q, .--nd 11, act on a point, the angles P Q and Q R are each 60°; P = 2 lbs., Q = 8 lbs., R = 2 lbs., find the resultant. — Ans. 10 lbs. 4. Seven forces act on a point so tln.t the angle between every consecutive two is 45°. The first force and the alternate forces C from the first are each 5 lbs. ; the second and alternate forces from the second are each 8 lbs., find the resultant of the whole. . — .4715. 8 lbs. 5. If two lines, AB, C A, represent two forces acting on a ' point, the one towards the point and the other from it, show how to find the resultant. G. Three forces of 119, 120 and 1G9 lbs. act on a point and Iceep it at rest ; show that the angle between the first and second is a right angle. 7. A cord is attached to two fixed points, A and B, in the eame horizontal line, and bears a ring weighing 10 lbs. at C, so A that A C B is a right angle ; find the tension in the cord. — Ans. J5721bs. 8. Two cords, A C = 44 inches, B = 117 inches, are attached to points A and B in the same horizontal lino and to a weight of 10 lbs. at C. The angle A C B is a right angle j find the pressures on A and B. — Ans. 3*52 and 9*30 lbs. 9. Prove that if two forces be represented by two diagonals of a parallelogram, their resultant will be represented by a line equal to twice one of the sides of the parallelogram. 2b v.— PARALLEL FORCES. 37. Take a bar of iron b c (Fig. 21), and attach it by the middle point d to a fine thread passed over a I pulley E capable of moving without friction. To the other end of the cord attach a mass of lead and a hook A, which will exactly balance the bar. Take a number of equal weights and suspend one of them from each of two points on the bar equally distant from d, and two others from the hook ; there will still be equilibrium. Take off the two weights from the bar and hang them one below the other at the point d : vrith this arrange- ment also there will be equilibrium. This is an illustration of the fact that two parallel forces, applied at two points in a body, produce the same effect as if they were applied together at the centre of the line which joins the points. Replace one weight at b, suspend two others at midway between c and d (Fig. 22), and hang thrtc from the hook at a ; in this case also there will be equilibrium. PARALLEL FORCEfl. 27 and attacii it asscd over a c. on. To tlie and a liook vC a numbor trom eacli of D, and two ccinilibrium. liang tlicm his arrange- Tliis is an rces, applied le effect as if of tlie line others at ' hang thrte here mil be If c" be taken as the point of support, such that D c"=^ D B, then it will be necessary when one weight B ■-^^ ^ig. 23. is placed at b to suspend four from c" and five from A. 38. These experiments, which might be multiplied aiid varied indefinitely, show that two parallel forces, acting in the same direction, can be counteracted by a single force parallel to them ; and consequently they may be replaced by such a force. Moreover, we learn that in magnitude the resultant is equal to the sum of the forces, and if one force be double or triple of the other, the point of application of the first must be twice or thrice nearer the resultant than that of the second ; in other words, the distances of the forces from the resultant are inversely proportional to the forces, Fi A C B Fig. 23. 89. Let F, f' be parallel forces acting towards the same direction at the extremities of a rod A b (Fig. 23). 28 PARALLEL FORCES. ! Let R be their resultant and let c be its point of applica- tion. Tlic principle explained above is expressed by the following equations : — R = F + F FAC = F'-BC If E be any point in the continuation of b A then F-AE + F-BE = R-CE This is the general equation for determining the position of the resultant, that is to say, it includes all cases. For example, let e coincide ^yith a, then F X + F-A B = RA C or F-x\B = RA C Let E coincide with n, tlien FA B + F X = RC B or FA B = RC B bo) act r *^ in fori in R=F-F' Composition of Parallel Forces acting towards OPPOSITE Directions. 40. Let F and f' be parallel forces, acting in opposite directions, and let f be the greater (Fig. 24). The force f may be de- composed into two forces ; one represented by b b' at B equal and opposite to f', and the other equal to F — f' at a point c, such that, (F— F)-CA = F-AB or F-AC = F'CB The forces at b will be in equilibrium, and an effective force remains at c equal to f — f'. This, then, is the resultant r of the forces f and f'. Therefore the resultant of two unlike parallel forces acting on a solid Fig. 24. 1 TAUALLEL FOP.CES. 29 t of applica- xpressed bj of BA then rmining tlio includes all then = Tf A C = TvCB fa TOWARDS r, in opposite lid let F be (Fig. 24). ay be de- .wo forces ; ed by b b' id opposite other equal a point c, n effective len, is the refore the on a solid body at points A and b is equal to their difTcrcnco, and acts in the direction of the greater, through a point c in the line a b produced on the side of the greater brce, such that F : F : : CB : C A 41. These cases may be collected under the follow- ing rule "which applies to any number of parallel forces in the same plane. To fir.d the magnitude of the resultant take the sum of those forces which act towards one direction from the sum of those which act in the opposite direction. 1 I I n Fi- 25. To find the position of the resultant draw a straight line cutting the directions of all the forces, take a point a on this line and find the distances of the forces from a measured along the line. Multiply each force by its distance from the fixed point A, before each product thus obtained write the proper sign and add all the products together. Let X be the distance of the resultant n from a. Multiply R by a: and write the result on one side of an equation, the sum of the preceding products being on the other. 42. We have yet to show how to determine the proper sign of each of the products. Imagine the line drawn from a to the force to be a rod jointed to the fixed point a, then if the force would tend to turn this rod in Jie direction of the hands of a clock make the sign o! the product + , but if the force would tend to turn the rod in the opposite direction, then make the ^V^i "V so PARALLEL FORCES. Sign — . TLns in Fig. 2G llio product is + F'a, in Fig. 27 it is — F-a, in Fig. 28 — Fa, '^nd in Fig. 2D wc liavo + F'a, -f- A— Fi;r. 26. A-^ cr Fig. 27. Fi-. 28. Fi-. 20. 43. Exami^h /. — Parallel forces of 2 lbs. and Gibs, respectively act at the ends of a rod IG inches loni^ ; find the magnitude and position of the force which will keep the rod at rest. Let B c be the rod, and let d be the point of applica- tion of tlie force required, then D is also the point of application of the (8) B f2)r ■\ — ^-4— i- (3) resultant of the given forces. The mairnitude of the resultant =6+2 = 8. To find the position of Fig. 30. the point d, we may proceed in several ways, thus : — First. Since the arms must be proportional to the forces, that is to say, as G : 2, if the rod be divided into 8 equal parts, b d will be equal to G of these parts and c d to 2. Each part will be 16 -f- 8, or 2 inches, therefore BD = 12 inches and CD = 4 inches. Second, Or we may let b d = a; then c d = (IG — ic), but by the above rule 2BD--G CD = or 2x — 6 (16 — .t) = .*. a; = 12 PAKALLEL FORCES. 31 s + F-aj in 1 in Fig. 29 TJdrdly. Wo may lot d be tlio fixed point of rcfcr- B DC r Fi-. 20. )S. and G lbs. :nclics long ; :e which will t of applica- )rce roquired, so the point on of tho the given itnde of the Ig + 2 = S. position of rs, thus : — lonal to the lividcd into 8 larts and c d ;s, therefore Ihes. (IG-a;), Fig. 31. cuce and apply the above rule, letting u d = a; then F X + F X C B = U X B D or2xO+Gxl6 = 8xa; .-. X = 12 44. The above rnle fails in* the case of two equal and parallel forces which- act towards opposite directions. Such a pair of forces has not a single resultant ; the forces form what is termed a couple. A couple can bo kept in equilibrium only by another couple Centre of Parallel Forces. 45. To find the resultant of a number of parallel i forces not in the same plane ; as, for example, when [the forces act at points in a solid body we must first find the resultant of two, then combine that resultant with a third, and so on. Let F, f', f", &c., be forces [acting at points a, b, c, &c., in a solid body. The resul- [tant r.i of f and f' will act at a point Lj in ae, such that A Li : B Li as f' : f. The resultant of Bi and f' will act at Lo in the line Li c, such that Li L2 : c 1-2 as f' : Bj. In this way we may proceed until we find Fig. 32. .. '.. iLra^libMrfMItt St I 32 PARALLEL FORCES. tlie point L at which acts the resultant r of all the forces. 4G. Now sui)pose the direction of the forces to be changed, the forces being of the same magnitude, remaining parallel and being applied at the same points as before. The resultants Ri, Rg, &c., though altered in direction; will still pass through the points Lj, Lg, &c., and the final resultant r will pass through the point l. Hence the resultant of a system of parallel forces acting at different points in a rigid hodij passes through a fixed pointy ihe position of ivhich is independent of the direc- tion of the forces. This point is termed the centre of the forces. Exercises. 1. Show how to find the resultant of two given parallel forces acting towards the same direction, the distance between the forces being given. 2. Forces of 8 lbs. and 12 lbs. respectively act at points A and B 12 inrhes apart; find the magnitude and position of their resultant. 3. Show how to find the magnitude and position of tho resultant of a number of parallel forces when the magnitudes and directions of tho forces are given and their several distances from a fixed point. 4. What is a couple ? IIow may a body be kept at rest when a couple is acting on it ? 5. The smrller of two parallel forces of 24 and 30 lbs. respec- tively is 5 inches from the resultant ; what is the distance between the larger force and the resultant ? — Ans. 4 in. 6. The resultant of two forces is OG lbs. ; the smaller force is -?,!j of the resultant and is distant from it 3 ft. 6 in.; find the otii or force and its distance from the resultant. — Ans. 3Glbs.; 2 ft. 11 in. 7. The perpendicular distance between two parallel fcroes df 1 fcaJ that 9. md tbat 98 iu 10. P Q. 6 ii). 11. of 11 ia to b tied tl e&d. 12. lon<7. .;.*»*' PARALLEL FORCES. 33 ant n of all e forces to be e magnitude, le same points hoiiirh altered nts Lj, Lg, &c., ^li tlie point L. I forces acting lirough a fixed t of the direc- \ the centre of en parallel forces ice between tlie i at points A and osition of their position of the magnitudes and III distances from lept at rest when |d 30 lbs. respec- listance between I smaller force is 6 in. ; find tha .—Ans. 30 lbs.; para,Uel fcroea cwt. and 20 lbs. respectively is 2 ft. 9 In. ; find the distance Hetweenthe smaller force and the resultant. — .4ns. 28 inches. §8. Two parallel forces which act at the extremities of a rod I ft \onrf have a resultant of 1 cwt. ; one of the forces is 35 lbs ; &id the distances of the points of application of the forces from tibut of the resultant. — Ans. 20 and 44 inches. 9. A B is a rod acted on at A and B by parallel forces P ijid Q. C is the point of application of their resultant R. Given that 11 = 154 lbs., Q = 99 lbs., AC = 5] ft.j find AB.— ^ns. !>8 in. 10. In the last example, given that A B = IG in., H = 104, and -= i; find P, Q, AC, and B C.—Ans. 39 lbs., 65 lbs., 10 in., 6 in. 11. At points equally distant on a rod 20 Inches long, weights of 1 lb., 2 lbs., 3 lbs., 4 lbs., and 5 lbs. are suspended. The rod ii to be supported by a single thi*ead ; at what point must it bo tied that the rod may remain horizontal ? — Ans. 13^ in. from the end. 12. A weight of 1.^ cwt. is carried by two men on a rod 8 ft. long. The weight Is hung from the middle; one man is 1 ft. from one end and the other 2 ft. from the other end of the rod ; find the weight borne by each. — Ans. *9 cwt. and 'G cwt. VI.— MOMENTS. 47. Suppose a rod o d, capable of turning about the %icd point o, io be acted on by a force f, the tendency of the force to turn the rod about o depends on the magnitude of the force and on its distance from the j^int 0. We might, for example, double this ten- iency, either by doubling the force, or by keeping the force the same and causing it to act at tNvice the dis- tince from o. Hence the tendency of the force to s^ 34 PARALLEL FORCES. turn the rod about o is measured by tlie product of the force by the perpendicular o d. 48. The product of any force f ly the perpendicular from a point on its direction is termed the moment of the force with respect to the 2^oint. Thus if AM represent the force f (Fig. 33), tbo moment of f about o is a m x o d. This product is numerically twice the area of the tri- angle having the line representing the force for base and the given point for apex. 49. We show therefore that the mo- ments of two forces are equal when we prove the equality of the triangles formed by joining the extremities of the lines representing the forces to the given point. 50. It is evident that the moment of a force about a point in its own direction is zero. 51. It is convenient to consider moments in one direction, as, for example, that of the hands of a clock as positive, and moments in the opposite direction as negative. The moment of a force is therefore a product, the sign of which is determined in the way described in § 42. Thus in Figs. 34 and 37 the moments are positive, ai in Figs. 35 and 36 negative. Fig. 33. HI A-^ A^ a ♦-A + a A icr. 34. Fiff. 35. 36 ris. 37. 5S mom (!• ahov the V, i (" the s -^t point (ii aboin is at ^^ point ■'I' ^1 wcig] and a \ii ^ mal horizi • Til - (ncgh B, am then < direct He Tli( TIk The PARALLEL FORCES. roduct of tlio perpendicular moment of the ?ig. 3 3), the lis product is '.a of the tri- presenting the point for apex. that the mo- 3 equal when ■ the triangles rcmities of the ;es to the given I force about a ►mcnts in one mds of a clock e direction as a product, the cribed in § 42. 3 positive, anil + GT 52. The following are the chief properties of moments : — (i.) The Slim of the moments of a mnnher of forces about any point taken icith their proper signs is equal to the moment of the resultant about the same point. (ii.) JVJien any number of forces are in cquiUbriwin, the sum of the moments ivith their proper signs about any point is zero. (iii.) When tu'o forces act on a particle the moments about a point in the direction of the resultant are equal. Example. A liorizontal beam A B is attached, by a joint at its middle point c, to a vertical beam c d. A weight of G cwt. is suspended from a, and a force r, applied at c in a direction making an angle of 45° with the horizon, keeps a e horizontal. Find r. There are three forces actinof on the beam a b (neglecting its weight). The weight at a, the force at B, and the reaction of the hinge at c. Let a c = a, then c B = a, and if we draw c e pernondicular to the B X Fig. 38. direction of p, angle c b e = 45°, and c e o. 72 va ITcncc the moment of p about the i)oint c = — s/2 Ihc moment of the weight about c == ~ 6 «. The moment of the reaction about c = (§ 50). Therefore the sum of the moments is -— — Gfl5 + 0=-0 \ p = GV2=G X 1-414! 8-4852 cwt. II Fi-. 37. SG 'I TAIIALLEL FORCtS. Exercises. 1. Show tliat when three forces act along the sides of a triangle, as in the figure, they cannot be in equihbi'ium. 2. Explain what is meant by the moment of a force about a point. State the principal properties of moments. 3. Show by the Principle of Moments that if three forces act on a body, and the directions of two of them pass through a fixed point, there cannot be equilibrium unless the direction of tho third also passes through the fixed point. 4. A metal rod A B is bent at its middle point C, so as to form an angle of 105°. It is then laid across a smooth peg, and weights P and Q are attached to the ends. When the rod has assumed the position of rest, it is found that a plumb-line at C divides the angle A G B into angles of 60° and 45°. Prove that P» : Q2 : : 2 : 3. VII.— CENTRE OF GRAVITY. 53. The attraction of the eartli wliicli causes a body to have weight, acts on every particle of the body; thus, if we take a stone and pound it into small frag- ments, the sum of tlie weights of the small particles will be equal to tliat of the whole stone. If one of these particles be attached by a fine thread to a fixed point o, the thread will take the direction of the vertical through o. If several of them be sus- pended from points near together, the threads will be parallel. When therefore the particles are united so as to form the body, we may regard their weights as a system of parallel forces. Suspend the body by a point a (Fig. 40), the re- sultant of the weights of the particles will be equal to th th( bO( wci Ban api of wir . S.11H follo^ 54 on a of iv] Ifth the be Wl ports centre two fc vcrtici snppo; G are hence, be sup CENTRE OF GRAVITY. Or-; ; along tlie sides ty cannot be in e moment of a .cipal properties tliree forces act tlirougli a fixed direction of tlio )int 0, so as to smooth peg, and hen the rod Las plumb-line at C 15°. Prove that |causes a body of the body; small frag- all particles a fine thread le direction of liem be siis- [rcads will be ire united so weights as a 40), the re- ll be equal to Fig. 40. their sura, and will liave the direction of tlie vertical through a. Suspend the body again from another point ; the weight of each particle will have the same magnitude and tlie same point of application as before, but the direction of the forces with regard to the body will be changed. The result is the same as if each force had been turned about its point of application ; hence the new line of suj)port or direction of the resultant will intersect the old one in the centre of the forces. If the body were composed of a plastic material, and pierced in the direction of the line of support in several different positions, all the lines of perforation would intersect in a common point. This point is termed the centre of gravity of the body. We are led therefore to the following definition : — 54. The resultant of a system of parallel forces acting on a ingid body passes through a fixed point the position of which is independent of the direction of the forces. If the forces he the weights of the several elements of , the body, the fixed point is termed the centre of gravity. When a body is suspended from a point a it com- ports itself as if its weight were concentrated at tlio centre of gravity g. We may consider therefore that two forces act upon it — the resultant weight along the vertical through g and the reaction of the point of support along the vertical through a. When a and G are in the same vertical line the body is at rest ; hence, if the C. G. be supported, the whole body will . be supported. ;38 CENTRE OF GRAVITY. hecl, one or several levers, Z, c, (Fig. C8), may be placed in the axle as in tho ivindlass. 82. When the axle is vertical, and the levers hori- zontal, the machine is termed a capstan (Fig. GO). i Toothed Wheels. 83. Several wheels and axles are frequently ccm- bined by means of spur wheels (Fig. 70). Several conditions must bo satisfied in each pair of wheels. The teeth on each iTlieel must be equal to one another^ [ GO WINDLAHB a:su capstan. and equally distant, ond the teeth of one wheel should be of the same size, and as far apart as those of the n:. CS. Fig. CU. rad woi; lie and reac tl icr I'lir a ?'((( pinio tl ir Ijniid TOOTUED "WHEELS. 61 ^\Leel in contact with it. The number of teeth will therefore be proportioncal to the circumftrences or tlie radii of the wheels. Fig. 70. Equation of equilibrium/or tiro-tootlicd wheels, Suppose tAYo sucli ^Yhc•els to be in equilibrium with a woiiilit v.\ at the axle of the larger, and Wo at the axle of tiic smaller. lA^t the radii of the wheels be \\^ and n,, and let the radii of both axles be r. Also let r be the reaction between the wheels at their point of contact. Tiie ecpulibrinm of the larger wheels gives wf?- ^^-^ v w^ The e(|ullibrium of the smaller . . . \\.:v=V'i\^ Wi I\i No. of teeth in larc^er. therefore Wo 1^2 ^'^- of teeth in smaUer. St. Sometimes the teeth are placed on a straight bar instead of awheel. Such a toothed bar is termed a rack^ and the wheel in contact with it is called a pinion. The jack (Fig. 71), used for lifting great weights through a small height, is a rack and j)inioii. The linndle by which the rack is raised is termed a winch. The instiumcnt is an example of a double wheel and 'I il i i If V?' I i 62 COMT'OUKD WHEEL AND AXLB. Figs. 7 1 and 72. axle ; the ^Yincll takes the place of the first wheel, the lower pinion being its axle. The upper pinion is the second axle and supports the rack. The large toothed wheel of the upper axle works in gear with the lower pinion. The two pinions are usually, as in the figure, of the same size. Tojind the equation of equilibrium. Let V be the power applied to the winch, w ^he weight of the rack and its load, a the radius of the wmch, h the radius of the large toothed wheel, v the radius of each pinion, and ii Uie mutual reaction between the first pinion and i\\?i large wheel ; then from the equilibrium of the first axle we have p'r/==R*?\ and from the eqn ilibriimi of the second axle we have w * r = R ' b. Therefore v • ah=^vf ' r^. 5. EXERCISES ON LEVERS. 63 Exercises. 1. A weiglit of 30 lbs. balances a -wciglit of 20 lbs. at the ei- trenntics of a straight lever 15 feet longj find the length of the arms. — Ans. 6 foot, 9 feet. 2. The arms of a lever are 7 in. and 9 in. in length, and the weight 3 lbs. is attached to the shorter arm ; find the povrcr. — Ans. 2^ lbs. 3. If one end of a bar rest on abeam and a weight of 501b.i. be suspended from it at one-fifth of its length from the beam, v.'liat power at the other end will support the weight, and what will be the pressure on the beam ? — Ans. 10 lbs., 40 lbs. 4. A beam 4c^ ft. long is supported horizontally by two props at its extremities and produces a pressure of 4 lbs. ou each prop; where must a weight of 3G lbs. be placed that the whole pressure on one prop shall be 10 lbs. ? — Ans. 9 inches from one prop. 5. When two weights, 15 lbs. and 5 lbs,., arc suspended at the ends of a lever, the fulcrum is 9 feot fxora the smaller weight ; where must the fulcrum be when the weights are each increased 5 lbs. ?—Ans. 4 ft. from ono end. 6. The C. G. of a wheelbarrow and its load, which weigh 100 lbs., is in a vertical line 18 inches from the centre of the wheel ; what power applied to the handles at a distance of 3 ft. 6 in. from the centre of gravity will just lift the barrow ? — Ans. 30 lbs. 7. A beam 12 ft. long balances about its middle point; about what point will it balance if a weight ecjjaal to twice that of the beam bo placed at one end ? 8. A beam A B 10 ft. long and weighing 56 lbs. balances about a point 3 ft. from A. When a weight is placed at B, the beam balances about a point 1*4 feet from B; find the weight. — Ans. 224 lbs. 9. The pressure on the fulcrum is 7, and the sum of the forces 13; find their distance from the fulcrum when the forces are 14i inches apart. — Ans. 6 in. and 20 in. 10. Find the tru3 weight of a substance which, when placed in II [i^ *' 1 '. 1 \f ti > H%r 111 64 EXERCISES ON LEVERS. ii 1^1 ( ' scale of a balance, seems to weigh 140 grammes, and in the other appears to weigh 154*35 grammes. — Ans. 147 grammes. 11. If in a balance one arm be "98 of tlio other, and a body placed in the scale of tlie shorter arm balance 14"7 kilogrammes in the otlier scale, find the true weight of the body. — Ans. 15 kilo- grammes. 12. The beam of a false balance is attached to one arm of a true Valance, and weighs 2 lbs. A body placed in one scale of the false balance requires a weight of 3 lbs. in the other, and then the whole weighs 7'0 lbs. ; find the weight of the body and the ratio of the arms. — Ans. 2'9 lbs. ; 30 : 29. 13. The weight of a steelyard is 1 lb., the movable weight also 1 lb., the point of suspension of the body 8 inches, and the C. G. cf the beam 3 in. from the fulcrum ; graduate the beam for weights from 1 to 12 lbs. 14. What efToct is made on tne graduations by increasing the movable weight ? 15. "With a wheel and axle a power of 8 lbs. sustains a weight of 12 lbs. ; what is the radius of the axle, that of the wheel being 24 in. ? — xins. 16 in. 16. The circumferences of wheel and axle are respectively 1 yard and 15 inches; what power will sustain a weight of 1 i tons ? — Ans. 12 5 cwts . 17. Find the pressure on the catch of a ratchet- wheel (Fig. 72) 12 inches in diameter when it is attached to an axle 5 inches in diameter, sustaining a weight of 60 lbs. — Ans. 25 lbs. 18. A capstan turned by two horses is used to draw in a boat; the levers to which the horses are attadisd are 12 feet long, and the radius of the axle is 18 inches. When each horse is pullinjT with a force of 7^ cwts., find the tension of the cord attached to the boat. — Ans. 6 tons. 19. A uniform bent lever, the weights of whoso arms are 5 lbs. and 10 lbs., rests with the shoi'tor arm horizontal, what weighi must be attached to the end of the short arn". that the lever may rest with the long arm horizontal ? — Ans. S7i lbs. 85 cular wood rountli tliroiij ally V the d Avhicli pulley (Fig. howcv replac( til en t PULLEVS. 65 nd in tb<) mmes. idy placed OS in the 15 kilo- of a true le of the and then 1 and the ight also he C. G. beam for asing the a weight leel being pectively eight of is- 72) nchea in a boat ; ug, and pulling ttached e 5 lbs. weighi er may The Pulley. 85. A T^iilley is a cir- cular disc of metal or wood, capable of turning round an axis passing through its centre. Usu- ally a groove is cut in the disc to keep a cord which passes over the pulley from slipping off 1 (Fig. 73). Sometimes, however, the cord is replaced by a strap, and then the edge is convex (Fig. 74). The pulley may be considered as a lover with equal arms, A o, V, (Fig. 75), so that forces p and p' in equi- librium, at the extremi- ties of the cord, are equal. Whether the two parts of the cord be parallel or not these forces are equal, so that if one end of the cord be attached to a dynamtter (Fipr. 70), the indication of Fig. 73. Fig. 74. }M \- if Hi' G<; PULLEYB. j,Jr vP« Fi;?. 75. the instrument will be the same as if a weight equal to p were attached to it directly. A pulley, the axis of which is fixed in space (Fig. 73), is termed a/urj J pulley, and in mechanics serves tlio purpose only of changing the direction of the power. If the axis be movable, the pulley is termed a viovahh pulley. Let such a pulley with a weight q attached, bo supported by a cord CBAD (Fig. 77), fixed at a point c, and having at the other extremity a force p. The cord b c exerts a certain pressure T on the fixed point c, and this point reacts with a force equal and opposite. The pulley is then supported by the three forces r, t, and q, which may be supposed to act at the points A b and o. That there may bo equilibrium it is necessary that the directions of these forces shall meet in one point (§ 20). Let i be the point in the direction of q, in v;hich p and t meet. Fig. 76. Fig. 77. Tlic and that if wi find Q. resui 2 p c is PULLEYS. C7 will be a weight attached A pulley, li is fixed •.73), is :iliej, and Jrves tho changing he power, movable, ermed a lley with iched, be a cord '), fixed 1 havinc: remit J n :ord B c pressure point c, reacts ual and nilley is by the , and Q, iipposcd may bo )f these be tho r meet. The triangles o i i and obi are equal in every respect, and therefore o i bisects the angle a i b. It follows that the forces r and t are equal, and, consequently, if wo know the angle i, and the weight q, we can find p, or if we know r and the angle i, we can find Q. When the cords a d, b c are parallel, q is the resultant of two equal parallel forces, and therefore 2 p = Q. In this case the pressure on the fixed point c is also equal to ^ q. Several movable Pulleys with separate strings, SG. In this system of pulleys each movable pulley has a string of its own, one end of which is attached to the beam, and the other end to the next pulley, the power being applied to the free end of the cord passing over the highest pulley (Fig. 78). If we neglect the weights of the pulleys, and take t, t', t" to represent the tensic'^ of each separate cord, then T = ^Q;T'=^T==iQ;T" = ^T' = ^Q and T" = P ; hence P = - Q. o If there be n movable pulleys, then P = -- -Q, If we are required to take into account the weights w w' w" of the pulleys, then T = ^ (W+ Q);T' = kw' + T);P = \ (W" + T) 2 2 therefore if there be 7i pulleys 'I ^1 'I Wa ^==^n Q + ^W 2«-i W' + ).»— a W". &c. iii- ilh! C8 PDLLEVS* 11 w Fig. 78. Several movable Pulleys in one block, the same string passing round all. 87. Ill tills case wo suppose the parts of the string hetwecn the pulleys to be parallel. The tension of the string is the same throughout, and is equal to r ; hence, if, as in the figure (Fig. 79), there are three movable pulleys, six parts of the string support iiie weight, m vj'ij, rn.T.Kvs. no .<;'*■ .'J:*' .f_--. same B'l 'I ^1 I string lof the licnce, )vablc sight, :''i"i?iii':ii:,ir:il hfi""' IP? Mt'iiiiillll' " m mu Fi2. 79. %. i:5# ■ : 6iW-"'!l ; .':.■ M Fig. 80- ^ ^('r 70 rULLEYS. each l)earing | ; hence p = -J q. If there be n strings from the lower block P =- i Q. n The weight of the pulley will be taken into account by adding it to Q. w> Several Movable Pulleys with Separate Strings, one end of each being attached to the Weight. Let P bo tlio power and W the weight. The tension in the strint^ which passes over the first pulley A is P, that in the string over B supports the weight of the pulley A, and the tensions in the two parts of the strinc; over A. If wo neglect the weights of the pulleys, the tension in the string over B is 2 P. Simi- larly, the tension in the string over C is 4 P, and so on. Now all the strings support the weight, consequently W = tho sum of the ten- sions -- P (1 + 2 + 22 + 2=^ + &c.) By summing this geometrical progression, wo obtain W = P ^ ^A (2'^ — l)j '?'' being tho number of strings attachcl to the weight. If tho weights of the pulleys w-^, w^, wg, are taken into account, we find the several ten- ay Wa, &C. B©£ Bions to be Ti = P To =- 2 P -h «'i To = 2T. + w, 4 P •{- 2 v; I + tP2 Honco W 2n-ip^ 2"-- to, -{- + 4- 2 P -f- n\ + 4P -i- 2-iUi 4.7P- w M— 1" -{■ 8P-I- itJi ■V 2 4 ^ 1 1, i hi m 76 THE WEDGE. (Fig. 84) Instead of to the body. The condition of equilibrium remains the same, namely, that F height "W base. The Wedge. 89. A wedge is a triangular prism used as a movable inclined plane, and is employed to sepa- rate bodies that are urged together by great pressures (Fig. 85). The resis- tances to be overcome act perpendicu- larly to the faces in contact with the body, and the force is usually applied perpendicularly to the base ; hence a section Fig. 85. of the wedge perpendicular to the faces sustaining the pressure is a triangle, the sides of which are perpen- dicular to the forces, and therefore, when there is equi- librium, proportional to the forces. Let p be the pressure necessary to keep the wedge in its place, R the pressure on each face, b the breadth of the base of tlie triangular section, / the length of one of the equal sides ; the equation of equilibrium is therefore R I In practice, howovor, the friction between the surfaces plays Bucli an important part, and is eg largo coraprtred ^vith the power, that tho above p-roportiou ca.nnot be appli xl with any degree of accuracy. Again, the power usually employed with tho wedge is not pressure but percussion, and we cannot accu- rately state the relation between the force of a blow and tho resistance it overcomes. Nevertheless one fact shown to b3 true THE SCREW. 77 when friction is neglected is also found to bo true in practice, viz. that the more acute the angle of the wedge the more powerful is the instrument. li J .1 i vys ho ny ith ;u- ho ue The Screw. 00. When an engineer wishes to make a road to the top of a very steep hill, lie carries the road round the hill, making it ascend gradually. In a similar manner a path may be made from the bottom to the top of a column by a winding shelf or ledge. k'ncli an arrangement would be a screw (Fig. 8G). A screw may he considered as an inclined j^^ane tvound round a cylinder. The projecting coils are termed the threads of the screw. They may be square as in Fig. 88, or triangular as in Fig. 87. The distance between the upper edge of one thread and the corrospondii.g edge of the next, measured on a line parallel to the a>'is, is termed the distance between the threads. The screw is usually connected with a concave cylinder termed a nut, ' on the interior surfaca of which a spiral cavity is cut, corresponding exactly to the thread of the screw which moves in it. 4i :■'« I\ Hi ill I ; 78 THE SCREW. •|llBllOBjillJiW!"=,, ,i«iiiui»iii's;i '';,'; Fig. 87. Fig. 88. The Condition of Equilibrium when a weight is supported by a Screw. 91. Let the weight w be supported by a force p, applied at the circumference of a screw in a (I'vfction perpendi- cular to a plane through the axis. Tlie arrangement is equivalent to an inclined plane acted on by a hori- zontal force (Fig. 84). The condition of equilibriam . . ,, . P lieiofht IS m this case .r- = W base. Now suppose the screw to be urn oiled from the cylin- der. The whole inclined plane thus formed will be similar to ihid portion of it which would go exactly on^e round the cyiindcr. Let abc he such a portion (Fig. 90). ".^ssss^sai :«*:.:4- THE SCREW. 79 m r-A, Q IS Then i is ilie circaniforcucc of the cyh'ndtT, bc the distaiiCv bet.Tcen the threads. Let ac = c, bc = d then -- = — W c Fig. 83. Fig. 00. Tlie screw is, however, rarely used without a lover. Let F be a power applied at the end of a lever, the length of which from the axis of the cylinder is Z, and let r be the radius of the cylinder. Let p be a force applied at the circumference of the cylinder, which would have the same effect as f, then bj the principle of the lever, r"r = f'Z or r = F- I r Substitute for p in the above equation, and remark that c = 2;rr, therefore Fl d d , ^ iL 27r/ d , F — hence -- 27rr W Wr c or the power is to the weight as the distance between the threads is to the circumference of the circle described by the power. We may, therefore, increase the mechanical force of the screw, either by diminishing the distance between the threads or by lengthening the arm of the lever, or by both. !',» . ii 1.1 3r' i lii^l *1 sMi I 'I 80 THE SCREW I'TirSS. The screw is commonly used to exert pressure, llie press represented in Fig. 1 is a familiar example. The Fig. 9i. upper beam supporls tlie nut which is fixed. The heaa c of the scrcWj moving in a socket fixed to the block D, is pierced for the insertion of levers. The screw is some- times associatctl wiili a spur wheel, and is then tcrmtd an e^idlcss 5"rew, because the teeth of tlie wheel succeed aud ropli-xe each otlier as thoy a^ vanc<^. so that they never arrive at the end of the screw. Ylj. 92. f -ma EXERCISES ON THE INCLINED PLANE. 81 The Exercises. 1. If tlie power represented by P act in a direction parallel to llie plane, and support a weigut W, solve the following : — I. W = 100 lbs., hcit^lit -= 3 ft., length =- -t ft.; find P.— , Ans. 75 lbs. 3. W = 122 lbs., hciw,^. -= 11 in., base »= E ft; find R.— Ans. 120 lbs. 3. P==351bs., lioight = 15in., length=27in.; find W.— Ans. 03 lbs. 4. P=R; find the inclination of the plane. — Ans. 45°. 5. P= 160 lbs., height = 32, base --= 255 ; find \Y—Ans. 1,285 lbs. G. W = 14,207 lbs., height =-- 72 ft., base = 1,295; find P. — Ans. 792 lbs. 7. A railway train, weighing 50 tons, is supported on an inclined plane, rising 1 ft. for every 50 ft. of length, by a rope attached to a stationary engine ; find the tension. — Ans. 1 ton. 2. Solve the following exercises, the direction of the power being horizontal : — 1. W = 15, height = 6, base=9; find P.— ins. 10. 2. W=36, heights 5, length = 13; find V.~Ans. 15. 3. P =3-5, baso = 24, length^ 25; find W.— 4tis. 12. 4. W=9, heights 28, base =^45; find R.— .I'/is. 10-G. 5. P = ^R; find the inclination. — Ans. 30°. G. P acting parallel to the plane, or 2 P acting horizontally, will support W; find the inclination. — Ans. 60°. 3. A weight oi .?53 lbs. is supported on an inclined plane rising 88 ft. in a length of 137 ft. by two equal forces, one being hori- zontal and the other parallel to the plane. Find the forces. — Ans. 02 lbs. (A horizontal force P will support P x base ■— height) A force P parallel to the plane will sujipcrt P x length -i- height •• P 88 -" 4. A vertical force of 20 Fba. and a horizontal furce of 84 Iba. S-'1 ■ I . * 'I 82 EXERCISES ON THE INCLINED PLANE. -1^ support a weight on an inclined plane, the liciglit of which is 21 and length 221. Find the weight.— ylns. 900 lbs. 5. A weight of 80 lbs. is supported on a smooth inclined plane, the inclination of which is 30°, by a string attached to a point in the plane. Find the tension of the string. 6. What horizontal force will support 100 lbs. on a plane inclined at an angle of 45° ? 7. The diameter of a sci-ew is 7 in., and the distance between the threads one-fourth of an inch ; what power applied at the circumference of the screw will support a weight of 110 lbs. ? — Ans. 1"25 lbs. 8. When the circumference of the screw is 12 in., and there are three threads to the inch, find the weight which will bo supported by a power of 10 lbs. — Ans. SGOlbs. 9. In a common press the diameter of the screw is 6 in., the distance between the threads ^ in., and the length of the lever, measured from the axis, is 4 ft.; what power will support a resistance of 352 lbs.? — A7is. '7^33. 10. If the circumference described by the end of the lever bo 10 ft., the power 10 lbs,, and there be three threads in 2 in., find the resistance supported. — Ans. 1,800 lbs. 11. If a screw be formed upon a cylinder whose length is 10 in. and circumference 4 in., how many turns must be given to the thread, in order that the power may bo one-eighth of the weight ? —Ans. 20. 'w 83 ■^*r i Work done by a Machine. 02. Wlicn a weight is lifted through a certain height, the product of the weight in pounds and the height in feet is a mea'iure of the work expended. 'L'hc work expended in hfting 10 lbs. to a heigiit of 5 ft. is repre- sented by the number 50. Now the following proposition is true of all machines if we neglect friction: — The work done by the power is equal to the work done hij the iveight. If therefore we imagine the weight to be lifted to a certain height //, and then find the space, ;?, through which the power will move, the equation vf h = r /? is always true. Wo will now apply this principle to determine the equations of equilibrium for the simple machines. 93. The Lever. Let a b be a lover, of which c is the fulcrum (Fig, 93), and let a power p at a balance Fig. 03. a weight q at b. Suppose the lever to move about the fulcrum, then a will describe an arc a a', and b an arc B b'; then by the above principle p'a a' = q-b n'. But the spaces a a' b b' are proportional to the arms a c and b c, therefore the lengths of the arms may be substituted for the spaces, thus P-AC = QBC 94. The Wheel and Axle. Let c be the circum- III tl !^T;- 84 ^VUllK. 'f» illl Fig. 94. fcrencc of the \vhcol, and c tliat of tlie axle. Imagine the wlicel to go round once, then evidently p ^viil descend through a space c, and q will ascend through a space c, and by equating the work done we have Qc = PC. Since the radii of circles arc proportional to the circumference, we may substitute a o, bo for c and c, and then we have Q-AO = r-BO 95. The Pulley. Let us first consider the single B ^, movable pulley with parallel strings. Suppose the weight to rise through a height h, then the string on the right will rise through a height h in con- sequence of the rise of the pulley, and also through a height It in consequence of the shortenincr of the string: on the left. There- fore the point of application of the power will move through a space 2/i : hence W7i = P-2/i or W = 2P In the arrangement represented in Fig. 78, if the weight be supposed to ascend 1 foot, it may be shown, by repeating the above reasoning, that the second pulley will ascend 2 feet, the third pulley 4 feet, and nth pulley reckoning from the lowest 2""^ ft., and therefore the power will move through 2" feet : hence W == 2" • P In the arrangements represented in Figs. 79 and Fig. 95. WOl:K. 85 80, if the weight ascend through h feet, each of the strings attached to the weight will be shortened h feet, and consequently, if there be G parts of the string, the power will move through Gh feet .-. W/i = P-O/t W »= G P Similarly, if there be n parts of the string to the lowei block \V = n P 96. The Inclined Plane. Tirst, suppose the force to act in a direction u parallel to the plane. Imagine the body to be moved from a to a— - B, then the weight I'ig- ^0. will have been lifted through a height b c, and the power F will have acted in the line of its direction through a space A d. Therefore, by Cfjuating the work done, WBC = F-AB. This result is the same as that obtained in § 88. The ratio of the power to the weight in this case of the inclined plane was first found by Stevens in the six- teenth century, by the fact that when a chain passes over the top of a smooth inclined plane, so that a part rests on the plane and a part hangs vertically, there ^^ill be equilibrium if the two extremities are in the same horizontal line. Secondly, let us suppose the force to be horizontal. Imagine, as before, that the body moves from a to b, then the weight will have been lifted through a height B c, and the power will have acted in the line of its direction, that is to say, horizontally through a space IMAGE EVALUATION TEST TARGET (MT-3) <. \° ^ ^ 'Qr Qf % 1.0 I.I 1.25 I l_ 1^ 1.4 IM [2,2 IM 1= 1.6 V <^ /^ 'cr^l Photographic Sciences Corporation # V <^ :\ \ ^9) V >, ^ '%^ <^ » 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 86 EXERCISES ON ^YOP.S. equal to A c. Hence, by equating the work done, we have W-B C = FA C. 07. TJie Screw. Let the circnmferonce described by tlie power be c, and let tlie distance between two tlireads be (/. Imagine the power p to act at the end of a lever and to move once round. The screw will rise tlirough a space equal to tlie distance between two threads, con- sequently, by equating the work done by power and weight, we have Exercises. '* Solve the following problems by equating the work done by the power and that done on the weight. 1. A wheel and axle is used to support a weight of 130 lbs. the circumference of the wheel is 8 ft. and that of the axle 6 in. ; find the power. 2. Sketch a system of pulleys by which a power, by acting through 64 ft., would raise a weight through one foot. If the power in this case is 10 lbs. : find the weight. 3. In the system of pulleys represented in Fig. 81, if the weight of A be 3 lbs., and tliat of B be 4 lbs., find the work done by the pulleys when the weight is raised ?..ft. 4. In the preceding example find the weight when the power is 20 lbs., the weights of the pulleys being taken into account. 5. The height of a plane is 5 "2 ft. and base 67"5ft. ; what weight will bo supported by a power of 13 lbs. acting along the plane ? — Ans. 1G'J"25 lbs. 6. When a power of 20 lbs. is apjdied to lift a weight through 2 in. the power descends through 3 in. ; find the weight. — Ans. 30 lbs. 7. With a wheel and axle a power of 14 lbs. balances 1 ton, if EXERCISES ON WORK. 87 no, we bed by breads a lever ■ougli a s, con- er and done by 130 lbs. xle 6 in. ; tlie power desccnclod tlirongb 80 in., through what height wouhi the weight be raised ? — Ans. 4 in. 8. When there are four movable pulleys in one bloclc, how much string passes through the hands in raising a weight 6 in. ? — Ans. 4 ft. 9. With a block of three movable pulleys, how much cord would bo required for a man to raise himself 30 ft. ? — Ans. 210 ft. 10. Show that, in order that the pulleys in Fig. 70 may revolve in the same time, the diameters of the lower block must be as the numbers 1, 3, 5, and those of the upper as the numbers 2, 4, 6. 11. The circumference of the circle described by the end of the lever of a screw is 9 ft., and there are two threads to the inch ; what power will support 14i lbs. with such a screw ? 12. The length of an inclined plane is to the height as 85 to 13 ; if the power be horizontal and equal to 52 lbs., find tho wcight.- -Ans. 8oG lbs. 3y acting If the 1, if the the work he power 3Count. t. ; what [along the through kveisht. — ■ 1 ton, if -Jk »a ] I DYNAMICS. I.— MOTION. 1. The examples of motion wliich constantly recui to our observation are very varied. Sometimes a body moves in a straight line, as when it falls freely to the earth. Sometimes it describes a curve, as when it is projected in a direction not vertical. Sometimes the body turns and retraces its path, as is the case with the ball of a pendulum. The velocity with which the body moves, and the nature of the forces which pro- duce motion, are other elements admitting of endless variation. It will evidently be necessary, therefore, to consider some of these circumstances of motion apart from the others. For example, it will be convenient to consider, first, the conditions of motion of a body independently of its size and shape, and, making an abstraction of these i)roperties, to consider the mo- tion of a very small particle of matter or a material point. 2. AVhen a moving point or particle is in motion, the line containing all its successive positions is termed BIOTION. 89 rocni \ body to the it is es the e with fch the 1 pro- ndless re, to apart enient body Ing an nio- [ ate rial lotion, termed the patJi of tlic point. The path may be a straight line or a curve. 3. The motion of a point is said to be uniform ^Yhtn the point passes over equal spaces in equal times. In the case of uniform motion the velocity of the moving point is constant, and is measured by the length of path passed over in a unit of time. This length is usually expressed in feet, and the time in seconcls. Frequently, however, other units are chosen ; thus, a train may proceed with a speed of 40 miles an hour, a ship may sail with a speed of 10 knots an hour. Velocity expressed in other units may, however, be readily reduced to feet per second. For example, 1 mile an hour = 1 ^^ ft. per sec. The velocity is variable when the lengths of the patli described in equal times are not equal. 4. The length of path described in a certain time divided by the time is termed the mean velocity for tliiit portion of the path. 5. Variable velocity at any instant is measured by th" mean velocity for an infinitely small space com- menced at that instant. It is the space the body would describe in a unit of time if from that particular instant the velocity remained constant. For example, a horse may travel fror^ one place to another with a variable velocity ; but we may say that at a particular instant he is running at a speed of 20 miles an hour. We mean that for a small distance he moves with a speed which, if maintained fur an hour, would carry him over 20 miles. 90 UNl^wnai VELOCITY. Uniform Velocity. 6. Let V ft. per second be the velocity of a moving point. If the velocity be uniform, the point will move through V ft. in every second of time ; hence in 2 seconds it will move through 2 vft., in o seconds through 3 vft., and so on. Consequently, if the point move for t seconds, it will j^ass through t v ft. Let s ■■= the space passed through in t seconds then s = \ t. 7. Suppose two straight lines be drawn at right angles, and v units of length bo measured on one and t units on the other, then these lengths may be taken to represent respec- tively the velocity and the time, and since the area of the rectangle of which these lengths form the sides is represented by the product of the two numbers v and ^, Fig. 97. J! this area may be taken to represent the space. Variable Velocity. 8. When the velocity is variable, the rate of increase or decrease must be known in order that the motion of the point may be determined. When the velocity varies uniformly, the increase or decrease of velocity per second is termed the accclera- tion. 9. The acceleration is usually denoted by/; so that y== the number of feet added to the velocity in every second of time. For example, if the velocity of a point at one instant be 30 ft. and a second later 4-0 ft. per second, then /"== 10 ft. per second. If at one instant the velocity by VARIABLE VELOCITY. 91 be 10 ft., and three seconds later 70 ft., the acceleration is CO ft. -f- 3, or 20 ft. per second. 10. From the preceding definitions it lollows that if V be tiie velocity at tlie commencement of the time, the velocity at the end of the 1st sec. =r v + y 2nd sec. = v + 2f Srd sec. = v + 2>f at time t == v -^ ft 11. Let us take a vertical line, as in Fig. 98, on which we can represent intervals of time, and at the points representing dilTerent instants of time draw horizontal lines proportional to the velocities at these instants, tlien by joining the extremities of these lines we obtain a fi-^urc the area of whicli represents the space passed over precisely, as in the case of a constant velocity. 1 2. Let the velocity at the commencement of tlie motion be 0, then at the end of a second the horizontal line will represent/* ft. at the end of 2 seconds 2yft., and so on. The space passed over in the time t will now be represented by a tri- angle. Now the area of a triangle is found by multiplying half the base by the height. .-. s = \ftK 13. If the initial velocity be v, the velocity at the end of t seconds will be V 4- ft^ and the space will be represented by Fig. 99, which con- sists of a rectangle, whose area is v tt and a triangle, whose area is \f^ .-. s = Yt +\ft\ II 1 i-j. 'j'd. 02 VARIABLE VELOCITY. 14. If the velocity be diminished uniformly by /ft. per second, wc have the velocity at the end of t seconds, v = v — ft^ and the space s, represented by the rectang, a b c d (= v t) — the triang beg (== -J/^") E C ,-, s = yt — yi\ Fig. 100. 15. The area of either of these figures may also be found by multiplying the mean breadth (i.e. half the sum of tlie upper and lower breadths) by the height ; hence the sjj ace passed over hu a particle moving with uniform acceleration is found hi/ multiplying half the sum of the initial and final velocities hy the time. 16. From the two formula? of motion V = Y ±ft s ^Yt ± \ff' we may obtain a third, which does not involve the time, thus : — Multiply the first by t and subtract twice the second, . , ^_±_ V + V But from the first, t =-\- — - — ; by equating these values of t, we obtain u'^ = V^ + ^f^- 17. We have now three equations of motion, v=_- V ±ft (I.) s = yt ± ift' (n.) v^^= Y' ± 2fs (ill.) Where v is the initial velocity, f the velocity added in every second, t the time daring which the body has been moving. veJ at sec fine 3 4. veloc 8 sec EXAMPLES OF UNIFORM ACCELERATION. 93 Examples. 1. A point moves from rest with uniform acceleration; the velocity at the end of three seconds is 18 ft. ; what is the velocity at the end of seven seconds ? The velocity gained in 3 sees. = 18 ft. The velocity gained in 1 sec. = V ^^ ^ ^* ,'. The velocity gained in 7 sees. = 7 x G == 42 ft. 2. A body moves at one instant with a velocity of 30 ft. per second, and 5 seconds later with a velocity of 50 ft. per second ; find the acceleration supposing it to be uniform. Velocity at the end of 5 sees. ^ 50 Velocity at the beginning =rr 30 ^l^ Gain in 5 sees. -- 20 Gain iu 1 sec. 4 hese 3. Find the space passed over in six seconds by a point which moves from rest and acquires a velocity of 30 feet. 'Eir^i metliod : — Initial velocity = Final velocity ==30 = 15 90 )Gen By adding and dividing by 2 we have the mean velocity . . . Therefore the space = 15 x 6 = Sixond mctliod : — Velocity gained iu 6 sees. . = 30 ft. .'. The velocity gained in 1 sec. = 5 ft. Thus/= 5. But S = ^/i3 .-. s= i X 5 X 36 = 90ft. 4. If a point start with velocity of 20 ft. per second and gain a velocity of 10 ft. per second, find the space passed through in 8 seconds. Hero V = 20, / = 10, and t -- 8 butS = Vt + Ift^ .-. S -- 20 X S + i X 10 X G4 -= 450 ft. u EXAMPLES CF UNIFOE:vI ACCELERATION. Or tnu3 :— Inidal velocity Final velocity = 20ft. = 20 + 10 X 8 = 100 = 60 Adding and dividing by 2 we obtain the mean velocity . . Therefore the space = 60 x 8 = 480 ft. 5. "Wluit ia the velocity of a body whicli comraenced to move with a velocity of 80 ft. per second, and has continued in motion for 7 seconds, the velocity rcj^'ularly diminisliiug at the rate of 8 ft. per second ? Decrease per second . . . . = 8 ft. „ in 7 seconds . . . = 56 .*. Final velocity =80 — 50 = 24. 6. Show by a diagram that the spaces passed through in successive seconds by a point moving from rest under the action of a force producing a uniform acceleration are as the odd numbers 1, 3, 5, &c. Let O LI be the line along which to measure intervals of time. Let a^ re- present the first second, a^ a 2 the next, ag fla the third, and so on. Let a^ i'^, ^2 '^'2» ^3 "^'s' ^^' represent the velocities after 1, 2, 3 seconds respectively. The figures tti Vj, a^ v^, ttg Vq, &c. represent the spaces passed over in the successive seconds ; but by drawing lines through the points v^ I'a v^, &c. parallel to O "vl, and through a^ a^ ag, &c. parallel to ON, we divide all the figures into trianglco equal to Oaii'i, the numbers of triangles in tlio successive figures being respectively 1, 3, 5, 7, &c., therefore the spaces have the propor- tion of these numbers. 7. A particle A starts from a point with a velocity of 20 feet per second accelerated uniformly by 10 ft. a second. Four seconds later another particle B starts after A with a velocity of 100 ft. per second, accelerated regularly by 20 ft. per second. Fig. 101. EXAMPLES OF UMFOUM ACCELERATION. 95 When B ovcrtalccs A, how long will A have moved, and bow fur will it be from the start inc; point ? Let t be the number of seconds thro;igh which A will have moved, and s the space, then {t — 4) will be the time during which B will have moved, and the space s will bo the same. The formula for A's motion will therefore bo s = Vi< + 4A<« or s -^ 2Gt + ^tf^ ' (I.) And the formula for B's motion s = V2 (t — 4) + 1 /, (< — 4)» or s = 100 {t — 4)+ 10 {t — 4;' = 10i2 4- 20t — 240 (II.) Equate these tAvo v;ilues of s, .'. 20 < + 5 i^ = 10 i' + 20 « - 2i0 .-. 5 fc2 =. 240 ^2 _ 4S == 16 X 3 t = 4v'3. ti Exercises. 1. A point moves from rest with uniform acceleration, and the velocity at the end of 4^ seconds vs 90 feet; find the acceleration. 2. Define the term acceleration. A body moves at one instant with a velocity of 30 feet per second, and 7 seconds later with a velocity of 79 feet per second. Supposing the velocity to bo uniformly accelerated, find the acceleration. 3. Explain clearly the meaning of the equation v = V + / i. 4. Find by tho help of a diagram the space passed over in 5 seconds by a body* which began to move with a velocity of 17 feet per second, and acquired a velocity 43 feet per second by the end of the fifth second, supposing tho acceleration to be uniform. — Ans. 150 ft. 5. Explain the meaning of the symbols in the equation s = Yt ± hft' and show by means of a diagram how it mr.y be obtained. G. A body starts with a velocity of 80 feet per second, and the velocity decreases regularly by 10 feet a second. What will be the velocity after 7 seconds ? v*^ 9G EXERCISKS ON UNIFOIIM ACCELERATION. 7. In the proceding example what will bo the space paaeed over ? — Ans. 315 ft. b. A point starts from rest witli an acceleration of 32 feet per second, liow far will it move in 8 seconds ? — uins. 1,024 feet. 9. The acceleration is 8 ft. per second, and the initial velocity 1 1 ft. per second ; find the space passed over in 8 seconds, and the velocity at the end of the time. — Ans. 344 ft. ; 75 ft. 10. The sjiaco passed over in 5 seconds is 105 ft., and tlic final velocity 35 ff . ; find tlio initial velocity and the acceleration. — Ans. 7 ft.; 5 ft. 11. A body moving from rest is observed to move over 6 ft. and 11 ft. res[>ectively in 3 consecutive seconds; show that the acceleration is 5 ft. 12. A body uniformly accelerated moves at the end of 2 seconds with a velocity which would carry it through 5 miles in the neict half hour; find the acceleration. — Ans. 7*3 ft. 13. A body moving with uniform acceleration describes 570 ft. in the eighth second ; find the acceleration. — Ans. 70 ft. 14. A body has described 3U2 ft. from rest in 7 seconds ; find the velocity acquired. — Ans. 112 ft. 15. A body has described 54 ft. from rest in 3 seconds ; find the time it will take to move over the next 120 ft. — Ans. 2'38 seconds. 10. A body moves over 70 ft. in the fourth second; find the acceleration. — Ans. 20 ft. 17. A body describes 354 ft. while its velocity increases from 43 to 75 feet per second; find the whole time of motion and the acceleration. — Ans. sees. ; 5'3 ft. 18. A body in passing over 135 ft. has its velocity increased from 7 to 53 ; find the whole space described from rest and the acceleration. — A71S. 137xVt ^^' 5 ^^'^ ^^• 19. Fi?id the numnrical value of the acceleration when in a quarter of a second a velocity is produced which would carry a body over 2 ft. in every third of a second. — Ans. 24. 20. A body moving from rest is observed to pass over 30 ft. and 44 ft. respectively in 2 consecutive seconds ; find the acce- leration and the time from rest. — Ans. 8 ft. ; 4 and 5 sees. 97 passed :eet per eet. velocity ds, and lie final ition. — VQT 6 ft. hat the seconda :he neirt 8 570 ft. 3s ; fi.nd find ns. 2-38 find the es Croni and the Lcreased and the en in a carry a 3r 3G ft. e acce- « II.— FORCES PRODUCING MOTION. 18. ITItlicrto we liave considered the motion of a particle under various circumstances independently of tlie forces which produce the motion. We have now to consider the relations between the forces and the motion which they produce. The science of Dynamics rests on certain fundamental principles termed the Laws of Motion. These laws were stated by Newton as follows : — J. Every lody continues in its state of rest or of uniform motion in a straight line, except in so far as it may he compelled hy impressed forces to change that state, II. Change of motion is proportional to the impressed force, and takes place in the direction of the straight line in n'ltich the force acts. III. To every action there is always an equal and contrary reaction; or the mutual actions of any two bodies are always e^ual and oppositely directed. Because the first law states the property of matter termed its inertia, that is, its inability to alter its state of rest or motion, it has been called the Law of Inertia. It may be paraphrased as follows : — When a hody is not acted on hy anyforce^ if it he at 98 FORCES PRODUCING MOTION. i rest it will reinain at rest, and if it he in rpoiion it will continue to move in a straight line icith uniform velocity. On the first two laws the theory of the motion of the heavenly bodies is based, and the nniform agreement of the deductions from these laws and observations in astronomy is one of the strongest confirmations of their truth. In the Second Law no distinction is made between bodies at rest and bodies already moving, but in all cases ch ^nge of motion is proportional to the impressed force, the change being in the direction of the force. Hence this law teaches us that when a force or forces act on a body in motion, the change of motion in mag- nitude and direction is the same as if the forces acted on the body at rest. We may, therefore, expand the Second Law so as to exhibit what is implied by negation in Newton's statement of it by dividing it into two parts, thus : — When a force acts vpon a body in motion, the change of motion is in the direction of the force, and is the same in magnitude and direction as if the force acted on the body at rest. IVhen pressure produces motion, the acceleration varies directly as the j^resswe, and inversely as the mass. The truth of the first part is shown by such con- eiderations as the following :— When a person is on board a boat ^vlHch is moving nniforinly along a stream, any movement he makes LAWS OF MOTION. 99 produces exactly the same effect as if the boat wcr2 at rest. When a stone is let fall from a point on land it falls in the direction of the vertical, and when a stone is let fall from the p.st of a ship in motion, it reaches the deck at a point vertically below the starting point. Now the stone falls from the mast to the deck in the same time whether the vessel be at rest or in motion ; again, if the vessel passes horizontally through any distance, 3 ft. suppose, dm'ing the fall the stone also passes through 3 ft. horizontally, that is, through the sam.e space as it would have passed through had it remained at the top of the mast. Vie conclude, therefore, that the liorizontal motion due to the velocity of the vessel, and the vertical motion due to the attraction of the earth, have each their full effect in their own direction, that is to say, in the resultant motion the stone is dis- placed horizontally in a certain time exactly as if its vertical motion did not exist, and it is displaced ver- tically in the same time as if its horizontal motion did not exist. This will be made still plainer by reference to the following experiments : — Take a board a b e (Fig. 102) which has a curved groove A B cut in it, so that the direction of the curve at B is horizontal. If a ball m be made to slide down this groove tlie direction of its motion is horizontal. If the groove be prolonged horizontally to the point m, and the surface of the crroove and ball bo so well gi polished that friction may be disregarded, after the ball has passed b it will move along b m with uniform velocity. At the instant ball m passes the point B, t^MBMI i1 100 FALLING BODIES. let anotlier ball 111! fall vertically along b e, and let B m be taken at sncli a length that m moves from b to m in Fig. 102. exactly the same time as m! moves from b to e. If now the horizontal groove be removed, and the ball be allowed to slide down the groove A b, and then to fall freely, it will neither arrive at the point m nor the point E, but at a point d, having passed through a vertical distance b e and a horizontal distance b m. It will arrive at the point d in exactly the same time as the ball m' falling free would pass from b to e or the ball m after descending through the arc A b would pass from B to M. Falling Bodies. 20. Tlie fall of bodies to the earth in various circum- stances offers remarkable illustrations of the precediig principles. n LAWS OF FALLING D0D1E3. 101 let B M to M ill E. If ball be to fall 16 point vertical It will as the the ball as from circum- eccdii'g When bodies of different material fall in the air, they do not usually pass through the same heights in the same time. A ball of lead and a scrap of paper fall through the air with very different velocities. The difference arises from the dilTerence in the resistance offered by the air, which varies with the form and dimensions of the body and with the velocity. If the bodies are made to fall in a tube from which the air Las been expelled, then the time of descent and the velocity acquired will be the same. Tlie motion of all bodies in vacuo is uniformly accelerated. It is usual to call the force producing the motion *' gravity," and to indicate the acceleration by g. Hence g is the number of feet added to the velocity of a body moving freely in vacuo for every second of time. This acceleration is not absolutely the same at all points on the earth's surface, but its variations are very small. It increases with the latitude of the place and decreases with the height above the sea. In London a velocity of nearly 32"2 feet is added in every second when a body moves in vacuo. We may therefore at once apply the forniuhe for uniform acceleration to the case of falling bodies. When the body falls from rest the equations of motion (§ 15) are V = gt s = ^ gl^ and by equating values of t from both equations v^ = 2 gs. WTien the body has an initial velocity v these equa- tions of motion are V = y ± gt s \t ± i gt 102 LAWS OF FALLING BODIES. and by finding tlie value of t from the first equation and substituting in the second In all cases in which the body moves in a vertical line the space passed through is equal to the mean velocity multiplied by the time. Exanvple 1. A body falling under the action of gravity has a velocity of 80 feet ; how far will it fall in 5 second3 ? Initial velocitv = 30 feet Final „ = 30 + 5 X 32 Mean Sps i> » u 110 5 550 feet. Example 2. A body is projected vertically upwards with a velocity of 100 feet per second; how far will it ascend in 3 seconds ? Initial velocity = 100 Final „ = 100 - 3 x 32 Mean Space }) if i} ^ J- 52 8 ■ 156 Example 3. To find the space passed through in the n^^ second: — Velocity at commencement of n*^ second = (n — 1) g. Velocity at end of nP^ second = n g. Mean velocity and space = h (2 ?i — 1) ^. Substituting successively 1, 2, 3, &c., for n, we find the spaces passed through in the successive seconds are -t 3 g, 5~, 7 5, &c., and generally the space passed through in the n*^ second ia the n*^ odd number multiplied by g. vi i' i I LAWS OP FALLING BODIES. 103 Example 4. A body is projected vertically upwards with velocity v ; to what height will it rise ? The velocity at any point is given by the equation v* = v' — 2 gs ; but at the highest point the velocity is 0. Hence, if h be the height o = v'^ — 2 <7 ^, hence h = tt Tlie velocity acquired in fulling from this height is given by the formula v^ == 2 <^3) &c- I'epi'C- sent the position of the par- ticle at the end of the successive seconds, and the curve joining these points represents the path. Fig, 103. i B W EXERCISES ON FALLING BODIES. 105 , and on to itally on of T B curve Exercises. In ilie f llowing exercises use 32 ft. as an approximation for 'ii e acceleration due to gravity. 0)1 the formula v = Y ± g t. 1. A body falls for 8 seconds; with wliat velocity is it mov!ii,j at the end of that time ? — A ns. 236. 2. If a body is let fall, how long will it take to acquire a velocity of 500 ft. per second ? — A us. 15|. sees. 3. A body is projected upward with a velocity of 80 ft. per BGCond; determine the velocity it will have at the end uf 3 seconds, and the number of seconds that must elapse before i 3 velocity equals its initial velocity.— .^-1 nj. IG ft. ; 5 sees. 4. A body is thrown downward with a velocity of ICO ft. for Becond ; find its velocity at the end of 5 seconds, and the nunil -.r of seconds in which a body that is merely dropped would accuiro that velocity.— .4 ns. 320; 10 sees. 5. A body A is projected downward with a velocity of 210 rt. per second; at the same instant another body B is project d upward with an equal velocity; how much faster will A Lo moving than B at the end of 6 seconds ? — Ans. 9 times. 6. A body is thrown upward with a velocity of 96 ft. y^^ second ; with v/hat velocity will it bo moving at the end of 4 seconds? — Ans» Downwards with a velocity of 32 ft. per second. 7- A body is dropped from a certain height h, at the gnme instant as another is thrown upward ; what initial velocity niuit the latter have that it may meet the former half way ? — Ans. 8. A body is at a given instant moving upward with a given velocity v\ show that it will be moving downward with an equol 2 V velocity after — seconds, and that it will reach its highest point 9 v after - seaonds, if i tli ;■;} I 106 EXERCISES ON FALLING BODIES. 9. A body h thrown np with a velocity a , aflter how long will it be descending with a velocity h g ? — Ans. a+ 6. On the formula S == Vt ± ^ g t* 10. How many feet will be described in 7 seconds by a body that moves freely from rest under the action of gravity ? — Ans. 784. 11. Through how many yards would a body falling freely from rest descend in 3 minutes ? — Ans. 172,800. 12. A body is projected downward with a velocity of 50 ft. per second ; how far will it fall in 4^ seconds ? — Ans, 549 ft. 13. A body is projected upward with a velocity of 150 ft. per second; how high will it have ascended in 64 seconds? — Ans. 299 ft. 14. If a body is thrown upward with a velocity of 9G ft. per Bt?cond, how far will it be from the starting point at the end of 4| seconds, and what wiU be the whole space it wiH have described ? — Ans. 95 ft. ; 193 ft. 15. A body is projected upward with a velocity of dg ft. p^r second ; determine the height of the body, and with what velo- city, and in what direction, it will be moving at the end of 4 seconds. — Ans. Height 4:g, moving downward with velocity g. 16. A body 13 projected upward with a velocity v ; show that 2v it will return to the point of projection after — seconds. H 17. A body falls for a time t, and has a velocity V at the beginning, and-u at the end of thu,t time ; find the space described. —^715. i (F + v) t. On tlie formula v'^ <= F' 4- 2gs, 18. If a body is thrown upward with a velocity of 36 ft. per Bccond, find its greatest height. — Ans. 20| ft. 19. If a body falls freely through 1,600 ft., find the velocity it acquires. — Ans. 320 ft. 20. A body is projected vertically upward with a velocity of 100 ft. per second; how long v/ill it take t ^ reach the top of a tDWcr 100 ft. high ? — Asis. 11 or 5 sees. i s s 1 ir it ac fo .^ A EXERCISES ON FALLING BODIES. 107 tliat 21 . A body is thrown upwards with a velocity of 80 ft. per second; how fr will it ascend ? 22. Two stones are di-opped from different heights and reach the ground at the same time, the first from a height of 81 feet and the second from a weight of 49 feetj find the interval between their starting.- -Ans. 4 sec. 23. A body is projected upwards with velocity of 60 feet per second ; what will be its velocity after it has passed over 50 feet ? —Ans. 20 ft. 24. A body is thrown downwards with a velocity of 10 feet per second ; what will bo its velocity after it has fallen through 75 ftet ? 25. A stone A is let fall from a certain point, and after it has fallen for a second, another stone B is let fall from a point 100 ft. lower down j in how many seconds will A overtake B ? — Ans. SI- seconds from the starting of A. 26. With what initial velocity must n. body be thrown down- wards that it may strike the ground, which is 60 ft. below the starting point, with a velocity of 104 ft. per second ? — Ans. 80 ft. 27- A body commences to move with a velocity of 80 ft. per second, and after it has passed over 100 ft. it has a velocity of 120 it: per second; find tho acceleration. the .'ibed. Mass. per dty it ity of of a 22. Hitherto we have considered a force producing motion as measured by the acceleration or the velocity it generates. It is, however, necessary to take into account the quantity of matter moved. When the same force acts on different bodies, it produces in each a 108 MASS. nniform acceleration ; but the acceleration will not bo the same in both cases. This difff' ncc arises from what is called the mass of the bodies. 23. Masti is a term for the quantity of matter in a body. It is not easy to give such a definition of mass as will lead us at once to a method of measuring it. We are compelled to measure mass by its effects. We assume that the attraction of the earth on all particles of matter is the same, and is not dependent on the nature of the matter attracted. This assumption Fccms to be justified by the fact that bodies of all kinds fall with equal velocity in the exhausted receiver of an ah'- pump. Hence we measure the mass of a body by its weight, and can only define the mass as a quantity proportional to the weight. If, then, at the same spot in the earth's surface one body is twice as heavy as another, the mass of the first is twice that of the Bccond. Suppose, however, the body is weighed by a spring balance (Fig. 2) at a certain place, and weighed again by the same instrument at another place nearer the equator, it will be found that the body is lighter at the latter place. It is found also that the acceleration due to the attraction of the earth is also less at the second place than at the first in the same proportion. This illustrates the fact that when the mass remains the same, the weight varies as the acceleration due to gravity. Consequently, writing m for mass, w for weight, and g for acceleration due to gravity, w varies as m when g is constant, w varies as g when m is constant ; J] i ij I LAWS OF MOTION. 109 i therefore w varies as m g when both vary ; and, there- fore, by a theorem of algebra By selecting a suitable unit of mass we may write w = M ^. Similarly, if a force f act on a mass m, and produce an acceleration/, F==./M. These two equations combined give For example, if a force equal to a weight of 12 lbs. moves a mass whose weight is Gl lbs., the acceleration is 82 X 12 -f- 64, or 6. Or we may reason thus : — Since the weight of G 4 lbs. produces an acceleration of 32 on a mass of G 4 ll)s., to find the acceleration w .ich a furce of 12 lbs. will pro- duce we have As G4 : 12 :: 32 :/ .•./=G. 24. The above relation between force, mass, and acceleration is expressed in the second part of the Second Law of Motion. The product M/is sometimes called the moving force , and this statement is then made thus — the moving force is proportional to the pressure. This law may be illustrated by the following arrangement: — Suppose two weights r and q, whose masses are respectively m and 772', to be connected by a fine in- extensible cord passing over a fixed pulley. If p == q there will be no motion ; if r be greater than q there 1 ^1 ■. -ft' ■*:« I U^ % I I i ■I \ i! 110 LAWS or MOTION. will be a pressure producing motion equal to the excess of p over Q or (p — q). The mass moved is m + m', hence / (m + m) = P — Q, but w ««= — and /«' = -^ 9 9 therefore f = a — 25. To find the tension in the cord. We may consider the tension t to be the upv.-ard force acting both on p and q ; the pressure producing the motion of p downwards will be p — t, and the pressure moving q upwards t — q, and the accelera- tions arc tlie same ; 1 r P— T hence /= . and/ therefore or dividing by g^ r — T T ^^ Q m m' P — T _ T — Q G Q consequently T 2PQ P ci Fig. 104. P+Q For example, suppose two weights connected by a cord passing over a fixed pulley to be 3 ozs. and 5 ozs. respectively; then,/= 32- 2 = 8, and if the weights are allowed to move from rest, the equations of motion become v = St and s = it^. The tension in 2x3x5 the cord will be 3 + 5 == 3|. If the 8 lbs. '■1. i ATTWOOD a MACHINE. Ill instead of being dlviJotl into unequal parts 3 and 5 were divided into eqnal parts, the tension of the string would be 4 lbs. ; hence we see that the tension is less when the parts are unequal than when they arc equal. That which is true in this particular case is generally true. •f lbs. Attwood's Machine. 26. The machine represented in the figure was devised by Attwood for testing experimentally the laws of motion and the results derived from the theory of falling bodies. It consists of an upright beam, supporting at the upper extremity a nicely constructed wheel turning on * a horizontal axis, and two equal weights r, r', connected by a fine silk thread which passes over a groove in the wheel. To diminish friction the axis of the larger wheel turns on friction wheels (Fig. 105). The pillar is furnished with a graduated scale, a movable ring n, and stage m. Another important appendage to the machine is a small clock with a pendulum beating seconds. Suppose the weights p and r' to be at first equal : they will then be at rest. Let the movable ring be in the position w, below the weight p and on this weight place a small bar;?, p will descend with accelerated velocity until it reaches the ring (iiy. Here the bar p will be lifted off the weight p by means of the ring (n), and p will move onward with a uniform velocity equal to the velocity at the moment when (p) was retained by the ring. Now place the stage in such a position that the weight p 112 AT'lWOOD'd 3IACI1INE. ' ! I i ATTWOOD S MACHINE. 113 m^^ may reach it in exactly one second after it lias passed the ring (/i). The distance between the stage and ring will be the space described by a body moving with a nniform velocity in a unit of time. Let us now suppose that the ring is exactly so far below tlio starting-point p as to lift off the bar when p has moved for exactly one second. Also let the stage be so placed as to stop p exactly one second after the bar has been removed. Then the distance n m\& the acceleration. Now the following relation is found to exist between this acceleration and the weights : if wc double the weight of the bar ;?, while we keep the sum of the weights p p' and j> the same, we double the acceleration, and if we take a bar n times the weight of the first, the second acceleration will be n times that of the first. If now we keep the bar the same, but double the sum of the weights, we shall find that the acceleration is then only half what it was before, and generally if we multiply \\\q, weight moved by n we shall divide the acceleration by n. Thus the acceleration varies directly as the pressure producing motion, and inversely as the mass moved. In every respect we find the results agree wutli the theory of Article 23. The chief advantage secured by this machine is that we may make the acceleration as small as we please, and thus render the motion slow enough to be observed without difficulty. Suppose, for example, that the equal weights are eadi 3r7 grammes and the bar 1 gramme, then, taking the acceleration due to gravity as 3:2-2, we have, by the tfn 114 MOTION PRODUCED BY PRESSURE. tlieory of Article 23, an accelercation equal to 32*2 x •q^Tt; J—-JL = -■• Now if we keep the bar on tlirongliout the time it is found that the space described in the first second is J ft., in the next second 3 x ^h. in the third 5 X ^U. and in the nth second the nth odd number multiplied by ^ ft. Again, by cutting off the pressure producing motion at the end of the successive seconds, and measuring the spaces passed through in the next second, it is found that the velocity at the end of the first second is J ft., at the end of the next 2 x J ft., and at the end of the third 3 x -J ft., and at the end of the nth second n X "I ft. Thus the results of the experiment agree entirely with those of the theory of falling bodies. Lot us direct attention again to tlie relation between the force acting on a body and the acceleration produced. This relation is given by tlie formula f = m'f, where f is the resultant force expressed in pounds, /the increase of velocity per second ex- pressed in feet, and m the mass of the body, the unit of mass being that mass to which a force of 1 lb. will give a velocity of 1 foot in a second of time. The formula shows that the acceleration is proportional to the force, for instance, if the force acting on a body be doubled or trebled, the rate at which the velocity increases will be doubled or trebled. If the forces acting on a body have no resultant, f = o and therefore / = o, hence the body is either at rest or moving with uniform velocity. EXAMPLES AND EXERCISES. 115 2-2 X )ar on scribed thodd motion ng the ! found sift., of tlie second agree iQ force relation it force )nd ex- )f mass ocity of to tlie bled or oubled and with Examples. 1. What velocity will bo communicated to a mass of 100 lbs. by a pressure of 10 lbs. in 10 seconds ? If the weight of 100 lbs. moved its mass, the velocity acquired would be lOg or 320. Since by the second Law the velocity produced is proportional to the force, the mass remaining the same, as 100 : 10 : : 320 : v, /. v = 32. 2. A mass of 8 lbs. hanging over the edge of a smooth table pulls along a mass of 28 lbs. placed on the table j find the acceleration and the space passed over in 3 seconds. Here the pressure producing motion is the 8 lbs. and the mass moved 8 + 28 = 36 lbs. If the weight of 36 lbs. moved the mass of 36 lbs. the accele- ration would be gf or 32 J but if the weight of 8 lbs. moves the mass of 36 lbs. the acceleration / is found from the following proportion : — As 36 : 8 : : 32 : / .•./== 7^. .-.s = 4/^2=^ >^ 71 X 9= 32 feet. Exercises. 1. A pressure of 2 lbs. moves a mass of 12 lbs. j find the acceleration. — Ans. 5^. 2. How far will a mass of 50 lbs. be moved by a pressure of 1 lb. in 12 seconds ?—Ans. 46-08 feet. 3. What pressure by acting for 6 seconds will produce a velocity of 120 feet per second in a mass of 12 lbs ? — Ans. 7a lbs. 4. If weights of 3 and 5 lbs. are connected by a string which passes over a fixed pulley, what will be the velocity ., ^nerated in a second of time ? — Ans. 8. 5. In the above example, how far will the heavier weight descend in 5 seconds? — A71S. 100 feet. 6. If the weights be respectively 15 lbs. ?ind 17 lbs., in how long will they move through 144 feet ? — Ans. 12 seconds. : i'i; lie MOMENTUM. ■ : ! 7. A weight of 2 lbs. hangs over the edi^e of a smooth table, and draws a weiglit of 50 lbs. laid on the table j what velocity will be acquired in 2J seconds? — Ans. 3j\ feet. 8. IIow far will either of the weights move in 5 seconds ?— Ans. IS^^^feet. 9. A steam-engine moves a train weighing CO tons on a level mad from rest, and acquires a speed of 5 miles an hour in 5 minutes. If the same engine move another train, and give it a speed oi 7 miles in 10 minutes, find the weight of the second train, supposing the resistance to amount to the same in both cases ?- -Ans. 85 y tons. MOMENTUM. 27. We have already shown that if p be a pressure producing the motion of a mass m, and if / be the velocity generated in a unit of time, then by taking a suitable unit of mass the relation of these quantities is expressed by the equation p =/m. The product of tho mass into the velocity at any instant is termed the momentum at that instant. Hence, when the acceleration i? uniform, /"m is the momentum at the end of the first second, and also the increase of momentum for every second. The above equation states that tlie momentum generated in a unit of time is proportional to the force. The momentum bears the same relation to the moving force that the velocity does to the acceleration ; the momentum varies with the time, and the moving force is the increase of momentum for a second of time. 28. When a moving body produces the motion of able, and y will be onds ? — n a level hour hi d give it ) second in both pressure be tlic aking a antities at any nt. I is tlie also the above 1 a unit moving )n ; the ig force :ie. )tion of another body, the greater tlie momentum of tlie first the greater tlie momentum generated in the second. 118 MOMENTUM. :( II [■•' \ I:; As an illustration, refer to the macliinc for driving piles, represented in Fig. 106. The block of iron a is raised to the top of the frame c c, and allowed to fall. It might lie on the head of the pile b for an indefinite length of time without producing any effect, but a momentum is generated in the fall which is destroyed in an exceedingly short time by the resistance of the pile. On the one hand we see that the greater the mass A, and the greater the height from which it falls, the greater is the effect, and on the other hand the greater the resistance of the pile the shorter the time during which it is overcome by the momentum of a. Whenever the resistance to be overcome is great, the action of the force must be limited to a proportionately short time. AVhen a hammer is made to fall on the head of a nail the momentum of the hammer at the instant of contact is spent in overcoming the cohesion of the particles of the wood. With a given momentum the shorter the duration of the shock the greater the effect of the force in overcoming this resistance. Sup- ' pose that the board is not firmly supported so that it yields under the blow (Fig. 1)7), the duration of the shock is prolonged, the resistance which may be over- come is lessened, and consequently the nail drives badly. It enters much better, however, if we place behind the plank some solid body which diminishes the recoil of the board. 29. We have considered hitherto forces which act for an appreciable time, but we now see that there may be forces which act for a very brief period, and yet in this short time produce or destroy a great momentum. WOMKNTDM. 119 Fig. 107. Fig. 108. These are termed impulsive forces. An impulsive force is 071 e which produces a finite chaiige of motion in an indefinitely short time. It is evident tliat such forces cannot be measured by the momentum generated in a unit of time ; they are estimated by the whole momentum generated. Suppose, for example, that a blow given to a cricket ball of weight w drives it from the bat with a velocity v, then w the momentum?; x — is the measure of the force. 9 I } M ' . 1 :.• ■: i 120 NEWTON S TIIir.D LAW. CO. The first and second Laws of Motion explain the action of external forces on a body. By their means wo are able to investigate the motion of a particle subjected to given forces. We have yet to examine the cases of motion which arise from the mutual action between or among two or more bodies. We shall be assisted in understanding the nature of this action by the following illustrations : — ■ (i.) When the block of the pile-driving engine (Fig- lOG), after falling from a strikes b, it exerts a force on tlic pile which causes it for a very short interval of time to overcome the resistance offered by the soil. At the same time the block itself is acted on by u "• rce upwards which brings it to rest, and these two forces are equal. (ii.) If two balls moving in the same straight line are made to strike one another, the force of the first on the second :s exactly equal to' that of the second on the first, the force being measured by the momentum. N Fig. 109. Let two balls m and n (Fig. 109) be approaching along the straight line a h vrith equal momentum, then it will be found that if they are not capable of rebounding, they will bo reduced to rest after impact, the action on one being exactly equal ant pposite to that on the other. Suppose, for example, that the balls weigh G lbs. and 8 lbs. respectively, and that their velocities aro 4 ft. and 3 ft. per second, so that the momentum ot NEWTON'S THIRD LAW. 121 each is 24 -r g^ then they will both be brought to rest by the impact. It is worthy of remark that in this case the C. G. of the two balls will not change its position during the motion ; for let c be the point at which the balls meet, then a c and h c will be. described in the same time, and therefore these lines must be proportional to the velo- cities of the balls. But since the momenta arc equal, the velocities must be inversely proportional to the masses, and consequently c divides the line ah into parts inversely proportional to the masses. Therefore c coincides with the C.G. of the two bodies, that is to say, the C.G. is the point at which the bodies will meet, and therefore has a fixed position. (iii.) If a cannon free to move on a smooth plane be discharged, the cannon will recoil, and its momentum will be exactly equal to that of the shot, but in the opposite direction. Let ivi be the weigh i. of the shot, and I'l its velocity on leaving the gun, and let iv^ be the weight of the gun, and v^ its velocity of recoil, then We need not write the g in the denominators, for they cancel. (iv.) Suppose a shell to be moving on a horizontal plane with a velocity v, and suppose it to burst in the line of its motion into two parts, whose weights are ii\ and ic^. Let the velocity of the first part after explosion be less than the original velocity by i'^, then the velocity of the second part will be greater than the original velocity by 1*2, such that u'l ' r, = u\, ' I'o. 122 V-' NEWTON S TIUHD T.AW. U ili The momentum produced in one direction by the exi)]osion will retard the motion of ti'i, and that pro- duced in the opposite direction will be equal to the former, and will increase the velocity of u'2. The sum of the momenta of tlie two parts will thus be the same before the explosion as after it. This we might have concluded also from the First Law of Motion, since by tliat law a body cannot of itself alter its momentum. These are illustrations of a law which is known as Newton's Third Law of Motion. To everi) action there is always an equal and contrary reaction. The momentum produced in any direction by the mutual action of two bodies being called action^ the momentum produced at the same time in the opposite direction is termed reaction. This principle was assumed in the case of statical forces in § 7, and also in § 15, p. 8, and § 25, p. 110, where the tension in a cord at rest or in motion was supposed to be the same at every point. !|t Exercises on Newton's Tiiikd Law. 1. A body A of weight 10 lbs. strikes a body B at rest, and weighing 100 lbs., witli a velocity of 100 ft. per second ; find the velocity of B supposing A to be brought to rest by the impact. 2. Two bodies, whose weights are 6 Ibi. and 10 lbs., and whose velocities are 50 ft. and 20 ft. per second respectively, approach, ani.1 after impact move on together; find the common velocity. 3. Three balls, weighing respectively 5 lbs., 7 lbs , and 8 lbs., lit" in the same straight line. The first is made to impinge on the second with a velocity of 60 ft. per second without rebounding ill II 111 EXERCISi:3. 12:3 Tho first and socouJ toge'liT inipln'^ro in the sr.mo way cu tlio third ; find the final velocity. — Ans. IT. ft. 4. A guu weighing 5 tons is charged with a sliot weighing 28 lbs. ; if the gun bo free to move, with what velocity will it recoil when the ball leaves it with a velocity of 100 ft. per second? — Ans. •25ft. 5. Two wooden balls weighing 12oz. and 10 oz. are niado to impinge on one another. One of the balls is furnished with a spike to prevent the rebound. If the velocity of the enuilli'r hn 12 ft, j^or second, what must that of the larger be that the motion may be destroyed by the impact ? f). A shell at I'est bursts into two pnrt.", the smaller of uhicli is one third of the whole j what will be the ratio of the iaiti^il velocities of the parts ? 7. A shell moving with a velocity of 50 ft. per second bursts ii'. tlie line of its motion into two parts, which weigh rospuctivi ly 30 lbs. und 02 lbs. The velocity of the larger piece is iucrca.sed in tho direction of motion by 30ft. per second; what is tlio velocity of tl:e smaller? — Ans. 12ft. in a direction opposite to the direction of motion immediately before the explosion. 8. Two bodies subjected to their mutual attraction, and to no other forces, start from rest. If their masses are as 5 to 2, and the acceleration of the larger is 40 ft. per second, find the aece- loration of the smaller. *■.<> ■'-' ,1 ■ •i T *' .■ -■ ^ «' \ ■ r * «. ■ 4 iJ' \ - ■t:' I. k x i I" i! ill 111 LONDON UNIVERSITY EXAMINATION PAPERS. -M- SrATllICULATION. 1. State tlio roiaMon between tlio power and weight on a smooth inclined p. ,uo when the power acts parallel to the plane. What power sustains a man standing on a slope ? (§ 88.) 2. Define the centre of gravity of a body. A uniform rod, three feet long, which weighs 3 lbs., has a 1 lb. weight hung from one end by a string one foot long : find the centre of gravity of the whole, and show where the rod i.'iust be supported in order that the whole may balance. (§ 54. The rod must be supported 13^ inches from one end, and when it is horizontal the C. G. will be 3 in. below this point.) 3. A cord without weight or friction passing round a single fixed pulley has a weight of 1,000 grains attached to one of its extremities, and one of 2,000 gi'ains to the other : when left free, how far will this latter extremity have descended in two seconds ? Ans. f = i :. s = ~ a. 4. Let A B and A C represent in direction and magnitude tv/o equal forces acting at A, the angle between their directions being 50°. Draw a line which shall represent in direc- tion and magnitude a force which will be in equilibrium with these forces ; and state the j^ q angle which its direction wiU malce with A B. 5. Explain why a man who has to carry a heavy box in one hand must throw his body on one side. (§ 66.) LONDON CNIVEUSIXY 125 :oN L smooth What rm rod, lit hung f gravity orted in must be t)rizontal a singlo 10 of its eft free, ccondfl ? C X m one If the forco of gravity, instead of acting vertically, wore to act horizontally from east to west, would this aflcct the position within a body of its centre of gravity ? (§ 53.) 6. What is the ratio between the power and the weight in a ecrcw which has ten threads to the inch, and is moved by a power acting perpendicularly to an arm at a distance of one foot from the centre ? — Ans. 5^^^. 7. How far will a body fall from rest in four seconds ? With what velocity must a ball be thrown vertically upward, in order to return to the hand after four seconds ? — Ans. 8 g, or 257"C ft., and 2 g, or 01 -4- ft. 8. A weight, W, is suspended from a movable pulley C, A and B being two fixed jnilleys. When the system is at rest, what are the forces which keep the point C in equilibrium ? and why is it necessary that the two weights W' should bo equal to one another? (§§18,10.) 0. Two men, A and B, carry a weight of 200 lbs. on a pole between them. If the men be 5 feet apart, and the wei'fut slung at a di^itanco of 2 feet from A, what part of tho w^eight will he boar, neglecting tho weight of the pole ? — Ans. 120 lbs. 10. Explain why a plumb-lino will not remain at rest except in one position. (§ 54.) 11. An arrow is shot vertically up'.vard v.'ith a velocity of 100 feet in a second when it leaves tho bow. IIov/ long will it be before it reaches tho ground again? — Ans. 200 -f- g, or 6] sees. 12. If another arrow, half tho weight of the former, be shot with the same /o7*ce vertically upward, how long will it be before it readies the ground again ? Point out vrhero you apply any of the laws of motion in ansvrering this question. (The moving force varies as the mass and as the velocity generated ; hence, if the force remain the same, and the mass be doubled, tlio velocity will be diminished by one half, and there- fore the time will be lessened in the same proportion.) 13. If a man has to raise a weight and has only one pulley at his disposal, show how he must apply it in order to obtain the utmost advantage. — Ans. Fig. 77, with parallel cords. 14. Two forces, not parallel, represented in magnitude and I , ;l'i ly: u''\V r :V I ;>''{'■; 'i: 126 MATRICULATION EXAMINATION rAPERS. !' i it ' direction by AP and BQ act at the extremity of a lover A B. Exhibit by means of a diagram the moments of these two forces round the fulcrum O, and state what condition will be satisfied when there is equilibrium (Fig. 60). 15. Show, from the property of the Centre of Gravity, that in a common balance it makes no difference in what part of the Ecale-pan the weight is put, whether in the centre or at the edge ($ 54). 16. A man who weighs 160 lbs. wishing to raise a rock, leans with his whole weight on one end of a horizontal crowbar 5 ft. long, which is propped at the distance of 4 in. from the end iu contact with the rockj what force does he exert on the rock, and what pressure has the prop to sustain ? — Ans. 2,3 10 lbs. 2,400 lbs. 17. A solid cube of wood rests by one of its sides upon an inclined plane in such a manner that the upper and lower edges of its base are horizontal. The inclination of the plane can be increased, and it is so rough that the cube will topple over before it will slide ; find at what incliniition this will take place (Fig. 49).— Ans. 45°. 18. How docs it appear that the force of gravity acts on all Bubfltances alike ? 19. A person inside clings to the roof of a railway carriage which then rushes horizontally over the edge of a precipice ; what change, if any, in his motion will result if he lets go his hold ? Give a reason for your reply. 20. A plummet, the string being held in the hand, is immersed in a current of water, and the string ultimately settles in a slant- ing position ; explain by a diagram the nature and action of the forces which determine the position of the string. — Ans. The force of the current is horizontal, the weight vertical, and their resultant has the direction of the string. 21. How would you find the C. G. of a straight, but not uniform rod? (§53.) 22. A straight lever, 6 ft. long and licavier toAvards one end, iti found to balance on a fulcrum 2 fb. from the heavier end ; but when placed on a fulcrum at the middle, it requires a weight of '(ill. LONDON UNIVliRSlTY 127 3 lbs. hung at the lighter end to keep it horizontal ; what ia the ^•eight of the lever ? — Arts. 9 lbs. 23. Enunciate the Second Law of Motion. A ball is projected in a horizontal direction from a rifle placed 1,000 ft. above the level of the sea. Find the elevation of the ball two seconds after the discharge, neglecting the resistance of the air. — Ans. 1,000 — 2g', or, y35-6ft. 24. Exhibit, by means of a diagram of which the construction is to be explained, the method of finding the resultant of the forces represented by the lines in Figure 21, the forces all acting at the point M, and lying in the plane of the paper. 25. Find the Centre of Gravity of a system consisting of two spheres, 8 oz. and 24 oz. in weight, connected by a rigid rod without weight, the distance between the centres of the spheres being 1 foot. — Ans. U in. from the smaller woiLjht. 2G. Mention some of the most important applications of the screw. 27. State the Third Law of Motion, and give a numerical application of it. Explain the " kick " of a gun. 2S. A boat is moored in a stream by two ropes, fixed to posts, one on each bank, and inclined to the direction of the current at angles of 30° and 45°. Draw a figure from measuring which tho proportion may be found between the strains on the two ropes. 29. A weight is to be raised by means of a rope passing round a horizontal cylinder 10 inches in diameter, turned by a winch with an arm 2.^ ft. long. Find the greatest weight which a man could so raise without exerting a pressure of more than 50 lbs. on the handle of the winch. — Ans. A little less than 300 lbs. 30. Define the C. G. of a body. A triangular board is hung by a string attached to cje corner. "What point in the opposite side will be in a lino with the string ? (§ 5G.) 31. When a body in motion is left wholly to itself, not being influenced by gravity or any other external force, what is the nature of the motion of translation of the body ? Give some evidence from observation or experiment of the truth of your Btatcment. (§§ 2, 3, 4.) I'Hr 128 MATRICULATION EXAMINATION PAPERS. i i il: Si' I 32. Explain by reference to a diagram wby a stone only falls 16 ft. during the first second, while yet the force of gravity generates in that time the velocity of 32 ft. per second. (Fig. 101.) 33. Sketch a system of pulleys, one fixed and two movable, in which one end of the string passing round each pulley is attached to the weight, and find the relation of the power to the weight in equilibrium. (Fig. 81.) 34. A stone after falling for one second strikes a pane of glass, in breaking through which it loses half its velocity. How far will it fall in the next second ? — Ans. g, or 33"2 ft. 35. It is required to substitute for a given vertical force, two forces, one horizontal, the other inclined at an angle of 45° to the vertical : determine the magnitudes of these two forces. — Ans. If P = the given force, the first = P, the second >/2"P. 36. How can the true weight of a body be determined by means of a balance vrith unequal arms ? (§ 77.) 37. A body falling with a uniformly accelerated motion, passes tiiroujGjli 10 ft. in the first tAvo seconds after startinc: : how fiir will it be from the starting-point at the end of the third second ? —Ans. 22| ft. 38. Show how to find the resultant of two forces, A and B, which act upon a given point in directions that make an angle a with each other, the force A acting towards a given point, and the force B away from it. (§ 25.) 30. Six smooth vertical posts are fixed in the ground at equal intervals round the circumference of a circle, and a cord, without weight, is passed twice round them all in a horizontal plane, and pulled tight with a force of 100 lbs. Find the magnitude and direction of the resultant pressure on each post. — Ans. 200 lbs. towards the centre. 40. A straight lever 20 inches long weighs 15 oz. Wliero must the fulcrum be placed in order that the lever may be in equilibrium when a weight of 16 oz. is hung at one end, and a weight of 9 oz. at the other ? — Ans. S\ in. from the 16 oz. 41. What is the pressure on the fulcrum when the lover is in equilibrium ? — Ans. 40 oz. LONDON UNIVERSITY 129 forco, two 45° to the . — Ans. If 'mined hy on, passes liow fiir 1 second ? A and B, 1 angle a oint, and I at equal , witliout lane, and tudo and 200 lbs. Wliero ly be in 1, and a )Z. er is iu 42. Draw a four-sided figure witli unequal sides, and describe (1) a geometrical, and (2) an experimental method of finding its C. G. (§ 67.) 43. A piece of uniform paper in the fonn of a regular hexagon has one of the equilateral triangles obtained by joining the centre to two consecutive angular points cut out. Determine the position of the C. G. of the remainder of the paper. 44. A heavy particle is dropped from a given point, and after it has fallen for one second another particle is dropped from the same point. What is the distance between the two particles when the first has been moving during five seconds ? — Ans. 144 ft. 45. A balloon has been ascending vertically at a uniform rate for 4*5 seconds, and a stone let xliU from it reaches the ground in 7 seconds ; find the velocity of the balloon, and tho height from which the stone is let flill. — Ans. If S = the height, and V the velocity, then S = V x 4*5 j also, S - | {/ 7*; hence, V = 5*4 X g, and S = 24'3 g. 46. In a system of one fixed and four movable pulleys, in which one end of each string is fixed to a beam, find the relation belween tho power and the weight (neglecting the weight of the pulleys) when one of the strings is nailed to the pulley round which it passes. Wliat is the forco exerted on the beam to which the strings are attached ? — Ans. Wlien the string is nailed to one pulley, that pulley becomes nothing n^re than a link, and the system, when used for raising the weight, is reduced to three movable pulleys, hence P = | W. 47. Two uniform cylinders of the same material, one of them 8 in. long and 2 in. in diameter, the other 6 in. long and 3 in. in diameter, are joined together, end to end, so that their axes are iu tho same straight lino ; find the centre of gravity of the com- bination. — Ans. 5||in. from the base of the shorter. 48. If two forces acting on a point are represented in magni- tude and direction by two sides of a triangle, under what circum- stances will the third side correctly represent their resultant ? Forces of 20 and 10 act along tlie sides A B and B C respectively ofanoquilateral triangle j find the magnitude of their resultant. rr If • i "I io'O I ISTO.— January. 1. Sliow Low to find tlio resultant of tliree given forces acting on a point ; and prove that, to produce equilibrium, their direc- tions must be in the same plane. 2. Find the ratio of the power to the weight for equilibrium on a bent lever of the first kind, -^hen the forces act at right angles to the arms. Supposing the arms make an angle of 1 20° with each other, and have the relative lengths of 1 and 5 ; find the magnitude and point of application of the resultant of the power and weight when the lever is in equilibrium. 3. Two heavy particles, vreighing respectively 3 and 5 ounces, are attached to the ends of a straight rod 8 inches long weighing 2 ounces; find the centre of giavity of the system. — Ans. 4"8 inches from the smaller weight. 4. A body of given mass is acted upon by a constant force; find the space described in a given time. If a particle move from rest through 40 5 feet in 4^- seconds under the action of a constant force, find the acceleration. — Ans» 4 ft. 1870.— June. 1. What is meant by the moment of a force a]30ut a given point ? How is its magnitude determined ? If several forces, which do not balance, act in the same plane upon diflfcrcnt points of a solid body one point of which is fixed, what condition must bo fulfilled in order tho.t the body may be in equilibrium ? Show that this condition cannot be fulfilled unless the fixed point is in the plane of the forces. 2. What condition must be fulfilled in order that a body may be in equilibrium upon a hard, smooth, plane surface ? A weight resta upon a smooth horizontal plane, and is acted on EXAMINATION PAPERS. 131 their direc- by a force, equal to the weiglit of G lbs., in a direction inclined obliquely downwards at an angle of 30° to the horizon. Find the ningnitude of the horizontal force required to provoTi', motion. 3. Find the relation between the po ^r and the weight in the Third System of Pulleys (in which each string is attached to the weight), when the weights of the pulleys are disregarded. Also, show what the relation would be if thg weights of the pulleys were taken into account. 4, Wha^ is meant by the " acceleration" due to a force ; and upon what docs its magnitudo depend ? If the velocity of a body increase from 12 to 13 fee*- ^^cr second while it moves over a distance of 5 feet, what is the acceleration ? Indicate the course of the reasoning upon which your calculation is based. — Ans. 2*5 ft. ♦ 1 1 ' ' 1 1 if 132 EXAMINATION TArEliS. * i hh 1871. 1. Show that aa the angle between two forces is increased, their resultant is diminished. Two forces of the magnitudes 5 and 1 1 act at angles of G0°, 90°, and 120° respectively : compare their resultants in the three cases.— ^Tis. V 201 : V 146 : V 91. 2. A weight of 50 lbs. is attached to a straight lever without weight, at a distance of 3 inches from the fulcrum, and is balanced ia one case by a power of 6 lbs., and in another case by a power of 16 lbs. : find in each case the pressure on the fulcrum, and also the distance between the points of application of the power and weight, v;heu they are applied — (a) on the same side as the fulcrum, (b) on the opposite sides of the fulcrum. Show how the answer to each part of the question would be affected if the lever weighed 9 lbs., and its centre of gravity were at the fulcrum. 3. Four heavy particles, of the relative weights 2, 3, 4 and 5, are j)laced at the corners A, B, C and D respectively, of a hori- zontal square board : find the common centre of gravity. Would its position change if the board were inclined ? 4. If a heavy body is thrown vertically up to a given height, and then falls back to the earth, show that, neglecting the resis- tance of the air, it passes each point of its path with the same velocity when rising and when falling. A ball is allowed to fall to the ground from a certain height, and at the same instant another ball is thrown upwards with just sufficient velocity to carry it to the height from Avhich the first one falls : show when and where the two balls will pass each other. 5. A mass originally at rest is acted on by a force which in l-308th of a second gives to it a velocity of 5^ inches per second : show what proportion the force bears to the weight of the mass. P _ 161 _ — vlns. Acceleration = 161 ft. .*. w ~ ' " ~ 5. • W 9 HYDROSTATICa 1.— introductio:t. 1. All matter with whicli wo avo acquainted presents itself in one of two forms, either solidf as iron, wood, stone, ovjluid^ as air, water, oil. When a knife is nsed to separate into parts a block of wood, however sharp the knifo may bo, it will not enter the block unless a certain pressure be applied to it ; but if a very thin plate of metal be placed in air or water, very little resistance will bo experienced to its motion in the direction of its plane. This difference distinguishes the two classes of bodies. The particles of a solid body cohere, so that a force is necessary to separate them ; the par- ticles of a fluid may bo separated by the slightest possible force. Hence a fluid differs from a solid by tlio absence of cohesion ai.d friction. Although tlicro is no fluid in nature bctv/ccn the particles of which there is an absolute absence of friction, never- theless these substances which we shall consider as fluid fulfil this condition so nearly as to make the i I ■ 2 INTRODUCTION. ! ^ I .n ij. III ^ ill ii.i conclusion3 founacd on it practically correct;. Ilcnce wo shall define ajluid as a collection of material par ' tides ivJiich move freely amongst themselves^ and conse* qucnth/ can he diqnaced ht/ the slightest force. 2. There arc two classes of fluids— gases and liquids. The particles of the former not only yield to tlio smallest pressure, but have a repulsive force among themselves. There is no such repulsive force "between the particles of a liquid. A cubic foot of any gas may readily bo con.prcs.^cd into half a foot, double the pressnro will reduce ifc to a quarter of a foot, and when the pressure is removed the gas returns to its original bulk ; but no ordinary pressure produces any sensible compression on water or any other liquid. Ilenco gases have been denominated elastic fluids, and liquids non-elastic. o. The perfect mobility of tho particles cf a fluid leads to special conditions of equilibrium. The gcienco -svluch treats of thcso conditions is termed Hydrostatics. 4. We have said that liquids cannot be com- pressed by any ordinary force ; and this is so far true, that both in tho theory of hydrostatics and in practice, they are assumed to bo perfectly incom- pressible. But under great pressure liquids are ejlightly compressible. By sinking a vessel filled with fluid in the ocean to the depth of 2000 metres, where every square centimetre supports a w^eight of i-OO kilogrammes, and every square inch nearly COOO lbs., it is found that tho volume of water is diminished one-tv/entieth. 5. If tho pressure of the atmosphere bo removed THE TRANSMISSION OP PRESSURE. 8 Ilcnce rial par- id consB' SC3 find ily yield vo force ivc force olo foot 3 half a quarter . the gas pressure r or any minated f a fluid X. The termed 3e com- i so far and in incom- ids are 1 filled metres, eight of nearly vatcr is k m from a given volume of water, its bulk increases by ••00005 of this volume. The increase in the case of . r.iercu vy would bo nearly '000005 of its original bulk. IT.— TUB TRANS:JISSI0N of PRESSUnB. 6. In mcclianics wc have considered the action of forces on rigid or SDlid bodies. We have found, for example, that when two parallel forces are applied at the ends of a lever, there will be equili- brium if one particular point in the lever be fixed. This is a property of solid bodies. When, however, the forces act on fluids, the result is different. Each point in the fluid must be directly sustained. Let us imagine a straight tube fall of water, v/ith a piston fitting one end. If a pressure of 1 lb. be exerted on the piston, the water will escape from the other end of the tube ; to prevent this, a pressure of 1 lb. must be exerted in the opposite direction by means of an air-tight piston like the first. There is here nothing new : if instead of pistons and fluid we had a rigid rod of wood or metal, the same relation between the forces would produce equilibrium. But let us bend the tube in any way we please (Fig. 1), and the two equal pistons, A and r, placed at the two ends, will be in equilibrium if acted on perpendicu- larly to their surfaces by equal forces. It is not necessary that the forces should here bo directly opposite. The pressure exerted by the piston A is R ■It \: mm ■■ I' ill m.: 4 TUE TRANSMISSION OF PRESSURE. trausmiUed througli the windings of tlic tube, and acts perpendicularly to tlio surface of the piston B, even \\hen tlio diroctioa is entirely dilTorent from i m Fig. 1. til at of A. It is to tills pressure at c that tlio force applied must be equal and opposite. 7. It must bo remembered that we are speaking of the equilibrium of the fluid in the tube, and not of the tube itself. We suppose the tube to bo fixed firmly on its support. The equilibrium of the solid tube and the equilibrium of the fluid are distinct things, and must be considered separately. 8. Take a closed vessel of water, and make in it an aperture B (Fig. 2), into which a piston with a surface of one square inch may be fitted, and exert a pressure of 1 lb. on this piston. \Ye may imagine that a tube passes perpendicularly from the aperture B, traversing the mass of the liquid, and meeting the surface of the vessel again perpendicularly at a point A. Since there is equilibrium the fluid surrounding the portion A B must exert everywhere a pressure upon this portion equal to that which would be exerted by the surface marked oS* if it were that of a rigid tube. Hence, the pressure at B will be trans- mitted to A, and, in order that it may be in equilibrium, a force of 1 lb. must be applied ex- ternally. THE THANSMISSION 0^ PRESSURE. 5 i As wo may repeat the same hypothesis for each point in the surface of the vessel, wo conclude tliat the pressure exerted on i; is transmitted by iho liquid equally iu every direction, so that every squaro 1 1 If Fis. 2. inch of surface is acted on by a normal pressure of lib. 9. Imagine a surface p q (Fig. 3), within the liquid, and suppose a tube constructed from b to a point A in the surface, so as to covei at that point an area, m n, equal to one square inch ; the sui face- gn n will sustain a pressure of 1 lb. Moreover, sinco tbe mass of the liquid itself is in equilibrium, it follows that m n must also sustain a pressure of 1 lb. in the opposite direction. This important principle of fluids may be enunciated thus : Aincssuve exerted at any ^oint loliatever in a Jluid in eq^idlihrlum is ti'ansmiited' undiminished to every point in the Jluid. 10. If there are two pistons, b and c, each equal to A, then when a pressure of 1 lb. is applied at a„ a pressure of 1 lb. must also be applied at b and c. Suppose the pistons united so that they present the eaino area to the fluid, they will sustain the samc^ i I t) THE TRANSMISSION OF PRESSUnn. pressure — that is to say, twice the pressure on A. A piston having tliree times the area of A will sustain thrice the pressure, and i^enerally tlie pressures sus- tained are proportional to the areas. 11. Suppose, for example, that the pistons arc fitted to two tuhes connected by a space filled with water (Fig. 4). Let tlic area of the smaller piston be one square unit, and that of the larger ten square units. When the smaller piston is pressed in with a force of 1 lb., it will be necessary to exert a pressure on the larger of 10 lbs., in order to prevent it from being moved. ]^ a and b be the areas, the pressure on the first is to llie pressure on the second as a : h. If the radius diameter or cir- cumference of one piston be n times the corres- ponding line of the other, the area of the first will be 7i' times the second, and the pressure on the first will therefore be ?t^ times that on the second.* Example. — If the circumference of the smaller be 2 in., and that of the larger 15 in., the pressure p on the former is to the pressure P on the latter as 2* : 15'. Let the pressure on the former be 1 lb., then that on the latter is 15^^ -r- 2', or 56-25 lbs. 12. The principle of virtual velocities applies to this as to every other mechanical power. Suppose, * It must be roniembered tliat the ratio of the areas is what is kere wanted, the area of a circle is the square of the radiuB multiplied by ir (or approximately, by y). THE BUAMAII TRESS. on A. sustain ?s sus- >ns arc ;d with piston that of When sscd in will 1)0 snre on )rdcr to movotl. a3, the to the s a : 5. or cir- corres- will be he first iller be re p on ttcr as 1 lb., 3lies to nppose, s what is le radius i for cxam[ amount of water le, the small piston to descend n in., tlie driven out of the smaller cylinder into the larger will be a X « cubic in. If t:ic larger pis'ton ascend m in., the air : -nt of water forced into it will be i x m ; hence rt X n but h X in P b A p = ax licnce p X 11 = I' X m The principle explained above is applied in the Bramah or Hydraulic Press. The Bramali Press. 13. Description of the Press. — Tiic whole instru- ment is represented in Fig. 5, and a section of it in Fig. 5. Pig. C. a and h are two pistons working through air-tight collars in strong cylinders filled with water, and connected by a tube c c. The small piston a ii If \m 1 I IF! '■t: 8 THE BRAMAII TRESS. worked by a lever represented on the right of the upper figure, and the larger piston b is attached to a movable platform, on which the substance to bo ^.' '^ii^^. 1 d Fig. 6. pressed is placed. At p and r are valves opening upwards. The tube d passes into a cistern of water. Action of the Press. — Suppose the cylinder filled with water, and a in its lowest position; on raising a, the atmospheric pressure forces water from tho reservoir through c?, and when a is afterwards forced down, the valve p closes, tho valve r is opened, a portion of the water is driven along c, and the cylinder h is made to ascend. By repeating this process any required compression of the sub- stance between the platforms may be produced. At TUE CRAMAU PKEb3, /there is a plug, "which can be unscrewed when the compression is completed. 14. The press is named from Mr. Bramah, "who invented the simple and beautiful contrivance repre- sented at r,i m in Fig. 6, for preventing the escape of the water from the cylinder. The ring or collar m is lined with leather, which is allowed to hang down like a flap in the cylinder. When water is forced upwards between the piston and the cylinder, it fills the fold of leather, and causes the edge to embrace the piston, so that the liquid cannot force its way higher than m. The greater the pressure, the more tightly will the flap close round the piston. TJiG relation hetii'een tlie power exerted on the press and iJic force at the end of the lever. 15. Let the diameter of the piston in the pump bo 1*5 centimetres, and the diameter of the piston of the press 20 centimetres. Let the force bo 18 kilogs., applied at 36 centimetres from the fulcrum, and let the piston of the pump be G centimetres from the fulcrum. Force at a distance of 30 cents, from fulcrum =13 »» »» ^ » ffi =iUi? .*. Pressure on area (1*5) -=108 M 1 =.7705=48 »» 202 =48 X 400 = 19,200 kilogs. IG. In the same way we may find the relation for any dimensions. Lot 7n and n be the arms of the lever, f the force acting on the lever, p the pressure exerted by tho i -i •i I i '■ i 'f, ,''':*-: 10 EXERCISES ON THE PROPEKTIES OP FLUIDS. 1 piston of the pump, a and h tlio areas of.tlio pistons. Then, by the principle of the lover, m the pressure on tho smaller piston = F. — 110 the pressure on one unit of area = F. — ' pressure on b units of area = F If R and r be the radii, m na IV h h a R2 j,3 Hence the pressure onh = F -^ -„ n T" Instead of a pressurcof 1 lb. on the piston (Fig. 4), a pound of water in the tube above would produce tho same effect. If the vertical tube holding this water bo made twice as long, but of half the sectional area, the pressure on a unit of area will be doubled, while tho same amount of fluid only will be used. By continually repeating this process, the small quantity of fluid may be made to support a very great weight. This principle, stated as follows, is sometim.es termed the Hydrostatic Paradox : — Any qiiantiti/ oj fluid, Jioicever small, may he made to sti2:)^ort any iccigldj Jioiccver larye. Exercises on the Pboperties of Fluids. 1. Wliy cTocs a liquid readily cliaugo its form^wlien its position is changed ? 2. "What properties of liquids are not possessed by solids ? 3. If a solid be placed inside a splicrc, and a pressure be exerted upon ILo solid by a piaton, how will the pressure bo transmitted ? EXEECISES ON THE PEOPL.TIES OF FLUIDS. 11 4. If the Bpliero bo filled with a liquid, and pressure bo csGvted by means of the piston, bow will the pressure act on tlw Bpliorc ? 5. If a vessel full of liquid be provided with two pistons, one of which is 20 square centimetres in area, and the other 100 square centimetres, and a pressure of 3 kilogs. be applied to the former, what pressiu'e will bo exerted on the latter ? Pressure on 20 square c. = 3 kilogs. „ on 1 square c. = ^, »» on 100 square c. = 3 X IOC = i: C. If one piston bo 10 centimetres in circumference, and tho other 80, what must bo the pressure applied to the smaller that the larger may support 1000 kilogs, ? Pressure on area of 80^ = 1000 fi »3 1 = 10= = _ 1000 C'l(iO 1000 X 100 = i: »» » *" ~ cioo 7. If tlio area of the smaller piston bo 12, and that of tho larger SCO, what pressure will the larger sustain when 5 kilogs. is applied to the smaller ? 8. The diameter of the smaller piston being 3, and that of tho larger 13, what pressure on tho smaller will produce a pressuro of too lbs. on tho larger ? — Ans. 25 lbs. 9. Tho circumferences of tho pistons are as 4 to 55, and tho pressure on tho smaller 4.00 grammes, find that on tho larger. — Ans. 75,G25 grammes, 10. In descending £0 centimetres, tho smaller piston raises a weight of 6G0 lbs. through a height of '5 centimetre. What pres- sure is applied ? — Ans. 20 lbs. 11. Two vertical tubes a and B are connected by a horizontal tube ; A is short, and is provided with a piston ; b is long, and has no piston. Tho diameter of a is 50 times that of c, what height cf water in b will support ICOO kilogs. in tho piston in A (a depth of a centimeter produces a pressure of one gramme on a square centimetre) ? — Ans. 4 metres. 12. Tho pressure on a plane area in the form of a sqiiarc, tho £iJo of which is a yard, is known to bo uuirorm, aud to be equal 'I' /£ nf (■' .• r - •'I ..•'■(■ i ■ i J ft- ■A I 12 PRESSURE OP A FLUID ARISING FROM ITS WEICnT. , 1; 1 to 2700 lbs. ; find the pressure at a point v;liea the unit of Icngtli is a sq. in. — Ans. 2-!^ lbs. 13. The pressure at any point in a rectangular surface 1"5 metres long and 1'3 metres broad, is 100 grammes, a centimetre being the unit of length. Find the pressure on the whole surface. — ilns. 1300 kilogs. 14. If the diameter of the pump of a Bramah press be 3 centi- metres, and that of the press 120 centimetres, wliat force will 1 kilog. applied to the pump produce on the press? — Ans. 1C03 kilogs. 15. If a power of 100 kilogs. j: applied to a lever at 30 centi- metres from the fulcrum, tlie piston of the pump being attached 6 centimetres from the fulcrum, what will be the force acting on the press iu t'ie last question ?—Ans. SOO.OOO kilogs. in v. III.— rPcESsunE or a fluid ArjsiNG feom its WEIGHT. 17. Liquids have weight like sol id 3 ; but tho particles of a liquid glide over each other without friction. Ilcnco, when a liquid is contained in a vessel, it is always found that the surface is in a horizontal plane. When the fluid is at rest, we see that this must be tho case. If part of the fluid surface were inclined (Fig. 7), we should have particles acted on only by their weight at rest, on an inclined plane free from friction, which cannot bo. Fig. 7. eating \m :-')i PRESSURE OF A FLUID ARISING FROM ITS WKIGHT. lo IS. When the liquid is contained in several vessels communicating with each other, it is still found that the surfaces in the different vessels all lie in the same horizontal plane. Thus, if tubes of various shapes be connected as in the figure (Fig. 8), the water rises in all to the same level. This fact is frequently expressed by saying that liquids maintain Fig. 8. their level. If a small vertical tube t communi- cating with the vessels terminate below tlie level of the surface, and be pierced with a small hole at the top, a jet of water rises from the orifice to the level of the surface. The force with which the water leaves the aperture is proportional to the depth of the aperture below the surface. 19. This principle is applied in many ways. Water is distributed over a city by subterranean V n II ,-*! ■i\ ' " -t >•- 1 ^, .1 § ■ H PRESSURE OF A FLUID ARISING FROM ITS WEIGHt. 1 I i ii i 111 pipes issuing from a large reservoir situated on an adjacent elevation. These pipes are terminated by stop-cocks, >Yliicli arc bebw the surface of the reser- voir, so that when they are turned on the water escapes. 20. Tlie law according to which the pressure produced by the weight of a fluid is transmitted in every direction frequently makes the effect of this weight very different from that of a solid. Suppose, for example, that a cylinder of ice is placed in a vessel of the same shape, the weight of the ice will exert a pressure on the base, but none on the sides of the vessel. The sides may be removed, while the mass Avithin retains the same posi- tion. When, however, the ice is melted, there is not only a pressure on the base equal to the weight, but also a pres- sure on the sides. If a hole be made in the vessel, the liquid will flow out, and the force with which it escapes will be proportional to the depth of the hole. This variation in the pres- sure may be illustrated by the following experiment. 21. Take a hollow cylin- der or tube of glass, one end of which may be closed by a disc of metal supported by a thread (Fig. 9). Apply the disc to the tr.^c, Fig. 9. prcs- PRESSURE OF A FLUID ARISING FROM ITS WEIGHT. 15 and i^lunge tlio closed end in a vessel of -water. At a certain depth, depending on tlio weiglifc of the disc, the thread may be dropped, and tho disc supported by tho fluid pressure only. If tho cylinder bo moved horizontally, no change takes place. This result illustrates the fact that the pressure at points in the same horizontal plane are equal. If a heavier disc bo taken, it must be plunged to a greater depth before it will bo supported, and it will bo found that the weight supported is proportional to tho depth. Suppose, now, wo take a very thin disc, make tho cylinder descend to a certain depth, then pour wTvter gently into the tube, wc shall find that when tho water in the tube reaches tho level of that outside, the diso will bo detached ; being equally pressed on both sides, it falls by virtue of its weight. Hcnco tho upward pressure on the disc is tho weight of tho liquid which would fill tho cylinder to tho level of the surrounding fluid. 22. Suppose tho area of the disc and C3dinder to be 40 square centimetres. The weight of a cubic centimetre is 1 gramme, and, with a base of 40 square centimetres, and height 1 centimetre, we get 40 cubic centimetres. Ilence, at a depth of 1 ccnti- mctrCj a disc weighing 40 grammes will be supported. At a depth of 2 centimetres tho pressure of the fluid v/ill be twice 40 grammeg, and so on. Thus, fur every square centimetre of area in tho section of tho cylinder there will be a pressure of as many grammes as there are centimetres in tho depth of the diso below the surface. 23. Tho principle that the pressure depends on rm ,1" -iiW- %0: It til I' I Si! 1 n IG rilESSURE CF A FLUID AIIISIKG FROM ITS WEIGHT. the depth, and not on the quantity of fluid, may be fiirtlier illustrated by the following series of experi- ncnts. Take a balance having a circular disc oi metal attached by a fine wire (Fig. 10) to one end of the beam, just heavy enough to balance the scale at the other. Take a cylindrical vessel open at both Fig. 10. ends, and place it with axis vertical on a tripod, so that the wire supporting the disc passes along the axis of the cylinder. Let the disc and cylinder be properly ground, so that when pressed together they form a vessel which will hold water. Place a certain weight, a pound for example, in the scale ; the disc will now be pressed against the cylinder, and, if water be poured gently into the cylinder, the disc will remain r. TEEssunc or a fluid arising from its wEiciir. 17 ly be peri- !C Oi ! end scale both tliot ds of perly rm a }iglit, now r be jni.iin in contact until the water rises to a certain hciglit. If more water be added, tho pressure will overcome the weight in tho scale, and will forco open the bottom. On a rod attached to tho stand mark tho height of tho water when the highest level is reached. It will bo found that exactly one pound of water will bo supported. Pour tho water out of tho cylinder into tho scale, taking away tho 1 lb. weight, replaco tho cylinder, and fill it again gently up to tho height h) tho water in tho scalo will bo just EuTiciei't to keep tho disc in contact with tho cylinder. "We see, therefore, that the pressure on tho disc is equal to tho weight of the column of water above it. Kcpcat the experiment with vessels B and c (Fig. 10) of different shapes, having the area of the base tho same. Water must be poured to the same height as before, in order that the pressure on the disc may bo 1 lb. We shall find, however, that with vessel E there is more than 1 lb. of water, and with vessel c less than 1 lb. Thus the pressure on a plane docs not depend on tho amount of water used ; it cicpends on the size of the plane and its depth below the surface. In all the cases tho pressure is equal to the weight of a column of fluid having the plane for base and the depth of the plane for height. Only in tho case of tho cylinder is this tlie samo as tho wcidit of tho water used. Thus we see that the pressure on the horizontal base of any vessel is tho v.'cight of tho flaid which a vessel having the samo base and vertical sides would contain, when filled to tho same level. !M:i M \'' S ! . -ii 18 THE PRESSUHE ON THE SIDES OF A VESSEL. I Example. — Whafc 13 tho ^pressure on tlio base of a cono -whicli lias tho vertex upwards and is filled with water to tho depth of 18 centimetres, tho radius of the base bein t> A cubic inch 1723 ounces. of 13 EXERCISES, 21 ExErvCiSES. 1. Find Iho pressure at tlio depth of 80 metres in a lake— ilns. 3,000 grammes on a squavo centimetre. 2. Find tlio pressure on a Bquaro inch at a depth of COO feet in a lake, the weight of a cubic foot being 1000 oz3. — -4ns. 41CGa ozs. 3. In the experiment illustrated in Fi^. 9. if tho section of the cylinder bo 30 square centimetres in area, and tho weight of tho disc 1080 grammes, to what depth must it bo sunk beforo tho thread may bo dropped ?-^Ans. 30. 4. Find tho pressure on a surface, 18 units by 15, nt a depth of 20, tho weight of a cubic unit of v.-ater being 10 lbs. — Ans. EJ,0001b3. 5. Two cubical vc:ccl3, tho cdgC3 of v/hich are as two to one, aro filled with water, compare tho pressure on tho bases.— -Ans. 8 to 1. C. A cubic inch of mercury v;cigh3 8 ozs., what will bo t];o pressure on tho bottom of a vessel, tho area of which is a square foot, tho depth below the surface being 5 inches ? — Ans. 300 lbs. 7. A plate of metal h held at tho end of a hollow glass cylinder C centimetres in diameter, wliilo the Ciliudcr is immersed verti- cally. It is found that tho plato is just supported at a depth of 35 ccutimetrcs; find the .weight of tho plato in grammes.— uljis. OCO grammes. 8. If tv;o vessels have equal bases and tho samo depth, but one wvlens upwards to tho mouth, and the other diminishes upwards to tho mouth, on which is the pressure on tho base the greater ? • 0. What will bo tho pressure on a rectangle 11 by GO inches, immersed so that a diagonal is vertical in a fluid a cubic inch of which weighs 1^ oz. ?—'Ans. 30,195 ozs. 10. What will bo the pressure on a vertical triangle immersed in water with the baso in the surface, tho base being 50 centi- metres, and height 30 ?— ilns. 7500 grammes. 11. The upper surface of a vessel filled with water is a squard whoso side is 8 metres, and a pipo communicating with the interior is filled with water to a height of 2 metres abovo tho surface ; find the weight which must bo placed on the lid of the vessel to prevent the water from escaping.— -^ns. 128,000 kilogs. 12. A house is supplied with water from a reservoir, 211 feet abovo the ground floor, by means of a pipe laid underground, what will bo tho pressure per square inch on a tap 25 feet abovo the 1 1 if ■" f4l LIM ^r » r llfl i\:n I in I 22 rr.EssuRE on a solid immersed ni a liquid. prouncl floor, supposing a cubic foot of v/atcr to weigh 1000 ozs. ?— Ans. 1500 ozs. 13. A vessel is filled with water, how will the pressure on the base bo aflccted if a piece of metal be dipped into the water ? 14. If the vessel be not full, how will the pressure on the base oc affected by dii^ping a piece of iron into the water ? W. ! lY.-TEE PRESSURE ON A SOLID IMMERSED IN A LIQUID. 29. When bodies aro placed in a fluid, it is a matter of common observation tliat they will sink or float, according as they are, bulk for bulk, heavier or lighter than the fluid. Even when a body sinks, it appears to bo of less weight in water than in air. For example, it is easy for a person to raise a mass of stono under water which ho is unable to lift above the surface. The pressures exerted by the fluid on the surface of the body have a resultant which acts in a direction opposite to that of the weight of the body. If the vreight of the body exceed the fluid pressure, the body sinks; if tho pressure bo greater than the weight, tho body ascends ; when the weight and the pressure are equal, the body rests. These different conditions may be made to succeed each other at pleasure, with the toy represented in Fig. 11. A hollow globe of glass, containing air and water, floats prrrsuRE ON A ^OLiD i:>i:.irn?ED in a liquid. 23 a ft A ,t8 Fi^. 11. on tlio surface of tbo water in a gLiss cylinder, over the top of which is stretched an elastic cover. In the lower part of the globe there is a very small hole, by which water can enter or escape. If the hand be placed on the clastic top, the air in the cylinder is compressed, and its pressure on the surface of the water is increased. This auimientcd pressure is transmitted through the water to the air in the globe, which is condensed. The consequence is that a small quantity of water enters the globe, increases its weight, and causes it to sink. When the pres- sure is removed, the bulb rises again. The pressure may bo so regulated that the bulb shall rest sus- pended in any part of the water. 30. The law according to which a fluid excrls a pressure on a body immersed is as follows : — WJicn a solid is immersed in a llqidcJf the resultant p'essiiro en it is the ivcight of its oim hiilh of U^uul, Lot A be a solid immersed in water (Fig. 12). Imagine the spaco A filled with water, and then suppose the mass A to become solid j ;" " it will remain suspended liko any (lij"^ other part of the fluid. Since its t weight acts downwards, it is evident I'is- 12. that there is an equal force acting upwards, counter- acting the weight. This forco is the resultant of all the fluid pressui.':s, acting in all directions, upward, ;i^ IS If t.''ik ni 1 24. SPECIFIC CEAVITT. downward, and latcrtally. Some of these pressures neutralize otliers ; but there must remain a balance of upward pressure, since the under surface is at a greater depth than the upper, and it is this resultant pressure which counteracts the weight of ^Jie imagined isolated mass. Therefore the upward pressure on A IS equal to the weight of a Since two forces in equilibrium must act in the samo straight line, the resultant must pass through tlio centre of gravity, y, of the mass. If we imaixine the ice to have become stone with- out altering its bulk or its surface, the pressure acting upon it will still bo the aame, namely, the weight of the same bulk of water as would exactly fill the space occupied by the stone. Thus the weight of the body is diminished by the weight of the water displaced. V.~SPECiriC GSAVITY. 31. Equal bulks of different substances have dif- ferent weights. If equal volumes be taken, lead is heavier than iron, iron is heavier than water. It is a problem of great practical importance to determine the relation between the weights of dif- ferent substances. For this purpose all substances are compared with water, and the relative weight of any substance is expressed by the number of times the weight of the given body contains tho weight of BrECIFIC Cr.AVITY. 25 1 an cqufil bulk of water. Tliis number of times is called tlio specific gravity of tlio body. 82. Tlie measure of the spcoiJiG gmvittj of a lochj is tltG ratio of the iceight of any volume of the hotly to the ii-eifjhi of the same volume of a certain standard siibstancG. Water by its nnivcrsal distribution and the ease with wliicli any required volume is obtained and purified, is best adapted for the standard substance. Ilcnco if s bo the specific gravity of a body, tlio weight of the body over the weight of an equal bulk of water =5. Let V be tho volume of tho body, and let w De the weight of a unit of water, then tho weight of tho same bulk of water =w;V, thus :— ■\7 — ^ = B or W =. loYs vN If the unit of volume be a cubic centimetre of distilled v;atcr at 4° «(;=one gramme; if tho unit of volume bo a cubic foot Z6'=1000 ozs. nearly. Thus, if wo know the speeiGo gravity of a body wc can find tho weight of any volume. Example 1. — What is tho weight of 43 cubio inches of copper the specific gravity of which is 8"S2. 48 X 1000 Weif^rht of 48 cubio inches of water = 1728 Hence, weight of 43 cubic inches of eubstanco, whoso specific gravity is 8*82 . 48 X 1000 X 8-82 1723 =215 ozs. m 1 1' hM : ■:* hit i-- i':.f. i,' 26 SPECIFIC »3nAvrn*. Example 2. — What is tho wclglifc of 250 cub'o centimetres of mercury, the relative weight being l3-o9G ? 1 cubic centimetre of water weighs 1 gramme, 1 cubic centimetre of a substance of spcciiic gravity 13'596 weighs . . . 13'oOG grammes, 2o0 cubic centimetres weigh 250 X 13'50G grammes, Or 3399 grammes. Example 3. — What volume of lead of specific gravity 11 '4 will weigh 190 grammes ? 11 '4 grammes of lead have the same bulk aa 1 gramme of water. Hence 11*4 p-rammes of lead measure 1 cu. cent. 1 gramme 100 grammes 19 l> »» r» Or ICii cubic centimetres. _1_ 11-4 190 11-4 >» J) Examph 4i. — A piece of marble of specific gravity 2*7 weighs half a ton. What is its size ? Haifa ton= 1 7,920 ozs. Volume of 1000 ozs. of water=: 1 cu. ft. 1 oz, „ = -001 „ half a ton „ =17920 x -001 cu. ft. :^17'92cu. ft. Volume of the same\ -i-.qo wciclit of a substance > = - — ^ssG'OG cu. ft. whose sp. gr. is 2*7 ) ^ * Example 5.- -Find the specific gravity of a mixture of 1 cu Co csntlmetro of a fluid of specific I ^1 ^2 ErECIFIC CRAVirY. 27 LO gravity 1*3, and two cubic ccntimclrc3 of cpccifio gravity 2*2. Weight of 1st, 1'3 grammes. Weight of 2nd, twice 2'2 or 4*4 Weight of mixture, hi grammes. Volume of mixture =3 cubic centimetres, and vreight of equal bulk of water =3 grammes. 5*7 Hence specific gravity of inixturc=-rr =1'0. Example C— To find the specific gravity of a mix- ture of given v lumes of any number of given fluids. Let Vj Vo V3 etc. bo the volume of the fluids, let $1 St> ^3 be their specific gravities. Then the vvxiL^ht of the mixture is lu (s-^-^ + s.Nc, + 53 V3 + etc.) .*. Weight of same volume of water is w (Vi + Vo + V3 + ctc.), and consequently tlio specific giavity of the mixture is Sy^^ -f sNc> + go Vo + etc. Vi + Vo + V3 +etc. Example 7. — To find the specific gravity of a mixture where the weight and specific gravity of the compounds arc known. Let Wi Wn tv^ etc., be the weights, let Si s^ b'.^ be the specific gravities. ^o^ 6^3 The volumes are — , ~^, etc. Hence, substituting in the last exercise, we have the specific gravity of the mixture ^J-L + v'n + v'.7 + etc. tC: u\ -+—•+•—+ etc. V. u ' s. 23 EXERCISES. EXEHCISES. 1. A body measuring 18 cubic centimetres floats in water with its wliolo bulk immersed ; find its wciglit. 2. Define the specific r;ravity of a body. If 130 cubic ccnti- metres of a body weigh 120 grammes ; find its specific gi'avity. — Ans, 'S, 3. If 200 cubic centimetres of brass havo a specific gravity of 7'5, find the weight. — Ans. 1500 grammes. 4. Having given thai; 10 cubic feet of etono weighing 21,050 0Z3., havo a specific gravity of 2*5, find the weight of a cubio foot of water.— ilns. 093 ozs. 5. What is the weight of 1 2 cubic inches of mercury, the specifio gravity of which is 13'5, if a cubic foot of water weighs 1000 ozs.— Ans. 9375 ozs. G. What volume of zinc, of specifio gravity 6*85, wiU weigh 109"6 cTaramcs ? — Ans. 10 cubic centimetres. 7. \Miat is the size of a block of stone, the weight of which is 1 cwt., and specific gravity 2-4?— .4?is.-74G cubic feet. 8. Three liquids, the specific gravities of which are respectively 1'2, '90, and l"4oG, are mixed in the proportions of 1-^ gals, of the first to 1\ gals, of the second, and 1^ gals, of the third j find the specific gravity of the mixture.— u4ns. 1*2. 9. What will be the weight in water of 100 cubic centi- metres of iron, the true wciijht being 781 grammes.— ilns. C31 grammes. 10. A cubic centimetre of alcohol weighs '70 grammes, what will bo the true weight of a body which floats in alcohol with 1000 cubic centimetres of its volume immersed? — Ans. 71/0 grammes. 11. What is the weight of a cubic centimetre of a body which floats in water with one-fifth of its volume above the surface ?— Ans. 'S grammes. 12. Find the weight of a cubic foot of mercury, the specific gravity of which is IG-HCS, supposing a cubio foot of water to weigh 1000 ozs.— J.ns. 813 lbs. 13. If a cubic unit of the standard substance weigh '35, what is the weight of 1000 cubic raiitsof substance whose specific gravity is 3 }—Ans. 1050. 14 The specific gravity of two liqui4;3 are resptsctively 1'3 and •S3, latter J 15. find th IG. of the mistur o3. gravitj and till now p( water ( that w] its own merse i water, shall th 31 princip^ appears placed. Archim cuse, ai wliethci TO FIND THE SrEClFIC GRAVITT. 29 ' •S3. A mcasuro of tlio former is added to thrco measures of tho latter J find the specific gravity of tho mixture.— yl lis. '9025. 15. Ten cubic centimetres of a substance weighs iV ki]op;3. ; find tho specific gravity of the substance. — Ans. 6*25. IG. Equal volumes of thrco fluids are mixed, tho specific gravity of the first is 1*55, that of tho second is 175, and that of the mixturo is 1*0 j find the specific gravity of tho third.--.4ns. I'o. VI.—TO FIND TEE SPECIFIC GRAVITY, %Yith 1 which ico?— ^pGciiic iter to ?hat is jravity '3 and 83. — To find tho relative weight or spcciric gravity of a body, we require to know its weight and tlie weight of an equal bulk of water. We vrill now point out methods of obtaining the weight of water equal in bulk to a given body. Wo have seen that when a solid is immersed in a fluid, it displaces its own volume of the fluid. Suppose that we im- merse the body gently in a vessel exactly filled with water, and weigh the water which overflows, we shall then have the required weight of vv^atcr. 34. A more convenient method is afforded by tho principle that the solid, when weighed in a fluid, appears to bo lighter by the weight of tho fluid dis- placed. This law is termed the principle of Archimedes. It is said that Hiero, King of Syra- cuse, applied to Archimedes for a test to provo whether a crown which had been made by hia orders iMa -1: V ■M ^^■t .*« ;* ! ■ lif- -i^ i M m r ■,i; ■t fi%i' so TO FIND THE SrECinC GRAVITY OF was all gold, or wl.ether the goldsmith had dis- honestly substituted a baser metal for a portion of the gold. While the philosopher was thinking of the sub- ject, he chanced to enter a bath filled with water, and noticed that, as he entered, the liquid flowed over. fli H^ ' ( rig. 13. ' t This observation suggested a solution of his problem. He took the crown, and a quantity of pure gold of the Bame weight, and immersed them successively in the same vessel, filled to the brim with water. As the crown displaced more water than the equal bulk of gold, he concluded that it was partly composed of a A SOLID HEAVIER THAN SVATEIS. 81 UgLtcr metal, and tlio king's suspicions were con- firmed. 35. The principle of Archimedes may be t'^^ted experimentally thus: — Two cylinders, the fir.t a hollow, and the second h solid, and of such a size as exactly to fill the first, are suspended from the scale of a balance and weighed. Tho cylinder h is then suspended from a and allowed to dip into a vessel of water. Tho opposite scale then goes down ; bat it is found that equilibrium is restored when tho cylinder a is filled with water A Solid heavier than Water. 36. Hence, to find the ' ^"^ecific gravity of a solid heavier than water, and not soluble in it, weigb tho body in air, then weigh it in water; tho specific gravity the true weight loss of weight in water. blem. Df the n the s the ilk of of a Example, — A piece of metal w^eiglis 18 kilogs. in air, and 15 in water; find its specific gravity. Tho loss of weigbt in water =18 — 15=3 kilogs. 18 Hence tho specific gravity =-^=G. 37. An instrument termed Nicholson's hydro- meter may be employed to compare the weight of a body with that of an cipal bulk of fluid. 82 TO FLND THE SrECIFIC GllAVITY OF ii < I '■ It consists of a liollo\v cylladcr c (Fig l-i), sup- porting a light cup i on a tliin stem above it, and a Lcavy cup h below it. Tho wbolo floats in Iho fluid. A certain wcigbt (suppose 10 grammes) is placed in the upper cup, and a mark is made on tbo stem, to indicate tbc surfiico of the liquid. The body is then placed in ilio cup ; tho hydrometer will descend, in cons'equenco of tho additional weight. Such weights arc then taken ofT, as will cause tho instrument to rise until the mark made is again at the surface. Tho K weights taken off will be tho y weight of tho body. For ex- Fig. 11. ample, let us suppose 7 grammes removed, then this is the weight of the body. Now remove the body to the lower cup ; the instrument ascends. Place such a weight on it as will bring it to the same level. The weight added is the loss of weight in water. Sup- pose 3 grammes added, then the specific gravity _7 ""3 *2i fc Q. To Find the Specific Gravity of a Solid Lighter than Water. 28. 1st Method, — If a wire gauze w be stretched over the lower cup in Nicholson's hydrometer, so as i lost A SOLID LIGUTEll THAN VrATER. sup- md a fluid, pposo ;n tlio rk is iicato . Tho .1 ihxi will CO of Such off, as cnt to ado is Tho be tho 'or cx- Dse 7 n this ody to such Tho Sup- - Lghter itched so as to prevent a light body rising, tho method last de- scribed will suit this case also. 2nd Method. — Tho specifio gravity of a solid lighter than water may also bo found by attaching a heavier body to it. Example. — A block of wood, weighing in air 8 lbs., is tied to a piece of iron weighing G lbs. ; tho wholo weighs in water 4 lbs., and tho iron only weighs 5 lbs. Find the specific gravity of the wood. Tho weight of water equal in bulk to tho iron , ~ G — 5 = 1 Tho weight of water equal in bulk to both iron and wood=(8 + G) —'1—10 •*. Weight of water equal in bulk to tho wood , =10 — 1=9. Q Hence tho specific gravity of tho svood=-^. And in every case tho specifio gravity of the lighter body weight of light body lossof weight of both —loss of weight of heavy body To find tho Specifio Gravity of a Liquid. 89. 1st IMJiod.— -Weigh a flask filled with tho liquid, then weigh tho flask filled with water ; deduct from each result tho weight of the flask, then tho specific gravity of the liquid is its weight, divided by tho weight of tho water. Example, — A flask, when empty, weighs 2 ozs., when filled with alcohol it weighs G ozs., and when W '1.1 \ ' S4 TO fUn'd the fc'rECiFic cr-wiTy of a gas. filled with water ifc wciglis 7 023. FiiiJ the specific gravity of the alcohol. The weight of the alcohol is G--2 = -l- cz3, Tlic weii-hfc of the water is 7 — 2 = 5 ozs. ,*. The specific gravity of the alcoliol=l-T-5=*8. 2nil MctJiocI. — Weigh a hocly of linown Avelght in the given liquid and in water. The weight of the body's bulk of the liquid is the loss of weight in the liquid. The weight of the body's bulk of water is the loss of weight in water. Hence the specific gravity __tlie loss of weight in the liquid tho loss of weight in water. Hxample. — A platinum ball, weighing 2 lbs., when suspended in olive oil, weighs 1*915, and in water weighs 1'907. Find the speciuc gravity of tho oil. Weight of oil displaccd=2-l"915=:-085 Wei'ihtof water CD J) =2-l-907=:-093 • ». Sp. gr. of tho oil = •085~-093=-915. To find the Specific Gravity of a Gas. 40. Use a flask with an aii'-tight stop-cock, ex- haust it by means of an air-pump, weigh it first empty, then full of the gas under consideration, and, finally, full of water. On deducting the first result from each of the other two, we have specific gravity of gas weight of gn?? contained by the flask weight of water contained by the flask. TO FIND THE SPECii'IC GRAVITY OP A SOLID BODY. 35 JExamph. — A flask v,'cigliin;5 10 grms. wViCn empty, weighs 10'G1G5 grms. when fjllcd with air, and 510 grms. when filled with >v^ater. Find tlio relalivo weight of air. The air which tills the flask weighs '0-105 grms. And tho water 500 gi.^m3. The rclativo weight is therefore •ClG5-i- 500 = -001203. To find the Specific Gravity of a Solid Body Soluble in Water. 41. "Weigh the hod J in air, and then in some liqnid in whieh ib is insoluble. The loss of weight is the weight of a quantity of this liquid having the same bulk as tho body. The weight of tho .same bulk of water is found by dividing tho result by the specific gravity of the liquid. Example. — A picco of sodium weighs 8 grms. in air, and 1-07 grms. in naphtha, the specific gravity of which is known to bo '84. Find tho specific gravity of tho sodium. The weight of the naphtha having the same volumo as the sodium=8— 1'07, or 6*93 grms. The weight of the same bulk of water = G-93-^-81, or 825. g Hence the specific gravity =- — = "07, o'25 Instead of definite weights, let us take general ,f '.it in II \fW 'hi so TO FIND THE SPECIFIC GRAVITY OF A SOLID P-ODT. symbols. Let tr<,=tho weight of the sodium in air, i:^„=tho wclglit in naphtha, and let S=the spcciHo gravity of the naphtha ; then the specific gravity — 7 -;^ — - = — - — . Hence ive may find the specific gravifij referred io water hy Jinding first the sj)ecific gravity referred to some other liquid, and then multi^plyiiij hi/ the sjK'cifio (gravity of this liquid. 42. "We may also find the specific gravity of a liquid by compariug tho amount disi^jlaced by a body made to floafc in the liquid and. in water. Fig. 15 represents such a body, termed a common hydrometer. It is usually made of glass, and consists of a tube A, a large bulb B, and a small one c, loaded £0 as to keep the stem vertical. - Let to be the weight of the in- strument, and (I the area of a section of the stem. Suppose tho instrument to sink to D in water, and to E iu tho other liquid, then if v bo tho volume below D, and S the speciSc gravity of the fluid, 1; ^ S is tho volume below E. But since the volume of the pfirb DEisfl.ED: /. |=i; + ». ED. A E DENSITY. 43. In tlie preceding examples we have not taken account of the weight of the air displaced by the solid weighed in it. The true weight of the solid is the weight iry air, together with the weight of the air clisplaced. Wo may find approximately the weight of the air displaced from the weight of the water dis- placed. At the ordinary pressure of the atmosphere 1 licrc of water weighs 1000 grms., and 1 litre of air 1*293 grms. ; hence the weight of air displaced is •001293 of the water displaced. Examiile. — A body weighs 8 hilogs. in air, and 5 in water ; find its specific gravity, taking into account the weight of the air. The weight of water displaced is very nearly 8 — 5, or 3 kilogs. This is not quite true, because 8 is not quite the true weight of the body. Hence the weight of the air is very nearly •001293 X 3000 grms., or 3-879 grms. Ilcuec the Bpecific gravity is if ■I '1 Density, 44. The density of a bo '^y is a term for the ratio of the quantity of matter in the body to the quantity of matter in an equal bulk of a standard substance. Thus, if water be the standard, the density / mass of the body v unass of an equal bulk of water./ Tho density, therefore, refers to the mass as tho specific gravity refers to the weight of a body. i; t w i 1 1 83 DENSITY. If we suppose fhe force of gravity constant, as is the case at any one particular spot on the earth's sur- face, the weight will be proportional to the mass, and the density and specific gravity of a body will be the same. The variation of the force of gravity docs not affect the density, but it docs affect the specific gravity. From the above formula we see that if the volume bo kept the same the density varies as the mass, and if the mass remain the same the density varies inversely as the volume. Hence, when both volume and mass vary, mass the density varies as Therefore d = CM V volume where c is a constant. It has been shown in "Mechanics" (p. 176) that writing M for mass, W for weight, and j for acceleration due to gravity, W varies as M when ^ is constant, W „ 3 M „ /. W „ My both vary .-. W == CMy, but from the above formula ]M = dV X a constant The units may be so chosen, therefore, that ^Y = ydV. PROBLEMS ON THE DENSITY OF LIQUIDS. 30 for Problems on tlie Density of Liquids, 45. If two liquids that do not mix meet in a bent tube, their heights abovo their common surface will be inversely proportional to their densities. Let A be the upper surface of the denser fluid of density d, B that of the lighter fluid of density rZ^, and let c be their common surface (Fig. IG). Let n = the height of A above the horizontal plane c D and 111 — height of B above c d. The pressure at c = 9(\^i\ tho pressure at d = gdli^ and these are equal, since they are in the same horizontal plane. .•. gdk = (/djii 1 1 Fig. 18. or 7i : 7i 1 - d d. Examples. 1. A body susprncTod by a thread from ono scale of a balance weigiis 3G"81 grammes; a vessel of water is placed so tlmt tlio body is suspended in tlio water, and then C'21 grammes must bo placed in tlio scale above the body to restore equilibrium ; find tbo specific gravity of tlio body. — Ans. 5"9. 2. Find tlio specific gravity of a pioco of load wliich weighs 47'48 grammes in air and 4-3'33 grammes in water. 3. The specific gravity of aluminiura is 2 '50, and a fragment when weighed in water loses 37S grammes. What is its weight ? ■Ans. 9"G7G8 grammes. 4. A body weighs 5 grammes ; when the body and 100 grammes of lead of specific gravity ]l"'i^l are weighed together in water, tho loss of weight is 9 5 grammes. What is tho specific gravity of tho body ?— .Ijis. C'59. , !,p. iO EXAMPLE?, 5. A cannon mado of metal of specific grivity S'j.- weighs 216 Icilogs. in air, and 210 kilogs. in water. Ta inero a flaw in the cannon ? What is its volumo ?-~Ans. liiere is a flaw j the volumo = G6,000 cuhic contimetres. C. A body weighs 25 grammes in air, 20 grammes in water, and 15 grammes in another liquid; find tho Epecific grrvity of the body and of tho second liquid. — Ans. 5 and 2. 7. A balloon weighs 2G3"525 grammes when empty, and 379'S95 grammes when filled with air at tho ordinary pressure. What is its capacity, tho weight of a litre of the air being 1-293 grammes ?--Ans. 90 litres. 8. Tho same balloon, when filled with gas under the same conditions of temperature and pressure, weighs 293 'GSr' What is tho specific gra\'ity of the gas referred to air as the standard?— Ans. "26 nearly. 9. A body floats with one-third of its bull: under water; what is its specific gravity ? Let Vi = weight of body ; then J of its bulk of water = V/, and tho weight of its own bulk of water is tl.oreforo 3 W. Tho specific gravity is therefore W -f 3 W = L 10. A piece of gold weighs 9'7 grammes, a flask full of water weighs 95 grammes ; the gold is put into the flask, and displaces a Email quimtity of the water, which flows over. Tho weiglit of tho whole is 10-1"2 grammes ; find the specific gravity of the gold.— Ans. 19-1. 11. Nicholson's Iv"*. dieter is nsod as follows : — 51*72 grammes is placed on the upp.r ' p, a fragment of metal is placed on tho cup, and it is found necessary to take off 11"85 grammes to make tho instrument float at the same level ; tho metal is then placed in the lower cup, and 2*03 grammes added above to restore tho former level. Find the specific gravity of the metal. — A71S. 7'31. 12. A globo of glass weighs in pure water G'9 grammes, in sea water C*2 grammes, and in air 13 grammes. Find the specifio gravity of tho sea water. — Ans. 1*115. 13. A cylindrical picco of cork of specific gravity '25 and length of 10 inches floats upright in water; how much of it will be immersed ? — Ans. 2*5 inches. 11. A piece of wood weighs 7 lbs. and a piece of iron weighs 7"8 lbs. ia air and 0*7 lbs. in water ; the wood and iron together wei.i^h 5*3 lbs. iu water. What is the specific gravity of the wood ? — Ans. §. EXAM PLCS, ^11 IP A body woicjlis 10 oz3 more :n air ^h..n in wn'cr , iT T ouHj foot of wakv weighs 1000 oz;i., v/hat :s tlic volumo of tho bouy in cubic feet ?—Ans. '01. lU. A cylinder flo^<-s in r fiuiil of epccific gravity 2"3 \v it', two- thivds of its bulk u'uvo tlio surlaco. What is tho speciiii; , t'^ivity of tho C3 ruder ? — A )ts. *S3. (Tho specific gravity refcrrod to tho liquid = J .*. referred to water =^ i x 2o.) 17. If tho cylinder bo removed to another fluid, and then floats with one-third of its bulk out, find the siiecillc gravity of the fluid. —Ans. 1*25. 18. A vessel containin:^ water is placed in ono scale of a balanco» and counterpoised by a weight placed in tho other scale. If 10 cubic centimetres of metal bo suspended in tho water so that rone of tho water is spilt, wiiat additional weight must bo placed in tho other scale to restore equilibrium ? — Ans. 10 grammes. 19. To ono scale of a balance a weight of 20 lbs. and specifio gravity 8"5 is suspended, and allowed to rest in water, what weights must bo put in tho other scale to restore equilibrium ? Sidulioa. 8'5 lbs. of metal lose 1 lb. in water. weigh 7'5 lbs. in water. lib. 20 lbs. i> }> a weighs 7-5 )> 85 20 X 7-5 "8-5 19 ==r^} 20. A rod of wood floats with one-fourth of itn . .Dec-':' of water, and a stone eiiual in bulk to one-tenth of tho rod, -. i^an placed on tho rod, sinks it until tho top of tho rod is level with U' bu) lace, Fi, i tho specilic gravity of tlio stono. — A)is. 2 0. 21. A body whoso specific g: ivity is unity is placed in water, how much weight does it lose ? 22. A vessel and water weigh 10 lbs., what will tho^ weigh when a body which floats with 5 "70 cubic iuchea immersed is placed in the water. — Ans. lO.jV lbs. 23. Two bodies, a and h, of difleront bulks wei,i;,'Ii tlio samo ia water, which will weigh the most in mercury ? — Ans. Tho smaller. 24. VVliich will weigh Iho inost in air ? — Ana. The lai'ger. 111 If I; •it -i 42 11; VII.— THE AIR AND GASES. 4G. Many of the properties of liquids arc also possessed by elastic fluids. For example, the air lias weight. To prove this, let us take a flask fitted with an air-tight stop-cock, and by means of an instrument to be hereafter described, lot us pump all the air out of it, and turn oil' the stop-cock. Let the flask be placed in the scale of a balance, and such weights in the other as will keep the beam horizontal. When the stop- cock is turned on, the air will be heard to rush in, and the flask "will descend. If, instead of the flask, a glass globe capable of containing a cubic foot be emptied and weighed, when the air is admitted it will be an ounce and a half heavier than before. Again, pressure is transmitted by liquids and gases according to the same laws ; for exam[)le, the pressure at a point in a gas is the same in all directions. 47. A Ycry simple experiment will prove the existence of atmospheric pres- sure. Fill a tumbler with water, and place a sheet of paper over the top, hold the palm of the hand against the paper, and invert the tumbler. Tlic hand may now be withdrawn without the water falling (Fig. 17). The atmospheric pressure sup- ports the water, the paper being used only to prevent the air from replacing them in the glass. 4S. If a glas;) cylinder be tiglitly closed at one THE AIR AND GASES. 43 } me end by an air-tiglit membrane, and the air be with- drav.'n from the interior by means of an air-pump, the pressure of the atmospliere, not being supported by the clastic force of the air inside the cylinder will burst the membrane. 49. Nothing in the history of ecicnco is more re- markable than that up to the time of Galileo (1G3G) the fact that the atmosphere exerted pressure was unknown. Efiecl.s produced by this pressure were observed on all sides, but they w^re set down to other causes. When the water rose in a pipe from which tho air had been withdrawn, as is the case with tho common pump, in apparent violation of tho laws of gravity and of the law that liquids maintain their levels, philosophers, unable to account for tho circumstance, satisfied themselves by considering it due to a freak of nature. They set it down as an axiom that "nature abhors a vacuum," that nature would not allow any part of the universe to be void of matter. Thus, they argued, when air has been withdrawn from the pipe of a pump, nature's abhor- rence of a vacuum compels the water to rise and fill tho vacant space. In 1G3G, however, some mechanics made a pump to raise water from a well, the surface of the water being 50 feet deep. They four' 1 that they could not make the water rise more than 32 feet. They applied to tho celebrated philosopher, Galileo, to solve the mystery. He could not explain it, but ho said one thing was certain — Nature's abhorrence of a vacuum did not extend over 32 feet. The subject was tiien investigated by Torricclli, ii ■■ 1 ■ . ';| I 4i TUE AIR AND GASE3. I il I I a pupil of Galileo. ITo argued that ■whr\fcvcr the cause miii^ht bo, since it was sufficient to support 32 foef> of water, if a heavier liquid were used, a column of less altitude should be supported. For example, Fig. 13. smco 32 feet of water is sustained, of mercury, which is 13| times as heavy as water, 32 feet divided by 13^, or 28 inches, ought to be su^iportcd. He conducted his experiment as follows :— Ho took a vessel of mercury and a glass tube over 30 MAGDEBUrvO HEMISmrnES. 45 inclics loTif^, open at one end and closed at the otlier. Having filled the tube ■'.vith mercury (Fig. 18), ho closed the end with his finf:;er, inverted the tube, plunged the covered end below the surface of tho mercury in the vessel, and removed his hand. The mercury in the tube subsided until it stood at the height of 28 inclics. Supposing tho surfaco of tho mercury in tho vessel to bo prolonged under tho tube, ho saw that the fluid outside the tube was only subjected to tho pressure of tho atmosphere, while that under the tubo sustained the column of mercury above it. Tho secret was now revealed to him, for he concluded at once that the pressure of tho atmosphere on an ar(?a equal to tho aperture of tho tube was equal to tho weight of the mercury in the tube. This conclusion was confirmed by carrying tho apparatus up a mountain. Pascal, a French divine, argued that when a portion of tho air was left below, the pressure of that which remained must be dimin- ished ; consequently, if the column of mercury wero really supported by this pressure, the height of tho column must vary with difTercnt distances above tho earth's surface. The experiment was tried, and tho mercury in tho tube was found gradually to sink during the ascent. \P ' I i\ m ■'M ■ in Magdeburg Hemispheres. 50, A few years later (in in54) a Professor in Magdeburg constructed an apparatus which not only li ' ;!' : i ■ i.' ' i ! f' f 1 1^1 .III 46 THE BAROMETER. Fig. 10. exhibited tlio force of tlie atmospheric pressure, but showed tliat this pressure acted equally in all directions. The apparatus con- sists of two hollow metallic hemi- spheres, the edges of which are made to fit exactly (Fig. 19\ To one of the liemlspheres there is i tube and stop-cock, so that when the two are placed together the air may be pumped out. When filled with air the hemi- spheres may be separated without diffi- culty, but when the air is exhausted, in whatever position they arc held, a powerful effort is required to separate them. The Barometer. 51. The apparatus used in Torricelli's experiment constitutes a barometer. Whatever the size of the tube, the pressure at the base of the column on a level with the surface of the liquid outside must be equal to the pressure of the air on the same area anywhere in that surface. Thus, if the area of a section of the tube be one square inch, the pressure on a square inch is the weight of the mercury in the tube ; if the section be a tenth of a square inch, the weight of the column of mercury above the surface is the pressure of the air on the tenth of a square inch, and ten times this weight is, of course, the pressure on a square inch. It is easy, therefore, to find by means of this experiment the pressure of the air. Suppose, for example, a tube is used having a section exactly one mg. catcd MAKRIOrTE'S LAW. •1.7 square centimetre in area, and suppose tlio column of mercury to stand at tho usual hciglit of 70 centi- metres ; tliero will then bo 7(j cubic centimetres of mercury in tlio tube. Now, the speciGc gravity of mercury is 13*500, hence 70 cubic centimetres of mercury weigh 13*590 grammes X 70 = 1033 grammes. A pressure of 1033 grammes on a square centimetre is termed a pressure of one atmosphere. If we take inches and ounces as the units, sup* posing the height of the barometer to bo 30 inches, and tho wcii^'ht of a cubic inch of water lOGO 172t the atmospheric pressure = 30 X 13*506 x 1000 ~ 1728 ozs. = M*7 lbs. rM i \\':) 1 ' M M Marriotte's Law. 52. On account of the elasticity of a pressure exerted by it changes with its temperature and with its volume. For example, taho a cylindrical vessel fitted with an air-tight piston (Fig. 20). When held vertical, tho piston will bo supported by the resistance of the air within the cylinder. An effort is required to force in the piston, and tho more the air is compressed the greater the pressure required to keep the piston from return- ' S-^y tho moj. Instead of a force applied by the hand, let us place a vreight on the piston. Suppose the pressure of the air as indi- Fig. 20. Gated by tho barometer to bo 15 lbs. on the squaro ^^3 A/. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 !r«= IM I.I 1.25 M 2.2 IM 1.8 U ill 1.6 ■J-V# C? / Photographic Sdences Corporation m ,\ #^ :\ \ ^\^ '^\%:' ^^ # % ^^ <> 'ife^ 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 o Si ! I 43 MARRIOTTE S LAW. inch. Suppose, for convenience of rccironing, wo take a cylinder A B, of 1 square inch in area, and 12 inches long. Before the piston is inserted the air in the cylinder sustains a pressure of 15 lbs. Put in the piston, and pla':o on it such weights as will cause it to descend to A, the middle of the cylinder. The air will have been compressed into half its bulk, and it will be found that the weight amounts to 15 lbs. Then double the pressure reduces the volume one half. If we make the piston and its load tv/ice 15 lbs., so that the whole pressure is thrice 15 lbs., the piston will descend to E, one-third of the height. When the pressure is four times 15 lbs., the bulk is one-fourth of that at a pressure of 15 lbs. A pressure of 12 atmospheres will reduce the air to ono-twelfth of ifs original volume. This relation of the pressure to the volume of a gas is termed Marriattc's Law. " If the temperature be kept the same, the pres- sure of a quantity of air or gas is inversely propor- tional to the space it occupies." n.'nis law w^as proved by Boyle and ^larriottc as follows : — A bent tube A B with open ends (Fig. 21), and one arm a longer than the other, is fixed to a graduated st?jid, and a little mercury is poured ir.Lo it. The surface C c' of the mercury will be in the same horizontal plane, and will be sub- jected in both tubes to the pressure of the atmosphere. The end B is now closed, and more mercury poured into A. The mercury rises to D in the arm B, compressing Tig. 21. arriottc a.s DELATION OF PRESSURE AND TEMPERATURE, 40 the air, and to d' in the arm A. Tlic prcssuro of tlio mercury in the two arms hclow the same level D d" balance each other, and the pressure at d" is therefore that of the atmosphere, added to tho weight of the column d' d". Suppose Pj be written for the original pressure on the air in B, and Po for the pressure after com- l)rL"Ssion ; suppose that the pressure of tho air would Bupport the mercury at a height /i, and let tho weight of mercury in one inch of the tube be iVy then the pressure of the atmosphere = Jiiv, and that of the atmosphero and tho column D'D" = (h + 'D'U')iv, Now, when the scale is applied to +ho spaces EC, BD, it is always found that ttti h + D'D' hcnco BD BC _Pi The relation between pressure nnd volume may be also expressed thus : — If the temperature remains the same, tho volume multiplied by the pressure re- mains the same. Thus, if V bo the volume when tho pressure is P, and V^ the volume when the pressure is pi, vp = yipi. : m m I V- . It il 'r m I Eolation of Pressure and Tomporaturo. 53. The pressure of a gas is aflccted by change of temperature. If the hcut syringe (Fig. 20), with a certain load 50 THE WEATHER-GLASS. on the piston, be placed in waini water, the piston will be forced up unless additional weight be added. If the pressure remain the same, an increase of temperature of 1° centigrade produces in a given mass of gas an expansion of -003663 of the volume it would have at 0° centigrade.* The Weather-Glass. 5-i. The pressure of the atmosphere not only varies with the height above the sea, but it changes at the same place within certain limits ; con- sequently, the height of the mercury in the barometer- tube is subjected to fre- quent variations. These variations at the level of the sea range from 785 to 731 millimetres, or from 31 to 28 inches. A change in the height of the barometer frequently coincides with a change of the weather. In our cli- mate it is a matter of ob- servation, that in rainy and stormy weather the barometer is usually below the average height, and in fine and dry weather the * For explanation and examples of tliis law, geo Orine'i " Heat," page 64. THE COMMON TUMP. 51 height of the barometer is usually above the average. Hence the barometer is used as a weather-glass. The form of instrument used for this purpose is represented in Fig. 22. It consists of a siphon ba- rometer, in the shorter leg of which is a float, which rises and falls with the mercury. The float is connected, by some mechanical contrivance, with a wheel, bearing a needle. When the pressure of the air varies, the sinking or rising of the float causes the needle to move over a graduated scale in front of the case enclosinfr the barometer. ? in I 1 1 The Common Pump, 55. The pressure of the atmosphere is taken ad- \antage of for the purpose of raising water above its level. The common pump is a machine by means of which this is eiTccted. A b is a cylinder, having its lower end closed with -^ valve B (Fig. 23) opening upwards, and connected by means of a pipe with the w^ater wdiich is to be raised. A piston contain- ing cither one or lwo valves opening upwards, is worked in the cylinder by means of a vertical rod and a handle. Suppose the piston to be in its lowest position, then, when it is raised, the valve B opens, and a partial vacuum is produced in the cylinder and pipe ; and the pressure of the atmosphere without being greater than the pressure within the pipe, the water rises, and it continues to rise until tho pressure within and without becomo equal. N m If f " i if J: ^ 52 THE LIFTING PUMP. "Wlien ilio piston descends the valve in the piston opens, and the air within the cylinder escapes. When the upward- stroke of the piston is complete, it is again de- j)ressed; the water passes through the valve in the piston, and on the next stroke it is discharged at the spout. If tlie valves and pipes be perfectly air-tight, the greatest height to which the water could be raised by a common pump, reck- oning from the surfiice of the water in the well to the bottom valve, is 33' 75 feet, when the mercurial barometer is 30 inches, for 30 inches X loS = 3375 feet, 13-5 being the specific gravity of mercury. The Lifting Pump. 66. Sometimes pumps are used to lift water to a great height. In this case the spout is connected with a vertical tube, and the water, instead of escaping by the spout, rises in the tube, and is prevented from returning by a valve. i ^! Yis. 23. IHE ARCHIMEDIAX SCREW. 53 The ArcTiimedian Screw. 57. Ono of tlio earliest maclilnes used for lifting water on record is tlie screw of Arcliimedes (Fig. 24). It consists 0^ a cylinder inclined to tlie vertical, and exactly fitted by a screw having the same axis. If a small solid body were placed at the bottom, and the screw turned round, each point of the screw would Fig. 21. pass beneath the body at the lower edge of the cylin- der, and the body would thus be raised gradually to the top. In the same manner, if water has access to the bottom, on turning the instrument it will bo gradually raised until it flows out at the top, Tho Porcing Pump. 58. The forcing pump differs from the common pump in the following respects: first, there is no spout or outlet to the upper part of the cylinder ; secondly, the piston A is solid ; thirdly, in addition to the pipe closed by the valve C, and communicating with the cistern, there is another pipe passing froia i I: m \ i}\ 54 THE FIRE-ENGINE. I ill Mil >i; the lo^Ycr part of tlie cylinder, " " "h a valve c' opening upwards, through which the water is to be raised. Snppose the piston to be in its lowest position ; when it is raised the valve 1c opens and c' closes, the PI column of water within the Ij cylinder and pipe is re- lieved of the pressure above A, and the pressure of the at- mosphere without being thus greater than that within the pipe, the water within rises, and continues to rise until the pressures within and without are equal. Let the piston be now made to descend, and the process repeated until the water has risen above the top of the pipe ; then when a the Fig. 25. descends, the water in cylinder, not being able to return on account of the valve c, is forced up the pipe, in which it is retained by the valve c'. The Fire-Engine. 59. When a continuous stieam is required, the valve leads into a strong air-vessel, out of which the pipe passes. After the down-stroke of the piston the air in the chamber is condensed, and its prestiure on the surface continues to lift the water during the up- stroke. The fire-engine consists of two force-pumps connected with an air chamber (Fig. 2u). T^ie THE SIPHON. bb pumps are worked by a lever, and are arranged so that when one piston descends and closes the valve below it, the other ascends and opens the valve communicating with the water in the cistern. ■ ^1 ■ n Fig. 20. Tli9 Siphon. CO. The siphon is a bent tube a n c (Fig. 27) open at both ends. Let the tube be filled with fluid and the shorter leg inserted into a vessel of fluid which it is required to empty, and the extremity c of the other leg closed by the finger. Let tlie level of the surface of the fluid meet the two legs of the siphon in g and d respectively. Provided the height of B above g d is not greater than that of a column of water, the weigh ♦" of which is equal to the atmospheric pressure, the wa*-er in d b will be supported by the atmospheric pressure. Let p be the atmnsphcrio I Ml'i !'■ 1 'I ! t , 4' m 1 If '' 56 THE SIPHON. I 5 i 4 } i I i La. pressure, tlie pressure at D=P, and is transmittecl along the tube. The press are at B=P — tbe pressure due to a column of water of hciglit D E. The pressure at G=P. The pressure at C=P4-the pressure due to a column of water of height f/ c. Consequently the pressure on the end c, which we have supposed closed, exceeds the atmospheric pressure by the pressure due to a column of water of height of g c, or C F — D E. If the end c be opened the fluid will descend. Since at every point of the tube there is a pres- sure in the direction A B c, as soon as the liquid at any point moves forward, the particles behind being re- lieved of the pressure in front immediately follow, and thus a continuous stream is produced, which will only cease when the surface of the fluid has descended to A, the extremity of the shorter le^ of the siphon. If the leg B c terminated at G, it is evident that there would be equilibrium. If the height d e were greater than that of the water-barometer, the liquid would descend in each tube to that height, leaving a vacuum at B THE ATR-ru:rp. 57 The Air-Pump. 61. The air-pump is a machine constructed for removing the air from a closed vessel, called a receiver. From the centre of a metal plate to which the receiver K is fitted (Fig. 28), so that no air can pass Fig. 28. hetween them, runs a pipe communicating with the barrel of the pump b by means of a valve e, opening upwards. In the barrel is an air-tight piston ii also furnished with a valve opening upwards. Suppose n to be drawn up, then a partial vacuum will be formed below the piston, and fresh air rushing in from r vrill open the valve at e, Now let H be made to descend, then the pressure of the air will close the valve e and open the valve in H, and the air in the barrel will be forced out, la \ \.. I J -i! I I ! k I i! 58 THE AIR- FUJI P. the mcantirao tho air in tlio receiver o::pands and refills tlio barrel. Now n again descending and ascending" expels another portion, and bo on till tho air in r becomes so rare that its clastic force is insufficient to open the valves. It is obvious that, although the air gets rarer and rarer at each stroke of tho piston, tho whole can never be exhausted. Tho pump which wo have described is called tho Single-barrelled, or Smcaton's Air-pump. Hawks- bee's Air-pump is fitted with two barrels instead of one (Fig. 29). Tho two pistons aro moved by a toothed wheel in such a manner, that one descends whilo the other ascends, and thus tho air can bo forced out with greater rapidity. 62. To find tlic density of tlic air in a receiver after any nimiber of turns of the icheel. Let R be tho volume of tho receiver and pipe, and 5 the volume of each barrel. Then the air v^hich occupied tho space R when n was at tho bottom of the barrel, fills the space R + Zj, when n ascenJs to the top. If therefore d = the density before the stroke, and di the density afterwards, we shall have d^ (U + h) = d*n. In the same manner it will be found if d^ be tho density after two turns, d,, the density after n turns, that (/g (R -f s) = d^n .-. ^3 (n + ly = ^^2 ^3 (R -i- &) =: cZg-R . • . dl(R + 5)3 = dK^ .'. dniU + hy = d'll\ THE sirnoN gavcc. 19 Tims it appears tliat the density of tho air de- creases in a geometrical progression. Tlie Siphon Gauge. This is a glass tube s (Fig. 29) sere wed to a pipe commauicatiDg with tho receiver. Ono end a is Fk. 29. closed, and a portion of the tube, rather more thaa a bf completely filled with mercury. T'li I ! i It •;.!3 GO THE DIYING-BELL. If it be not more tlian 28 inclics in lengtli, tlie tube a h ■will at first remain filled ; but as the ex- haustion proceeds, the mercury v/ill sink in a h and rise in h c, and the difference in the heights of tho mercury in a h and h c will always measure tho pressure in the receiver. Thu? if wo call the pressure of tho air which supports 76 centimetres of mercury one atmospliere^ and let n centimetres = the difference in the heights, the pressure of the air in the rccoiver is jr-;ths of ono ^ 7G atmosphere. If d be tho density of the air outside, and di the density in the receiver, since density is proportional to the pressure, d^ 'I d heiirht oi barometer. The Diving-Bell. 63 T!:e diving-bell is a strong iron vessel, which is immersed in water with its mouth downwards. Tho air within prevents the water from filling thu vessel j but as it descends the pressure increases, and the air is compressed. Let us suppose that the pressure of the air at the surface is equal to the weight of a column of water 33 feet high, then when the bell has descended 33 feet, the air within will be compressed to half its original bulk. The pressure of the air inside would support a column of mercury twice as high as the mer- curial barometer at the surface. At a depth of twico 33 feet the air would be compressed to one- third its original bulk, and so on. EXEKCISES. Gl I |M: To prevent tlio advance of tlio viator in tho bell, and to supply tlio workmen with fresh air, a tube, connected with a force-pn^ip at tho surface, passes to the bell. There is a.oO a tubo through which tho workmen can allow tho foul air to escape by turning on a stop- cock. The bell is suspended by chains or ropes. Exercise. — A cylindrical divin.c^-bell 8 feet long descends until tho water in the bell is SO feet below the surface, when the height of a water-barometer is 32 feet. If no air has been supplied from above, to what height has the water risen in tho bell ? The pressure at the surface =tho pressure of 32 ft. of water. ,1 „ a depth of 80 ft. = the pressure of 112 ft. of water. With a pressure of 32 ft. of water, the air occupies 8 ft. of the cylinder. .'. With a pressure of 1 fb. of water, the air would occupy 8 X 32 ft. ; a.nd with a pressure of 112 ft. of water, -yX^, or 2| ft. I t 4 U " > 1 ExEr.cisES. 1. \Vlicn tlie lioislifc of tho barometer is 76 centimetres, find the height to which water will rise in a tube from which tho ju'r has been withdrawn, the specific gravity of mercury ijcing 13'503. —■Ans. 1033 ceutiraetrc3. 2. What would bo tho height of tho barometer if a liquid of Bpccllio gravity 2 '5 wore used, when the mercurial barometer Btantis at 76 centimetres "^—Ans. 413"3 centimetres. C. When the water barometer stands at 03 feet, wliat would bo 1 1 J! *■ , <. C2 EXEPwCISES. the heiglifc if alcoHol of specific gravity •825 vvera used instead of water ? — Ans. 40 feet. 4. What would be the effect of malcing a small aperture at the highest point of a siphon ? 5. What difference is made in the height of the column of mercury in a barometer when the tube is inclined ? C. How would the action of a siphon bo affected by carrying ib up a mountain ? 7. If when the barometer stands at 30 inches, the air in the glass cylinder represented in Fig. 23 is half exhausted; find the pressure on a square inch, taking the specific gravity of mercury as 13 "5, and the weight of a cubic foot of water as 1000 ozs. 8. If the weight of a cubic inch of mercury be 7 '8 ozs., what ia the pressure of the air on a square inch when the mercury stands at29'5 inches? — Ans. 230"1 ozs. 9. What would be the height of the water baromcler in the above case, the specific gravity of mercury being 13"5 ? — Ans. S98-25 inches, or 33 feet 2^ inches. 10. A barometer is partly filled with water and partly wiJ.li mercury, and the height of the water is three times that of the mercury ; find the height when the pressure of the atmosphere is 1020 grammes on a square centimetre, the specific gravity of mer* cury being 13 '593. Solution. Let x be the height of the mercury ; then 3 a; =:tlie height of the water, and 13"59G x = the height of water equal to x centims. of mercury. Now, a water-barometer would stand at 1020 centimetres. .-. 3.r + 13 50Ga? = 1020 .'. 03 = 01 "5 centimetres. 11. IIow will a fall in the barometer affect the action of a Eiphon ? 12. How will a change in the barometer affect the pressure on the base and sides of a cubical vessel filled with water? 13. Find the greatest height over which a liquid of specific gravity 3-5 can be carried by means of a siphon, when the barometer stands at 7G0 millimetres. 14. If a vessel of water containing a floating body be placed under the receiver of an air-pump, and the air gradually exhausted, vrhat will bo the effect on the floating body ? Solution. The weight of the floating body is equal to the weight EXERCISES. 63 eil instead of aerture at tlio 10 column of of fluid (air and water) displaced; lienco, as tlio surrounding air 19 withdrawn — that is to suy, as the air displaced by the body is (diminished — the water displaced must increase. When the receiver is completely exhausted, the body will sink, until the weight of the water displaced is alone equal to the weight of the body. 15. To what height would the mercury rise in a Torricellian tube, if it be immersed in water until the surface of the mercury in the cistern is 45 inches below the surface of the water, supposing the barometer to stand at 30 inches, and the specific gravity of mercury to be IZo.—Ans. 33| inches. IG. The receiver of an air-pump is five times as largo as the barrel; after how many strokes will the density of the air be diminished by nearly a half ?— -.4ns. 3. 17. In the above example, how much air has been removed after sis strokes? — Anc Taking the original volume as unity, the air removed is 18. "What would bo the cfTcct of making a small holo in the trp of a diving-bell ? 19. What is the pressure on a square centimetre in a dlvlno^-^>c- 1; ii f \ MATIUCULATION. 1. Explain the difficulty of opening a lock-gate when the water is at a different level within and without the lock ; also, wliy the force required to open the ga' > is not proportional directly to the difference of level. 2. The weight of water is 770 times that of air; at what depth in a lake would a bubble of air be compressed to the density of water, supposing the law of Marriotte to hold good throughout for the compre?sion ? Solution. — Let the pressure at the surface be P grammes on a square centimetre, then the pressure, P', at a depth of x centi- metres is {x + P) grammes. Let V be the volume of the bubble at the surface, and V that at depth x, then by § 52 V P=V' P', but by hypothesis V=770 V, therefore by substituting for P' and V we obtain a;=7C9 P centimetres. If we take feet and ounces as our units x=1Ux'7(jO P feet. 3. A body weighs in air 1000 grains, in water 300 grains, and in another liquid 420 grains ; what is the specific gravity of the latter Hquid ?—Ans. '8285. 4. Sketch the common pump, describing its action, and stating the limitation to which this is subject. 5. A mercurial barometer is lowered into a vessel of water, so that the surface of the water is finally sis inches above the cistern of the barometer. What land of change will take place in the reading of the column of the instrument ? Give a reason for your reply 6. Draw a diagram representing (in section) a siugle-barrel air-pump when the piston is being lowered. MATRICULATION EXAMINATION PAPERS. 65 and stating Explain tlio action of such a pump. If the receiver hold 61 grains of air, and 8 grains be removed by the first stroke of the piston, how much will be removed by the second stroke? — Ans. 7 grains. 7. If a bottle filled with air bo tightly corked, and lowered into tlio ocean, the cork will be forced in at a certain depth. Why is this ? and what will take place if the bottle bo filled with water instead of air ? 8. Explain why a siphon cannot be used to convey water from a lower to a higher level. 9. A solid weighs in vacuo 100 grains, in water 85 grains, and in another fluid 83 grains ; what is the specific gravity of this Cuid ?^Ans. S. 10. Explain the action of the ordinary lifting pump ; find the limit of its action ; and show that this will be diiTereut at the top of a mountain and at the level of the sea. 11. If a barometer were carried d^wn in a diving-bell, what would take place ? Give a rough quantitative result. 12. A solid, soluble in water but not in alcohol, weighs 31G grains in air, and 210 in alcohol ; find the specific gravity of the Eolid, that of alcohol being 0'85. — Ans. 21G25. 13. A tapering tube is bent in the middle at an acute angle, and being filled with water is inverted with the two ends completely immersed beneath the surfaces of the water in two vessels, the narrower leg being held vertical, and its end being only just below the surface. Point out the mode in which the water will flow j and show when the flow will cease. 14. Describe and explain the barometer; and state on what principle It is used to determine the height of a mountain. Might it also bo used to determine the depth of the sea ? 15. Define clearly what is meant by specific gravity. Is there any difference in specific gravity between 4 lbs. of iron and 2 lbs. of the same metal ? A body whose specific gravity is 3"5, weighs 4 lbs. in water* What is its real weight ? — Ans. B'G lbs. IG. Describe, by means of a diagram, the common pump. State the limit to its action and the cause of the same. 17- Explain the action of the siphon. Is there any limit to the tcight of an embankment over which water may bo carried by means of a siphon from a higher to a lower level ? I' HI ' »:» in Hi 11 ifi H r ; i If I \f ,.!■: :l ■!' lii'li- I G6 LONDON UNIVERSITY 18. If as mucli ad Jitlonal air were forced into a close J vessel as it previously contained when in communication -willi tlio atmo- spliere, what would be the pressure on a square inch of the internal surface ? 19. Describe the forcing-pump, and state why it cannot in general be replaced by a common pump. 20. State precisely what information is gained by knowing the specific gravity of a body. Show how to find the specific gravity of a solid lighter tian water, and not soluble in that liquid. 21. At what depth in a lake is the pressure of the water, in- cluding the atmospheric pressure, three times as great as at the depth of 10 feet, on a day when the height of the liijuid column in a water-barometer is 33 feet G inclies? — Ans. 97 feet. 22. A lump of beeswax, weighing 2895 grains, is stuclc on to a crystal of quartz weighing 795 grains, and the whole, vvhen sus- pended in v;ater, is found to weigh 390 grains ; find the specific gravity of beeswax, that of quartz being 2'G5. — Ans. '905. 23. A barometric tube of half an inch intcrnil diameter is filled in the usual way, and the mercury is found to stand at the height of 30 inclies. A cubic inch of air having been allowed to pass into the vacuum above the mercury, the column is found to be depressed 5 inches. What was the volume of the oi'iginal vacuum ? 22 Solution. — The volume of 5 inches of the tube = '^- ^ = 1 cubic inch nearly. Let X be tlie volume of the vacuum, then aj 4- 1 is the volume occupied by the air. This air exerts a pressui'O = 5 inches of mercury. But the air occupied 1 cubic inch when under a pressure of 30 inches of mercury. Hence, by applying the formula of page 49, (x + 1)5 = 30 .*. X = 0. 24. A bottle holds 1500 grains of water, and when filled with alcohol it weighs 1708 grains; but when empty it weighs 520 grains ; what is the specific gravity of alcohol ? — Aits, '792. 25. Describe the construction and action of the common pump. 20. A tube closed at one end is filled with mercury, and the open end dipped below the surface of mercury in an open vessel. If the tube bo inclined at an angle of CO' to the vertical, what is BIATRICULATION EXAMINATION PAPERS. 67 the greatest length the tube can have so as to remain full of mercury, the height of the barometer beinij at the time 30 inches ? —Ans. CO inches. 27' A piece of cupric sulphate weighs 3 czs. in vaciio, and 1'80 ozs. in oil of turpentine ; what is the speciiic gravity of cupric sulphate, that of turpentine being 0*88 ? — Ans. 2i%-. 28. If the height of the barometer rises from 30 inches to 3025 inches, what is the increase of pressure (in ozs.) upon a square foot ? — the weight of a cubic foot of water being taken to be 1000 ozs., and the specific gravity of mercury 13-5G. — Ans. 282-5 ozs. 29. Describe an experiment which proves that the upward pressure of a fluid on any substance immersed in it, is equal to the weight of the fluid displaced by the substance. Give a sketch showing the arrangement of the appp.ratus. 30. If a bladder containing 300 (;ubic inches of air under a pressure equal to that of 30 inches of mercury be sunk 2 10 feet below the surface of water, the barometer standing at 23"5 inches, to what volume wiL the air in the bladder be compressed? [Specific gravity of mercury =13"C.] — Ans. 37 '2 cubic inches. 31. A piece of metal weighs 211"6 grains in vacuo, 187 '32 grains in water, and 182'37 grains in a solution of sodic chloride ; find the specific gravity of the solution. — Ans. 1*2. 32. Describe the common barometer, and point out the principle on which its action is based. What change iu the atmospheric pressure on a square inch is indicated by a fall of one inch in the height of the barometric column? (A cubic inch of water weighs 252'7 grains, and the specific gravity of mercury is 13"6.) — Ans. A decrease of pressure of 343672 grains, or nearly half a pound on the square inch. 33. If two liquids that do not mix meet in a bent tube open at both ends, show that at rest their heights above the common surface of contact are inversely proportional to their specific gravities (§ 45). 31. The specific gravity of mercury is 13 6, and the height of the mercurial barometer is 30 inches. What is the greatest height to which water can bo raised by means of the common pump ?— Ans. 34 feet. 35. Desciibe and explain the construction and action of the double-barreled air-pump. Ill k it 1 : f; ! I ( ' n Mm \l IW^ 1 ' 68 EXAMINATION PAPERS. 36. Describe the forcing-pump, and explain its action. If tlie area of the piston is 3*5 square inches and a power of 77 lbs. is employed in forcing it down, find the pressure of the air within the air-chamber, and the height to which water can be raised above the piston when the surface of the water in the chamber u 1 foot above the piston. ^One cubic foot of water weighs C2-3 lbs.) 37. Al vessel in the shape of a jiyramid 5 feet high, and with base 4 feet square is filled with water. Find the pressure upon the base and account for it being greater than the total weight of water in the ver.sel. 38. Describo the barometer-gauge of the air-pump j and explain how the degree of exhaustion of the air in the receiver is deter- mined by it. OPTICS. Defjiitions. 1. Light is the agent which acts npon the organs of vision so as to produce the sensation of sight. The light by which an object is seen always comes from the object to the eye. Bodies are said to be luminous when they affect the organs of sight without the aid of other bodies. Thus, the sun, the stars, a candle, a lamp are luminous ; trees and stones are non-luminous. The light by which non-iurainous bodies are seen is not emitted by them, but is received from some luminous body, and is slinply scattered or dispersed by them : if the luminous body disappear they cease to be visible. 2. Light passes through some bodies, and not through others. When the light is transmitted so perfectly that objects tan be seen through the inter- vening body, the latter is said to be transparent. When light passes through a body, but, as in the case of ground glass, not so that objects may be distin- ^jruished through it, the body is said to be translucent. Bodies through which no light can be transmitted are called opaque. The space through which the pheno- mena of light take place is called a medium. The medium may be a vacuum or a transparent substance, :l ! H i ' i m ! ! 1 :i "^c !v^ I ^ 'iilil S OPTICS. such as air, water, glass. A ' '^dium is homogeneous when its chemical coinpositio nd density are the same in all its parts. Propagation of Light. 3. In a homogeneous transparent medium light is propagated in straight lines. If a cylindrical tuhe be placed in line with a candle, the light may be seen when the tube is straight, but not when the tube is curved. The same is true if the ends of the tube be closed by glass plates and the tube be filled with water. Eay— Pencil— Focus. 4. A rai/ is the smallest conceivable portion of light, and is represented by the straight line along which it is supposed to be propagiUed. A collection of rays is called a Pencil of Ligld. A parallel pencil is one whose component rays are parallel to one another. A converging pencil is one in which the rays all proceed to a single point, and a diverging pencil one in which the rays all proceed from the same point. The point to or from which the rays of a pencil proceed is called \\\Q focus of the pencil. Velocity of Light. 5. The velocity with which light is propagated is so great that to ordinary observers on the surface of the earth the interval tahen by light to pass from one point to another is inappreciable, yet by means of astrononucal observations, as well as by elaborate and delicate experiments, this velocity has been determined. Light travels from the sun to the earth in 8 minutes 16 seconds. The velocity varies with the medium. Thus, it is not the same in water as in air. Light incident on a Surface. 6. When a ray of light meets a surface at a point, and the perpendicular (or as it is sometimes termed the normal) to tlic surface is drawn from the point, the angle between the ray and the perpendicular is termed the ancjle of incidence. When rays come in contact with an opaque and polished surface, their direction is changed, and the rays are bent back or rejlec^'d. The angle between the ray after reflection and the per- pendicular to the surface is iermQd the angle of rejlcction. When a luminous ray meets the surface of a trans- parent medium it passes into the medium, but not along the same straight line. It is bent or refracted at the surface. The angle which the ray after refraction makes with the perpendicular to the surface is termed the angle of refraction. Usually reflection and refrac- tion occur together, for when light falls on the surface of a body some of it is scattered or dispersed, making the body visible from all sides, although the light may reach it only in one direction. A portion of the light is absorbed, another portion is reflected, and if the body be transparent a large portion passes into it. We shall, however, consider only the reflection and refrac- m iil 3 1, ~w i- Ol'TlCS. tion of liglit, and siiall suppose polislicd surfaces to reflect all the light incident on them, and transparent surfaces to transmit all. Law of Beflection. 7. Light is reflected according to the following aws ted raj/, am I the per- Pig. 1. 2)oint of incidence lie in equal to the angle of le perpendicular to the rface bisects the angle between the incident and reflected rays. Thus, if A li be a ray incident at n on the surface of which M N is a section, and n n the normal at i), then A B, B D, and n c lie in the same plane, and the angle a b d is equal to the angle d u o. Experiment to illustrate the Law of Reflection. 8. To prove the law experimentally, a graduated circle, of which the plane is vertical, is mounted on a stand. Two tubes, movable round the centre, and having their axes directed along radii of the circle, slide LAWS OP REFLECTION. on the arc, and are capable of being adjusted at any angle to the vertical through the highest point a. At the centre of the circle is placed a small plane mirror m, which is exactly perpendicular to the plane of the circle. In making an experiment, a pencil of solar light 8 is directed along the tube to the mirror m. Here the luminous pencil undergoes reflection, and takes a direction M E, which is ascertained by moving the tube e until the light can bo seen as a bright spot in the mirror m,, by looking along the tube. On reading off the numbers of degrees in the arcs A E and a s, we find that these arcs are equal. Hence the angle of reflection a m e is equal to the angle of incidence a m s. The first part of the law is proved by the arrangement of the apparatus, since the plane of the rrys s m and m e is parallel to that of the graduated circle, and consequently perpendicular to the plane of the mirror m. Fig. 2. : I' I J I' I il: ! I I "'iP* ! 11 OPTICS. MIRRORS. 9. Bodies having polished surfaces in which objects can be seen by reflection are called mirrors. Mirrors are divided, according to their form, into plane, con- cave, convex, spherical, parabolic, &c. Images* 10. Before discussing the properties of mirrors, it is necessary to consider wha^ is meant by the image of an object. If an object be placed before an opaque screen having a small aperture about the size of a pin- hole in it, and a she.*; of paper be held at a short distance beyond the screen, a picture or image of the object will be seen on the paper. Suppose the screen and paper to form opposite sides of a box, as in the figure, then the rays from the upper end of the object will pass straight through the hole, and will illuminate a point a on the back of the box with their own colour; the rays from the lower extremity of the object will do the same at a point b above the former, and rays from intermediate points will fall in a similar manner between a and n. When the hole is small the rays from any one point of the object can fall only on the point of the screen which is opposite to it, and hence the smaller we make the aperture the more distinct will be the picture of the object. If we enlarge the small hole the picture will become more indistinct, and the colours fainter, and when the hole reaches a certain fcize, the light from different points of the object will Fig. 8. PLANE MIRIIOIIS. 7 fall on the same part of the card, and no image will be formed. Images in every way similar to tlie above, but much brighter and more distinct, may be produced with a larger aperture by means of mirrors and lenses, the nature of which we proceed to explain. To explain the farmation of images with a plane mirror. 11. When objects are placed befora a plane mirror, images of the objects are produced i)y the reflection of the light which passes from the objects to the mirror. Let o be a point in the object, and a b a right section of the mirror by a plane through o. Luuiinous rays proceed from this poin<- in all directions, and some of them fall on the mirror. Let us select one of these, o m ; let M N be the normal at ji, and M R the direction of the reflected ray. Draw I from 0, perpendicular ^^' * to the mirror, and continue r m to meet o i in i. Because mn is parullel to oi, the angle n m n is equal to the angle i, and om n to the angle o. But by the law of reflection rmn = omn, therefore, also, i =«o, and the triangles m o c, m i o, having a common side and the angles of the one respectively equal to the angles of the other, are equal ; consequently o c == i c. Thus the ray m r proceeds as if it came from a point i at the back of the mirror, at the same distance from it as the point o, and in the same ucrpendicular. A \ \\ % • OPTICS. similar demonstration might be applied to any other reflected ray ; hence all the rays after reflection pro- ceed as if they came from the point i. i is therefore the image of o. Similarly, every other point of the object has its image as far behind the mirror as the point is in front. The images of all the points in the body form the image of the body. To draw the paths of the rays hy which the image is seen. 12. Of all the reflected rays appearing to proceed from one point of the image only a small pencil will enter the eye of the observer. In order to determine this pencil we must first determine the image i of the point O by the preceding law ; then draw lines from the point i to the eye of the observer. Let a b be the points of the mirror in which the extreme line inter- sects the mirror ; then the pencil of rays which comes Fig 5. Fig. 6. PLANE MIRRORS. VI originally from o will be reflected at a b, and reach the eye of the observer e. A similar pencil may be drawn for every point of the object. It will be seen that only the part a b of the mirror is concerned in the reflection of the light to the eye. Limits of visibility of the image. Let IK, II., be straight lines from a point i of the image throngh the limits m and n of the mirror, then it is evident that the eye of the observer must be within the angle k i l, in order that it may see the image i. (Fig. 6.) I Continued Beflections at Plane Mirrors. Parallel Mirrors. 13. Let MN, p Q, be sections of two parallel mirrors, bisected by a plane perpendicular, and let o be a point of the object in this plane, o has an image i^ to the left of M N, and an image i' to the right of p q. Light from o, after reflection by m n, proceeds as if it came from i^ and some of this light falls on p q. Fig. 7. This will form an image of i^ to the right of pq ; similarly an image i" of i' will be formed as far to the left of MN as i' is to the right. Tbese images again in 10 OPTICS, iii H 0^_^'.. VR •■": c ; :. Fig. C4. 39. Descartes, to whom is due the dis- covery of this law, employed a spherical glass vessel exactly lialf full of water, and caused a small ptncil of light to enter the water at tlie centre. He then found that if the pencil passed into sphere at l (Fig. 24), and out at r, the ratio of lb, th€ distance of l from the vertical through the centre to R T, the distance of r from the same vertical was constant. The law is best illustrated experimentally by means of a vessel formed of a ring of glass and two circular glass plates. The vessel is filled with water up to the centre. At the back of the glass is a rod l m (Fig. 24), movable about the centre o, so that it can be made to take the direction of the incident ray, and a rod o n also movable about the centre, to indicate the direction of the refracted ray. The length of the rod, om, is exactly equal to that of o N. A single beam of light is made to enter the vessel at a point, l, and to pro- ceed to o, and the rod lm is moved to take the direction of the beam. A screen is then placed below REFRACTION. M. on. ' es, to edis- law, lerical xactly (Yater, small ht to ter at [e then if the i into of L8, centre al was means ircular to the ig.24), ade to od ON rection OM, is f light o pro- ^e the below the vessel, and the rod o n is made to take the direc- tion of the refracted ray, by causing it to point to the spot which is illuminated on the screen. A movable scale then gives the distances of m and n, ^rom the vertical through o. It is found that whatever may bo the angle made by the rod lm, the ratio of these distances is always the same. The limiting angle of rrf faction. 40. Let m be the refractive index for two given media, the light passing from the rarer to the denser, and let A o (Fig. 23) be a ray incident obliouely at o. By the construction of § 37, draw o c, the refracted ray. Then it has been shown that — '■ == m. or o d = — . OD m Now let A be inclined more and more to the vertical, so that it approaches the surface ; then o c increases, and becomes more and more nearly equal to the radius cf the circle, and then o d approaches the radius -r- m. Thus, the limit of o d is -^ , and the limit of the 7n anele BON is such as will give — == — . Let us ° ° BO m denote this angle by d. If the light proceed from n along B o, then it will emerge alo. ig o a, provided the angle b o n be less than the limit 6. If the angle B N be greater than 0, the light will not emerge, but will be turned back at the surface according to the laws of reflection. The ratio — is equal to the sine of tho bo B 32 OPTICS. II ! i' 'ii % I s •■' angle BON (§ 35) ; consciuently, \i m be the refractive index for Iho rarer to the denser of two media, tijc ancrlo wliose sine is canal to — is termed ihe limiting ° 7/1 angle of rt fraction^ or the limiting angle of total ve- il ectiou, Jt is sometimes also termed the critical angle. Vox instance, suppose the refractive index for air t(. 4 ^Yatcr to be -- . Constrnct the right angle doi*, with b a radius 1, and willi centre, o, describe a circle, cut off o n -=^ o, and draw d b perpendicular to o n (Fig. 23). The angle v, o n will be the limiting angle of refraction. No light which passes through water to the surface, and is incident at an angle greater than bo n, can pass out of the water. It will t)e totally reflected. \ 1 . Let o be a luminous point under water (Fig. 25) ; by the above construction find the limiting angle (,' Fig. 25. '• I TOTAL BEFLKCTION. 33 fraclivc ilia, tlic limit ill f/ otal re- crilical or ail' t( 0, ci.l off Fig. 23). cf Faction. > surface, , can pass Tig. 25); r anglo (.' (about 48'5*), ami tlirongli o draw o m, o m', eacli milking an angle witli tlie vertical throngh o ; then rays of light passing from o to the surface within the angle M o m' ^vill emerge, but any ray o u falling without the angle will be totally reflected in the direc- tion R T. The effect both of refraction arnl of total reflection may be easily exhibited by taking a glass tumbler of water with a spoon in it, and holding it nbovo tli«» ovo (Ficf. 2G), so as \.^ <:r'<^ the j^urOi'^r of the Tig. 26. water from below. The image of the spoon will be seen as in a mirror, by reflection. The part in the water will be seen by rays which are refracted so that \i 34 OPTICS. it will not appear as the direct continuation of tlie part withont the water. EjBTects of Refraction. To find the relation letiveen the true and apparent depths of an ohjecf under u-ater. 42. Let L be a luminous poiut in a dense medium, as water. Let l a and l b be two rays from l, and let A c, n D, be their directions after passing into the air, and let the continuations of c a and c b meet in l' ; then the eye which rf reives the pencil of which l a and L B are the extreme rays, sees the point L as if it were at i/, hence l' is iha image of l. To find the position of l' when the pencil is vertical, let us consider only the ray l a c. Let K li be the perpendicular to the surface, and let it meet the surface in n. When L A is very near to l k, l' is on L K. is the sine of the angle between c a and the Fig. 27. NA AL' •mal ; NA is the sine of the angle between l a and the A Ju normal. i)nt by the law of rofraciion -r—rr -^ *--r = Wj the refractive index for air AL A L into water; therefore a l = m • a l'. witl the Let I the direj the glasj facet EE inci< SUCCESSIVE RKFUACTIONS. 35 If A be very near n then n l =- m * n l'. Thus, if d be the true depth of a stream, and d' its apparent dcptli, then d == d''m. Since m is greater than 1, an object in water will not seem to be so deep as it really is. For example, a fish under water will seem to be nearer the surface than it is in reality. A sportsman aiming at the fish as he sees it will fire above it ; he mup.t aim lower than the image of the fish in order to hit it. 43. If a straight stick be placed in water, the image of the immersed part will be above the part itself (§ 42), consequently thestick will appear bent, as in the figure. Fig. 28. Passage of a ra>/ ilirovgh a transparent plale with parallel faces. 44. When a ray passes through a transj)arent plate with parallel faces, it emerges in a direction parallel to the direction of incidence. Let M N be the plate, s a the incident ray, a b the direction after refraction at the first surface of the glass. Because the sur- faces are parallel, the angle KB A, which is the angle of incidence at the second Fig. 29. 36 OI'TICS. I I ii surface, is equal to the angle i a b : and since if t]»e light were reversed it would proceed along the same path, the angle after emergence ebd^sag, and therefore a s and b d are parallel. In the same way it may be shown that if the ray passes through any number of plates with parallel faces, some of which are repeated, its directions in the same medium are always parallel. For example, if a ray pass through a series of plates as follows, the parts in 1 and 4 will be parallel, and so will be the parts in 2 and 5. 1 2 toater glass 3 air 4 wafer f/Iass Drviafion of a raij in passinrj throvrjh a prism. 45. A medium honnded by two faces which intersect is termed a [)risn» (Fig. 31). A platio perpendicular to the two intersecting faces is termed a principal plane. Pig. 30. Fig. 31. LBK6B8. 37 Let OD be a ray from a luminous point o in the principal plane, wlr'ch cuts the prism in the section ABC. The ray is bent towards the normal at d, and on emergence it is bent again away from the normal at K. The ray is received by the eye h as if it came from o'. The angle between the emergent ray k ii and the incident ray od is the angle of deviation.. It is found that the deviation is always towards the thickest part of the prism. The deviation is a minimum, that is to say, it is the least possible when the angle of incidence and the angle of emergence are equal (Fig. 31). i^ i! LENSES. 46. Any portion of a refracting medium which has t.wo opposite surfaces, either both spherical, or one spherical and the other plane, is termed a lens. 47. The combination of spherical surfaces, either with each other or with plane surfaces, gives rise to six kinds of lenses, sections of which are represented in Fig. 33 ; four are formed by two spherical surfaces, and two by a plane and a spherical surface. Ffg. 32. wj m OPTICS. A is a double convex, li is a plano-convex, c is a con- verging concavo-convex, d is a double concave, e is a plano-concave, and f is a diverging concavo-convex. The lens c is also called the converging meniscus, and the lens f the diverging meniscus. 48. From § 45 we may at once conclude that -vvhcn a ray passes through a lens the de- viation is always tcf^ards the thickest part (Fig. Go). Hence a pencil of light may be mad«^ converging by a lens which is thickest in the middle, bat not by a lens which is thinnest in the middle.. Consequently the first three lenses, which are thicker at the centre than at the borders, are termed con- verging ; tlic others, which are thinner in the centre, are di- verging. In tlie first group, the double convex lens only need be considered ; and in the second, the double concavo, as the properties of each of these lenses apply to all those of the same group. 49. The straight line which passes through the centres of the two spherical surfaces of the lens is termed the principal axis. the as par fort thr( that but Fig. 33. a con- E is a 'oiivex, IS) and lay at 1 a ray he de- ds the Hence 2 made Inch is Jilt not nest in tly the ich are han at lI con- icli are trc di- ip, tlie f need 1 the ivo, as these of tlic I ^vhich brcs of Ices of Incipal LENSES. 89 50. In every lens there is a point such that for every ray which passes through tlie point the direction after emergence is parallel to the direction before incidence. This point is termed the optical centre of the lens. When the thickness of the lens is small, as is usually the case, the distance between the two parallel parts of the ray is also small. We may, there- fore, .without sensible error, suppose a ray passing througli the optical centre to proceed without refraction, that is to say, in the same straight line. Converging Lenses. Tlie principal focus, 51. When a small pencil of rays parallel to the axis passes through a double convex lens, it converges to a point /(Fig. 34), which is termed the principal focus Fij. 31.. of the lens. The distance cf of the principal focus from the lens is termed the focal length of the lens. The focal length is always the same for the same lens, Int it varies with the radii of the two surfaces and Tvith the refractive index. The fcoal length may be expressed in terms of these quantities, but it will suffice ii 40 OPTICS. for our present purpose if we suppose the focal length always to be det<;rmined experimentally. Since rays may be incident on either side, it is evident that there are two principal foci, one on each side of the lens, and (neglecting the thickness of the lens) equidistant from the optical centre. If rays from a luminous point at a great distance pass through the lens, they will meet at the principal focus. Hence we may determine the focal length of a convex lens by forming with it a distinct image of tho sun on a sheet of paper, and then measuring the distance of the image from the lens. If rays from a luminous point placed at the principal focus pass through the lens, they will emerge parallel and will not converge to form an image of the point. It is evident, therefore, that any ray whatever which passes through the principal focus will after refraction be parallel to the principal axis. To find the focus of a small pencil of rays parallel to one another^ hut not to the principal axis, 52. Let be the optical centre, and p or f' the prin- cipal focus of the lens. To determine the focus, it will 'ig. n:^. fl CONVEX r.KNaKS. 41 be sufficient to find the point of intJTsection of any two rays. The most convenient rays for tine purpose are the ray s c, wliich passes through the centre, and which W8 may suppose to continue in the ?anie straight line, and the ray f' m, which passes through the prin- cipal focus f', and cons.^quently after refraction is parallel to the principal axis. Let / be the point of intersection of these rays, then / is the focus of the pencil. Since /m f'c is a parallelogram, m/= c f'= c f, therefore / is on the line k f/ drawn through the principal focus / perpendicular to the axis. Any other ray parallel to s c will after refraction pass through /; hence the above proposition affords an easy way of determining the direction of any ray pu after refraction (B^ig. 36). The rule will be — draw a line 9 o through c parallel to p R, and let it meet f k in/, then r/ is the direction of the refracted ray. To find the hnage of a luminous point on the principal axis. 53. Let c be the optical centre of the lens, and let f be the principal focus on one side, and f' the prin- cipal focus on the other side. Also let fk be a perpendicular to the axis at f'. rig. 30. ■ , I 42 OPTICS. 1st. Let p be furtlicr from llie lens tlian f. Tlio image of p may be determined by finding the point o. intersection of any two rays from p which pass throug] the lens. One of these rays may be p o, or that whicli passes through the centre. Let p r be any other ray. Through the centre c draw m c/ parallel to p r, and let it meet f'k in /. Then by § 52 p r passes through /, and the image of p will be found by producing the line R/to meet the axis in i. Now the triangles pro, c/f', have these sides respectively parallel, and consequently proportional ; hence, as c f' is by hypothesis less than p c, therefore /f is less than c r, and the point i lies beyond f'. It is evident, therefore, that when the point p is beyond the principal focus on one side, its image is beyond the principal focus on the other side. The image is therefore real. 54. If p be distant twice the focal length from the lens, that is, if c p = 2 c f, it may be easily demon- strated by means of the above construction that c I == 2 c f' ; hence, in this position, the conjugate foci p and I are equidistant from the lens. 2nd. Let p be nearer to the lens than the principal focus. As in the previous case, let p r be any ray, let F^. 37. CONVEX LENSES. 43 il Tlio oint 0. irougl which jr ray. and let he line luently 58 than it I lies en the idC) its er side. om the lemon- n that ite foci 'incipal ay, let MC, drawn through the centre c parallel to pr, meet f'k in the point/. Then n/is the direction of the ray PR after refraction (§ 52). Let /r be continued to meet the axis in i. As the sides of the triangles r p c, /c f', are parallel, they are also proportional ; but as c f' is by hypothesis greater than c p, therefore r'f is greater than c r, and i lies beyond p on the same side of the lens. As the light from p does not really pass through I, it is a virtual image of p. Tliese results may, therefore, be summed up thus : — With a converging lens, when a luminous point is at an infinite distance, its image is on the opposite side of the lens at the principal focus. As the luminous point moves nearer to the lens, its image moves further away. When the point is at a distance equal to twice the focal length, the image is also at a distance equal to twice the focal length on the opposite side of the lens. (Fig. 33.) When the point reaches the principal focus on one side, the image has passed to an infinite distance on the other. Finally, when the point is nearer the lens than the principal focus, the image is further from the lens than the point, but on the same side. ill To, find hi/ a geometrical construction the image of a point not on the principal axis. 55. Let c be the optical centre and f the principal focus of the lens, and let p be the luminous point. To determine the position of the image of p, it will be sufficient to find the point of intersection of any two rays which proceed from p and pass through the lens. The two most convenient rpjr^ for the purpose are, the j. i 44 orxjcB. I ray passing throngli llio centre anJ the ray parallel to the axis. Now, tlio ray v c fiom p through the centre will proceed in the straight line r u (§ 50), and th« ray pe, which proceeds from p prrallel to the prin- cipal axis, will after re- fraction pass through f (§ 51). Hence, if we draw p E parallel to the axis, and then draw the line EF meeting the con- tinuation of PC in I, then ^^S- ^^- I is the image of p. Image of an object with a Converging Lens. 56. Let o be the optical centre, and f the principal focus of the lens, and also let a b be the object. 1st. Let A D be at a very great or infinite distance. All the rays from a c to the lens will be sensibly parallel, and will therefore converge to the principal focus. The imacre will therefore be reduced to a point. 2nd. Let a b be at a definite distance from the lens greater than ihe focal length. The image of a b may Fig. 89. llel to centre (1 th« prin- ,cr rc- llgll F if we to the iw tlic 16 con- I, then 3X18. •incipal stance. ensibly •incipal to a he lens B may CONVEX LBNSBB. 45 be constructed as follows: — Draw from A the ray which passes through the centre o ; and also draw the ray A E to the lens parallel to the axis. The first of these Fig. 40. rays will proceed in the straight line ao, and the second will pass through the principal focus f ; hence, if the line E F be drawn and continued to meet A o in «, then a is the image of A. Similarly, by drawing b o through the centre, n a parallel to the axis, and g f through the focus, meeting bo in i, we obtain h, the image of b ; a 5 is therefore the image of a b. It is evident that the image is real and inverted, and smaller than the object, when the latter is at a distance greater than twice the focal length, and larger than the object when the latter is at a distance from the lens between once and twice the focal length. 3rd. When the object is at the principal focus, the image is at an infinite distance. 4t]i. Let AB be nearer the lens than the principal focus/! The image of a is obtained, as in the previous case, by drawing a o through the centre a e parallel to the axis, and B f through the principal focus. In this case, however, it will be found that f e and o a do not i6 or. C8. moot on the side of the lens opp^ite to a, but at tho point a, on the same side as a. In the same way we Fi^. 41. may find h the image of r.. Hence we see that the image is virtual, erect, and larger than the object. Diverging Lenses. TJie principaifoms, 57. Wlicn a small pencil of rays parallel to the axis passes through a double concave lens, the rays diverge after refraction by § 45. They do not therefore meet llie axis in their passage, but if their directions Ic Fig. 42. CONCAVE LENSES. 47 at tho vay we continued bftckwarda, these will meet the axis in a point F. This point is the principal focns. hat the jct. the axis diverge re meet iions \^^ Tojind the focus of a pencil of rays parallel to one another J but not to the principal axis. 58. Let r be the optical centre, f the principal locus on one side, and F'that on the other side. Let f k be perpendicidar to the principal axis at the point v. To determine the focus of the pencil it will be sufficient Fig. 43. to find the point of intersection of any two rays. Let one of the rays be a c, which passes through the optical centre, and let the other be the ray nR, which, if con- tinued in the same straight line, would pass through f'. By § 57, after emergence, this ray n r will have the direction r d parallel to the axis. Continue d r to meet ac in/, then /is the focus of the pencil. Since /r f' c is a parallelogram, /r = c f' = c f, therefore/ is on the line f k. iji To find the image of a point on the axis formed by a diverging lens, 59. Let c be the optical centre, and f the principal focus of the lens. Let fk be perpendicular to the axis u u 48 OPTICS. at p, and let p be the luminous point. To detemnno the image of p it will be sufficient to find the point of intersection of any two rays from p through the lens. Let one of these be pc, which passes through tlie P ^Plf Fig. 44. centre o. Let pr be any other ray, and let a line through c parallel to pr cut fk in /, then p r after refraction will have the direction of jTr (§ 58). Let fn cut PC in i, then i is the image of p. It is evi- dent that p and i are both on the same side of the lens, and that i is nearer the lens than p. Images of objects loith a diverging lens. 60. Let AB be the object, and let o be the centre and F the principal focus of the lens. From A draw a o to tlie centre and A e parallel to the axis. Join fe, and let a be the point in which it intersects ac, then a is tlie imnge of a (§ 50). By a similar construction, 5, tlio image of b, may be found. It is evident, therefore, lli.it the imnjo is virtual; for CONDITIONS OF Vl.SlIill.nV. 49 the liglit after refraction does not "o tlirongh It, tliat it is always erect, and always siiiuller than the olject. Fig. 43. We sec, tlierofore, tliat double concave lenses, like convex mirrors, give only virtual images, and that donble convex lenses, like concave mirrors, give real images when the objects are beyond the principal focus, but virtual images when the objects are within the focal length. B' The conditions of the vislhilihj of images. 61. Take a double convex lens and a sheet of white paper for a screen. Hold the lens before a lighted candle, or before a window in a room, at the distance of a few feet.* Place the paper a short distance from the lens, so that the lens is directly between the candU and paper. On moving the lens backwards and for- wards, a position will be fotind in AvJiieh a distinct inverted image of the candle or win;]ow v. ill be formed * As such a lens is oiisily td^taiucd, aiul the experiment is siuiplo and iu>>li"uctive, the ruiidc;r tIio%lJ follow it out. 50 orTics. on the screen. On placing the eye at a distance of about nine inches behind tlie paper, the image will be distinctly seen on it ; and if the paper be taken away while the eye is in a straight line with the image, lens, and object, the image will be still visible at the place at which it appeared on the screen. The distance nine inches is here mentio d because the eye cannot see objects distinctly at a distance less than nine inches, and for ordinary vision rather more than nine inches is a convenient distance. The rays proceed to the eye from the image in the same way as from an object having the same position jind dimensions. Hence, in order that the image may be seen the eye must be beyond it and at a distance greater than nine inches. For instance, to find the position of the eye for distinct vision in Figs. 39 and 40, we must continue the line o f beyond a b, and place the eye at a distance from a J, which shall represent the ordinary distance of distinct vision. If the same lens be placed neir any object, as, foi example, a page of print, the image cannot be received on a screen, but will be found on the same side as the object (Fig. 41) ; and, on placing the eye near the lens, the image will be seen erect and magnified. The rays from the object are refracted by the lens, and reach the eye with the same inclination as if they came from an object having the position and dimensions of tlie image. Hence, if a converging lens having a focal length of about three inches be used to look at a page of print when the lens is held at a distance of a foot from the eye the print appears inverted, and when the MAQNiryiNG rOWER OF A LENS. 51 lens is held close to the eye, and at a distance of an inch from the paper, the print appears erect. Now, with a diverging lens the image is always \irtual ; consequently, however near or distant the object may be, it can be seen in an erect position when the lens is held near the eye (Fig. 45). The magnifying 2')ower of a lens. 62. A lens magnifies an object when the image of the object seen at the distance of most distinct vision subtends a larger angle at the eye than the object would subtend at the same distance. The ratio of the angle under wliich the image is seen to the angle under which the object is seen measures the magnifying power of the lens. This ratio is found to be nearly equal to the ratio of the observer's distance of distinct vision ( t^mes as great as it is at the surface of the earth, nrid approximately the time which a heavy body would occupy in falling from a height of 1G9 feet to the surface of Jupiter. 6. Ten weights, each of 20 lbs., are to be lifted to a height of 8 feet from the ground. Show how a system of pulleys might be arranged so that, disregarding fric- tion and the weight of the pulleys, all the weights could bo lifter' together by exerting a force equal to one of them. Show that the distance through which this force would have to act would be the same as when the weights are ruit^ed one by one by the same power. examixation papers. 11 ball And f tlie it is the om a to a 'stem fric- coukl ne of force C. 7. Explain Iiow it may be proved experimentally that the resultant pressure of a liquid on a body immersed in it is an upward pressure etj^ual to the weignt of the liquid displaced. How may this property be employed to compare the specific gravities of two liquids ? 8. Describe fully ^ ii experimental method of deter- mining the weight of a cubic inch of water. 9. 'A litre of air at 0°o, and under 760 mm baro- metric pressure, weighs l'29o gramme. Show how to find the weight of 73 litres at the same temperature, and under 1,000 mm pressure. D. 10. When a ray of light falls upon a rotating mirror, eliow that the reflected ray turns twice as fast as the mirror. 11. A beam of light, on passing obliquely from air into water, is bent away from the surface of the water; and a straight stick, with one end inmiersed obliquely in water, appears to be bent towards the surface of the water. Show that these arc illustrations of the same law of Refraction. 12. Explain how to draw a figure to represent the formation of a real magnified image of a small object by a lens. If a real image five times as high as the object is to be thrown on a screen at a distance of 36 inches from the object, show what must be the focal length of the lens employed. >•!*-. ■>)-•• w^^mmm. 12 EXAMINATION PAPEK8. JoNE, 1873. A. 1. A heavy plummet is immersed in a stream, the string being held by a person standing on the bank. The string is found to settle in a sloping position. Show by means of a sketch the three forces which keep the plummet in equilibrium. 2. Give examples of bodies in stable, unstable, and neutral equilibrium. If a body be in stable equilibrium, how is its centre of gravity affected by a small displace- ment of the body ? 3. Two men carry a block of iron, weighing 176 lbs., suspended from a uniform pole 14 feet long : each man's shoulder is 1 foot 6 inches from his end of the pole. At what point of the pole must the heavy weight be sus- pended in order that one of the men may bear |ths of the weight borne by the other ? B. 4. State the Three Laws of Motion, and give examples of each. A rifle is pointed horizontally with its barrel o feet above a lake. When discharged, the ball is found to strike the Vvater 40' » feet oIT. Find approximately the velocity of the bal;. 5. Find the tension on a rone which draws a carriap^e vof 8 tons weight up a smooth incline of 1 in 5, and causes an increase of velocity, in a second, of 3 feet per second. 6. If, on the same incline, the rope breaks when the EXAMINATION PAPERS. 13 carriage has a velocity of 48*3 feet per second, how far will the carriage continue to move up the incline ? 7. Explain the principle which enables us to find the Specific Gravity of a body by weighing it in water. A piece of metal of specific gravity 9*8 weighs in water 56 grains. "What is its true weight? 8. Describe an experiment, showing that the pres- sure of water at any depth below its surface is exerted upwards as well as downwards. 3. In a vessel not quite full of water, and closed at the top by a membrane, a small glass balloon, with an opening under water, contains just sufficient air to make it float. Explain the principle on which the bal- loon sinks when the membrane is pressed down.. Will it rise or sink with increase of pressure, when the open- ing is above water ? D. 10. Enunciate completely (in two statements) the law of Reflection of Light. Employ it to find the positions of the images of a bright point placed be- tween two parallel plane mirrors. 11. A beam of light issues from a given bright point 3 feet above the surface of still water, and fallii;g obliquely on the surface, is divided into two parts, one of which is reflected and the other refracted. Find the position of the point of incidence and its distance from the bright point, so that the reflected and refracted beams may be at right angles to each other. [The index of refraction from air to water is ^.] 12. An object is moved from a considerable distance on the principal axis of a convex lens up to the lens. I'i EXAMINATION PAPERS. Find the corresponding changes in the position and size of the image. ineaning January, 1874. A. the Parallelogram of Forces. Explain the f the terms employed in yonr statement. Apply it to shov,' that if four forces acting on a p(nnt be represented by the sides of a rectangle taken in order, they will be in equilibrium. 2. A substance is weighed from both arms of an un- equal balance, and its apparent weights are 9 lbs. and 4 lbs. Find the ratio between the armr. 3. Find the relation between the power, P, and weight, W, in the sys- tem of pulleys represented in the figure, neglecting the weight of the strings, of the rod A P C, and of the pulleys. If the pulleys are 1| inches in dia- meter, and the strings parallel and attached at given points. A, B, and C, to the rod supporting the weight, to what point of the rod should the weight, W, be attached, so that the horl.-jontal direction of the rod may be maintained ? B. 4. When a body changes its rate of motion under the action of a constant force show that the space des- cribed in any time is the same as the space deseribetl by a I the sa If a feet it will of 40 5. I pound of the Give a G. I than tl load fall thr 7. A grains, specific 8. A^ the last 9. A twenty fourteei forced volume p res sun State 10. 1 still wa !* EXAMINATION TArEUS. 15 by a body moving uniformly with the mean velocity fur the same time. If a body is projected upwards with a velocity of 120 feet in a second, what is the greatest height to which it will rise, and when will it be moving with a velocity of 40 feet per second ? 5. I suddenly jump off a platform with a twenty - pound weight in my hand. AVliat will be the pressure of the weight upon my arm while I am in the air ? Give a reason for your reply. 6. In Attwood's machine one of the boxes is heavier than the other by half an ounce. AVhat must be the load of each, in order that the overweighted box may fall through one foot daring the first second ? C. 7. A body weiglis in air 80 grains, in water 5G grains, and in another liquid 46 grains. Find the specific gravity of this liquid. 8. "What becomes of the weight which the body, in the last question, appears to lose ? 9. A cylindrical bell four feet deep, whose content is twenty cubic feet, is lowered into water until its top is fourteen feet below the surface of the water ; and air is forced into it until it is three-quarters fail. "What volume would the air occupy under^ tiie atmospheric pressure — the "Water-barometer being at 34 feet ? State the principles on which your answer is based. D. 10. A bright point, G inches above tlK surface of still w\ater, is reflected from the bottom of the vessel, 16 EXAMINATION PAPEUS. I which is 2 feet deep, as well as from the surface of the water. Show how tu find the positions of the images formed by the reflections ions. ( M = - I 11. Enunciate by oid of a sketch, and in two state- ments, the Law of Refraction when light passes from a rarer to a denser medium. Also point out what is meant by the Critical Angle. 12. A simple lens is used as a magnifier. Sketch the relative positions of the object (an arrow) and its image. Tlic same lens is used as in photography. Sketch the relative positions of the object and its image. MATRICULATION EXAMINATION. June, 1874. A. 1. Explain the meaning of the words Composition and Resolution of Forces ; and show how forces may be compounded and resolved. A point is acted upon by a force whose magnitude is unknown, but whose direction makes an angle of G0° with the horizon. Tlie horizontal component of the force is known to be 1*35. Determine the total force, and also its vertical component. 2. Assuming the truth of the Parallelogram of Forces, show how to find the position and magnitude of tho resultant of two parallel forces, P, and Q, acting at different points of a rigid body. EXAMINATION rAl'EllS. 17 c 8. Two uniform heavy rods, A C, B C, rigidly connected together, are capable of turning round a hori- zontal axis at C. Find the me- chanical conditions which determine the position of equilibrium. B. 4. A balloon is carried along by a current of air moving from east to west at the rate of six miles an hour, — having no motion of its own through the air ; and a feather is dropped from the balloon. What sort of a path will it appear to describe, as seen by a man in the balloon ? 5. Suppose that at the Efpiator a straight hollow tube were thrust vertically down towards the centre of the earth, and that a heavy body were dropped through the centre of such a tube. It would soon strike one side. Find which, giving a reason for your reply. 6. What is meant by saying that the accelleration produced by gravity is represented by the number ^2-2? From a point in a smooth inclined plane a ball is rolled np the plane with a velocity of 10 1 feet per second. How far will it roll before it comes to rest, the incli- nation of the plane to the horizon being 30° .^ Also, how far will the ball be from the starting-point after five seconds from the beginning of motion ? C. 7. A wine bottle, which below the neck is perfectly IMAGE EVALUATION TEST TARGET (MT-3) // 'Q>.. / f/i 1.0 I.I 1.25 li^ 11121 1^ 2.2 1^ IM \^ Ilia Ill 1.8 U III 1.6 Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 iV ^q\' ■1? :\ \ V..X ^..%> > 6^ <^ "ply it to determine tho conditions of equilibrium in the single movable pulley \vith inclined strings, and in the screw. 3. A ladder rests against the side of a house and is inclined at 60° to the ground. The pressure of the ladder against the wall being equal to a force of 60, and the friction at tho same placo equal to a force of 40, find the iwcs sure and friction at tho poitjt where the ladder rests on the ground. Find also the weight of the ladder.— il)js. 40 and CO ; 80. 4. What is the average velocity of a body during the first second of its fall under gravity ; also during tho first two seconds ? ShoAv how, by knowing the average velocity, you can find the whole space fallen through. 5. A body whose mass is 100 grammes, is thrown vertically upwards with a velocity (;f 080 centimetres per second. What is the energy of the bociv — (1) at the moment of propulsion — (2) afterhalf a second — (lij after one second ? (N.B.i/ = 980). G. What is meant by Specific Gravity in general, and par- ticularly according to the French system of measurement ':' Describe, in detail (with a sketch), the method of obtaining tho specific gravity of a liquid by means of the specific-gravity bottle. 7. If you have a deep tank or ocean of water, the pressure will be greatest at the bottom, and will be diminished, by a definite amount, for every foot you rise above the bottom. Does this law hold for atmospheric air ? If not, in whi:! direction does it differ in this case, and what is the cause of the difference ? 8. Ten cubic centimetres of air are measured off at atmospheric pressure. When introduced into the vacuum of a barometer they depress the mercury, which previously stood at 76 centimetres I' I ' I: I ! LI' ^1 s; !!! 30 EXAMINATION rAl'ERS. and occupy a volume of 1 5 c.c. By how mucli lias the mercurial column been depressed ? — Ans. 25|- . y. A piece of wood floats in a beaker of water with nine-tenths of its volume immersed. When the beaker is put under tho I'eceiver of an air-pump, and the aii withdrawn, how is the im- mersion of the wood affected ? 10. A plane mirror, in the shape of a circle, revolves about ^ vertical diameter. A fixed horizontal ray of light falls upon its centre and is there reflected. Prove generally that if the mirror move through any angle the reflected ray will appear to have moved through double that angle. 11. An object 6 inches long is placed sj'^mmctrically on the axis of a convex spherical mirror, and at a distance of 12 inches from it. The image formed is found to be 2 iuchcs long. What is the focal length of the mirror ? — Ans. 3 in. 12. If the refractive index of a ray of light in passing from air to water be ^, and in passing from air to glass f , find, by aid of a diagram, what it will be for the ray when passing from water to gli' S3. — Ans. ^. ]\r ATKICUL ATIO X EXAMINATION. June, 1878. 1. Explain what is meant ' ^ the moment of a force. Prove that when two forces act upon a point the algebraical sum of their moments about any point in their plane is equal to the moment of their resultant about the same point. 2. State tho principle of the Triaugle of Forces. Employ it to construct an inclined plane on which a horizontal force of 3 lbs. will support a weight of 4 lbs., and find tho resistance of the piano. 3. Two bodies start together from rest, and move in directions at right aii;Tles to each other. One moves uniformly with a velo. city Oi' 3 feet per second, the other moves under the action of a constant force. Determine the acceleration duo to this force, if tho bodies at the end of 4 seconds are 20 feet apart. — Ans. i. EXAMINATION PATEKS. 31 nercurial ne-tentbs mder tlio 3 the im- s about ^ 3 upon its tbe mirror ,r to have illy on tlie I 12 inches ,Dg. What ssing from find, by ssing from ^> )N. rce. Provo ical sum of qual to the Employ it tal force of •Gsistance of in JIvoclious with a velO' action of a this force, if , — Ans, 2. 4. The space passed over in any time may be represented by an area. Explain clearly the meaning of the statement, and under what conditions it is true. Show how to employ it to determine the space passed over by a body in 10 seconds after it starts from rest : ; has its velocity increased by 1 foot per second at the beginning of each second. 5. A mass of 488 grammes is fastened to one end of a cord which passes over a smooth puUoy. "What mass must be attached to the other end in order that the 488 grammes may rise through a height of 200 centimetres in 10 seconds {g = 1)80) ? — Ans. 492. 6. Describe the hydrostatic balance, and show how to employ it to determine tbe specific gravity of a solid heavier than water. 7. A wooden vestiel, 6 inches square and 6 inches in height, with a neck 2 inches square and 3 inches in heiglit, is full of water ; find the position of the centre of gravity of the water. Also find the pressure on the base of the vessel. — .4ns. 3/^ ins. from the base. Pressure = the weight of 324 cubic inches of water. 8. A piece of copper and a piece of silver fastened to the two ends of a string passing over a pulley, hang in equilibrium when entirely immersed in a liquid whose specific gravity is 1*15. Determine the relative volumes of the masses, the specific gravities of silver and copper being respectively 10*47 and 8-89. 9. If the height of the water-barometer be 1,033 centimetres, what will be the pressure on a circular disc whose radius is 7 centimetres, when sunk in water to a depth of 50 metres ? — Ans. 771033 grammes. 10. Give a sketch and explain the principle of action of the Forcing Pump. Also show by what modification a continuous stream may be obtained by means of it. 11. Enunciate the law connecting the volume and pressure of a given weight of gas. When the height in the mercurial baro. meter changes from 29"55 inches to 30"33 inches, what is the change in the weight of 1,000 cubic inches of air, assuming that 100 cubic inches of air weigh 31 grains at the former pressure, and that the temperature remains constantly at O ° C ? — Ans. I !i ( '; o'J EXAMINATION PAPERS. 12. A man, 6 ft. high, sees bis image in a plane mirror hung vertically. The top of the mirror being G ft. from the ground, determine its smallest length in order that the man may see bis full-length image in it. 13. Show how to find the position of the image of an aiTow placed in front of a concave spherical mirror. Explain when it is an erect, and when an inverted image. mateiculatio:n" examixatiox. January, 1879. 1. Define Stable, Unstable, and Neuti-al equilibrium, and state the conditions of stability of a body resting on a horizontal plane. 2. A telescope consists of 3 tubes, each 10 inches in length, eliding within one another, and their weights are 8, 7, and 6 ounces. Find the position of the Centre of Gravity when the tubes are drawn out to their full length. — Ans. -fj inches from the middle point. 3. State the condition in order that three or more forces acting on a bar which is free to turn about a fixed axis may not produce motion about that axis. A uniform bar 2 ft. long and weighing 3 lbs. is used as a steelyard, being supported at a point 4 inches from one end. Find the greatest weight which can be weighed with a movable weight of 2 lbs., and find the point from which the gradations should be measured. — Ans. 16 lbs.; 8 inches beyond the weight, an inch being taken to half-a-pound. 4. Show by the aid of a sketch, exhibiting the reEolution ox forces, lioiv a ship can sail at right angles to the direction of the wind. 5. A stone dropped into a well reaches the water with a velo- city of 80 ft. per second, and the sound of its striking the water is heard 2/^ seconds after it is let fall. Find from these data the velocity of sound in air (g = 32). — Ans. 1,200 ft. per second. 6. A mass of G ounces slides down a smooth inclined plane whose height is half its length, and draws another mass from rest whii the 7. a boi the I 8. It is and 1 i;:ich i base ( 7 ozs. 9. 1 Solid a of any 10. ; in orde siphon and a where wuderg(| 11. plane si 12. internal the diffj asthepJ 13. nil Tor. jugate fd of a poi reflectini EX-VMIXATION PATERg. 33 ror huug ground, y see Ws in arrow wlien it ,,and state horizoutal s in lengtli, fld 6 ounces, le tubes are the middle rest over a dist;ince of 3 ft. in 5 seconds, along a horizontal tublo wljich is level with the top of the plane, the string passing ovtr the top of the plane. Find the mass on the table. — Ans. 394 czs. 7. Th3 resultant vertical pressure of a fluid on the surface of a body immersed in it is equal to the weight of fluid displaced by the body. How is this proved by experiment ? 8. A vessel, shaped like a portion of a cone, is filled with wator. It is one inch in diameter at tbe top, and 8 iucues at the bottom, and 12 inches high. Find the pressure in pounds per square inch at the centre of the base, and also the whole pressure on the base (a cubic foot of water weij^hs 1,000 ounces). — Ans. Nearly 7 ozs. U^} ozs. 9. How would you show that atmospheric air is heavy, like s.ilid and liquid bodies ; and how would you find the exact weight of any volume of it ? 10. Explain the action of the siphon. "What are the conditior.s in order that it may work properly ? If the shorter arm of the siphon for emptying a vessel of water be 32 feet long, and a bubble of air occupies 6 inches of the tube at the end where the water enters, find the changes which the bubble of air undergoes while the process of emptying goes on. 11. State, and show how to prove, the Law of Refraction by plane surfaces. 12. An arrow, pointing towards the observer, is seen by internal reflection in an isosceles right-angled prism. Explain the difference in, and give a sketch of, the images seen, according as the prism is three-sided, or a four-sided Wolliis/.on prism. 13. Explain the formation of images by a concave cylindrical mi ror. Find the relation between the distances of the two con- jugate foci from the mirror. What is the position of the imago of a point which is at the distance of the diameter from the reflecting surface of the cylinder ? 1 A ar III m anc tha Loi ior Li\ < con < ran of .stu (( Me (( to 1 me: kiK to I Foolscap Svo, cloth, price Ss. 6d. GROOMBRIDGE'S GUIDES. A GUIDE TO THE MATPJCULATJON EXAMINATION OF THE UNIVERSITY OF LONDON. COXTAIN'IXa Advice and hints on all the Subjects, the Method of Ex- amination, Number of ]\Iarks given, Directions for lilling up the Schedule, Entering Name, Writing tlie Papers, and much Useful Information to every Cundidate, whether working with or without a Tutor, WITH SPECIMENS OF EXA:\IINATI0N PAPERS. " The subject is dealt with very thorou^lily and comprehensively, useful hints and details hein^j: furnished on ;ill jKiiiits, with advice as to the course of reading that should bo undertaken." — !