Plate 1. In the triangle ABC, let AB express the time in which a body is falling, and BC the velocity which it has acquired at the end of the fall; let AF, AD, be parts of the time AB; and through F, D, draw FG, DE, parallel to BC. Because the triangles ABC, ADE, are similar, AB is to AD as BC to DE; but AB and AD express times of descent, and BC expresses the velocity acquired in the time AB; therefore, since (Prop. XXIII. Cor.) the velocities are as the times, DE expresses the velocity acquired in the time AD. In like manner GF, any other right line parallel to BC, expresses the velocity acquired in the time AF. Therefore the sum of the lines which may be supposed drawn parallel to CB in the triangle ADE; that is, the whole triangle ADE, will represent the sum of the several velocities with which the falling body moves in the time AD. For the same reason, the triangle ABC will represent the sum of the velocities with which the falling body moves in the time AB. Since therefore it is manifest, that the space which a body passes through in any moment of time is as the velocity with which it moves at that moment; and consequently, that the spaces through which it passes in any times whatsoever, are as the sums of the velocities with which it moves in the several moments of those times; the spaces passed through in the times AD, AB, are to each other as the triangles ADE, ABC. But the triangle ADE (El. VI. 19.) is to the triangle ABC in the duplicate ratio of the homologous sides AD, AB, and also of DE, BC: that is, the spaces are as the squares of the times, and also as the squares of the last acquired velocities. consequently the spaces described are in the compound ratio of the times and the velocities. EXP. Let there be two pendulums, one of which vibrates twice as fast as the other, a ball let fall from such a height above the ball of the shorter pendulum as to reach it in one vibration, must fall from four times this height, to reach the longer pendulum in one of its vibrations. COR. 1. Hence, if the forces are variable, the spaces described are as the forces and squares of the times ;-or as the squares of the velocities directly, and the forces inversely. For by the Prop. (calling S, V and T, the space, velocity and time) S is as T x V, and (by Prop. A. p. 14.) V is as F x T.. S is as F x T', and S is as Va COR. The times in which bodies fall from unequal heights, and their last acquired velocities, are as the square roots, or in the subduplicate ratio of their heights. Since TT is as S, T will be as✔ S; and since VV is as S, V will be as ✔ S. COR. 3. If the time of the fall of a body be divided into equal parts, the spaces through which it falls in each of these parts, taken separately, will be as the odd numbers 1, 3, 5, &c. The spaces being as the squares of the times or velocities, if the times be as the numbers 1, 2, 3, 4, the spaces will be as 1, 4, 9, 16; whence, in the first time the space will be as 1, in the second time, the space passed over will be as 3, in the third, as 5, &c. SCHOL. SCHOL. Since S is as T., and as in the first second of time a body freely descending by the force of gravity falls through 161 feet, we easily find the space described in any given number of seconds; for S 16'1 x T'. Thus in 5" a body will fall through 402 feet; for 16.1 X 25= 402. Again, the spaces, fallen through in the 1st, 2d, 3d, seconds, are 16'1; 161 × 3; 16'1 × 5 respectively. PROP. XXVII. The space which a body passes over in any given time from the beginning of the fall, is half that which it would pass over in the same time, moving with the last acquired velocity. Fig. 10 For the triangle ABC (by Prop. XXVI.) expresses the space passed over in the time AB, Plate 1. when the motion is uniformly accelerated; the last acquired velocity is expressed by BC; and the rectangle of AB, BC, rightly expresses the space passed through in the time AB with the equable velocity BC: since therefore the triangle ABC is half of the rectangle AB, BC, the proposition is manifest. PROP. XXVIII. The motion of a body thrown upwards is uniformly retarded by gravitation: the time of its rise will be equal to that in which a body falling freely acquires the same degree of velocity with which it is thrown up; and the height to which is rises will be as the square of the time, or first velocity. The same force which accelerates a falling body, acting in an opposite direction upon one thrown upwards, must retard it: and, since the action of gravitation is uniform, in whatever time it generates any velocity in a falling body, it must in the same time destroy the same velocity in a rising body: through whatever space the falling body must pass to acquire any velocity, the rising body must pass through the same to lose it: and whatever ratio the spaces bear to the velocities and times in one case, must take place in the other: the effect of gravitation in rising bodies being in all respects the reverse of its effect upon falling bodies. SCHOL. As the force of gravity near the surface of the earth is constant, and known by experiment, and as the spaces described by falling bodies vary as the squares of the times, (T) or as the squares of the velocities (V): hence every thing relating to the descent of bodies, when accelerated by the force of gravity; and to their ascent, when they are retarded by that force, may be deduced from the foregoing propositions. (1.) When a body falls by the force of gravity, the velocity acquired in any time, as T", is such as would carry it uniformly over 2 F T in 1"; where F = 16'1 feet. EXAM. The velocity acquired by a falling body in 6′′ = 32°2 × 6, or such as would carry it uniformly through 193 feet in 1". (2.) The (2.) The space fallen through to acquire the velocity V is 4F. For S: F:: V2 : 2F(* EXAM. If a body fall from rest till it acquire a velocity of 20 feet per second, the space V9 From these three expressions, V2FT; S= and S FT (Cor. 1. Prop. XXVI.) 4F any one of the quantities S, T, V, being given, the other two may be found. EXAM. I. To find the time in which a body will fall 400 feet; and the velocity acquired. = Since S FT... T= √F= 400 ·· T = √ F = √16-1=5′′ nearly, and V being equal 2FT 161 feet = velocity acquired. EXAM. 2. If a body be projected perpendicularly downwards, with a velocity of 20 feet per second, to find the space described in 4" The space described in 4" by the first velocity is 4 x 20, and the space fallen through by the action of gravity is 161 x 4', therefore the whole space described is 337.6 feet. EXAM. 3. To what height will a body rise in 3" which is projected perpendicularly upwards with a velocity of 100 feet per second? 2 The space described in 3" by the first velocity is 300 feet, and the space through which the body would fall by gravity in g" is 16'1 x 3' = 144'9 feet, therefore the height required is 300 1449 = 155°1 feet. SECT. II. Of the Laws of Gravitation in Bodies falling down inclined Planes. DEF. IV. An inclined plane, is a plane which makes an acute or obtuse angle with the plane of the horizon. PROP. XXIX. The motion of a body, descending down an inclined plane, is uniformly accelerated. In every part of the same plane, the accelerating force has the same ratio to the force of gravitation acting freely in a perpendicular direction, and is therefore (El. V. 9.) equally exerted exerted in every instant of the descent; whence, (as was shewn concerning bodies falling freely, Prop XXIII.) the motion must be uniformly accelerated COR. Hence, whatever has been demonstrated concerning the perpendicular descent of bodies, is equally applicable to their descent down inclined planes, the motion in both cases being uniformly accelerated by the same power of gravitation. PROP. XXX. The force, with which a body descends by the attrac tion of gravitation down an inclined plane, is to that with which it would descend freely, as the elevation of the plane to its length; or as the sine of the angle of inclination to radius. Let AB be the length of an inclined plane, and AC its elevation, or perpendicular Plate 1. height. If the force of gravitation with which any body descends perpendicularly be Fig. 11. expressed by AC, and this force be resolved into two forces, AD, DC, by drawing CD perpendicular to AB; because the force CD is destroyed by the re-action of the plane, the body descends down the inclined plane only with the force AD. And (El. VI. 8. Cor) AD is to AC, as AC is to AB; that is, the force of gravitation down the inclined plane is to the same force acting freely, as the elevation of the plane to its length, or as the sine of the angle of inclination ABC is to the radius AB. COR. I. Hence, the force necessary to sustain a body, on an inclined plane, is to the absolute weight of a body, as the elevation of the plane to its length for the force requisite to sustain a body, must be equal to that with which it endeavours to descend; which has been shewn to be to that with which it would descend freely, as the elevation of the plane to its length. COR. 2. If H be the height of an inclined plane, L its length, and the force of gravity be represented by unity; the accelerating force on the inclined plane is represented by H For by the Prop. the accelerating force is to the force of gravity (1) as H is to L.. the accelerating force = H H COR. 3. Hence varies as the sine of the angle of inclination. COR. 4. If a body fall down an inclined plane, the velocity V generated in T" is such as would carry it uniformly over x 2 FT feet in 1" where, as before, F is equal 16.1. H For (by Prop. A. p. 14.)the velocity varies as the force and time, (i. e.) as H H × T, and the velocity generated by the force of gravity in one second is 2 F, therefore V = -분 H X 2 FT. 8 Ex. Ex. If LH:: 2 : 1, a body falling down the plane will, at the end of 4", acquire a 32.2 × 4 = 64.4 feet per second. velocity of H COR. 5. The space fallen through in T" from a state of rest, is x FT2 for (Prop.. XXVI.) the spaces described vary as the squares of the times. Ex. I. If H = feet. the space through which a body falls in 5′′ is× 16.1 × 15 = 2011 Ex. 2. To find the time in which a body will descend 40 feet down this plane. Since S = FT2, therefore T =✔HF=✔ =2.2 seconds. V2 COR. 6. The space through which a body must fall, from a state of rest, to acquire a L V2 velocity V, is X For (Cor. 1. Prop. XXVI.) S is as therefore the space H 4F through which the body falls by the force of gravity, is to the space through which it falls down the plane, as the square of the velocity directly, and as the force inversely in the former case, is to the same in the latter; and if F 161) be the space fallen through by gravity, and 2 F is the velocity acquired in 1"; hence FS:: 2 2 F : x V2 and S = H I L V2 Plate 1. HX 4F Ex. 1. If L=2 H, and a body fall from a state of rest till it had acquired a velocity of 40 feet per second, the space described is Ex. 2. 402 I 64.4 If a body fall 40 feet from a state of rest down this plane, to find the velocity acquired. V2 = 4 FS× = 64 4 × 40 × = 1288, and V = 35·8 feet per second. <== 1 PROP. XXXI. The space described in any given time by a body descending down an inclined plane, is to the space through which it would fall perpendicularly in the same time, as the elevation of the plane to its length. Let AC represent the force with which a body would fall perpendicularly: CD being drawn from C perpendicular to AB; AD, as was shewn (Prop. XXX.) will represent the force with which the body descends down the inclined plane AB. And, since the spaces through which a body falls in any given time must be as the forces which move them, the space through which the body will fall down the inclined plane AB, is to that through which it will fall perpendicularly in the same time, as the force AD, to the force AC. But AD is to AC (El. VI. 8. Cor.) as AC the elevation to AB the length of the plane; therefore the 众 |