Ex. A particle is projected vertically upwards with a velocity of 100 feet per second, to find (i) its height at the end of 3′′, and (ii) the time when it is at a height of 140 feet above the point of projection. When a foot and a second are taken as the units of space and time the numerical value of g = 32.2, (Art. 43), and if u be the velocity of projection, and s the height at time t after projection, we have s = ut — gť3, (Art. 70). For the first part of the example t=3, u= 100; .. 83.100-32.2. (3) 155.1 feet = height of the particle at the end of 3 seconds. For the second part of the example s = 140, u = 100; and we have to find t from the quadratic equation Substituting the numerical values of u, g, s, we get after reduction, a double result, which is to be explained thus, at the end of 2".13 the particle is at a height 140 feet in its ascent, and at the end of 4".08 it is again at the same height of 140 feet on its descent, after having reached its highest point and then descending. 75. We subjoin a few interesting problems which can be solved by the principles already explained. PROB. Two bodies P, Q are connected by an inextensible string which passes over a smooth fixed pully; to determine the motion of each body, and the tension of the string. Let P, Q represent the masses of the bodies; T the tension of the string, the mass of which we will neglect, and suppose P> Q. Now moving force on P downwards = Pg - T, Qupwards = T-Qg; .. accelerating force on P downwards = P P A (i). Now the string being always stretched and inextensible, the velocity of P downwards and of Qupwards will be always equal, and therefore the rate of change of their velocities, i.e. the acceleration of the two bodies must be equal; Pg-T T-Qg Q whence T= 2PQ (ii), which gives the tension of the string,-and further substituting this value of T in either of the expressions (i), we get the acceleration on P downwards and on Q upwards Also velocity of P and Q after time t from rest = ...... space described P. M. S= P-Q 1P-Q 14 COR. 1. By taking P-Q as small as we please we may make the motion as slow as we please, and so capable of being measured, by which means the value of g might be obtained from observation. This is substantially the principle of Atwood's machine, which will be described hereafter, (Art. 82). COR. 2. If at any instant a part (R) of one of the bodies (Q for instance) were suddenly detached, there would be no instantaneous change of the velocity of either body, but the P-Q+R acceleration would become the string would become P+Q-RI, and the tension of 2P(Q-R) ་ P + Q − R · 9• 76. PROB. Two bodies P, Q are in motion, connected by a string which passes over a smooth fixed pully; another body R is suddenly attached to Q;—find the change of velocity and the impulsive strain on the string. Let P, Q, R be the masses of the bodies, and suppose R to become attached to Q by a string P connecting them suddenly becoming tight. Let V A R be the velocity of P and Q at the instant before this takes place, and 'the common velocity of the three the instant after, X, X, the impulsive strain on the strings PAQ, QR, respectively; then for the motion of the three bodies we have (Art. 46) PV' = PV – X1, QV' = QV+X1- X2, 2 whence by adding (P+Q+R) V' = (P + Q) V, Equations (i), (ii), (iii) determine the three quantities required to determine the change of motion completely. MOTION ON AN INCLINED PLANE. 77. PROB. A heavy body Q is drawn up a smooth inclined plane by another body P, which descends vertically; P being connected with Q by an inextensible string passing over the vertex of the plane. Let P, Qbe the masses of the bodies, T the tension of the string, and a the inclination of the plane to the horizon, R the pressure of Q on the plane. P Then resolving the motion of parallel to the plane and perpendicular to it, the weight of Q is equivalent to a force Qg sin a down the plane, Qg cos a perpendicular to the plane, the latter force is balanced by R, the pressure of the plane, And since the string continues stretched, the velocities of P and Q in these two directions are always equal and therefore the accelerations upon them are equal, that is, and acceleration on P downwards and on Q up the plane equation (ii) gives the tension of the string, and (iii) gives the acceleration from which the velocity acquired and space passed over in any time may readily be obtained. COR. The preceding problem may be varied by supposing to move on a smooth hori zontal table. The student may either investigate the motion in this case independently, or deduce the results from the present Art. by making a = 0. P 78. A heavy body descends freely down a smooth inclined plane; to find the time of motion and the velocity acquired. Let P be the mass of the body moving down the plane BA, the inclination of which to the horizon is a, R the pressure on the plane. Then resolving the forces on P parallel to the plane and perpendicular to it, P B C A |