PP'g and a x (W 1+A' + A) (P+P)(A® +1)* the actual tension, N 2. PP'S (Vi+ A' + A) (P+P')(A®+1) the pressure on the pulley. If o be the inclination of the plane to the horizon, we have A = tan. 0, and +A? = sec. 0. Hence gPP or o (P + P') sec. 6 (sec. 0 + tan. 0) (1 + sin. 6) x (1 + sin. 6)? P + P which somewhat simplifies the expressions. auda = and c = If 0 = 0. Then we get gPP .(a) P + P for the tension and pressure when the body P is drawn along an horizontal table by P' descending from its edge. Again, if P draw P along an horizontal plane situated below the pulley, then we have Y=0, A = 0 and B = 0 and equation (2) becomes 2gP(a - 1) as P+P. (X-a) Also Hence (X - a) = 4* – nie P.'(2-2).(18-m*) (P+Pa' - Pm 1 X 2P)* - 2PP'lm's + P"*} Also ds' = da. Hence (P + PP)* - 4moPP'+ 2PP'lm' o=P'g {(P + P')a? – P'mo} Imta a = 2 Xo which give the tension of the string and pressure on the pulley in this case. 2r= Let m = 0. Then gPP P + P PP2 and P + P which confirm the results marked (a). 601. Letr be the radius of the given circle, t the given time, & the altitude of the required diameter; then by the question 9 ť 2r 9. F = 2 602. Let zb be the length of the string to which W is attached, and x the space descended through in the time t; then the length of the pendulum being 26 - x, the time of one oscillation is 26 Hence N being the number of oscillations swung in the interval t, we have be Hence dx Х ✓ (26x - x*) 1 X vers.x. P+W 2(P-W) the number of vibrations required. It is evident that N is independent of the length of the string. Let P = 00. Then 1 N= or the number of oscillations performed in the time 2 through 2b is however great may be the power. 603. Let x denote the ratio required; then W and w being the weights of the wheel and axle R, r, their radii, we have pRT qro wr WR? pR+qr+ + 2 2 2pr - 2922 29x2 2p+W+(29 + W)x? Now s being the given space described in the time t, we get 2(29 + w). (px — qxP) = (p – 29x) (2p + W + 2q + wx") :: (29 + w) x + 29(2p + W).c = (2p + W)p :. 28 + 29. 2p+W P 29 + w 29 + w 2p+W 604. Let a be the height of the cone, r the radius of its base; then the distance of the centre of oscillation from the axis of rotation is (Venturoli,) S FicdM CdM M L = MAM = fx?dM JedM x being the distance of the molecule dM from that axis. But dM = 2Tydz 2XT da s retao Х The centre of oscillation of a conical surface cannot possibly be in the base. |