mass of the oscillating body, k the distance between the centre of gravity and suspension ; then the length of the pendulum is S Mk But in this case L= kar - men ge being the radius of the circle, and c the chord, and I the length of 613. Generally for n pulleys, let Q, Qy.... be their weights; then, the weight sustained is W 2" 1 and the moving force is w P 2 where P = P + e + Q3 +..... W Now the inertia of W= Q of Q, = 2(2"-1) (2— 1)'.IQ, 2 2.(2-1) &c. = &c. Х 3 Х Hence, supposing Q, = Q = &c., the inertia arising from the rotation of the pulleys, is Q x1 + 38 + 78 + 15 X &c.) 2(2" - 1)* Q. 22+2-3 X 2=+*+36 +8 2(2"-1) Also the inertia from the descent of Qas Qg, &c. is Q 2an 3. 2+1 + 3n + 5 3 Q x (2.84 +22+1 - 6 X 2 + + on (2* 1) 6(2"-1) M + 18) = by supposition. 6(2"-1)2 Hence the accelerating force of P is 6P (2"-1) - 6W (2-1). M and that on W is 6P (2" — 1)– 6W W = X .+ P't F = F = Again, let T, T,....T, be the several tensions required, and I, J, .... I, the corresponding inertias ; then we have P-T, = I, F = PF, :: T, = P x (1 – F) : T: = T: - 1, F and so on. The pressure on each axis is double the tension of the string that goes round it. 614. Let o denote the angular distance required, measured from the West, then by the composition of forces, it readily a -1 and ..0 = sin. Ő Hence it seems that at sea, when the ship is sailing iu a direction oblique to the wind, the position of the vane is no certain criterion as to the quarter from which it blows. It would also appear, at first sight, that the magnet is affected in like manner by the motion of the vessel; but a moment's reflection is sufficient to be convinced of the contrary. The magnet consisting of two equal arms, the action of the ship’s motion upon the one is counteracted by its opposite action upon the other. 615. Generally, w being the weight of the cylinder, and P the power or weight which puts it in motion, e the radius of the base, and r the distance, then the accelerating force on p is p=20,w = 133. 6, v = l,ę = 1, r= 10, and it will be found, after substituting and reducing, that t = .03214, &c. seconds. 616. Generally, required the length of a pendulum that would oscillate seconds at the distance of n radü of the Earth from its centre. If F be the force which accelerates the pendulum, whose length is L, then the time of an oscillation is got from (see Bridge, vol. II.) L TOC Hence, if I be the length of a second's pendulum at the surface of the earth, whereg is the accelerating force, we have If n = 2. Then inches = 9.8 inches. Within the surface F = g. Ř whence p is easily found. 617. Letr be the distance of the particle dM from the axis of suspension, a the length of the rod, D its density at the lower extremity, L the distance between the point of suspension and centre of oscillation; then Sr’dM (see Vince, p. 127.) L= 618. Generally let AC (Fig. 102,) be any curve whatever revolving uniformly round the vertical aris AB, and suppose the body P descending along the curve by the force of gravity; required the velocity of the body at any given epoch t. Let the 1 PM = y, AM = x, BC = B, AB = 6, and V the known velocity of the point C; then resolving the centrifugal force RP into the tangential and normal ones PQ and RQ, we have PQ = PR X cos. RPQ = PR X dy ds dx and RQ = PR X sin. RPQ = PR X ds Again, the velocity at C being V, that at P is |