605. Generally let R,R', &c., be the radii of the wheels, T, ', &c., those of the axles; then x denoting the weight to be raised each time, and P the power, the moving force of Pis rr. P-XX RR' .... Pt F = P - x. and the mass moved reckoning the inertia is R’R'R"&c. wherein w, w', w" &c. are the weights of the several pairs of wheels and axles, and d, d', d" &c., the distances of their centres of gyration from the respective axes of rotation. Hence the accelerating force is rr'.... X(RR'R"....) x RR'R" &c. PM - xm RR'R" &c. PM +x+wd + w'd?R? + &c. M = RR'&c, and m = rr' &c. Now n being the number of ascents, Q the whole quantity raised through the space s in the whole time T, we have Q nx = Q and .. n = .....-X.rr..... D being substituted for wds + w'd'R + &c. 2PM:- 3m.ro PM.x2 23 m PM +D+ (PMR + D+x) and we finally obtain 3m PM+3mD-PM 2PM(PMR +D) 22+ 2m which being resolved will give x, and therefore the quantity required, when for M, m and D we substitute their values, viz., RR', rr', wd + w'd" 2m 606. Let either weight be denoted by P and the additional one by p, and put P + p = Q, also let l be the length of the whole string, a, b the lengths of the parts of it to which P, p are attached previously to the motion, and suppose the weight Q to have descended x feet. Then since the weight of the strings may be represented by their lengths the moving force is Q-P + b + x - (a - x) = Q - P + b -a + 2x and the mass moved is P + Q +7 :. the accelerating force is by hypothesis N 2gM 29 x + X N N 2 6a 29 x ✓ (Mx + x2) x Q ♡ (Q-P+b-a)x + x2 P+Q+b+a Let b = a = 0. Then the velocity resulting from the actions of P, Q alone is 29 d': v (Q-P)x + r? PER and therefore that which is due to the weight of the string is v – v', which being put = 0 according to the question gives (Q-P+b-a).x+x® Q-Px+xs P+Q Q :: x= 2(P6- Qa) a + b which gives the part of the descent required. d = 607. The distance required (d) is that of the centre of oscillation from the axis of suspension, that is S fr’dM SædM 0 But M = x x x tan. 2 Let x = a, the altitude of the triangle; then 3 a. 608. Let r be the radius of the base of the paraboloid, W its weight, and I the length of the chain, (whose weight is represented by its length), a that part of it which is unwound at the commencement of the motion, and a + x that part unwound after the body has moved t seconds; then since the accelerating force is (a +x) W F = which being integrated between t = 0 and t = t", will give t" in terms of x, and therefore x in terms of t", the value required. 609. ? тр denote the ratios of the num- N = r,N 610. Let W be the weight of the given cylinder, r the radius of its base, 8 the required inclination of the plane to the horizon, and I its length, then the moving force is W sin. 0 and the mass is W + W 611. Letr be the radius of the circle; then D being the distance of the required centre from the axis, we have D= M M being the mass, and a the distance of the particle dM from the axis of gyration. But . M = 2yd.x = 2dr s 2x*dx / ma— M 2x (ne – zey? and D = M M 2 612. Generally if S denote the moment of inertia, M the 2 G VOL. II. |