605. Generally let R, R', &c., be the radii of the wheels, r, r', &c., those of the axles; then x denoting the weight to be raised each time, and P the power, the moving force of P is P+ x + wd2 + w'd'R +w"d"RR" + &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 = P.RR'. P.R R2x+wd2+wd'2R2+&c. and the force which accelerates the ascent of x is .. PM-xm x RR'R" &c. PM+x+wd2+w'd2R2+ &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 = min. D being substituted for wds + w'd'2R2 + &c. 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' 606. Let either weight be denoted by P and the additional one by p, and put P+pQ, also let / 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 + 1 .. the accelerating force is Let ba=0. Then the velocity resulting from the actions of P, Q alone is v'= √ p2 Q √(Q−P)x + x2 and therefore that which is due to the weight of the string is 607. The distance required (d) is that of the centre of oscillation from the axis of suspension, that is fx2dM Let x = a, the altitude of the triangle; then 608. Let r be the radius of the base of the paraboloid, W its weight, and 7 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)r (a+x)r2+ W r2 F = r2 which being integrated between t = 0 and t = t", will give t“ in terms of r, and therefore x in terms of t", the value required. 609. Let T12 ....... Τρ denote the ratios of the num 2.... ber of teeth in the pairs of wheels and pinions, and N1, N, N, their respective numbers of revolutions in any time t; then 610. &c. .... TnX m. Let W be the weight of the given cylinder, r the radius of its base, the required inclination of the plane to the horizon, and its length, then the moving force is 611. Let r be the radius of the circle; then D being the distance of the required centre from the axis, we have M being the mass, and x the distance of the particle dM from the axis of gyration. But 612. VOL. II. Generally if S denote the moment of inertia, M the 2 G |