and substituting in equation (6) we get v* in terms of X, which gives the velocity of the body moving along the straight line or inclined plane. The velocity along the curve is ds v=uX which is therefore known. As a particular case, let CA be the common parabola, its vertex being C and axis parallel to DB. Then the equation to the curve being X'=pY + (a – X) P da m X dX p_ P a X (dX) and equation (11) becomes d's X (D + 18-2(1 + ) A' + 1 p' + X p'YA' + 1)(1-1+2) which together with r, .p' would still give a very complex expression for e Let us farther particularize by making m = b, or by supposing the pulley fixed in the curre, and consequently b not <. In and (/X) ) which will also produce a very complicated result. These several computations being rather laborious than difficult, we leave to the student, to whom such exercises may be useful. The question may be still farther simplified by supposing one of the paths a straight line perpendicular to the horizon, and the other any curve whatever. ... In this case, A = tán 2 and X = 0 and a' = 0. Also equation (8) gives fX'= m - (1 + 1' – ^) d(fox) =1 da and dX'=0. Hence equation (7) becomes ds dX + (dfX) (12) ds“ das * = FX - B - + - 1 {PfX - B – P'(x - 1)} 02 = 2g (13) dae ds which give the velocities of P and P' when P is at that point of the curve whose abscissa is X. Let the curve which P describes be the common parabola whose equation is X'=py. X P X + (a - X) P Xo'. 2X P + P x 1.(4X4 + 1) {X.pm – Xo+p(a-X)*** m=b, pm = pb = a* and :. v (p+a+X") and we get P Pt 29 P.(X' – a*)-P.(1-1) P x 2.(4X + 1) (a-X) (aX + X' + p)? P.(X – 2*) – P.(a -- 1) (aX+X* +p) Let X = a; then 1 = 1 and v' = 0, as it ought from the hypothesis. Again, let X = a; then a = 0, and we have Pla' - ") + PI (po +2a2) 29 = 2gp 592. Let AD = x, AP=2, BC = AC Şa; then resolying the vertical tendency of P, viz. atIP a into directions tangential and normal to the curve, and then again the normal part into vertical and horizontal components, we get the pressure sustained by the curve in a vertical direction expressed by dx2 P = X dza a+IP a 2a+3 P. a a But since by the question za = 2ax + 2 d.x® 2ar + x2 atx 2ax + x2 atx Hence that part of P which remains unsupported by the re action of the curve is P 2ax + mono a + x or P. (a+x) = Pa = Wa, that is P is supported by W. Hence the weight P is supported in all positions. a = P. a 3 593. Let a be the height of the given cone, r the radius of its base; then since the moving force is P, and the momentum of inertia is (see Venturoli) 1 774 a = Mr? 10 10 M being its whole mass; therefore the accelerating force is Pro 10P 3M + 10P s= F = 594. Let R, r, be the radii of the wheel and axle respectively; then the moving force is qr. р R and the mass moved including the inertia p2 M R R? 2R mp2 + + 2 M and m being the quantities of matter in the rims of the wheel and axle, .. the accelerating force is – 9• F MR2 mr2 + 2R? – qrR 2R p + 29r2 + MR2 + mr and v=gFt = &c. 595. Let x be that portion of the chain which has descended in the time t; then the moving force in the direction of gravity is X + (1 x) nl+m-n. n m m d? x nl + m - dt ml 2nlr + m-nx? dt2 ml dir 1. dt = 7 ml x 2nl.x +m-n.3 ml dix 2nl x + x2) ✓ |