Х m Hence 12 (2+1-^)-(m-a') cos. 28 dá 1+-a 1-(m-a) cos." Also x = sin. B. {V1-(m-a) cos." B – NAP-(m-a)" cos." B} and x'=-sin.ß{78-(m-a')'cos.°8'- 12+1=2)-(m-a')'cos.°8"} Hence substituting in equat. 3 we get 2g X 18 - (m - a)* cos."B {P sin. B ( P — (m – a) cos* B – N12-(m-a)'cos.' B) + P sin.B'(7 (m – a)* cos.98 a)'cos.°8 – (1+1-2) - (m-a'y cos."B"} which gives the velocity of P'; and this being found, that of P is obtained from v'ds ds' =uX -a')B Х 1+ľa 1 - (m-a cos." From these results a multitude of particular consequences may be deduced. (1). Let a = a' = m. Then 29 x {P(-a) sin.B – P' (1-2) sin. B'} (P sin. B - P sin. ) (1+1'-a)? X (1+1'-) as it ought to be; being the common velocity of two bodies along two inclined planes of the same altitude. which gives the common velocity of two bodies when one of them, hanging vertically from the edge of a table, or any other horizontal plane, causes the other to move in a straight line along the table or plane. In this case if P = P'; then 1 or the velocity is such as would be acquired in falling freely through 1-a 2 (3.) Let a = d'em, B = * = ß. Then 2 P - P P+P which gives the velocities of two bodies hanging freely over a fired pulley, after having moved through 2 – n. B = 0. 2 Then 2gP(1-2) P x {2+-) — me cos.B} (2+1'-ay which is easily made to coincide with the expression marked (2). v2 = P't The problem may be still farther generalized, by supposing AC, A'C' (Fig. 100,) any curve lines in the same plane with the pulley. In this case we also have 29 (Px – P'sc') (3) P + P. ds2 22 ds'? s, s' being the arcs described, and the other symbols retaining their former signification. Now to resolve this case after a simple manner, let the coordinates at P, P of the curves originating in C, C' be X, Y; X', Y'. Also supposing the motions to commence at the points I, I', let the co-ordinates at those points be a, B; a', B'; and at the points A, A' a, b; a', 6. Moreover let the equations of the curves be expressed by Y=f.(X), and Y' = f* (X'); and make ID = I, ID = I PD = 1, P'D = a. This being premised, we have 1+1=1+a'. (5) (6) dX2 + DY dX2 + dY'2 dX' + d f'X') Again, a=N (m-Y)2 + (a - X) a'=(m-Y')'+ (a'-X) in=n(n-fX)2+(a-X) and 1+1'-=(m—f'X') + (d' - X') :: vim- f'X')* +(a'-—X')=l+l'- (m-fX)+(a – X)......(8) } · (7) Hence X' may be expressed in terms of X, and :. dX' and Alf'X') in terms of X. Substituting these functions of X in (6) and (7), and their resulting values in (6), we shall have v'' in terms of X and constants. Moreover v will be found from ds v=o. ds'' Ex. 1. Let the curves be two parabolas, C, C' being their vertexes, and p, p, their principal parameters; then dse Х X* = pY, X" = p'Y' X2 P 2XdX Also d. (X) = 2XX and a(foX') = р p' Hence equation (7) becomes 4X2 1+ p2 dX ds'? 4X2 X2 1+ p Х (m X. p р which being developed gives X'4 – (2m - pp'.X'– 2a'po x X + (m + a'?)p" = 1' X? P a biquadratic equation wanting its second term. This equation dX being resolved will give X'in terms of X, and therefore in X 2 ds terms of X. Hence 2 and x', and therefore u' may be ex ds'? pressed in terms of X, which will give the velocity required. From this example it is perceptible how very difficult it is to investigate the velocity practically of one body which preponderates over another when they both describe the most simple of curves. If their paths be both ellipses, or even circles in the general sense, the difficulty will be still greater, and so on for all curves whose equations are more complex. This difficulty is much diminished when the path of one of the bodies is a straight line, as in Ex. 2. Let A'C' (Fig. 100,) be a straight line whose equation is Y' = AX X = and X'= mA+a' + mA ta' A being a constant quantity, and AC any curve whose equation, as before, is expressed by X=fX. In this case equation (8) becomes v (m – A.X')2 + (d' - X')=1+1'-(m--fX)+ (a – X)" =lt1-a ::m - AX)2 + (a' -X)2 = (1 + ľ– 1). Hence X'? MA + á 2. (?+?' – 1)2 – mo – a'2 A: +1 + A: + 1 12 – 2(1+1) A% + 1 — B m? + a'2 – (1 +19 and As + 1 A% + 1 we get – 2(1 9 X + A: +1 Hence dX da a-(l+7) X (10) dX dX 1-21 + 1)) W(D+ A +1 which is a function of X alone. Now (dX)' + (dfX) ds (dX')* + A'(dX')* dX D + 18 - 211 + ) 1+ X . (11) (a – 1+)'(A' + 1) which is a function of X. Also x=fX-B, x' B 12-2(1 + ) A% + 1 2 = D dsa 19 df X |