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and x'=—sin.ß′ {√ l'a — (m—a')3cos.3ß'— √ (l+l'-x)2-(m—a')'cos.''} Hence substituting in equat. 3 we get

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P' sin.'(√

√ λ2 — (m—a)2cos.' B) +

(m a')3 cos.3ß′ — √↓ (1+l'—x)3 — (m—a')' cos.''} which gives the velocity of P'; and this being found, that of P is obtained from

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as it ought to be; being the common velocity of two bodies along two inclined planes of the same altitude.

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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

or the velocity is such as would be acquired in falling freely through

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which gives the velocities of two bodies hanging freely over a fixed pulley, after having moved through I-x.

T

(4). Let a = m, a' = 0, B =
B = = = =, B ==
6' 0.

2

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which is easily made to coincide with the expression marked (a).

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

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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 1, I', let the co-ordinates at those points be

a, B; a', B';

and at the points A, A'

a, b; a', b'.

Moreover let the equations of the curves be expressed by
Y =ƒ. (X), and Y' = ƒ' (X');

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(5)

(6)

(7)

:: √(m−ƒ'X')2+(a'—X')2=1+l' — √ (m·~ƒX)2+(a — X)a ..... (8)

Hence X' may be expressed in terms of X, and .. dX' and d(ƒ'X') in terms of X. Substituting these functions of X in (6) and (7), and their resulting values in (b), we shall have vo in terms of X and constants. Moreover v will be found from

ds v = v. ds'

Ex. 1. Let the curves be two parabolas, C, C' being their ver texes, and p, p', their principal parameters; then

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2

'+ (a' — X')2=1+1 — √ (m— X2)*+ (a− X')2 — — √√(m.

which being developed gives

X'4

Р

(2m-p')p'.X' -'2a'p'2 × X' + (m2 + a2)p'2 =

p' . {l+l'— √(m- X2
X2 ) ' + (a - x) · }

Р

X)

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a biquadratic equation wanting its second term. This equation

being resolved will give X' in terms of X, and therefore

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dX'

in

dX

x and x', and therefore v may be ex

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 equa

tion is

Y' = AX'

A being a constant quantity, and AC any curve whose equation, as before, is expressed by

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In this case equation (s) becomes

√↓ (m − A.X')2 + (a' − X')2 = 1 + 1' — √√ (m −ƒX)2 + (a - X)2

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•. (m − AX')2 + (a' — X')2 = (1 + l'′ − a)o.

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(mA+a')e m2 + a22 — (1 + 1)2 +
+
(A2+1)2

√ {(mA+a)

and by way of abridgment if we put

A2+1

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x = ƒX - ß,

1 + (dƒX ) 3

(λ − 7+7')2(A2 + 1)

P

and x' = f′ — AB − A√ (D + 12 = 2(1 + ?^^)

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. (11)

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