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Now by the question

k =

:. L = 5, and the two points of suspension are given bg

+ ✓

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2572

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k

2

T = 2r or

2

Hence it also appears that I cannot be less than 2r.

When, moreover, L = 27, k = r, or the point of suspension is in the circumference.

588. Let x, y, be the co-ordinates of the point required, referred to the tangent at the vertex of the parabola, whose latusrectum is a;

then

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Again g being the radius-vector corresponding to x, y, it is evident that the direction of reflection will be along p; and since by the question the body must strike the vertex of the parabola, therefore the time through p with the velocity acquired before impact continued uniform, must equal the time through y by the force of gravity. Hence

2y 2016-Y

9 b being the length of the axis of the parabola.

:: p= 2 N y.(6-y) But p = y + y by the property of the parabola, :: y

= 4by

Q2

4y?

16

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a 80

a

a

86 ..y =

+

(a -86) 20

400
8b-

I
N(166* — 4ab

a”)
20

10 Hence it is evident that 166 4ab must be = or > am

1+5 or ó = or > a X

8

589. Generally let 6 be the inclination of the cycloid to

1 the horizon; then the axis being a, the semi-base is circum

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2

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a tao x

N 4+2.

4 Again, if o denote the inclination of this chord to the horizon, it may easily be shewn that

N(4 + 7"). sin. Q = a sin. O and :. sin. p =

x sin. 8.

(4 + g?) Hence the time required is

2

2. 을 x N 4+ **} g sin.

2

2

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590. Let W denote the weight of the given inclined plane, & its required elevation; then since the moving force acting I plane is

P X cos. A that parallel to the horizon is

P. sin. 0.

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x sin. 28

Hence the accelerating force is
P sin. 0 cos. 0

P
P +W 2(P+W)
and since F = max. when the velocity = max. therefore

sin. 20 = max. =1= sin.

.:: 0 = * = 45°.

591. In any system of bodies P, P', &c., acted upon by the accelerating forces, F, F, &c., at the end of the time t, let u, u &c., be the velocities impressed by these forces, or such as the bodies would have moved with at the end of the time t, had the system been free; also let v, u', &c., be their actual velocities, and 0, ', &c., the forces which really accelerate them in their general

Then since by the connexion of the parts of the system there is occasioned no loss of vis viva, we have P.q.v+P.0.0 + &c. = P. Fu + P Fu' + &c

dv

du'

&c. dt dt

courses.

But Q =

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+ P

dt

v'du'

udu

Pu'du + &c. = P. +

* &c. dt

dt dt .. Pul + P've + &c. = Pu + P'z" + &c. there being no correction since the bodies are supposed to begin to move at the same instant. This expression will serve to resolve every dynamical problem that can be proposed.

In such as relate to systems situated on the surface of the earth, or as are impressed by the force of gravity, we have

u = 2gx, u = 2gx', &c. X, a being the spaces described at the end of the time t, the spaces descended through being positive, and those ascended through negative. Hence

Py8 + P'v' + &c. = 2g (Px + PX + &c.)...... (1)

x, a' &c. being positive or negative according as the body that describes it is descending or ascending, To apply this formula in the resolution of the problem, we have

Pue + Wv = 29 (Px - Wx')

Buto = v'.ds

s and s' being the spaces described by P and W in the vertical and horizontal directions respectively in the time t. Hence

29.(Px

Wx')
W+P

ds2

ds'?

But x = 0 and ds = ds' x cos. O
when 6 is the inclination of the string to the horizon. Hence

2gPx
W + P cos.? 8

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which gives the velocity of W generated from B to C.
Hence also

ds
ข ะ บ x =cos. O xove
ds'

sin. A sin. B
P

(W

Hence when A=

, the velocity of P is o.

This problem may be generalized thus : Let two bodies P, P connected by a string passing over a fixed pulley at D, (Fig. 99,) move along the inclined planes CA, C'A'; P ascending and P descending; then having given the inclinations

Of the planes to the horizon B, B ; the distance of D from the horizon, viz. m; the altitudes of the planes, viz. a, a'; and the lengths of the parts of the string between P and D, P and D, viz. I, l', when the motion commences; required the velocities acquired v, u' in the time t, or when these lengths shall be a, a'.

First we have
a ta'=l+l.....

(2) Moreover by (1)

Pue + Pu'? = 2g (Px Px')..... (3) and s, s' being the spaces described in the time t, viz. IP, I'P, we have 0:u' :: ds : ds'

ds

ds'

ds
and Pu?. + P'v' = 2g (Px – P'x'.)

ds

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Again, it is easily seen from the figure that

X = s X sin. B 2

x = $' x sin. B' and that

s= 1-(m-a) cos." B N72-(m-a) cos." B and s' = 7"-(m-a'je cos.”B = Na-(m-a')'cos.' B' ds ada

29-m-a'' cos.

X ds' a'da'

1-(m-a) cos.'

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But a ta' = 1+

and a' increases as a decreases .. da = - da'

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