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720. Let A be the area of the orifice of the cylinder, h its 'altitude, and r the radius of its base; then if the water descend through h - x during the time t, the moving force at the end of that time which causes pressure and an increase of velocity at the orifice is

#ro (h 2) and the upward accelerating force is :: Tor? (h x)

h-X 8

g P + 7r.22

h+3

Hence the force which actually accelerates each particle of the fluid downwards through the orifice is

h

h F =g+g

29 h + x

htu

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2.

Hence the velocity at the orifice is due to the altitude

hex
h fox

ich and is :: 219

ht it

Hence the velocity at the descending surface is

2A

√gh x ✓

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h to

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2A Ngh

h Nhx+x2 + x hxx? 1.

+ C} 2A Ngh

✓ hata2 and which being taken between x = 0 and x = h gives

ap?

x {h 2 + h l. (V 2 + 1)}
24 gh.
Trh

. { 2 + 1. (1 + 2)}

2AN & for the whole time of emptying the cylinder.

T

(1)

Again, the force which accelerates the descent of P is

h-X

dos 9

ht-3 dt2 where ds is the element of space described by P in the time dt. Hence (see eq. a) d's

h-2 'dix? 4Ah

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Finite !

vel. of P

0

= 0, when vel. of descending surface x = h) ::

C=h-h.l. (h)
ds
7284

(h - 1. (h) - x + h . lx)
dic

4Ah

and integrating again, we get

{h-(1 – 1.K)x x*.+ hal.x – hä}

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for the space described by P during the discharge of the water.

721. Let o be the given inclination of the tube to the horizon, a the altitude of the orifice of the vessel, B the observed distance at which the water strikes the plane from the vertical line passing through the orifice; then since the water describes a parabola, whose equation is (see p. 253,)

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where y and x are the vertical and horizontal co-ordinates originating at the orifice, and v the velocity of the issuing fluid.

Hence

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But if h be the required altitude of the fluid above the orifice, we have

2gh whence by substitution, &c.

BP

1
Х
4 cos.0

a + B tan.

Х

722. Let p be the parameter of the generating parabola, B the radius of the base of the paraboloid, y that of the required orifice, a the given altitude of the water above the vertex, and s the given space descended through in the given time T. Then at any variable altitude x y above the orifice, the velocity at the

р orifice is

29.(

r

P and that of the descending surface is my?

2g (2 прс

р Hence p

xdx yN 29 x

р x (x +

y? . 3y 2g

P

р Hence

2p
x (a

y?
3yo N 29

P

р 2y?

Xa р

P

) رود

y?

Х

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2p

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T =

which being reduced and resolved, will give the required value of y.

723. Let ß be the required altitude of the hole, a the altitude of the cone ; then since the equation to the parabola described is (see p. 253,)

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But v = 2g. (a – B)

Hence by the solution of a quadratic equation, Band :: the required position may be found.

Let x

724.

be
any

variable altitude of the water, A the area of the orifice, « the length of the axis, and ß the extreme ordinate ; then since the area of the descending surface is

my* the velocity of this surface is

A

х N 29

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Hence

'a’x dx

X

A 29 an integral which can be taken only between certain limits, as between x = 0, and x = 00. See Ihewell's Dynamics, p. 15.

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725. Let za be the altitude of the water when it is in equilibro, 2b the distance between the axes of the two legs; then the centre of gravity of the water in the legs in the state of equi

librium is the middle point of the line joining the bisections of the water in the legs. Make this point the origin of the vertical and horizontal co-ordinates y, x; then supposing the water to have descended down one of the legs, it will rise as much in the other, and the line which joins the bisections of the axes of the fluid in the legs, being divided in the inverse proportion of the altitudes of the water, will give the centre of gravity corresponding to this position. Hence if m denote the distance to which the centre of gravity of the water has been depressed in one leg and elevated in the other, it is easily shewn that

y : x :: m : 6 we also easily get

V mi + 6 x m = x x + yo

+ b)
boy?

= x2 + ya

:: (6. y

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i. by = 2 or the required curve is the common parabola, whose principal parameter is 6.

726. Let y = pr" be assumed as the equation to the required parabola. Also let a be the common altitude of the paraboloid and cylinder.

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