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Hence the attraction of the shell is

udu = S {du

{2rx - x* £2r xu}

udu
✓2rx – 2* 2(r - 1)u
2
rx =

(r
13 3

(r x) see Hirsch's Tables, p. 94. Let u=0. Then a = 0 and C=

W 2rx - 22 -7x)

(r-x) and if the ordinate corresponding to x be y, we get {y}, " – 4 tỷ

{ys † 2(r— x)u } a=U

yi

3(1-x)" 3(r – x)2• Again, let u = 2r

2r

+ a=2r-r 2r 7. r + 2x}

3(r - x)2 3(r — x)2 Let u = x. Then

+

X; then

1

Х

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x®)

9

5

2
a = 2

+
3(r - x) 3(r — x)2

4 73 + rex 3rr + 2
:. F =a á = 2r 2x

(r – x)3 which is the force which accelerates the body at any point of the descent. Hence by meaus of the formula

dir vdv = Fdx, and dt =

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the velocity at that point may be found, and also the time in reaching it.

640. Owing to the diurnal rotation of the earth, the body will have a centrifugal force in the direction of that I to the axis which passes through the top of the tower; this will consequently counteract, in some degree, the effect of gravity, and cause the body, during its descent, continually to recede from the tower in the meridian towards the equator. So that in northern latitudes

the body will strike the ground southward of the tower, and vice versá. For a full discussion of this subject see Laplaca Bulletin de Sciences, No. 75. Also see Emerson's Algebra, prob. 198.

641. Let y be the radius of any section parallel to the plane passing through the centre, x the distance of its centre from the centre of the sphere, and y' the radius of any circle concentric with the former ; then since the attraction to the centre of any particle within a sphere o distance, that attraction will be duly measured by the distance, and when resolved into two directions, one I to, and the other parallel to the plane, we have the pressure of this particle upon the plane, measured by x. Hence the pressure of the whole circle whose radius is y is

2x sy'dy taken between y' = 0, and y'=y; or

xy* = **. (78 – 2*) r being the radius of the earth.

Again, the pressure of the whole hemisphere is

# S xdx (p2 – x2) taken between x = 0, and x = r; that is

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But since the density of the earth is supposed uniform, and consequently its weight at the surface proportional to its magnitude x r, that weight is measured by

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HYDROSTATICS.

3

642. The pressure against the curve is measured by the product of the part pressed, and the depth of its centre of gravity in the fluid, (Vince, Prop. VII.)

Let be the abscissa required, and y the corresponding ordinate, and a, b, the whole abscissa and corresponding ordinate of the parabola ; then since the areas of the lower and upper parts of the parabola are respectively

y.

4
and (as – yr)

3 and the distances of their centres of gravity from the vertex are Syxdr

Szédx Jyder fæddx integrated between x = 0 and x, and x = x and a respectively; or they are 3

3 5 :: the depths of these centres in the fluid are

X, and a

(6 – 2)

-X, and.

3

5

5

Hence

y2 ( - ): (+8–y:)~ (++) :: * :

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3

282 - 3y :: y3 x (B2 y): (B3- y) x

:in: m

5 which gives the equation B2 5m + 2n

2023 Х

y? +

= 0. mt n

3

m + n 3(m + n) whereby to determine y and x.

nB3

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643. Let x be the depth of the required section, and a the whole depth or length of the cylinder; then the depths of the centres of gravity of the upper and lower portions are

,
and x +

am

a-X

Also the surfaces pressed are as

x and a - X
:. by Vince, Prop. VII., and the question, we have

x+a
X 2 : x (a - x) :: 1:1

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2

2

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644. Let x be the depth of the heavier fluid, a - x that of the lighter; and suppose y the depth of that column of the heavier fluid which has the same weight as the column of the lighter fluid ; then since the pressure against the surface o surface pressed x depth of centre of gravity of the surface x the specific gravity of the fluid, we have

P:P :: (a – 2)" X1: X X + P and P being the pressures on the upper and lower surfaces. Hence, by the question

(a − x) = x. (x + 2y) n

a-T

2

But since the weight of a fluid cc volume x specific gravity,

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645. Let . be that part of the axis a which is immersed, and suppose r the radius of the base of the cone; then since the base of the part immersed is

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646.

Let a be the axis of the parabola, « the depth of the required ordinate y; then since the pressure a surface pressed x depth of its centre of gravity x density of the fluid, we have

2y X X X X = max.

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