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diminished without disturbing the equilibrium of the fluid, the pressures at G, H......M, N, P will be equally increased or diminished.

COR. 2. If the fluid be acted on by no accelerating force, the pressures on ABC, abc must be equal; therefore pressure at F= pressure at G... pressure at N=pressure at P: or, the pressure is the same at all points in a fluid at rest acted on by no accelerating force.

6. Let the forces P, Q, R, &c. be in equilibrium when applied to pistons A, B, C, &c. fitting cylindrical apertures in the sides of a vessel filled with fluid. Let a, b, c &c. be the areas of the pistons, and suppose the fluid to be acted on by no accelerating force. Then since the fluid is at rest, the pressures on an unit of the surface of each of the pistons must be equal,

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7. Let the fluid be incompressible; p, q, r, &c. the distances of the pistons A, B, C, &c. from fixed points in the axes of the cylinders in which they play; p + Sp, q + Sq, r + dr, &c. their distances from the same points after they have been moved in any manner. Then since the volume of the fluid in the vessel remains the same,

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Sp, dq, dr, &c. are the virtual velocities of the pistons

A, B, C, &c. to which the forces P, Q, R, &c. are applied.

SECTION II.

ON THE EQUILIBRIUM OF NON-ELASTIC FLUIDS ACTED ON BY GRAVITY.

ART. 8. THE specific gravity of a body is the weight of an unit of its volume.

9. The density of a body is the quantity of matter in an unit of its volume.

10. Let W, M, V be the number of units of weight, mass, and volume contained in the weight, mass, and volume of a given body, S its specific gravity, D its density, g the force of gravity; then

S=weight of one unit=gD

M= mass of V units= DV

W = weight of V units = SV=gDV.

11. When a fluid acted on by gravity is at rest, the pressures are equal at all points in the same horizontal plane.

Let P, Q (fig. 4.) be any two points in the same horizontal plane in the interior of a mass of fluid at rest. The equilibrium of the fluid will not be disturbed if we enclose a part of it in a tube PMQ of uniform bore, having its branches MP, MQ symmetrical with respect to a vertical line. Then since the columns MP, MQ are symmetrical, and similarly situated with respect to the direction of gravity, they will balance when the pressures at P and Q are equal. But if the pressures at P and Q be unequal, the fluid will begin to move towards that end at which the pressure is the least, and the equilibrium will be destroyed; therefore in order that the fluid may be at rest, the pressures at P and Q must be equal.

12. To find the pressure at any point in a mass of fluid

at rest.

Let the vertical prism AEF (fig. 5.) be a portion of a fluid at rest; then (Art. 4.) the equilibrium of AEF will not be disturbed if we suppose it to become solid. Now AEF is kept at rest by its own weight, and the pressures on its sides and ends. The pressures upon its sides act in a horizontal plane; its weight, and the pressures upon its ends, act in a vertical line; therefore the latter must be capable of maintaining equilibrium separately; therefore pressure on DEF = weight of prism of fluid AEF+ pressure on ABC. Let m, p be the pressures at A, D respectively, p the density of the fluid, ABC =K, AD=8; then the pressure on ABC=mê, the pressure on DEF=pк, and the weight of AEF=gp≈к; therefore pk=gрk+mk, therefore p=gpz+m.

COR. 1. Let A be a point in the open surface of the fluid, then m=0, and p=gpx.

COR. 2. Since fluids press equally in all directions, and the pressure is the same at all points in the same horizontal plane, the pressure on a small area of any plane is ultimately equal to the pressure on an equal area of the horizontal plane that

intersects it.

13. The surface of a fluid at rest is a horizontal plane.

Let A, P (fig. 6.) be any two points in the surface of a fluid at rest, AB, PQ vertical straight lines intersected by a horizontal plane in B, Q; p the density of the fluid. Then, (Arts. 11, 12.) gp. PQ = pressure at Q = pressure at B=gp. AB; therefore PQ=AB, therefore A and P are in the same horizontal plane.

14. The common surface of two fluids that do not mix is a horizontal plane.

Let A, P (fig. 7.) be any two points in the common surface of two fluids that do not mix; BAC, QPR vertical straight lines intersected by horizontal planes in B, Q, and in C, R; p, σ the densities of the upper and under fluids respectively. Then, (Art. 12.) pressure at A— pressure at B=gp. AB,

also

pressure at C-pressure at Ago. AC,

.. pressure at C-pressure at B=g. (p. AB+o. AC)

in like manner

present a fews at R

CA 22.

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pressure at R-pressure at Q=g (p. PQ+o. PR), and, (Art. 11.) pres. at Q=pres. at B, pres. at R=pres. at C, .. p. PQ+σ. PR=p. AB+σ. AC,

and σ. PQ+o. PR=σ. AB+σ. AC

•. (σ—p). PQ = (o − p) . AB,

.. PQ=AB .. A and P are in the same horizontal plane.

COR. Hence the surface of stagnant water exposed to the atmosphere is a horizontal plane.

15. If two fluids that do not mix, meet in a bent tube, the altitudes of their surfaces above the horizontal plane in which they meet, are inversely as their densities.

Let PAQ (fig. 8.) be a bent tube containing two fluids of different densities; AP, AQ the portions of the tube occupied by the lighter and heavier fluids; p, σ the densities of the fluids in AP, AQ. Let the planes of the surfaces of the fluids, and the plane in which they meet cut a vertical in H, K, C.

The pressure of the fluid in AP at A=gp. HC,

and the pressure of the fluid in AQ at A=go. KC. When the fluids are in equilibrium these pressures must be equal

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COR. Let II be the pressure of the atmosphere at the surface of each fluid.

Then

pressure of fluid in AP at A=II+gp.HC,
pressure of fluid in AQ at A=Пl+go. KC,

and as before these pressures must be equal;

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16.

σ HC

To find the pressure of a fluid on any surface.

Let BPC (fig. 9.) be the given surface. Draw AK vertical cutting the surface of the fluid in A, through H, K draw horizontal planes cutting the surface BPC in the curves PM, QN.

Let P be the pressure on MPB, S the area of MPB, p the density of the fluid, X the depth of the centre of gravity of BPC below the surface of the fluid, AH=x, HK-8a, therefore. ultimately pressure on MQ=d, P. Sa, area MQ = d ̧S. Sx. But, ultimately,

=

pressure on MQ=gp.AH.MQ=gp.x.d2S.8x,

.. d„P=gp.x.d ̧S .. P=gp. fx. dr S,

and the pressure on the whole surface BPC=gp. fx.d2S, the integral being taken between the limits corresponding to the highest and lowest points in the surface.

But X. (area BPC)=x. d,S between the same limits;

... pressure on BPC=gpX. (area BPC);

or, the pressure of a fluid on any surface is equal to the weight of a column of the fluid whose base is equal to the area of the surface, and altitude equal to the depth of the centre of gravity of the surface below the surface of the fluid.

COR. When the surface BPC is a plane, the pressures are all perpendicular to BPC, and consequently parallel to each other; therefore the resultant of the pressure on PBC is equal to the whole pressure, and acts in a direction perpendicular to BPC.

17. The centre of pressure of a plane surface immersed in a fluid is the point in which the resultant of the pressure of the fluid meets the surface.

To find the centre of pressure of any plane surface.

Let ABC (fig. 10.) be the surface, OY the line in which its plane cuts the surface of the fluid. From O draw OX in the plane ABC perpendicular to OY, and let X, Y be the co-ordinates of the centre of pressure referred to the axes OX, OY.

Then since the pressures are parallel to each other, we shall have, (Whewell's Mechanics Art. 84.)

X.(pressure on ABC) = moment of pressure on ABC round OY, Y. (pressure on ABC) = moment of pressure on ABC round OX, Draw MP, NQ parallel to OX; HP, KQ parallel to OY; PT perpendicular to the surface of the fluid meeting it in T.

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