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

and σ. PQ+o.PR=σ.AB+σ.AC

.. (σ—p). PQ = (σ− 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;

16.

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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. Sx, area MQ = d2S. Sx. But, ultimately,

pressure on MQ=gp. AH.MQ=gp.x.d.S.dx, .. d. P-gp.x.d, S. .. P=gp. ↓x.d2S, and the pressure on the whole surface BPC=gp. fx.dS, 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.

Let OH=x, HK,=dx, OM=y, MN=dy, TMP=0, the density of the fluid= Therefore we have ultimately = p. pressure on PQ=gp. PT.PQ=gp. sin 0.x. dx.dy; moment of the pressure on PQ round OY

=gp.MP.PT. PQ=gp.sin 0. x2. Sx.dy;

moment of pressure on PQ round OX

=gp. HP. PT.PQgp.sin 0.xy. Sx.dy;

... pressure on ABC=gp.sin 0. Sy x ;

moment of the pressure on ABC round OY=gp.sin 0 . sa syæ2 ; moment of the pressure on ABC round OX=gp. sin 0.fx sxy; the integrals being taken between the limits corresponding to the boundary of the surface.

.. X. Sx Sy∞ = fx Sy∞2, Y. Sa Sy∞ = Sx Sy xy.

COR. 1. A physical plane ABC, one side of which is exposed to the pressure of a fluid, may be kept at rest by a single force equal and opposite to the pressure of the fluid applied at its centre of pressure.

COR. 2. If ABC were a plane lamina of very small uniform thickness, moveable round the axis OY, the values of X and Y would be those of the co-ordinates of its centre of percussion.

18. To find the vertical pressure of a fluid on any surface.

Let ABR (fig. 11.) be a vertical cylindrical column of fluid in a mass of fluid at rest, meeting the surface of the fluid in ABC, and the given surface in PQR. The equilibrium of ABR will not be disturbed if we suppose it to become solid. Then since the vertical pressure on PQR, and the weight of ABR are the only forces that act vertically on ABR, the vertical pressure on PQR is equal to the weight of ABR, and acts in a vertical through the centre of gravity of ABR.

The vertical pressure upwards on PQR when it forms the under surface of the solid APQ, is equal to the vertical pressure downwards on PQR when it forms the upper surface of the solid PQD.

For the pressures on any portion of PQR are the same in either case, and they act in the same line but in opposite directions; therefore the vertical pressures are equal and act in opposite directions. Consequently the whole vertical pressures on PQR are equal and act in opposite directions.

Hence the vertical pressure of a fluid on any portion of the interior of the vessel in which it is contained, is equal to the weight of the superincumbent column of fluid.

19. To find that part of the pressure of a fluid on any surface which acts in a direction perpendicular to a given vertical plane.

Let ABR (fig. 12.) be a cylindrical column of fluid in a mass of fluid at rest, perpendicular to the given plane. Let it meet the given surface in PQR, and the vertical plane in ABC. Suppose ABR to become solid. Then since the resolved part of the pressure on PQR perpendicular to ABC, and the pressure on ABC are the only forces that act on ABR in a direction perpendicular to ABC, the resolved part of the pressure on PQR perpendicular to ABC is equal to the pressure on ABC, and acts in a line passing through the centre of pressure of ABC.

20. When the fluid sustains a pressure arising from the weight or elasticity of a lighter fluid resting upon its surface, we must suppose the pressure of the lighter fluid removed, and the depth of the heavier increased, so that the pressure at any given point beneath its original surface may remain unaltered. The amount of the pressure of the fluid on any surface estimated in a given direction, and the line in which its resultant acts may then be determined as in the preceding Articles.

21.

To find the resultant of the pressure surface of a solid immersed in it.

of a fluid on the

Since any portion V of a fluid at rest may become solid without disturbing the equilibrium of the fluid, the resultant of the pressure of the fluid on the surface of V after it has become solid, must be equal and opposite to the weight of V. But the fluid will exert the same pressure on the surface of any other solid of the same form as V, and occupying its place. And

B

the weight of

of gravity.

acts downwards in a vertical through its centre Hence the resultant of the pressure of a fluid on the surface of a solid immersed in it is equal to the weight of the fluid displaced, and acts upwards in a vertical through the centre of gravity of the fluid displaced.

The solid may be either wholly or partly immersed, and the fluid of uniform or variable density.

Hence the resultant of the pressure of a fluid on the interior of the vessel in which it is contained is equal to the weight of the fluid and acts downwards in a vertical through its centre of gravity.

22. To find the conditions of equilibrium of a solid suspended in a fluid by a string.

Let GN, FM (fig. 13.) be verticals through the centres of gravity of the solid, and of the fluid displaced by it, EL the direction of the string by which the solid is suspended, T the tension of the string, W the weight of the solid, V the volume of the fluid displaced, p its density, and therefore gpV the weight of the fluid displaced, or (Art. 21.) the resultant of the pressure of the fluid on the solid. Now W acts downwards in NG, gp V acts upwards in FM; hence in order that the solid may be kept at rest by T acting in EL, EL must be vertical, and in the same plane with FM, GN; T=W-gp V acting upwards, or gp V - W acting downwards, according as W is greater or less than gp V; and if EGF be drawn perpendicular to GN in the plane GFN, W.GE=gpV.FE.

COR. 1. W acting downwards in NG, and gpV acting upwards in FM may be resolved into a single force W-gp V acting downwards in NG, and a "couple" gp V. FG in the plane MGF tending to make the solid revolve in the direction GFM; hence if any forces acting on the solid, can be resolved into a single force W-gp V acting upwards in GN, and a "couple" gp V. FG in the plane MGF tending to make the solid revolve in the direction MFG, they will keep it at rest.

COR. 2. When a solid floats in equilibrium, it is kept at rest by its own weight acting downwards in a vertical through

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