it would yield in that direction, and motion would ensue, which is contrary to the hypothesis.

This is called the principle of equal pressures.

It follows, from the principle of equal pressures, that if any point of a fluid in equilibrium, be pressed by any force, that pressure will be transmitted without change of intensity to every other point of the fluid mass.

This may be illustrated experimentally, as follows:

P

D

Let AB represent a vessel filled with a fluid in equilibrium. Let C and D represent two openings, furnished with tightly-fitting pistons. Suppose that forces are applied to the pistons just sufficient to maintain the fluid mass in equilibrium. If, now, any additional force be applied to the piston P, the piston Q will be forced outwards; and in order to prevent this, and restore the equili

Fig. 134.

brium, it will be found necessary to apply a force to the piston Q, which shall have the same ratio to the force applied at P that the area of the piston has to the area of the piston P. This principle will be found to hold true, whatever may be the sizes of the two pistons, or in whatever portions of the surface they may be inserted. If the area of P be taken as a unit, then will the pressure upon Q be equal to the pressure on P, multiplied by the area of Q.

The pressure transmitted through a fluid in equilibrium, to the surface of the containing vessel, is normal to that surface; for if it were not, we might resolve it into two components, one normal to the surface, and the other tangential; the effect of the former would be destroyed by the resistance of the vessel, whilst the latter would impart motion to the fluid, which is contrary to the supposition of equilibrium.

In like manner, it may be shown, that the resultant of all the pressures, acting at any point of the free surface of a fluid, is normal to the surface at that point. When the only force acting is the force of gravity, the surface is level. For

small areas, a level surface coincides sensibly with a horizontal plane. For larger areas, as lakes and oceans, a level surface coincides with the general surface of the earth. Were the earth at rest, the level surface of lakes and oceans would be spherical; but, on account of the centrifugal force arising from the rotation of the earth, it is sensibly an ellipsoidal surface, whose axis of revolution is the axis of the earth.

Pressure due to Weight.

155. If an incompressible fluid be in a state of equilibrium, the pressure at any point of the mass arising from the weight of the fluid, is proportional to the depth of the point below the free surface.

Take an infinitely small surface, supposed horizontal, and conceive it to be the base of a vertical prism whose altitude is equal to its distance below the free surface. Conceive this filament to be divided by horizontal planes into infinitely small, or elementary prisms. It is evident, from the principle of equal pressures, that the pressure upon the lower face of any one of these elementary prisms is greater than that upon its upper face, by the weight of the element, whilst the lateral pressures are such as to counteract each other's effects. The pressure upon the lower face of the first prism, counting from the top, is, then, just equal to its weight; that upon the lower face of the second is equal to the weight of the first, plus the weight of the second, and so on to the bottom. Hence, the pressure upon the assumed surface is equal to the weight of the entire column of fluid above it. Had the assumed elementary surface been oblique to the horizon, or perpendicular to it, and at the same depth as before, the pressure upon it would have been the same, from the principle of equal pressures. We have, therefore, the following law:

The pressure upon any elementary portion of the surface of a vessel containing a heavy fluid is equal to the weight of a prism of the fluid whose base is equal to that surface,

and whose altitude is equal to its depth below the free surface.

Denoting the area of the elementary surface, by s, its depth below the free surface, by z, the weight of a unit of the volume of the fluid, by w, and the pressure, by p, we shall have,

We have seen that the pressure upon any element of a surface is normal to the surface. Denote the angle which this normal makes with the vertical, estimated from above, downwards, by, and resolve the pressure into two components, one vertical and the other horizontal, denoting the vertical component by p', we shall have,

ន

Fig. 135.

But scoso is equal to the horizontal projection of the elementary surface s, or, in other words, it is equal to a horizontal section of a vertical prism, of which that surface is the base. Hence, the vertical component of the pressure on any element of the surface is equal to the weight of a column of the fluid, whose base is equal to the horizontal projection of the element, and whose altitude is equal to the distance of the element from the upper surface of the fluid.

The distance z has been estimated as positive from the surface of the fluid downwards. If < 90°, we have cosp positive; hence, p' will be positive, which shows that the vertical pressure is exerted downwards. If o > 90°, we have coso negative; hence, p' is negative, which shows that the vertical pressure is exerted upwards (see Fig. 135).

Suppose the interior surface of a vessel containing a heavy fluid to be divided into elementary portions, whose areas are denoted by s, s', s'', &c.; denote the distances of these

elements below the upper surface, by z, z', z', &c. From the principle just demonstrated, the pressures upon these surfaces will be denoted by wsz, ws'z', ws''z", &c., and the entire pressure upon the interior of the vessel will be equal to,

w (sz + s'z' + s'z' + &c.); or, 20 Χ Σ(82).

Let Z denote the depth of a column of the fluid, whose base is equal to the entire surface pressed, and whose weight is equal to the entire pressure, then will this pressure be equal to (s + s' + s" + &c.) Z; or, wZ. Es. Equating these values, we have,

w.Σ(sz)

= wZ. 2(s), .'. Z =

Σ(sz)
Σ(8)

(143.)

The second member of (143), (Art. 51), expresses the distance of the centre of gravity of the surface pressed, below the free surface of the fluid. Hence,

The entire pressure of a heavy fluid upon the interior of the containing vessel, is equal to the weight of a volume of the fluid, whose base is equal to the area of the surface pressed, and whose altitude is equal to the distance of the centre of gravity of the surface from the free surface of the fluid.

EXAMPLES.

1. A hollow sphere is filled with a liquid. How does the entire pressure, on the interior surface, compare with the weight of the liquid?

SOLUTION.

Denote the radius of the interior surface of the sphere, by r, and the weight of a unit of volume of the liquid, by w. The entire surface pressed is measured by 4r2; and, since the centre of gravity of the surface pressed is at a distance below the surface of the liquid, the entire pres

sure on the interior surface will be measured by, the expression,

Hence, the entire pressure is equal to three times the weight of the liquid.

2. A hollow cylinder, with a circular base, is filled with a liquid. How does the pressure on the interior surface compare with the weight of the liquid?

SOLUTION.

Denote the radius of the base of the cylinder, by r, and the altitude, by h. The centre of gravity of the lateral surface is at a distance below the upper surface of the fluid equal to th. If we denote the weight of the unit of volume of the liquid, by w, we shall have, for the entire pressure on the interior surface,

If we suppose hr, the pressure will be twice the weight.

If we suppose r = 2h, we shall have the pressure equal to of the weight.

If we suppose h = 2r, the pressure will be equal to three times the weight, and so on.