38

LEVEL SURFACES OF EQUAL DENSITY.

s.c. is opened, and the equilibrium is therefore stable; vice versa if w'-w is negative.

Various experiments have been devised for illustrating the instability of the equilibrium of two liquids, when the upper liquid has the greater density; for instance, by taking a tumbler full of water, closed by a card, and inverting it over a tumbler full of wine, so as to fit accurately; when the card is removed or slightly displaced so as to allow a communication between the two liquids, the wine will gradually rise into the upper tumbler and displace the denser water; the card may even be replaced by a handkerchief or a piece of gauze.

Ice again is less dense than water and rests in stable equilibrium on the surface; but if water is run on the surface of the ice, the horizontal plane form becomes unstable, and the ice becomes bent into waves.

By the application of heat variations of density are produced in a liquid which destroy the equilibrium, and set up convective currents; as seen exemplified in the Gulf Stream, and in boilers and kettles; also in the winds, and particularly the Trade Winds.

If however the heat is propagated uniformly in a downward vertical direction, the alteration of density does not interfere with the stable equilibrium of the liquid, provided the liquid expands with a rise of temperature, as is generally the case.

When the density of a fluid varies continually, the above arguments show that the fluid comes to rest under gravity so that the density increases in the downward vertical direction (exemplified in the air by the denser layers of fog), and so that

"the surfaces of equal density are horizontal planes."

GENERAL EXERCISES.

GENERAL EXERCISES ON CHAPTER I.

39

(1) Define a fluid, a viscous fluid, and explain how viscosity is measured.

Prove that in a perfect fluid the pressure is the same in all directions about a point, but in a viscous fluid only when the fluid is at rest.

Show how to distinguish between a soft solid and a very viscous fluid, and give examples of each.

From what property of a fluid does it follow that any portion of it may be considered solid. (2) Show that a solid whose faces are portions of spheres is the only one possessing the property that, if immersed in any fluid whatever, the resultant pressure on each face reduces to a single force.

(3) A hollow cone, whose axis is vertical and base downwards, is filled with equal volumes of two liquids, whose densities are in the ratio of 3:1; prove that the pressure at a point in the base is (3-3/4) times as great as when the vessel is filled with the lighter liquid.

(4) A vertical right circular cylinder contains portions of any number of fluids that do not mix; show that in equilibrium the fluids arrange themselves in horizontal strata and that the density cannot diminish on descending into the fluid.

Further, show that of all arrangements of the fluids in horizontal strata, geometrically possible, the one actually taken by nature corresponds to the minimum average pressure on the surface of the containing vessel.

40

GENERAL EXERCISES.

(5) Water is poured into a vertical cylinder, whose , weight is equal to that of the water which it will contain, and whose centre of gravity is at the middle point of its axis.

Find the position of the centre of gravity of the cylinder and water when the water has risen to a given height within it; show that the whole distance traversed by the centre of gravity while the cylinder is being filled is to the height of the cylinder as 3-2/2:1; and that when in its lowest position it is in the surface of the water.

If mercury is poured into the water, find when the C.G. of the water and mercury is in its lowest position.

(6) A large metallic shell which is spherical and of small uniform thickness is quite full of water.

A small circular part of the shell is cut out at some distance below the top of the sphere, and provided with a hinge at the highest point of the aperture.

Given W and W' the weights of the shell and the water contained in it; prove that the water will not escape, unless the centre of the aperture and the top of the shell subtends at the centre of the shell an angle greater than

CHAPTER II.

HYDROSTATIC THRUST.

27. Let us apply immediately the mechanical Axiom or Corollary of § 5, derived from the Definition of a Fluid, "The Stress on any plane in a Fluid at rest is a Normal Pressure," to the solution of an important Hydrostatical Question, the determination of the

Thrust of Water against a Reservoir Wall.

First, suppose the wall of the reservoir a masonry dam with a vertical face, AB representing the elevation of the face in a plane perpendicular to its length (fig. 20); and draw any plane BC through B the foot of the face AB to meet in C the surface of the water AC (which we have shown to be a horizontal plane).

Consider the equilibrium of the water in ABC; and, merely to fix the ideas, it is convenient to suppose this water solidified or frozen; we shall often make use of this supposition hereafter.

The forces maintaining the equilibrium of ABC are the force W of its weight acting vertically downwards, the thrust P of the wall acting horizontally, and the thrust R of the water on the plane BC, acting perpendicular to BC.

42

HYDROSTATIC THRUST

Then, denoting the angle ABC by 0, and resolving parallel to the plane BC,

P sin 0 W cos 0, or P= W cot 0.

=

We notice that if 0=45°, then P= W, so that the thrust P on the wall is given by the weight of the isosceles prism ABC.

Now if AB=h, the depth of the water in feet, then AC=h tan 0; and if I denotes the length of the wall, and w the weight in lb of a cubic foot of water or the liquid, then W = wlh2 tan 0,

an expression independent of 0, as it should be.

The average pressure on AB is P/hl=wh, the pressure in lb/ft2 at depth 1h feet.

But the pressure at any point of AB being proportional to the depth, as represented by the ordinate of the straight line Ab, the thrust P on AB is not uniformly distributed over AB; but the distribution of thrust P, as represented by the pressure p, is uniformly varied, as represented by the ordinate of Ab; the thrust P being represented by the area A Bb.

or

Resolving vertically to determine R,

R sin 0 W = {wlh2 tan 0,

=

R={wlh2 sec 0;

also if the vertical through G, the centre of gravity of ABC, meets BC in K, then K will be the point of application of the resultant thrust R; and CK=3CB.

Therefore also H is the point of application of the thrust P on AB, when HK is horizontal, and therefore AH=AB; the three forces P, R, W which maintain the equilibrium of ABC meeting in K.