axis. And in one such plane, which we may imagine carried round the axis in the direction of the force, the fluid pressure will increase in simple proportion to the angle at a rate per unit angle (§ 55) equal to the product of the density at unit distance into the force at unit distance. Hence it must be remarked, that if any closed line (or circuit) can be drawn round the axis, without leaving the fluid, there cannot be equilibrium without a firm partition cutting every such circuit, and maintaining the difference of pressures on the two sides of it, corresponding to the angle 27. Thus, if the

D

axis pass through the fluid in any part, there must be a partition extending from this part of the axis continuously to the outer bounding surface of the fluid. Or if the bounding surface of the whole fluid be annular (like a hollow anchor-ring, or of any irregular shape), in other words, if the fluid fills a tubular circuit; and the axis (A) pass through the aperture of the ring. (without passing into the fluid); there must be a firm partition (CD) extending somewhere continuously across the channel, or passage, or tube, to stop the circulation of the fluid round it; otherwise there could not be equilibrium with the supposed forces in action. If we further suppose the density of the fluid to be uniform round each of the circular lines of force in the system we have so far considered (so that the density shall be equal over every circular cylinder having the line of their centres for its axis, and shall vary from one such cylindrical surface to another, inversely as the squares of their radii), we may, without disturbing the equilibrium, impose any conservative system of force in lines perpendicular to the axis; that is (§ 506), any system of force in this direction, with intensity varying as some function of the distance. If this function be the simple distance, the superimposed system of force agrees precisely with the reactions against curvature, that is to say, the centrifugal forces, of the parts of a rotating rigid body.

701. Thus we arrive at the remarkable conclusion, that if a rigid closed box be completely filled with incompressible heterogeneous fluid, of density varying inversely as the square of the distance from a certain line, and if the box be movable round this line as a fixed axis, and be urged in any way by forces applied to its outside, the fluid will remain in equilibrium relatively to the box; that is to say, will move round with the box as if the whole were one rigid body, and will come to rest with the box if the box be brought again to rest: provided always the preceding condition as to partitions be fulfilled if the axis pass through the fluid, or be surrounded by continuous lines of fluid. For, in starting from rest, if the fluid moves like a rigid solid, we have reactions against acceleration, tangential to the circles of motion, and equal in amount to or per unit of mass of the fluid at distance r from the axis, & being the rate

of acceleration ($ 57) of the angular velocity; and (see Vol. II.) we have, in the direction perpendicular to the axis outwards, reaction against curvature of path, that is to say, 'centrifugal force,' equal to w'r per unit of mass of the fluid. Hence the equilibrium which we have demonstrated in the preceding section, for the fluid supposed at rest, and arbitrarily influenced by two systems of force (the circular non-conservative and the radial conservative system) agreeing in law with these forces of kinetic reaction, proves for us now the D'Alembert ($230) equilibrium condition for the motion of the whole fluid as of a rigid body experiencing accelerated rotation: that is to say, shows that this kind of motion fulfils for the actual circumstances the laws of motion, and, therefore, that it is the motion actually taken by the fluid.

702. In § 688 we considered the resultant pressure on a plane surface, when the pressure is uniform. We may now consider (briefly) the resultant pressure on a plane area when the pressure varies from point to point, confining our attention to a case of great importance; that in which gravity is the only applied force, and the fluid is a nearly incompressible liquid such as water. case the determination of the position of the Centre of Pressure is very simple; and the whole pressure is the same as if the plane area were turned about its centre of inertia into a horizonal position.

In this

The pressure at any point at a depth z in the liquid may be expressed by

where p is the (constant) density of the liquid, and the (atmospheric) pressure at the free surface, reckoned in units of weight per unit of area.

Let the axis of x be taken as the intersection of the plane of the immersed plate with the free surface of the liquid, and that of y perpendicular to it and in the plane of the plate. Let a be the inclination of the plate to the vertical. Let also A be the area of the portion of the plate considered, and x, ỹ, the co-ordinates of its centre of inertia.

Then the whole pressure is

The moment of the pressure about the axis of x is

Spydxdy = Ap+Ak❜p cos a,

k being the radius of gyration of the plane area about the axis of x. For the moment about y we have

[Spxdxdy = Apx + p cos a ffxydxdy.

The first terms of these three expressions merely give us again the results of § 688; we may therefore omit them. This will be equivalent to introducing a stratum of additional liquid above the free surface such as to produce an equivalent to the atmospheric pressure.

If the origin be now shifted to the upper surface of this stratum we have

Pressure = Apỹ cos a,

Moment about Ox = Ak3p cos a,

Distance of centre of pressure from axis of x =

k2

y

But if k, be the radius of gyration of the plane area about a horizontal axis in its plane, and passing through its centre of inertia, we have k2 = k12 + y2.

Hence the distance, measured parallel to the axis of y, of the centre of pressure from the centre of inertia is

y

and, as we might expect, diminishes as the plane area is more and more submerged. If the plane area be turned about the line through its centre of inertia parallel to the axis of x, this distance varies as the cosine of its inclination to the vertical; supposing, of course, that by the rotation neither more nor less of the plane area is submerged.

703. A body, wholly or partially immersed in any fluid influenced by gravity, loses, through fluid pressure, in apparent weight an amount equal to the weight of the fluid displaced. For if the body were removed, and its place filled with fluid homogeneous with the surrounding fluid, there would be equilibrium, even if this fluid be supposed to become rigid. And the resultant of the fluid pressure upon it is therefore a single force equal to its weight, and in the vertical line through its centre of gravity. But the fluid pressure on the originally immersed body was the same all over as on the solidified portion of fluid by which for a moment we have imagined it replaced, and therefore must have the same resultant. This proposition is of great use in Hydrometry, the determination of specific gravity, etc.,

etc.

704. The following lemma, while in itself interesting, is of great use in enabling us to simplify the succeeding investigations regarding the stability of equilibrium of floating bodies :

Let a homogeneous solid, the weight of unit of volume of which we suppose to be unity, be cut by a horizontal plane in XYX'Y'. Let be the centre of inertia, and let XX, YY' be the principal axes, of this area.

Let there be a second plane section of the solid, through YY, inclined to the first at an infinitely X' small angle, 0. Then (1) the volumes of the two wedges cut from the solid by these sections are equal; (2) their centres of inertia lie in one plane perpen

N

-P

dicular to YY'; and (3) the moment of the weight of each of these, round YY', is equal to the moment of inertia about it of the corresponding portion of the area multiplied by 0.

Take OX, OY as axes, and let @ be the angle of the wedge: the thickness of the wedge at any point P, (x, y), is 0x, and the volume of a right prismatic portion whose base is the elementary area dxdy at Pis Oxdxdy.

Now let [ ] and () be employed to distinguish integrations extended over the portions of area to the right and left of the axis of y respectively, while integrals over the whole area have no such distinguishing mark. Let v and v' be the volumes of the wedges; (x, y), (x', ') the co-ordinates of their centres of inertia. Then

v = [[[xdxdy] - v' = 0 (Jjxdxdy),

whence v-v=0 ffxdxdy = o since O is the centre of inertia. Hence v=v', which is (1).

Again, taking moments about XX',

And (3) is merely a statement in words of the obvious equation [ffx.x0dxdy]= 0 [ƒƒx2.dxdy].

705. If a positive amount of work is required to produce any possible infinitely small displacement of a body from a position of equilibrium, the equilibrium in this position is stable (§ 256). To apply this test to the case of a floating body, we may remark, first, that any possible infinitely small displacement may (S$ 30, 106) be conveniently regarded as compounded of two horizontal displacements in lines at right angles to one another, one vertical displacement, and three rotations round rectangular axes through any chosen point. If one of these axes be vertical, then three of the component displacements, viz. the two horizontal displacements and the rotation about the vertical axis, require no work (positive or negative), and therefore, so far as they are concerned, the equilibrium is essentially neutral. But so far as the other three modes of displacement are concerned, the equilibrium may be positively stable, or may be unstable, or may be neutral, according to the fulfilment of conditions which we now proceed to investigate.

706. If, first, a simple vertical displacement, downwards, let us suppose, be made, the work is done against an increasing resultant of upward fluid pressure, and is of course equal to the mean increase of this force multiplied by the whole space. If this space be denoted by z, the area of the plane of flotation by A, and the weight of unit bulk of the liquid by w, the increased bulk of immersion is clearly Az,

and therefore the increase of the resultant of fluid pressure is wAz, and is in a line vertically upward through the centre of gravity of A. The mean force against which the work is done is therefore wAz, as this is a case in which work is done against a force increasing from zero in simple proportion to the space. Hence the work done is wAz3. We see, therefore, that so far as vertical displacements alone are concerned, the equilibrium is necessarily stable, unless the body is wholly immersed, when the area of the plane of flotation vanishes, and the equilibrium is neutral.

707. The lemma of § 704 suggests that we should take, as the two horizontal axes of rotation, the principal axes of the plane of flotation. Considering then rotation through an infinitely small angle

round one of these, let G and E be the displaced centres of gravity of the solid, and of the portion of its volume which was immersed when it was floating in equilibrium, and G', E' the positions which they then had; all projected on the plane of the diagram which we suppose to be through I the centre of inertia of the plane of flotation. The resultant action of gravity on the displaced body is W, its weight, acting downwards through G; and that of the fluid pressure on it is W upwards through E corrected by the amount (upwards) due to the additional immersion of the wedge AIA', and the amount (downwards) due to the extruded wedge B'IB. Hence the whole action of