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 w3r 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 radical 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. In this 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 horizontal position. 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 p, 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 ≈, ÿ, the co-ordinates of its centre of inertia.

Then the whole pressure is

Sfpdxdy=ff (po+py cos a) dxdy
=Ap+Apy cos a.

The moment of the pressure about the axis of x is

Spydxdy=Apoÿ + Ak2p 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

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

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 XX'Y'. Let O 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 II', 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

Y

-P

X

dicular to '; 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, Or as axes, and let O 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 P is 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 a and a' be these areas, v and v′ the volumes of the wedges; (x, y), (x', ÿ') the co-ordinates of their centres of inertia. Then v=0[fxdxdy]=ax

- v′=0 (Sfxdxdy)=a'x',

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

Again, taking moments about XX',

and

Hence

vy 0 [fxydxdy],

-v'y' = (JJxydxdy).

vý-v'y=0fxydxdy.

But for a principal axis Exydm vanishes. whence, since v=v', we have

y=y', which proves (2).

Hence vý-v'y' =0,

And (3) is merely a statement in words of the obvious equation

[ffx.x0dxdy]=0 [ssx2.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 (§§ 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 wAz2. 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

gravity and fluid pressure on the displaced body is the couple of forces up and down in verticals through G and E, and the correction due to the wedges. This correction consists of a force vertically upwards through the centre of gravity of A'IA, and downwards through that of BIB'. These forces are equal [§ 704 (1)], and therefore constitute a couple which [704 (2)] has the axis of the displacement for its axis, and which [§ 704 (3)] has its moment equal to Owk2A if A be the area of the plane of flotation, and k its radius of gyration (§ 235) round the principal axis in question. But since GE, which was vertical (G'E') in the position of equilibrium, is inclined at the infinitely small angle 0 to the vertical in the displaced body, the couple of forces W in the verticals through G and E has for moment Who, if h denote GE; and is in a plane perpendicular to the axis, and in the direction tending to increase the displacement, when G is above E. Hence the resultant action of gravity and fluid pressure on the displaced body is a couple whose moment is

(wAkWh)0, or w (Ak2- Vh)0,

if V be the volume immersed. It follows that when Ak2> Vh the equilibrium is stable, so far as this displacement alone is concerned.

Also, since the couple worked against in producing the displacement increases from zero in simple proportion to the angle of displacement, its mean value is half the above; and therefore the whole amount of work done is equal to

w(Ak2- Vh)02.

708. If now we consider a displacement compounded of a vertical (downwards) displacement z, and rotations through infinitely small angles 0, 'round the two horizontal principal axes of the plane of flotation, we see (§§ 706, 707) that the work required to produce it is equal to

≥ w[Az2 + (Ak2 — Vh) 02 + (Ak2 — Vh) 0′2],

and we conclude that, for complete stability with reference to all possible displacements of this kind, it is necessary and sufficient that

709. When the displacement is about any axis through the centre of inertia of the plane of flotation, the resultant of fluid pressures is equal to the weight of the body; but it is only when the axis is a principal axis of the plane of flotation that this resultant is in the plane of displacement. In such a case the point of intersection of the resultant with the line originally vertical, and through the centre of gravity of the body, is called the Metacentre. And it is obvious, from the above investigation, that for either of these planes of displacement the condition of stable equilibrium is that the metacentre shall be above the centre of gravity.

710. We shall conclude with the consideration of one case of the