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dicular to '; and (3) the moment of the weight of each of these, round ', 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 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]=ox -v=0(Jfxdxdy)=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',

vý=0 [fxydxdy],

and

Hence

-v'y'=0 (xydxdy).
vy-v'y=0xydxdy.

whence, since v=v', we have

But for a principal axis Exydm vanishes.

ÿÿ', which proves (2).

Hence vý-v'y' =0,

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 (§§ 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

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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 to the vertical in the displaced body, the couple of forces Win 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

(wAk2 - Wh)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 + (Ak22 — 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

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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

equilibrium of a revolving mass of fluid subject only to the gravitation of its parts, which admits of a very simple synthetical solution, without any restriction to approximate sphericity; and for which the following remarkable theorem was discovered by Newton and Maclaurin :—

711. An oblate ellipsoid of revolution, of any given eccentricity, is a figure of equilibrium of a mass of homogeneous incompressible fluid, rotating about an axis with determinate angular velocity, and subject to no forces but those of gravitation among its parts.

The angular velocity for a given eccentricity is independent of the bulk of the fluid, and proportional to the square root of its density.

712. The proof of this proposition is easily obtained from the results already deduced with respect to the attraction of an ellipsoid and the properties of the free surface of a fluid.

M

A

P

We know, § 538, that if APB be a meridian section of a homogeneous oblate spheroid, AC the polar axis, CB an equatorial radius, and P any point on the surface, the attraction of the spheroid may be resolved into two parts; one, Pp, perpendicular to the polar axis, and varying as the ordinate PM; the other, Ps, parallel to the polar axis, and varying as PN. These components are not equal when MP and PN are equal, else the resultant attraction at all points in the surface would pass through C; whereas we know that it is in some

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B

n N

such direction as Pf, cutting the radius BC between B and C, but at a point nearer to C than ʼn the foot of the normal at P. Let then

Pp=a.PM, and Ps=ß.PN,

where a and ẞ are known constants, depending merely on the density (p), and eccentricity (e), of the spheroid.

Also, we know by geometry that Nn (1-e) CN.

Hence; to find the magnitude of a force Pq perpendicular to the axis of the spheroid, which, when compounded with the attraction, will bring the resultant force into the normal Pn: make pr=Pq, and we must have

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Now if the spheroid were to rotate with angular velocity w about AC, the centrifugal force, §§ 39, 42, 225, would be in the direction Pq, and would amount to

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the whole force on P, that is, the resultant of the attraction and centrifugal force, will be in the direction of the normal to the surface, which is the condition for the free surface of a mass of fluid in equilibrium.

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This determines the angular velocity, and proves it to be proportional

top.

713. If, after Laplace, we introduce instead of e a quantity € defined by the equation

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the expression (1) for w2 is much simplified, and

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(2)

(3)

When e, and therefore also e, is small, this formula is most easily calculated from

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of which the first term is sufficient when we deal with spheroids so little oblate as the earth.

The following table has been calculated by means of these simplified formulae. The last figure in each of the four last columns is given to the nearest unit. The two last columns will be explained a few sections later:

:-

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