Resultant pressure on a plane area. Loss of apparent weight by immersion in a fluid. 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, y, the coordinates of its centre of inertia. Then the whole pressure is [Spdxdy = √(P+py cos a) dxdy The moment of the pressure about the axis of x is Spydxdy = Apy + Ak3p cos a, k being the radius of gyration of the plane area about the axis of x. For the moment about У we have [Spxdxdy = Ap+ p cos a ffxydxdy. The first terms of these three expressions merely give us again the results of § 746; 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, by § 283, k = k‚2 + y2. 1 Hence the distance, measured parallel to the axis of y, of the centre of pressure from the centre of inertia is k2/ÿ; 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. 762. 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 equi apparent And Loss of single weight by immersion librium, even if this fluid be supposed to become rigid. the resultant of the fluid pressure upon it is therefore a force equal to its weight, and in the vertical line through its in a fluid. 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. Analytically, the following demonstration is of interest, especially in its analogies to some preceding theorems, and others which occur in electricity and magnetism. If V be the potential of the impressed forces, -dV/dx is the force parallel to the axis of x on unit of matter at xyz, and pdxdydz is the mass of an element of the fluid, and therefore the whole force parallel to the axis of x on a mass of fluid substituted for the immersed body, is represented by the triple integral extended over the whole surface. Let dS be an element of any surface at x, y, z; λ, μ, v the direction-cosines of the normal to the element; p the pressure in the fluid in contact with it. The whole resolved pressure parallel to the axis of x is P1 = fλpdS =SSpdydz, the same expression as above. The couple about the axis of z, due to the applied forces on any fluid mass, is (§ 559) Edm (Xy - Yx), dm representing the mass of an element of fluid. This may be written in the form – SSSpdxdydz (y dV dx dv dy the integral being taken throughout the mass. Loss of apparent weight by immersion in a fluid. Lemma. which is the couple due to surface-pressure alone. 763. 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 YY', inclined to the first at an infinitely 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 perpendicular 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 Ox, 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, ÿ), (x', ÿ') the co-ordinates of their centres of inertia. Then v=0[[[xdxdy]= añ0 - v' = 0 (Лfxdxdy) = a'x'0, whence v-v0ffxdxdy=0 since O is the centre of inertia. But for a principal axis (§ 281) Exydm vanishes. Hence [[[x.x0dxdy]= 0 [ƒƒx2dxdy]. equilibrium body. 764. If a positive amount of work is required to produce Stability of any possible infinitely small displacement of a body from a of a floating position of equilibrium, the equilibrium in this position is stable (§ 291). To apply this test to the case of a floating body, we may remark, first, that any possible infinitely small displacement may (§§ 26, 95) 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 stable, or may be unstable, or may be neutral, according to the fulfilment of conditions which we now proceed to investigate. placements. 765. If, first, a simple vertical displacement, downwards Vertical dislet 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 VOL. II. 21 in vertical displacement. Work done increasing from zero in simple proportion to the space. Hence the work done is wAz. 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. Displacement by rotation about an 766. The lemma of § 763 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 inflotation. finitely small angle round one of these, let G and E be the axis in the plane of 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 |