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CHAPTER VIII.

EQUILIBRIUM OF VESSELS CONTAINING FLUID.

SECTION I.
Equilibrium.

THE system consisting of a vessel and the contained fluid, will be subject to the action of three forces, viz. the weight of the vessel, the weight of the fluid, and the reaction of the surface upon which the vessel is supported. The conditions of the equilibrium may therefore be ascertained as in ordinary statical problems.

1. A cubical vessel, half filled with fluid, rests upon a horizontal plane; to find the greatest angle through which the plane can be turned before the vessel overturns, the lowest edge of the cube remaining horizontal, and all sliding being prevented.

It is evident that the angle may be as great as 45°, for, under this condition, the centre of gravity of the fluid contained in it, and of the vessel itself, will lie in the vertical line through the middle point of its lowest edge. It may easily be ascertained that this is the greatest elevation. For, supposing the elevation to be greater, in which case a portion of the fluid will have escaped, let ABCD (fig. 26) be a vertical section of the vessel through the centre of gravity G of the fluid, at right angles to the inclined plane. Let AK be the section of the surface of the fluid, which will be a horizontal line. Join A, G, and produce the line AG to M, which will be the middle point of BK. Draw GH at right angles to BK: join GB.

But, if 0 be the inclination of the inclined plane to the horizon, the tangent of the acute angle between the vertical through B and the inclined plane will be equal to cot 0. Hence, supposing the weight of the vessel itself to be inconsiderable, we must have, that equilibrium may be preserved,

a condition necessary à fortiori, if the weight of the vessel be not inconsiderable, since the centre of gravity of the vessel lies in a vertical line to the left of B. But, since AB is greater than BK, tan GBH must be greater than unity, and, O being supposed to be greater than, cot must be less than unity:

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thus the condition (1) is impossible. It is evident therefore that, whether the vessel be supposed to have weight or not, the plane may have any inclination up to 45°, but no greater.

2. The common intersection of two planes is a horizontal line, and between them rests an indefinitely thin hollow hemisphere, with its rim in contact with one of them, which is vertical, and the space between the internal superficies of the hemisphere and the vertical plane is filled with a homogeneous fluid: to find the pressure upon each of the planes and their greatest inclination for which the equilibrium is possible.

Let fig. (27) represent a section of the hemisphere and the two planes, made by a plane through the centre C of the sphere, perpendicular to the two planes; AE and AF denoting the sections of the two planes.

The plane AF will exert upon the hemisphere a force R, passing through C: also the fluid, which acts normally upon the hemisphere at every point, will produce upon it a resultant force P passing through C. The hemisphere is subject to the action of no other force beside these two, except the horizontal resultant S of the pressures of AE upon the rim of the hemisphere: the force S must therefore also pass through C.

H2

Letr = the radius of the sphere, 0 = LPCA, a = LEAF. Then, for the equilibrium of the hemisphere, we have, resolving forces horizontally,

But P sin 0, P cos 0, will be equal respectively to the pressure of the fluid upon the vertical plane and the weight of the fluid, παργ3. Hence the equations (1) and (2)

that is, to wgpr3 and become

whence

S + παργ3 = R cos a,

παργ3 = R sin a,

R = παργ3 cosec a,

S = παρ3 (cot a -3).

Since S cannot have a negative value, the greatest value of a consistent with equilibrium is given by the equation

cot a = 3, α = tan-1.

3. A cylindrical vessel, the weight and thickness of which are inconsiderable, is placed with its base upon an inclined plane, and prevented from sliding by the roughness of the plane; to find the height to which it may be filled with fluid without oversetting.

If r be the radius of the cylinder, tan a the inclination of the plane to the horizon, and e the length of the portion of the axis of the cylinder which is immersed in the fluid,

Stability of Vessels filled with Fluid.

1. A thin uniform hemispherical bowl of given weight, partly filled with fluid, is placed with its axis vertical upon

the highest point of a sphere: to ascertain the nature of the equilibrium in regard to stability, the bowl being supposed to be subject to a rolling but not to a sliding motion upon the supporting sphere.

Let A' (fig. 28) be the highest point of the supporting sphere, A the point of the bowl which, when it is in its position of equilibrium, is in contact with A'. Let O be the centre of the sphere of the bowl, O that of the supporting sphere. Let Q be the point of contact when the bowl is displaced from its position of rest by a motion of rotation.

Let P denote the weight of the bowl and W of the fluid which it contains. Join OO' which will pass through Q, and produce OA to meet O'A' produced in B. The centre of gravity G of the bowl, which is supposed to be of uniform thickness, will coincide with the middle point of OA, and the weight W of the fluid will act vertically downwards through 0.

Let OQ = r, O'Q=r', LAOW = 0, 200W = ψ. Then the moment of P and W about Q, tending to disturb the bowl still further from its position of equilibrium, is equal to

Wr sin 4 + P (r sin 4 – Ersin 0)

= (P + W)rsin 4 – Pr sin 0.

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But, since the arc QA must be equal to the arc QA' by the

nature of the displacement, which is due to rolling, we have

In order that the bowl may tend to return to its position of

rest, this expression must be negative: hence we must have

Suppose 0 to be a very small angle: then approximately this inequality becomes

which is the condition for stability with regard to small displacements of rotation.

2. A cone of given density is scooped out so that the hollow part is the inscribed paraboloid, the vertex of which bisects the axis of the cone. To determine how much fluid of given density must be poured into it, that it may float in a state of neutral equilibrium when placed, with its vertex downwards, in another given fluid.

Let r denote the radius of the base of the cone, h its altitude; x the distance of the vertex A (fig. 29) of the paraboloid from the surface PQ of the fluid which it contains, x' the distance of the vertex O of the cone from the surface P'Q' of the fluid in which the cone is immersed; V the volume of the fluid contained in the paraboloid and V' that of the fluid displaced by the immersion of the cone; y the radius of the circle PQ, y' that of the circle P'Q'; p the density of the solid substance of the cone, σ of the fluid poured into it, and o' of the fluid in which it is immersed. Let H be the centre of gravity of the paraboloidal volume PAQ, H' of the conical volume P'OQ'; M the metacentre of the fluid PAQ and M' of the fluid displaced, P'OQ'. Then, in case of a slight angular displacement of the cone, the fluid PAQ will act on the solid portion of the cone vertically downwards through M, and the fluid P'OQ' will act upon it vertically upwards through M'.

If the equilibrium be neutral, the moment of Vo to twist the solid matter of the cone about its centre of gravity must be equal to that of V'o' to twist it in an opposite direction.

Let denote the distance of the centre of gravity of the solid matter of the cone from its vertex: then, the volume of the whole cone KOL being wrth, the volume of the paraboloid KAL being wrh, the distance of the centre of gravity of the