Also HG must be perpendicular to PQ, and therefore since

FE must be perpendicular to PQ, that is, FE is the normal at E to the hyperbola. The problem is therefore reduced to that of drawing normals from F to the curve.

Let xy=c2 be the equation of the curve referred to AB, AC as axes, and let

< BAC=0, AB = 2a, AC 2b............(a).

=

Let x, y, be the co-ordinates of E; the co-ordinates of F are a, b, and the equation of the normal at E is

And if this pass through F, the co-ordinates of which

are a, b,

or

(b − y) (x cos 0 − y) = (a — x) (y cos ◊ — x),

x2 − (a + b cos 0) x = y2 — (a cos 0 + b) y ..............(B).

The equations (a) and (B) determine all the points of the hyperbola, the tangents at which can be lines of floatation.

Also (8) is the equation to an equilateral hyperbola, referred to conjugate diameters parallel to AB, AC; the points of intersection of the two hyperbolas are therefore the positions of E.

To find x, we have

x − (a + b cos 0). x2 + (a cos 0 + b) c2x − c2 = 0,

an equation which has only one negative root, and one or three positive roots, and there may be therefore three positions of equilibrium or only one.

If the densities of the liquid and the prism be p and σ, we have, since the area PAQ

Suppose the prism to be isosceles, then putting a = b, the equation for a becomes

x-c-a (1 + cos 0) (a3 - c2x) = 0;

from which we obtain a=c, which gives y = c, and makes BC horizontal, an obvious position of equilibrium, and also

the isosceles prism will therefore have only one position of equilibrium, unless

2

a cos2=> c;

and, since pc2 = σa2, this is equivalent to

Ex. 4. Determine the position of equilibrium of a balloon of given size and weight, neglecting the variations of temperature at different heights in the atmosphere.

If the temperature be constant, the pressure of the air at

II gz

= € II being the

k

atmospheric pressure at the level from which the height is measured.

The air displaced consists of a series of strata of variable density, and if z be the height of the lowest point of the balloon, the distance from that point of any horizontal section (X) of the balloon, and h its height, the weight of a stratum of the air displaced is

The form of the balloon being given, X is a known function of x, and if W be the weight of the balloon and of the gas it contains, the height z will be determined by equating W to the expression we have obtained for the weight of the air displaced.

50. A homogeneous solid floats, wholly immersed, in a liquid of which the density varies as the depth; to find the depth of its centre of gravity.

Let a, c, be the depths of the highest and lowest points of the solid, Z the area of a horizontal section of the solid at a depth z, and μz the density;

με

the weight of the liquid displaced = ["guzZdz.

α

Let be the depth of the centre of gravity of the solid, and V its volume, then

Vz

Zzdz;

therefore the weight of displaced liquid = guzV, and if ρ be the density of the solid, its weight = gp V; hence p = μž, or the solid floats in such a position that the density of the liquid at the depth of the centre of gravity of the solid is equal to the density of the solid.

51. If a solid float under constraint, the conditions of equilibrium depend on the nature of the constraining circumstances, but in any case the resultant of the constraining forces must act in a vertical direction, since the other forces, the weight of the body, and the fluid pressure, are vertical.

If for instance one point of a solid be fixed, the condition of equilibrium is that the weight of the body and the weight of the fluid displaced should have equal moments about the fixed point; this condition being satisfied, the solid will be at rest, and the stress on the fixed point will be the difference of the two weights.

As an additional illustration, consider the case of a solid floating in water and supported by a string fastened to a point above the surface; in the position of equilibrium the string will be vertical, and the tension of the string, together with the resultant fluid pressure, which is equal to the weight of the displaced fluid, will counterbalance the weight of the body; the tension is therefore equal to the difference of the weights, and the weights are inversely in the ratio of the distances of their lines of action from the line of the string, these three lines being in the same vertical plane.

52. For subsequent investigations, the following geometrical propositions will be found important.

If a solid be cut by a plane, and this plane be made to turn through a very small angle about a straight line in itself, the volume cut off will remain the same, provided the straight line pass through the centroid of the area of the plane section.

To prove this, consider a right cylinder of any kind cut by a plane making with its base an angle 0.

Let be the distance from the base of the centroid of the section A, SA an element of the area of the section and V the volume between the planes. Then

Now the centroid of the area A is also the centroid of all sections made by planes passing through it, as may be seen by projecting the sections on the base of the cylinder; it follows therefore, that, z being the same for all such sections, the volumes cut off are the same.

In the case of any solid, if the cutting plane be turned through a very small angle about the centroid of its section, the surface near the curves of section may be considered,

N

without sensible error, cylindrical, and the above proposition is therefore established*.

* The following form of proof may also be given.

Let ACB, the cutting plane, be turned through a small angle (0) about a line Cx, and let dA be an element of the area.

B

B'

Then the algebraical value of the additional volume cut off is equal to foyda, and, if this vanishes, fydA=0, which is the condition that the centroid of A should lie in the axis of x; and, taking C as the centroid, any plane through C will satisfy the same condition.