CENTRE OF PRESSURE.

71

(13) If a quadrilateral area be entirely immersed in water, and a, B, y, d be the depths of its four corners, and h the depth of its centre of gravity, show that the depth of its centre of pressure is

(a+B+y+8)

1

(By+ya+aß+as+Bô+yồ).

(14) Assuming that any quadrilateral is dynamically equivalent to six particles +1 at each corner, -1 at the intersection of the diagonals, and +9 at the C.G., find the depth of the centre of pressure in terms of the depths of the corners and of the intersection of the diagonals.

(15) If a plane regular pentagon is immersed so that one side is horizontal and the opposite vertex at double the depth of that side, prove that the depth of the centre of pressure of the pentagon is a(29+3/5)÷48,

where a is the depth of the lowest vertex.

(16) If the area be a regular pentagon with one side in the surface and the plane vertical, then the vertical line through the centre of pressure, terminated at the surface, is divided by the centre of the polygon in the ratio

2+sin 18°: 6(1+sin 18°).

(17) A regular hexagon is immersed in a fluid so that its plane is vertical and its highest side in the surface of the fluid; show that the depth of the centre of pressure below the surface is 3/3a, where a is the length of a side of the hexagon; so that the depth of the C.P. is to the depth of the C.G. as 23 to 18.

72

CENTRE OF PRESSURE.

(18) Show that the centre of pressure of a regular polygon of n sides, wholly immersed in a uniform liquid, and in a vertical plane, is vertically below the centre, and at a distance

r2(2+cos 2π/n)/12h,

where h is the depth of the centre and r is the radius of the circumscribing circle.

(19) An elliptic lamina is just immersed in a homogeneous liquid, the major axis being vertical; prove that, if the eccentricity be, the centre of pressure will coincide with the lower focus.

(20) A triangular area is immersed in liquid with one side in the surface; the ellipse of largest possible area is inscribed in it; show that the depth of the centre of pressure of the remainder of the triangle 18/3-5π

is

36/3-12π

of the depth of its lowest point.

(21) A rectangle is immersed in n fluids of densities p, 2p, 3p, ..., np; the top of the rectangle being in the surface of the first fluid, and the area immersed in each fluid being the same; show that the depth of the centre of pressure of the rectangle 3n+1 a 2n+12'

is

where a is the depth of the lower side.

Deduce the position of the centre of pressure of the rectangle when the density of the liquid varies continuously as the depth.

(22) Prove that the depth of the centre of pressure of a parallelogram two of whose sides are horizontal

and at depths a, b respectively below the surface

VERTICAL COMPONENT OF THRUST.

73

of a liquid whose density varies as the depth below the surface, is

3 a3+a2b+ab2+b3

4 a2+ab+b2

(23) If the centre of gravity G is fixed, and the centres of pressure, when a given line in the area is horizontal and vertical, are respectively K1, K2; then, when the line is inclined at an angle ✪ to the horizontal, the centre of pressure is at K, where GK meets K1K, in D, so that

and

2

K1D cos 0 K2D sin 0,

2

GK=GD(cos + sin 0).

41. Component Vertical Thrust of a Liquid under Gravity against any Curved Surface.

Suppose BC is a portion of a curved surface, for instance, a plate on the bottom of a cistern, boiler, or ship; considering as before the equilibrium of the superincumbent liquid BCcb by resolving vertically, then the component vertical thrust on BC is equal to the weight of the superincumbent liquid BCcb, and acts vertically upwards or downwards through the c.a. of the superincumbent liquid.

In fig. 28, BC represents a portion of a curved vessel containing the liquid, and the vertical component of the thrust on BC is downwards.

But suppose, as in fig. 29, that BC is a portion of the curved bottom of a ship; then the vertical thrust on BC is upwards; and now the superincumbent liquid must be taken as the fictitious quantity required to fill up the cylinder BCcb to the level bc of the water outside.

74

VERTICAL COMPONENT OF THRUST

So also if BC is the underside or soffit of an arched bridge; in a flood the upward hydrostatic thrust of the water may be sufficient to blow up the bridge, especially if the parapet is solid and the road is walled to prevent flooding.

Fig. 28.

Fig. 29.

Fig 30.

It might happen, as in fig. 30, that BC was part of a curved surface like the ram of a warship, such that the vertical thrust of the liquid on it was partly upwards and partly downwards; in this case it will be necessary to cut up BC into two parts by the line DFE, along which the tangent planes to the surface BC are vertical, and then to determine the downward vertical thrust on DCEF, and the upward vertical thrust on DBEF separately, by the preceding methods.

42. The Hydrostatic Thrust in a Mould.

Consider for example the upward and downward thrust on the two parts of a mould used in casting an object like a bell (fig. 31).

The bell-metal being poured in it to the level a of the top of the mould, the upward thrust on the upper half of the mould, tending to lift it and to allow the metal to escape, is equal to the weight of metal which would occupy the volume BACcab, and to counteract this upward thrust the mould must be fastened down, or weighted down with corresponding weights.

ON A CURVED SURFACE.

75

The downward thrust on the lower part of the mould, called the core, will be equal to the former upward thrust, plus the weight of the bell.

By making the bell very thin, we have the paradoxical result, that a very small amount of liquid metal is capable of exerting a very large upward thrust, especially with metal of high density.

h-a

α

Fig. 31.

In fig. 31, the external shape of the bell is shown as a cylinder, a cone, a hemisphere or hemispheroid, and a paraboloid; if a denotes the height of the bell, and h the head of metal above the bottom of the mould, then the upward thrust on the mould is in these cases equal to the weight of a cylinder of metal whose base is the base of the bell, and altitude h-a, h-fa, h-fa, h-ja respectively; this follows at once from the well-known theorems of Mensuration, due to Archimedes, that

"The volume of a cone is, of a hemisphere or hemispheroid is, and of a paraboloid is of that of a cylinder on the same base and of the same altitude."

The quasi-hydrostatic thrust of spherules may be similarly illustrated by a funnel filled with lead shot, inverted and resting on a table.