In all cases, the total pressure will exceed the weight of the liquid.

3. A right cone, with a circular base, stands on its base, and is filled with a liquid. How does the pressure on the internal surface compare with the weight of the liquid?

SOLUTION.

Denote the radius of the base, by r, and the altitude, by h, then will the slant height be equal to

√h2 + p2.

The centre of gravity of the lateral surface, below the upper surface of the liquid is equal to h. If we denote the weight of a unit of volume of the liquid, by w, we shall have, for the total pressure on the interior surface,

wxr2h + &w=rh√h2 + r2 = w=rh(r + {√h2 + r2).

4. Required the relation between the pressure and the weight in the preceding case, when the cone stands on its

5. What is the pressure on the lateral faces of a cubical vessel filled with water, the edges of the cube being 4 feet, and the weight of the water 62 lbs. per cubic foot?

Ans. 8000 lbs.

6. A cylindrical vessel is filled with water. The height of the vessel is 4 feet, and the radius of the base 6 feet. What is the pressure on the lateral surface?

Ans. 18850 lbs., nearly.

Centre of Pressure on a Plane Surface.

G

F

156. Let ABCD represent a plane, pressed by a fluid on its upper surface, AB its intersection with the free surface of the fluid, G its centre of gravity, O the centre of pressure, and s the area of any element of the surface at S. Denote the inclination of the plane to the level surface, by a, the perpendicular distances from 0 to AB, by x, from G to AB, by p, and from S to AB, by r. Denote, also, the entire area AC, by A, and the weight of a unit of volume of the fluid, by w. The perpendicular distance from G to the free surface of the fluid, will be equal to p sina, and that of any element of the surface, will be r sina.

Fig. 136.

From the preceding article, it follows that the entire pressure exerted is equal to wAp sina, and its moment, with respect to AB as an axis of moments, is equal to

WAP sina xx.

The elementary pressure on s is, in like manner, equal to wsr sina, and its moment, with respect to AB, is wsr2 sina, and the sum of all the elementary moments is equal to

w sina Σ(872).

But the resultant moment is equal to the algebraic sum of the elementary moments. Hence,

wAp sina x x = w sina Σ(sr2);

and, by reduction,

x=

Σ(87)
Ap

(144.)

The numerator is the moment of inertia of the plane ABCD, with respect to AB, and the denominator is the moment of the area with respect to the same line. Hence, the distance from the centre of pressure to the intersection of the plane with the free surface, is equal to the moment of inertia of the plane, divided by the moment of the plane.

If we take the straight line AD, perpendicular to AB, as an axis of moments, denoting the distance of O from it, by y, and of s from it, by 7, we shall, in a similar manner, have,

The values of x and y make known the position of the centre of pressure.

EXAMPLES.

1. What is the position of the centre of pressure on a rectangular flood-gate, the upper line of the gate coinciding with the surface of the water?

SOLUTION.

It is obvious that it will be somewhere on the line joining the middle points of the upper and lower edges of the gate.

Denote its distance from the upper edge, by z, the depth of the gate, by 27, and its mass, by M. The distance of the centre of gravity from the upper edge will be equal to 7.

From Example 1 (Art. 132), replacing d by l, and reducing, we have, for the moment of inertia of the rectangle,

But the moment of the rectangle is equal to,

hence, by division, we have,

MI;

2 = 11 = (27).

That is, the centre of pressure is at two-thirds of the distance from the upper to the lower edge of the gate.

2. Let it be required to find the pressure on a submerged rectangular flood-gate ABCD, the plane of the gate being vertical. Also, the distance of the centre of pressure below the surface of the water.

SOLUTION.

EG F

D

Fig. 137.

B

pressure

Let EF be the intersection of the plane with the surface of the water, and suppose the rectangle AC to be prolonged till it reaches EF. Let C, C', and C", be the centres of of the rectangles EC, EB, and AC respectively. Denote the distance GC", by z, the distance ED, by a, and the distance EA, by a'. Denote the breadth of the gate, by b, and the weight, a unit of volume of the water, by w.

The pressure on EC will be equal to a2bw, and the pressure on EB will be equal to a''bw; hence, the pressure on AC will be equal to

{bw (a2 — a'3) ;

which is the pressure required.

From the principle of moments, the moment of the pressure on AC, is equal to the moment of the pressure on EC, minus the moment of the pressure on EB. Hence, from

which is the required distance from the surface of the

water.

3. Let it be required to find the pressure on a rectangular flood-gate, when both sides are pressed,

the water being at different levels on the two sides. Also, to find the centre of pressure.

SOLUTION.

Denote the depth of water on one side by a, and on the other side, by

a', the other elements being the same as before. The total pressure will, as before, be equal to,

B

C

Fig. 188.

4. A sluice-gate, 10 feet square, is placed vertically, its upper edge coinciding with the surface of the water. What is the pressure on the upper and lower halves of the gate, respectively, the weight of a cubic foot of water being taken equal to 621 lbs.? Ans. 7812.5 lbs., and 23437.5 lbs.

5. What must be the thickness of a rectangular dam of granite, that it may neither rotate about its outer angular