3. To find the centre of pressure of a triangle immersed vertically to any depth beneath the surface of a fluid, the base of the triangle being horizontal and its vertex being the point of the triangle which is nearest to the surface.

Let c denote the depth of the vertex of the triangle below the surface, h the altitude of the vertex above the base. Let x be the depth, below the vertex, of any line y drawn parallel to the base of the triangle and intercepted by the other two sides. Then, c + being the depth of the elemental strip y dx below the surface, we have, supposing ≈ to be the depth of the centre of pressure below the vertex,

z. f (z + 0) y dz = f (x + c) zxy dx;

x.

0

or, since y varies directly as x,

x. f (x + c) x dx = f (x + c) x2 dx,

0

0

Since the line y is always bisected by a line passing through the vertex of the triangle and the middle point of its base, it is obvious that the centre of pressure must lie in this bisecting line. Its distance from the vertex, ifl denote the length and the inclination of the bisecting line to the horizon, will be equal to

3h + 4c
3h+ 4c
4h + 6c 4h + 6c

4. To find the centre of pressure of a square AB (fig. 5) immersed to a given depth below the surface of a fluid, one of its diagonals OC being vertical.

Let the diagonal CO produced meet the surface KDL in the point D, and let OD = h; let OA, OB, produced indefinitely be the axes of x and y. Let x, y, be the coordinates of a point P of the square, and x + dx, y + dy, of a point p indefinitely

near to it; then the depth of the elementary square, of which Pp is a diagonal, below the surface of the fluid, being

and its area being dx dy, we have, a denoting a side of the square AB,

The coefficient of ≈ in this equation is equal to

and the right-hand side of the equation is equal to

is immersed vertically in a fluid, the prime radius vector being coincident with the surface; to find the depth of the centre of pressure.

The required depth being denoted by x, we must determine ≈ from the equation

ffrd@dr.r sin 0 = ffrde dr. r2 sin2 0,

r sin being the depth of the element rde dr below the surface. Indicating the limits,

Now

S

(1 + cos 0)* sin2 0 do.

0

(1 + cos 0)3 sin Ode = {(1 + cos 0)} = 4.

Also

(1 + cos 0)* sin3 0d0 = (1 + 4 cos 0 + 6 cos2 0 + 4 cos30 + cos10) sin20d0

= 1 (1 − cos 20) d✪ + 4 sin2 0 d sin 0 + 3(1)

cos 40) do

+ 4 (sin2 0 – sin* 0) d sin 0 + sin2 0 cos1 0 d0,

Ꮎ

0

Ꮎ

π

sin' 0 cos Ꮎ ᏧᎾ

sin2 6

sin2 0 cos1 0d0;

cos (1 sin' 0) do

cos 0d0,

0

["cos' Ode = "(cos' 0 sin @) + 3 ["cos' 0 sin2 0 d0

Ꮎ

[" (1 + cos 20) d0 = }π ;

0

sin3 0d0

x. 4 = a.

π 21π

= π +

6. An elliptic area PCD, bounded by two conjugate semidiameters CP, CD, and the intercepted arc, is immersed vertically in a fluid, CD coinciding with the surface; to find the centre of pressure.

Let CP and CD be taken as the axes of x and y respectively. Let a denote the angle between CP and CD; let CP = a, CD = b.

Then the area of an elementary parallelogram, the sides of

which are dx, dy, parallel to the axes of coordinates, will be dx dy sin a, and its depth below the surface will be x sin a. Hence x and y are to be found from the equations

ffdx dy sin a .x sin a = ffdx dy sin a.x sin a. x,

or, reducing and indicating the limits,

and y ffdxdy sin a. x sin a = ffdx dy sin a.x sin a.y,

разг

Again ["f" xy dx dy = } ["22 (a2 – x2) xdx

12

b2

a

0

13

2

a2

= {a2b2.

Hence, from (1), ≈.} a3b = ¦ πα3b,

0

7. A rectangular flood-gate is moveable about its lowest edge. as a horizontal axis; having given the depth of the water on each side, to determine the moment of the couple which will keep it at rest.

Let c, c', represent the altitudes of the surfaces of the water on the two sides of the gate above its lowest edge: let b denote the breadth of the gate. Then, a being the depth of any point of the gate below the surface on one side, the moment exerted by the water on this side to produce rotation about the axis is equal to

The opposite moment due to the fluid on the other side will accordingly, putting c' instead of c, be equal to

Hence the moment of the required couple will be equal to gpc (c'3 ~ c3).

8. To find the centre of pressure of a rectangular plank immersed vertically to any depth within a fluid, the two ends of the plank being horizontal.

If a be the depth of the upper and b of the lower end of the plank, the depth of the centre of pressure will be equal to

Stevin: Euvres Mathématiques par Albert Girard, p. 496. 9. A circular area is just immersed in a fluid; to find the depth of the centre of pressure below the centre of the area.

If r denote the radius of the circle, the required depth will be equal to r.

10. To find the position of the centre of pressure of an equilateral triangle having one angle in the surface of the fluid and one side vertical.

Let 7 represent the length of a side of the triangle, and h the distance of one of its angles from the opposite side; then, the