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angle of the triangle opposite to the vertical side being taken as origin of coordinates, the axis of x being taken horizontal so as to bisect the vertical side, and the axis of y being taken vertically downwards,

= h, y = hl.

. 11. To find the centre of pressure of a segment of a parabola cut off by a chord passing through the vertex, the tangent at which lies in the surface of the fluid, and inclined at an angle of 45° to the axis.

The axis of the parabola and the tangent at its vertex being taken as the axes of x and y respectively, and I representing the latus rectum,

% = , = 0. 12. A parabolic area (fig. 6), cut off by a single ordinate, the abscissa of which is equal to the latus rectum, is immersed vertically in fluid, its axis being inclined at an angle of 30° to the horizon, and its vertex being at a distance equal to the latus rectum below the surface: to determine the position of the centre of pressure. Let the axis of the parabola be taken as the axis of x,

the axis of x, and the tangent at its vertex as the axis of y, and let l denote the latus rectum. Then 19

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3 5 V3 Tel.

y =l. 13 V3

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13. An area in the form of a sector of an equilateral hyperbola, bounded by the axis, a line through the centre, and the curve, is placed in a fluid with the axis in its surface: to find the depth of the centre of pressure.

. If a be the semi-axis of the hyperbola, and a the angle between the two radii vectores of the sector, the required depth will be equal to

tan 2a + log (1 – tan a) – log (1 + tan a) ia.

cos a . (sec 2a)} – 1


14. Two equal parabolas, with the same vertex, having their axes at right angles to each other, are immersed vertically in a fluid, the axis of one of them coinciding with the surface: to determine the position of the centre of pressure of the area which is common to both.

Let l represent the latus rectum of each parabola, then, the axis of x extending vertically downwards in coincidence with the axis of one parabola, and the axis of y coinciding in the surface of the fluid with the axis of the other parabola,

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15. A flood-gate ADO (fig. 7) moves upon a vertical axis A0, the area ADO, on one side of the axis, being the quadrant of a circle, and, on the other side, a rectangle ABCO of the same altitude: to determine the width AB of the parallelogram so that the gate may just open by the pressure of the water when it has risen to the top.

If a represent the radius of the circle, and x the width of the rectangle,

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16. A hollow cube, just filled with heavy fluid, is held with one diagonal vertical : to find the centre of pressure of one of the lower faces.

If a denote the length of each edge of the cube, and if the lowest point of the cube be taken as the origin of coordinates, the axes of coordinates being two sides of a lower face, the centre of pressure of this face is given by the equations

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17. The axis of a cylindrical vessel, containing a known quantity of fluid, is inclined at a given angle to the horizon: to determine the centre of pressure of its base.

If r represent the radius of the cylinder, ce the known quantity of fluid, and a the inclination of the axis of the cylinder to the horizon, the distance of the centre of pressure from the geometrical centre of the base is equal to


40% tan a


Heterogeneous Incompressible Fluids.


If the fluid be heterogeneous, let p denote the unit of pressure at any elemental area K, then, the notation of the preceding section being retained, we shall have, instead of the formulæ there given,

Σ (p Κα) (pKy) že

y Σ (DK)

Σ (DK) If, with the increase of the depth, the density of the fluid and therefore the unit of pressure vary discontinuously, we must subdivide the fluid into a series of horizontal strata, such that the law of the variation of the density and unit of pressure shall be continuous throughout each of them. We must then take the sum of the values of each of the expressions

(pK), (pKx),
E (pKx), E (pky),

Σ () for the several strata to which the plane surface is exposed, as constituting its value in the preceding formulæ.

1. To find the centre of pressure of a parallelogram OC (fig. 8) immersed vertically in a fluid with one angle 0 at the surface, the density of the fluid being supposed to vary as the depth.

Let a, B, be the inclinations of the sides OA, OB, to the vertical; let 0A = a, OB = b. Let Pp be any elemental

a parallelogram in the area OC, its sides dx, dy, being parallel to the axes of x, y, respectively, that is, to the sides 0A, OB, produced indefinitely. Then, h being the depth of P below the surface,

h = x cos a + y cos B,

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p = jkg (x cos a + y cos B). Thus, dx dy sin (a + B) being the area of Pp, SI (z cos a + y cos By dx dy = S S (x cos a + y cos B)xdx dy. Se

(oc cos a + y cosß) dy = bxo cos'a + b’x cos a cos ß + }68 cos'ß, SS (z cosa

(x cosa+y cosß) dxdy=b(žao cos’a+ ja’bcos a cosß + jab'cosos), ST. (cos a +

cos 8) = da.


6 (ja* cos” a + jaob cos a cos ß + ļaob? cos” B): hence 2.(fa cos’ a + jab cos a cos ß + fb cos' B)

- (ja’ cosa + jab cos a cos ß + cosB) a, ū.(4a cos” a + bab cos a cos ß + 462 cos' B)

= a.(3a' cos” a + 4ab cos a cos ß + 262 cos® B). Similarly y.(46% cosa B + 6ba cos ß cos a + 4a cos” a)

0(362 cos' ß + 4ba cos ß cos a + 2a cos? a). 2. To find the centre of pressure of a rectangular board immersed vertically in a vessel containing three fluids of different densities, the length of the board being just equal to the whole depth of the fluid and being divided by the three fluid strata into three equal parts.

Let a be the length of each portion of the board, and let p, p,p', be the densities of the three fluids, reckoning from the highest. Then, at a depth x below the edge of the higher extremity of the board,

p = gpx, in the highest fluid ;
p.= gpa + gp' (x a), in the middle fluid;
p = gpa + gp'a + gp(20 – 2a), in the lowest fluid.

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Hence, m denoting the breadth of the board, we have, putting mdx for K,

2*( pm)= gom | da = goma :

x dx {:

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S. (pa + p'a – 2p"a + p”a) a da



22,34(pKx) = gm

- gma(p + kp' + $p'): and therefore

2.3 (pKx) = {gma' (13p + 10p' + 4p"). Hence

13p + 10p' + 4p"
2 = fa.
5p + 3p' + P

p" 3. A semicircular area is just immersed in a fluid, its diameter being horizontal ; the density of the fluid varies as the depth: to determine the centre of pressure.

The centre of pressure will be in the vertical radius of the semicircle, and, a denoting the radius, at a distance from the centre equal to

32 a

4. A segment of a parabola, cut off by a double ordinate to the axis, is immersed in a fluid the density of which varies as

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