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axis of the prism, passing through G its centre of gravity. Let W be the weight of the prism, h the length of its axis, P the density of the fluid. Draw GE perpendicular to RD, GF perpendicular to SD. Take PD = & AD, QD = & BD; and bisect PQ in H. Then H is the centre of gravity of the fluid displaced; Therefore (Art. 22. Cor. 2.) GH is GH is perpengph. AD. BD. sin D = W.

dicular to AB or PQ; and

=

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and PG QG, for PH = QH, and GH is perpendicular to PQ;

.. PD2 – QD2 – 2 ED. PD + 2 FD . QD = 0;

.. AD2 – BD2 – 3ED. AD + 3 FD. BD = 0 ;

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The last term of this equation is negative, and therefore one root of it is negative; but the nature of the question excludes all negative values of AD and BD. Hence, there cannot be more than three positions of equilibrium as long as the same edge is immersed. All values of AD greater than RD, and of BD greater than SD are likewise inadmissible.

127. Two equal rods RD, SD, (fig. 71.) meeting each other at right angles, float with the angle D immersed; to find their positions of equilibrium.

Let G be the centre of gravity of the rods; P, Q, the middle points of the portions immersed; GR perpendicular to DR; GS perpendicular to DS; GH perpendicular to PQ; HN perpendicular to RD; 2c the sum of the lengths of the

immersed portions, when the weight of the fluid displaced is equal to the weight of the rods; RD=e; PD=a; QD = b. Then, a + b = c; and the centre of gravity of the fluid displaced must be in GH, it must also be in PQ, therefore H is the centre of gravity of the fluid displaced;

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The equation to PQ referred to the axes DR, DS, is

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the co-ordinates of G are e, e', therefore the equation to GH is b (y − e) = a (x − e); and H is the intersection of GH and PQ,

.. (b2 + a2) DN = ab2 + (a2 −ab) e, .. (b2 + a2) a = cb2 + (a−b) ec,

.. {(c − a)2 + a2 } a = c (c − a)2 + (2 a − c) ec,

.. 2a3 − 3ca2 + (3c − 2e) ac − (c − e) c2 = 0,

one root of this equation is c, the other two are

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128. To find (M) the metacentre of the prism RDS (fig. 70), the prism being inclined in the plane RDS.

The moment of inertia of the plane of floatation round an axis through its centre of gravity, perpendicular to RDS

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and the volume of the fluid displaced = 1h. AD. DB. sin D ;

... (Art. 26.) .h. AD. DB. sin D. HM = AB3. h,

12

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129. To find the metacentre of a cone floating with its axis vertical.

Let DC (fig. 72.) be the axis of the cone, meeting the plane of floatation in C, CA the radius of the plane of floatation, H the centre of gravity of the fluid displaced. Then DH = 2 DC; the moment of inertia of the plane of floatation

round a horizontal axis through its centre of gravity C =

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AC1; 4

π

AC,

3

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130. A conical vessel partly filled with fluid, floats in the same fluid with its axis vertical; to find whether the equilibrium of the vessel is stable or unstable.

Let DM (fig. 73.) be the axis of the cone making a very small angle with the vertical; C, c the points in which it cuts the plane of floatation and the surface of the fluid within; H, h the centres of gravity of the fluid displaced, and of the fluid contained in the cone, when the axis of the cone was vertical; M, m the points in which verticals through the centres of gravity of the fluid displaced, and of the fluid within ultimately intersect DM; G the centre of gravity of the cone. The weight of the fluid displaced, the weight of the cone, and the weight of the fluid contained in the cone, act in parallel lines through M, G, m, respectively. And the pressures of the exterior and interior fluids will tend to diminish or increase the inclination of DC according as (weight of fluid bda); mG is greater or less than (weight of fluid displaced); MG.

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weight of fluid bda: weight of fluid displaced = cD3 : CD3; therefore the equilibrium of the cone will be stable or unstable

according as eD' (GD-1D) is greater or less than

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131. An An open vessel containing fluid, is made to revolve round a vertical axis with the angular velocity a; to find the form of the surface of the fluid.

Let the axis of revolution be made the axis of x; and let z be measured downwards. The forces on the fluid at any point P whose co-ordinates are x, y, ≈, are, g acting downwards, and a2 √(x2 + y2) acting in the direction of a perpendicular from P on the axis of revolution; this force may be resolved into a2 √(x2 + y2) or a2x, in a direction parallel to the axis of x, and a3y, in a direction parallel to the axis of We have then,

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√(x2 + y2)

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y.

X = a2x, Y = a2y, Z=g; .. d2p = pa2x, dyp = pa3y, d2p=pg;

·· p = p { } a2 (x2 + y2) +gx} + C.

Let the surface of the fluid cut the axis of ≈ at the depth c below the origin; and let II be the pressure of the atmosphere.

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At any point in the surface of the fluid p = II, therefore the equation to the surface of the fluid is

0 = a2 (x2 + y2) + 2g (≈ − c) ;

the equation to a paraboloid generated by the revolution of a

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132. A hollow parallelopiped OF (fig. 74.) just filled with fluid, revolves round the edge OC, which is vertical, with the angular velocity a; to find the pressure on the side BF, and the centre of pressure of the side BF.

Let OA = a, OB = b, OC = c; OA, OB, OC, the axes of x, y, z, respectively; p the pressure at the point (x, y, z).

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Draw HPR, KQS, parallel to BD; MP, NQ parallel to BE; and let BM = x, BH = %.

The pressure at P = p { } (x2 + b2) + g*};

the pressure on PQ = p { } (x2 + b2) + gx} MN. HK ult.

the

=

pressure on KP p { } ( {} x3 + b2 ∞) + gxx}. HK ult.

the pressure on KR = p { } (} a3 + b2 a) + gxa}. HK ult. the pressure on BR = p { } (} a3 + b2 a) ≈ + { g÷2 a} ;

the pressure on BF = p { } (} a3 + b2 a) c + {gc2a}.

The moment of the pressure on PQ round BE

=

p { } a2 (x2 + b2) + gx} x . MN. HK ult.

the moment of the pressure on KP

= p { } a2 ( } x2 + 1 b2x2) + 1⁄2 g≈x2}. HK ult.

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