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the moment of the pressure on KR

= p { } a2 ( — a2 + 1 b2 a2) + 1 gza2} . HK ult.

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= p { // a2 (x2 + b2) + gx} ≈ . MN. HK ult.

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Hence if X, Z, be the co-ordinates of the centre of

of BF, referred to the axes BD, BE,

pressure

X { } a2 ( }} a2 + b2) + 1⁄2 gc} = 1 a2 († a3 + 1 b2 a) + 1 gca.

Z { } a2 ( } a2 + b2) + { gc} = { a2 (} a2 + b2) c + } gc2.

133. Two plates of glass meeting in the vertical Cy (fig. 75.), and making a very small angle e with each other, are immersed in water; to find the figure of the water elevated between them by capillary attraction.

Let one of the plates meet the surface of the fluid between them in PQ, and the plane of undisturbed surface in Ca.

=

Draw PN parallel to Cy. Let CN= x, PN Y,

the distance between the plates at P = ex;

H

Therefore (Art. 48.) exy =

nearly, the equation to a rect

g

angular hyperbola of which Ca, Cy, are the asymptotes.

134. A wire, the area of a section of which = K, can just sustain a weight W without breaking; to find the greatest pressure that can be applied to a fluid contained in a hollow cylinder of the same substance as the wire, without bursting it, a being the radius of the cylinder, and e its thickness.

Let ML (fig. 25.) be a portion of the cylinder; MK, HL perpendicular to its axis; MH, KL parallel to its axis; p the pressure of the fluid. The area of the section MH = e. MH,

therefore it can sustain a tension

e.

MH

W; and (Art. 49.)

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2e

A pressure

-

ка

W might be applied to a hollow sphere of

the same radius and thickness without bursting it.

135. To find the time of emptying a vertical prism or cylinder through a small orifice in its base.

Let A be the area of the base of the prism, the area of the orifice, the depth of the orifice below the surface of the

M

fluid at the end of the time t from the beginning of the motion. Then, (Art. 58.)

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if the depth of the orifice below the surface of the fluid was a when t = 0.

0 = C-24a;

.. √(.2g)k. t = 2A(√ a−√x);

and the whole time of emptying

=

A

g

136. To find the time of emptying a hollow sphere through a small orifice in its vertex.

Let a be the radius of the sphere, κ the area of the orifice, a the depth of the orifice below the surface of the fluid at the end of the time t from the beginning of the motion.

The area of the surface of the fluid at the end of the time t

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137. To determine the motion of a fluid oscillating in an inverted syphon PDB (fig. 76.) of uniform bore.

Let P, Q be the extremities of the column of fluid at the end of the time t from the beginning of the motion; A, B, the extremities of the column of fluid when at rest; a, ẞ the angles between AP, BQ and the vertical MN; AP = 8, K the area of section of the tube. The moving force on the fluid = gpk.MN ἔρκ..

= gpk(AP cos a + BQ cos B) = gp (cos a + cos ẞ)AP.

a

The mass of the fluid = pê.ADB, and, since the bore of the tube is uniform, every part of the fluid moves with the same velocity, therefore the effective accelerating force at any point tending to make the fluid return to its position of equilibrium

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138. A weight is raised by a rope wound round the axle of an undershot wheel; to find the velocity of the wheel.

Let K be the area of each float-board; u the velocity of the wheel; v the velocity of the stream; a the radius of the wheel; b the radius of the axle; W the weight. The relative velocity of the stream is vu, and, therefore, the force with which it impels the wheel = pK (v – u). And when the velocity of the wheel is uniform, this force is in equilibrium with the weight W, .. bW = a 1 p K (v – u)2.

The work performed by a water wheel is measured by the product of the weight lifted multiplied by the velocity of the

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a

The wheel would be kept at rest by a weightpKv2, therefore the work performed by the wheel is a maximum when the weight lifted is of the weight that would keep the wheel at

rest.

139. To find the position of the rudder of a ship, when the effect of the rudder in turning the ship is a maximum.

Let AP (fig. 33.) be the keel of the ship, PE perpendicular to the rudder. The resolved part of the resistance on the rudder estimated in a direction perpendicular to AP,

∞ (cos APE). sin APE. (Art. 69. Cor. 2.);

And this is a maximum when

0 = (cos APE)3 -- 2 (sin APE)2. cos APE,

or sin APE = //3.

The theory of resistances (Art. 68.) is the same as that given in the Note, Page 188, of Mr. Moseley's Hydrostatics. The correctness of the application of the theorem (Art. 53.) in this case, appears doubtful.

140. Example of the comparison of the specific gravities of two fluids.

A glass flask being filled with mercury at 20,6, the mercury appeared to weigh 1340,893 grammes; when filled with water at 20,5, the water appeared to weigh 98,7185 grammes; the weight of the air contained in the flask = 0.1186 grammes; therefore

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