Question #c9625

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Question #c9625

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Answer 1 · 2015-01-31T13:43:28

618lbforce

Explanation:

The size of the pool is unimportant, as long as it can hold the cube. Only the depth matters, because this determines the pressure (=force per unit area).

It's a pity this problem was stated in old english measures.
I'll convert to metrical units, and convert the answer back.

$1 f t = 0, 305 m$ $1ft=0,305 m$
$1 f t^{2} = {0.305}^{2} = 0.0930 m^{2}$ $1ft^2=0.305^2=0.0930m^2$

The pressure goes up by $\approx 10 k N / m^{2}$ $~~10kN//m^2$ for every meter down.
( $N$ $N$ =Newton is the metrical measure for force)

Top of the cube
Is at $9 \cdot 0.305 m = 2.745 m$ $9*0.305m=2.745m$
this gives a pressure of $27.450 k N / m^{2}$ $27.450 kN//m^2$
Multiply by area: $F = p \cdot A = 27.450 \cdot 0.093 = 2.554 k N$ $F=p*A=27.450*0.093=2.554kN$

The sides
Since pressure rises linearly with depth, we may use the average pressure which will occur halfway up the cube.
Depth halfway is $9.5 \cdot 0.305 m = 2.8975 m$ $9.5*0.305m=2.8975m$
Pressure will be $28.975 k N / m^{2}$ $28.975kN//m^2$
Multiply by area: $28.975 \cdot 4 \cdot 0.093 = 10.782 k N$ $28.975*4*0.093=10.782kN$

Answer
$F = 2.554 + 10.782 = 13.336 k N \approx 13.3 k N$ $F=2.554+10.782=13.336kN~~13.3kN$

Conversion:
$13.3 k N = 618 l b$ $13.3kN=618lb$ force.

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Question #c9625

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