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Q10 P

Expert-verifiedFound in: Page 439

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A swimming pool has dimensions $30.0\text{}m\times 10.0\text{}m$**** and a flat bottom. When the pool is filled to a depth of $20.0\text{}m$** **with fresh water, what is the force exerted by the water on (a) the bottom? (b) On each end? (c) On each side?**

(a) The force exerted by the water on the bottom is ${F}_{b}=5.88\times {10}^{6}N\left(down\right)$ .

(b) The force exerted by the water on each end is $F=196kN\left(outward\right)$ .

(c) The force exerted by the water on each side is $F=588kN\left(outward\right)$ .

The pressure in a fluid is the force per unit area exerted by the fluid on a surface:

$P=\frac{F}{A}$

Where, P is the pressure, A is area, and F is the force exerted on the fluid.

The pressure on the bottom due to the water is

${P}_{b}=\rho gY\phantom{\rule{0ex}{0ex}}=1.96\times {10}^{4}\text{}Pa$

The force exerted by the water on the bottom is then

${F}_{b}={P}_{b}A={P}_{b}\left(L\times B\right)\phantom{\rule{0ex}{0ex}}=\left(1.96\times {10}^{4}Pa\right)\left(30.\text{0 m}\times 10.0\text{m}\right)\phantom{\rule{0ex}{0ex}}=5.88\times {10}^{6}N\text{}\left(down\right)$

Hence, the force exerted by the water on the bottom is ${F}_{b}=5.88\times {10}^{6}N\left(down\right)$ .

Pressure varies with depth. On a strip of height dy and length L, the force

$F=PdA\phantom{\rule{0ex}{0ex}}=PLdY\phantom{\rule{0ex}{0ex}}=\rho gYLdY$

Which gives the integral as below.

$F=\underset{0}{\overset{h}{\int}}\rho ghLdY\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\rho gL{h}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\rho gh\right)Lh\phantom{\rule{0ex}{0ex}}={P}_{average}A$

On each end,

$F={P}_{average}A\phantom{\rule{0ex}{0ex}}=\left(9.8\times {10}^{3}\text{}Pa\right)\left(20.0{m}^{2}\right)\phantom{\rule{0ex}{0ex}}=196kN\left(outward\right)$

Hence, the force exerted by the water on each end is role="math" localid="1663669149245" $196kN\left(outward\right)\phantom{\rule{0ex}{0ex}}$ .

On the side,

$F={P}_{average}A\phantom{\rule{0ex}{0ex}}=\left(9.8\times {10}^{3}\text{}Pa\right)\left(60.0{m}^{2}\right)\phantom{\rule{0ex}{0ex}}=588kN\left(outward\right)$

Hence, the force exerted by the water on each side is $F=588kN\left(outward\right)$ .

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