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Expert-verified Found in: Page 439 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # A swimming pool has dimensions $30.0\text{}m×10.0\text{}m$ 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 ${F}_{b}=5.88×{10}^{6}N\left(down\right)$ .

(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 $F=588kN\left(outward\right)$ .

See the step by step solution

## Step 1: Pressure:

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.

## Step 2: (a) The force exerted by the water on the bottom:

The pressure on the bottom due to the water is

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

The force exerted by the water on the bottom is then

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

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

## Step 3: (b) The force exerted by the water on each end:

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,

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

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

## Step 4: (c) The force exerted by the water on each side:

On the side,

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

Hence, the force exerted by the water on each side is $F=588kN\left(outward\right)$ . ### Want to see more solutions like these? 