Chapter 15, Problem 8P

(a)

To determine

The force exerted by the water on the bottom.

Answer to Problem 8P

The force exerted by the water on the bottom is $5.88\times {10}^6\text{ N, down}$.

Explanation of Solution

Write the equation for pressure on the bottom due to water.

  $P_b=\frac{F_b}{A}$

Here, $P_b$ is the pressure on the bottom due to water, $F_b$ is the force exerted by the water on the bottom and $A$ is the area of the bottom.

Rewrite the above equation for $F_b$.

  $F_b=P_{b}A$        (I)

Write the equation for the pressure on the bottom due to water.

  $P_b=\rho gh$        (II)

Here, $\rho$ is the density of water, $g$ is the acceleration due to gravity and $h$ is the depth of water.

Conclusion:

Substitute $1000{\text{ kg/m}}^3$ for $\rho$ , $9.80{\text{ m/s}}^2$ for $g$ and $2.00\text{ m}$ for $h$ in equation (II) to find $P_b$ .

  $\begin{array}{c}{P}_b=\left(1000{\text{ kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)\left(2.00\text{ m}\right)\\ =1.96\times {10}^4\text{ Pa}\end{array}$

Substitute $1.96\times {10}^4\text{ Pa}$ for $P_b$ and $\left(30.0\text{ m}\right)\left(10.0\text{ m}\right)$ for $A$ in equation (I) to determine $F_b$ .

  $\begin{array}{c}{F}_b=\left(1.96\times {10}^4\text{ Pa}\right)\left(30.0\text{ m}\right)\left(10.0\text{ m}\right)\\ =5.88\times {10}^6\text{ N, down}\end{array}$

Therefore, the force exerted by the water on the bottom is $5.88\times {10}^6\text{ N, down}$.

(b)

To determine

The force exerted by the water on each end.

Answer to Problem 8P

The force exerted by the water on each end is $196\text{ kN, outward}$.

Explanation of Solution

Pressure varies with depth.

Write the equation for the force on a strip of height $dz$ and length $L$.

  $dF=PdA$        (III)

Here, $dF$ is the force on a strip of height $dz$ and length $L$ , $P$ is the pressure on the strip and $dA$ is the area of the strip.

Write the equation for $P$.

  $P=\rho gz$

Write the equation for $dA$.

  $dA=Ldz$

Here, $L$ is the length of the strip and $dz$ is the height.

Put the above two equations in equation (III).

  $\begin{array}{c}dF=\rho gzLdz\\ =\rho gLzdz\end{array}$

Integrate the above equation over the entire height of the strip.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{h}{\int }}dF}={\displaystyle \underset{0}{\overset{h}{\int }}\rho gLzdz}\\ F=\rho gL{\left[\frac{z^2}{2}\right]}_0^h\\ =\frac{1}{2}\rho gL{h}^2\\ =\frac{1}{2}\rho ghLh\end{array}$        (IV)

Write the equation for the total area.

  $A=Lh$        (V)

The pressure at the top of the pool is zero.

Write the equation for the pressure at surface of the pool.

  $P_t=0$        (VI)

Here, $P_t$ is the pressure at surface of the pool.

Write the equation for the average pressure.

  $P_{\text{avg}}=\frac{P_t+P_b}{2}$

Here, $P_{\text{avg}}$ is the average pressure.

Put equations (II) and (VI) in the above equation.

  $\begin{array}{c}{P}_{\text{avg}}=\frac{0+\rho gh}{2}\\ =\frac{1}{2}\rho gh\end{array}$        (VII)

Put equations (V) and (VII) in equation (IV).

  $F=P_{\text{avg}}A$        (VIII)

Put equation (II) in equation (VII).

  $P_{\text{avg}}=\frac{1}{2}{P}_b$        (IX)

Conclusion:

Substitute $1.96\times {10}^4\text{ Pa}$ for $P_b$ in equation (IX) to find $P_{\text{avg}}$ .

  $\begin{array}{c}{P}_{\text{avg}}=\frac{1}{2}\left(1.96\times {10}^4\text{ Pa}\right)\\ =9.80\times {10}^3\text{ Pa}\end{array}$

The length of each end will be $10.0\text{ m}$.

Substitute $10.0\text{ m}$ for $L$ and $2.00\text{ m}$ for $h$ in equation (V) to find $A$ .

  $\begin{array}{c}A=\left(10.0\text{ m}\right)\left(2.00\text{ m}\right)\\ =20.0{\text{ m}}^2\end{array}$

Substitute $9.80\times {10}^3\text{ Pa}$ for $P_{\text{avg}}$ and $20.0{\text{ m}}^2$ for $A$ in equation (VIII) to find $F$.

  $\begin{array}{c}F=\left(9.80\times {10}^3\text{ Pa}\right)\left(20.0{\text{ m}}^2\right)\\ =196\times {10}^3\text{ N}\frac{1\text{ kN}}{1000\text{ N}}\\ =196\text{ kN}\end{array}$

The force will be directed outward.

Therefore, the force exerted by the water on each end is $196\text{ kN, outward}$.

(c)

To determine

The force exerted by the water on each side.

Answer to Problem 8P

The force exerted by the water on each side is $588\text{ kN, outward}$.

Explanation of Solution

Equation (VIII) gives the force exerted on each side of the swimming pool by water.

Conclusion:

The length of each side will be $30.0\text{ m}$ .

Substitute $30.0\text{ m}$ for $L$ and $2.00\text{ m}$ for $h$ in equation (V) to find $A$.

  $\begin{array}{c}A=\left(30.0\text{ m}\right)\left(2.00\text{ m}\right)\\ =60.0{\text{ m}}^2\end{array}$

Substitute $9.80\times {10}^3\text{ Pa}$ for $P_{\text{avg}}$ and $60.0{\text{ m}}^2$ for $A$ in equation (VIII) to find $F$.

  $\begin{array}{c}F=\left(9.80\times {10}^3\text{ Pa}\right)\left(60.0{\text{ m}}^2\right)\\ =588\times {10}^3\text{ N}\frac{1\text{ kN}}{1000\text{ N}}\\ =588\text{ kN}\end{array}$

The force will be directed outward.

Therefore, the force exerted by the water on each side is $588\text{ kN, outward}$.