Chapter 14, Problem 10P

(a)

To determine

The force exerted by the water on the bottom of the pool.

Answer to Problem 10P

The force exerted by the water on the bottom of the pool is $\underset{\_}{5.88\times {10}^6\text{N downward}}$.

Explanation of Solution

Given that the dimension of the pool is $30.0\text{m}\times \text{10}\text{.0}\text{m}$, the depth of the pool is $2.00\text{m}$.

Write the expression for the pressure at the bottom of the pool due to the water in it.

  $P_b=\rho gz$                                                                                                                   (I)

Here, $P_b$ is the pressure at the bottom of the pool due to the water, $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $z$ is the depth of the pool.

Write the expression for the force exerted by the water on the bottom of the pool.

  $F_b=P_{b}A$                                                                                                                   (II)

Here, $F_b$ is the force, and $A$ is the area of the bottom of the pool.

Use equation (II) in (I).

  $F_b=\rho gzA$                                                                                                               (III)

Conclusion:

Substitute $1000{\text{kg/m}}^3$ for $\rho$, $9.80{\text{m/s}}^2$ for $g$, $2.00\text{m}$ for $h$ and $\left(30.0\text{m}\times \text{10}\text{.0}\text{m}\right)$ for $A$ in equation (III) to find $F_b$.

  $\begin{array}{c}{F}_b=\left(1000{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)\left(2.00\text{m}\right)\left(30.0\text{m}\times \text{10}\text{.0}\text{m}\right)\\ =5.88\times {10}^6\text{N}\end{array}$

The force on the bottom of the pool is directed downward.

Therefore, the force exerted by the water on the bottom of the pool is $\underset{\_}{5.88\times {10}^6\text{N downward}}$.

(b)

To determine

The force exerted by the water on each end of the pool.

Answer to Problem 10P

The force exerted by the water on each end of the pool is $\underset{\_}{\text{196}\text{kN outward}}$.

Explanation of Solution

Given that the dimension of the pool is $30.0\text{m}\times \text{10}\text{.0}\text{m}$, the depth of the pool is $2.00\text{m}$.

Equation (I) gives the expression for the pressure at the bottom of the pool due to the water in it.

  $P_b=\rho gz$

The pressure varies with depth. One a strip of height $dz$ and length $L$, the force is given by the expression,

  $dF=PdA$                                                                                                                (IV)

Write the expression for the elemental area of the strip.

  $dA=Ldz$                                                                                                                  (V)

Use equation (I) and (V) in (IV).

  $dF=\rho gzLdz$                                                                                                          (VI)

Integrate equation (VI) with limits from $z=0$ to $z=h$ to find the force.

  $\begin{array}{c}F={\displaystyle \underset{0}{\overset{h}{\int }}\rho gzLdz}\\ =\frac{1}{2}\rho gL{h}^2\\ =\left(\frac{1}{2}\rho gh\right)Lh\end{array}$                                                                                                    (VII)

Write the expression for the average pressure of water in a region with height $h$.

  $P_{\text{av}}=\frac{1}{2}\rho gh$                                                                                                          (VIII)

The product $Lh$ represents the area $A$ of the surface over which the force is measured. Thus, equation (VIII) can be modified as,

  $F=P_{\text{av}}A$                                                                                                                 (IX)

For each ends of the pool, the length is $10.0\text{m}$ and height is $2.00\text{m}$.

Conclusion:

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

  $\begin{array}{c}{P}_{\text{av}}=\frac{1}{2}\left(1000{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)\left(2.00\text{m}\right)\\ =9800\text{Pa}\end{array}$

Substitute $9800\text{Pa}$ for $P_{\text{av}}$ and $\left(10.0\text{m}\times \text{2}\text{.00}\text{m}\right)$ for $A$ in equation (IX) to find the force at each end of the pool.

  $\begin{array}{c}F=\left(9800\text{Pa}\right)\left(10.0\text{m}\times \text{2}\text{.00}\text{m}\right)\\ =1.96\times {10}^5\text{N}\\ =1.96\times {10}^5\text{N}\times \frac{1\text{kN}}{1000\text{N}}\\ =196\text{kN}\end{array}$

The force on each end of the pool is directed outward.

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

(c)

To determine

The force exerted by the water on each side of the pool.

Answer to Problem 10P

The force exerted by the water on each side of the pool is $\underset{\_}{\text{588}\text{kN outward}}$.

Explanation of Solution

Given that the dimension of the pool is $30.0\text{m}\times \text{10}\text{.0}\text{m}$, the depth of the pool is $2.00\text{m}$. It is obtained that the average pressure exerted by water in the pool is $9800\text{Pa}$.

Equation (IX) gives the force exerted by water at a given part of the pool.

  $F=P_{\text{av}}A$

For each side of the pool, the length is $30.0\text{m}$ and height is $2.00\text{m}$.

Conclusion:

Substitute $9800\text{Pa}$ for $P_{\text{av}}$ and $\left(30.0\text{m}\times \text{2}\text{.00}\text{m}\right)$ for $A$ in equation (IX) to find the force at each side of the pool.

  $\begin{array}{c}F=\left(9800\text{Pa}\right)\left(30.0\text{m}\times \text{2}\text{.00}\text{m}\right)\\ =5.88\times {10}^5\text{N}\\ =5.88\times {10}^5\text{N}\times \frac{1\text{kN}}{1000\text{N}}\\ =588\text{kN}\end{array}$

The force on each side of the pool is directed outward.

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