Chapter 10, Problem 14P

(a)

To determine

The total force and the absolute pressure acting on the bottom of swimming pool is to be found depth of 2 m and bottom surface of 22.0 m by 8.5 m.

Answer to Problem 14P

The total force acting on the bottom of swimming pool is $2.27\times {10}^{7}N$ .

Absolute pressure acting on the bottom of swimming pool is $1.21\times {10}^{5}Pa$ .

Explanation of Solution

Given:

Cross-section of bottom, $A=22.0\times 8.5m^2=187m^2$

Depth of swimming pool, $d=2.0m$

Atmospheric pressure, $P_a=1.013\times {10}^{5}Pa$

Density of water, $\rho ={10}^3{\text{kg/m}}^{\text{3}}$

Formula used:

Absolute pressure, $P=P_a+\rho gh$

Force, $F=PA$

Calculation:

Substituting values in formula for absolute pressure,

  $\begin{array}{l}P=1.013\times {10}^5+({10}^3)(9.8)(2)\\ P=1.21\times {10}^{5}Pa\end{array}$

Substituting values in formula for force,

  $\begin{array}{l}F=(1.21\times {10}^5)(187)\\ F=2.27\times {10}^{7}N\end{array}$

Conclusion:

The total force acting on the bottom of swimming pool is $2.27\times {10}^{7}N$

Absolute pressure acting on the bottom of swimming pool is $1.21\times {10}^{5}Pa$

(b)

To determine

The pressure acting at the side of pool near the bottom.

Answer to Problem 14P

The pressure acting at the side of pool near the bottom is

  $1.21\times {10}^{5}Pa$ .

Explanation of Solution

Given:

Atmospheric pressure, $P_a=1.013\times {10}^{5}Pa$

Density of water, $\rho ={10}^3{\text{kg/m}}^{\text{3}}$

Formula used:

Absolute pressure,

  $P=P_a+\rho gh$

Calculation:

Substituting values in formula for absolute pressure,

  $\begin{array}{l}P=1.013\times {10}^5+({10}^3)(9.8)(2)\\ P=1.21\times {10}^{5}Pa\end{array}$

Conclusion:

The pressure acting at the side of pool near the bottom is $1.21\times {10}^{5}Pa$ .