Chapter 6.6, Problem 39E

(a).

To determine

To Calculate: The hydrostatic force on the shallow end of a pool.

Answer to Problem 39E

The total force on the shallow end of the pool is $5625lb$ .

Explanation of Solution

Given Information:

• The depth of the shallow end is $3ft$ .
• The width of the pool is $20ft$ .

Concept Used:

• The hydrostatic pressure of a liquid of density $\rho$ at a depth $d$ from the surface of the liquid is $P=\rho gd$ .
• The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

Assume the density of water is $\text{ρ}g=62.5lb/f{t}^3$

The pressure in the liquid at any point that is at a depth $x$ from the surface of water is

  $P_x=\rho gx=62.5x$

The area of an infinitesimally thin strip of thickness $dx$ on the shallow end of the pool with width $w$ at the depth $x$ is

  $dA=width\times thickness=wdx=20dx$

Therefore, the force on this thin strip of area is

  $dF=P_{x}dA=62.5x\cdot 20dx$

  $=1250xdx$

Now the total force on this shallow end is the integral of $dF$ from $x=0$ (surface of water) to $x=3$ (depth of shallow end ). So

  $F={{\displaystyle \int }}_0^{3}dF={{\displaystyle \int }}_0^{3}1250xdx$

  $={1250\frac{x^2}{2}]}_0^3$

  $=1250\times \frac{3^2-0^2}{2}=5625$

Therefore, the total force on the shallow end of the pool is $5625lb$ .

(b).

To determine

To Calculate: The hydrostatic force on the deeper end of a pool.

Answer to Problem 39E

The total force on the shallow end of the pool is $50625lb$ .

Explanation of Solution

Given Information:

• The depth of the deeper end is $9ft$ .
• The width of the pool is $20ft$ .

Concept Used:

• The hydrostatic pressure of a liquid of density $\rho$ at a depth $d$ from the surface of the liquid is $P=\rho gd$ .
• The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

Assume the density of water is $\text{ρ}g=62.5lb/f{t}^3$

The pressure in the liquid at any point that is at a depth $x$ from the surface of water is

  $P_x=\rho gx=62.5x$

The area of an infinitesimally thin strip of thickness $dx$ on the deeper end of the pool with width $w$ at the depth $x$ is

  $dA=width\times thickness=wdx=20dx$

Therefore, the force on this thin strip of area is

  $dF=P_{x}dA=62.5x\cdot 20dx$

  $=1250xdx$

Now the total force on this deeper end is the integral of $dF$ from $x=0$ (surface of water) to $x=9$ (depth of deeper end ). So

  $F={{\displaystyle \int }}_0^{9}dF={{\displaystyle \int }}_0^{9}1250xdx$

  $={1250\frac{x^2}{2}]}_0^9$

  $=1250\times \frac{9^2-0^2}{2}=50625$

Therefore, the total force on the shallow end of the pool is $50625lb$ .

(c).

To determine

To Calculate: The hydrostatic force on a side of a pool is $48750lb$ .

Answer to Problem 39E

The total force on the side is $48750lb$

Explanation of Solution

Given Information:

• The depth of the shallow end is $3ft$ .
• The depth of the deeper end is $9ft$ .
• The length of the pool is $40ft$
• The width of the pool is $20ft$ .

Concept Used:

• The hydrostatic pressure of a liquid of density $\rho$ at a depth $d$ from the surface of the liquid is $P=\rho gd$ .
• The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

Assume the density of water is $\text{ρ}g=62.5lb/f{t}^3$

The pressure in the liquid at any point that is at a depth $x$ from the surface of water is

  $P_x=\rho gx=62.5x$

The length of the side $l$ at a depth $x$ is

  $\begin{array}{l}{l}_x=40\text{ for }0\le x\le 3\\ and\\ l_x=60-\frac{40}{6}x\text{ for }3\le x\le 9\end{array}$

The area of an infinitesimally thin strip of thickness $dx$ on the shallow end of the pool with width $w$ at the depth $x$ is

  $\begin{array}{l}d{A}_x=width\times thickness=l_{x}dx=40dx\text{ for }0\le x\le 3\\ and\\ d{A}_x=width\times thickness=l_{x}dx=\left(60-\frac{40}{6}x\right)dx\text{ for }3\le x\le 9\end{array}$

Therefore, the force on this thin strip of area is

  $\begin{array}{l}dF=P_{x}dA=62.5x\times 40dx=2500xdx\text{ for }0\le x\le 3\\ and\\ dF=P_{x}dA=62.5x\times \left(60-\frac{40}{6}x\right)dx\text{ for }3\le x\le 9\end{array}$

Now the total force on the side is the integral of $dF$ from $x=0$ (surface of water) to $x=9$ (deepest point). So

  $F={{\displaystyle \int }}_0^{9}dF={{\displaystyle \int }}_0^{3}2500xdx+{{\displaystyle \int }}_3^{9}62.5x\left(60-\frac{40}{6}x\right)dx$

  $={2500\frac{x^2}{2}]}_0^3+3750\frac{x^2}{2}-{\frac{2500}{6}\times \frac{x^3}{3}]}_3^9$

  $=2500\times \left(\frac{3^2-0^2}{2}\right)+3750\left(\frac{9^2-3^2}{2}\right)-\frac{2500}{18}\times \left(9^3-3^3\right)$

  $=11250+135000-97500$

  $=48750$

Therefore, the total force on the shallow end of the pool is $48750lb$ .

(d).

To determine

To Calculate: The hydrostatic force on the bottom of a pool.

Answer to Problem 39E

The total force on the of the pool is $300000lb$ .

Explanation of Solution

Given Information:

• The depth of the shallow end is $3ft$ .
• The depth of the deeper end is $9ft$ .
• The length of the pool is $40ft$
• The width of the pool is $20ft$ .

Concept Used:

• The hydrostatic pressure of a liquid of density $\rho$ at a depth $d$ from the surface of the liquid is $P=\rho gd$ .
• The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

Assume the density of water is $\text{ρ}g=62.5lb/f{t}^3$

The pressure in the liquid at any point that is at a depth $x$ from the surface of water is

  $P_x=\rho gx=62.5x$

If the height of an infinitesimally thin strip on the bottom of the pool along the width of the pool is $dx$ , then the thickness of the strip is

  $dt=\frac{40}{6}dx$

The area of this infinitesimally thin strip of thickness $dx$ on the bottom of the pool in the horizontal planewith width $w$ at the depth $x$ is

  $dA=width\times thickness=wdt=20\times \frac{40}{6}dx=\frac{800}{6}dx$

Therefore, the force on this thin strip of area is

  $dF=P_{x}dA=62.5x\cdot \frac{800}{6}xdx$

  $=\frac{50000}{6}xdx$

Now the total force on this end is the integral of $dF$ from $x=0$ (surface of water) to $x=3$ (depth of shallow end ). So

  $F={{\displaystyle \int }}_3^{9}dF={{\displaystyle \int }}_3^9\frac{50000}{6}xdx$

  $={\frac{50000}{6}\cdot \frac{x^2}{2}]}_3^9$

  $=\frac{50000}{6}\times \frac{72}{2}=300000$

Therefore, the total force on the bottom of the pool is $300000lb$ .