Chapter 7.5, Problem 25E

a.

To determine

To find: the fluid force working against each end of the tank.

Answer to Problem 25E

Total fluid force against each end of the tank is $209.7\text{3lb}$ .

Explanation of Solution

Given information:

Dimension of the base is $2\times 4$ ft

Height of the tank is $2$ ft

Formula used:

The work to move an object from $x=a$ to $x=b$ is:

  $W={\displaystyle \underset{a}{\overset{b}{\int }}F(x)dx}$ .

Total force on an end = pressure $\times$ area of a thin horizontal strip

Calculation:

Force on an end = $62.4\times 2\Delta y$ .

As, it is filled to within 2 inches at the top.

Hence, depth of the water = $\frac{11}{6}\text{ft}$ .

The total force is

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{11}{6}}{\int }}2\left(62.4\right)ydy}=2\left(62.4\right){\displaystyle \underset{0}{\overset{\frac{11}{6}}{\int }}ydy}\\ =124.8{\left(\frac{y^2}{2}\right)}_0^{\frac{11}{6}}\\ =62.4\times \frac{11}{6}\times \frac{11}{6}\\ \approx 209.73\text{lb}\end{array}$

b.

To determine

To find: if the tank is sealed and stood on end so that one of the square ends is the base then what happens to the fluid forces on the rectangular sides.

Answer to Problem 25E

The fluid force is doubled under the given condition.

Explanation of Solution

Given information:

Dimension of the base is $2\times 4$ ft.

Height of the tank is $2$ ft.

Formula used:

The work to move an object from $x=a$ to $x=b$ is: $W={\displaystyle \underset{a}{\overset{b}{\int }}F(x)dx}$ .

Calculation:

Sides are twice as length as the ends. Therefore, the force is doubled to be $419.4\text{7lb}$ .

The depth is doubled to $\frac{1}{2}\left(\frac{11}{6}\right)\text{ft}=\frac{11}{3}\text{ft}$ .

The force on side is,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\frac{11}{3}}{\int }}2\left(62.4\right)ydy}=2\left(62.4\right){\displaystyle \underset{0}{\overset{\frac{11}{3}}{\int }}ydy}\\ =124.8{\left(\frac{y^2}{2}\right)}_0^{\frac{11}{3}}\\ =62.4\times \frac{11}{3}\times \frac{11}{3}\\ \approx 838.93lb\end{array}$

As, $838.9\text{3lb}=2\times 419.47\text{lb}$ .

Hence the force is doubled.