Chapter 7, Problem 10RWDT

To determine

To compare: Both the tanks in terms of the pressure at the bottom and the total force on the bottom when both tanks are full.

Answer to Problem 10RWDT

The pressure at the bottom of the second tank is twice the pressure at the bottom of the first one whereas the force on the bottom of both the tanks is same.

Explanation of Solution

Given information:

There are 2 tanks, each hold 6 cubic feet of water.

Dimension of first tank: 2 feet wide, 3 feet long, and 1 foot tall.

Dimension of other tank: 3 feet wide, 1 foot long, and 2 feet tall.

Concept used:

  $\text{Pressure }=wh$ , where $w=$ weight density $\left(\text{in}\frac{\text{lb}}{{\text{ft}}^{\text{3}}}\right)$ and h is the depth in feet.

Weight density of water is $62.4\frac{\text{lb}}{{\text{ft}}^{\text{3}}}$ .

  $\text{Force}=\text{Area×Pressure}$

For the first tank,

  $\begin{array}{c}\text{Pressure }=\text{ 62}\text{.4}\times 1\\ =62.4\frac{\text{lb}}{{\text{ft}}^{\text{2}}}\end{array}$

For the second tank,

  $\begin{array}{c}\text{Pressure }=\text{ 62}\text{.4}\times 2\\ =124.8\frac{\text{lb}}{{\text{ft}}^{\text{2}}}\end{array}$

Now,

For the first tank,

  $\begin{array}{c}\text{Force at the bottom }=\left(2\times 3\right)\times 62.4,\text{ where Area at the bottom =}\left(2\times 3\right){\text{ft}}^2\\ =374.4\text{ lb}\end{array}$

For the second tank,

  $\begin{array}{c}\text{Force at the bottom }=\left(3\times 1\right)\times 124.8,\text{ where Area at the bottom =}\left(3\times 1\right){\text{ft}}^2\\ =374.4\text{ lb}\end{array}$

Conclusion:

The pressure at the bottom of the second tank $\left(124.8\frac{\text{lb}}{{\text{ft}}^{\text{2}}}\right)$ is twice the pressure at the bottom of the first one $\left(62.4\frac{\text{lb}}{{\text{ft}}^{\text{2}}}\right)$ .

The force on the bottom of both the tanks is same $\left(374.4\text{ lb}\right)$ .