Chapter 19, Problem 48A

A hole of area $12{\text{cm}}^2$ is made in the bottom of a barge 1.5 m below the freshwater surface. A board is held over the hole from inside the barge to stop water from leaking in. Show that the force necessary to hold the board in position is 18 N.

To determine

To show: Force applied on the board to stop leaking of water from the hole is 18N.

Explanation of Solution

Given:

Area of the hole = $12{\text{cm}}^2$

Depth of the barge = $1.5\text{m}$

Formula used:

The pressure is,

  $P=\rho gh$ .

Here, $\rho$ the density of the water, $g$ is the acceleration due to gravity, and $h$ is the height of the hole below the fresh water.

Calculation:

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$ , 9.8 m/s² for $g,$ and $1.5\text{m}$ for $h$ .

  $\begin{array}{c}P=\left(1000\text{kg}/{\text{m}}^3\right)\left(9.8\text{m}/{\text{s}}^2\right)(1.5\text{m})\\ =14700\text{kg}/\text{m}\cdot {\text{s}}^2\end{array}$

Also, the pressure is

  $P=\frac{F}{A}$

Here, $A$ is the area of the hole.

Rearrange the equation for $F$ .

  $F=PA$

Substitute $14700\text{kg}/\text{m}\cdot {\text{s}}^2$ for $P$ and $12{\text{cm}}^2$ for $A$ .

  $\begin{array}{c}F=\left(14700\text{kg}/\text{m}\cdot {\text{s}}^2\right)\left(12{\text{cm}}^2\right)\\ =\left(14700\text{kg}/\text{m}\cdot {\text{s}}^2\right)\left(12{\text{cm}}^2\right){\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\\ =\left(\text{17}\text{.64kg}\cdot \text{m}/{\text{s}}^2\right)\left(\frac{1\text{N}}{\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ \approx 18\text{N}\end{array}$

Conclusion:

Therefore, the force required is $18\text{N}$ .