Chapter 14.2, Problem 8PP

To determine

The total push force of water on the diver.

Answer to Problem 8PP

The push force of the water on the diver is $3.03\times {10}^5\text{N}$ .

Explanation of Solution

Given info:

Depth of the diver, $h=10\text{m}$

Pressure experienced by the diver, $P=202\text{kPa}$

Surface area of the diver is, $A=1.50{\text{m}}^2$

Formula Used:

Consider the conversion factor.

  $\begin{array}{c}202\text{kPa}=202\text{kPa}\times \frac{1000\text{N}/{\text{m}}^2}{1\text{kPa}}\\ =2.02\times {10}^5\text{N}/{\text{m}}^2\end{array}$

Write the expression to calculate the total push force of the water on the diver.

  $F=P\cdot A$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}F=\left(2.02\times {10}^5\text{N}/{\text{m}}^2\right)\left(1.50{\text{m}}^2\right)\\ =3.03\times {10}^5\text{N}\end{array}$

Conclusion:

Thus, the total push force of the water on the diver is $3.03\times {10}^5\text{N}$ .