Question

The bottom of a rectangular swimming pool is 16 m × 20 m. If the atmospheric pressure above the swimming pool changes from 737 to 761 mm of mercury, determine the amount by which the force on the bottom of the pool increases? (Assume the density of mercury is 13.6 ✕ 103 kg/m3.)

Answer 1

First, let's calculate the increase in the pressure:

$761-737=24\text{ mmHg}$

Now, let's convert this difference to Pascal:

$24\text{ mmHg}=24\cdot133.322\text{ Pa}=3199.73\text{ Pa}$

Then, let's find the area of the bottom of the pool:

$A=16\cdot20=320\text{ m^^b2}$

Having the difference of pressure and the area, we can find the difference of force:

$\begin{gathered} P=\frac{F}{A}\\ \\ F=P\cdot A\\ \\ F=3199.73\cdot320\\ \\ F=1023913.6\text{ N} \end{gathered}$