In Figure, an open tube of length L=1.8mand cross-sectional area A= 4.6$\mathit{c}{\mathit{m}}^{\mathbf{2}}$cmis fixed to the top of a cylindrical barrel of diameter$\mathit{D}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{}\mathit{m}$and height $\mathit{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{}\mathit{m}$. The barrel and tube are filled with water (to the top of the tube).Calculate the ratio of the hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel. Why is that ratio not equal to $\mathbf{1}\mathbf{.}\mathbf{0}$? (You need not consider the atmospheric pressure.)

1. Length of open tube, $L=1.8m$
2. Cross sectional area of tube, $A=4.6c{m}^{2}or4.6\times {10}^{-4}{m}^2$
3. Diameter of cylindrical bar,$D=1.2m$
4. Height of the barrel,$H=1.8m$

We can find the hydrostatic force on the bottom of the barrel from pressure due to the complete water column. Then we can find the gravitational force on the water contained in the barrel from the pressure on the water in the barrel. From these two values, we can easily find the ratio of the hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel.

Formulae:

Force applied on a body in terms of pressure, F = pA (i)

Pressure applied on a fluid surface, p = pgh (ii)

Area of the bottom of the barrel is given by:
$$\begin{gathered} A={\mathrm{\pi r}}^2 \\ =\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2(\because \mathrm{Radius}=\frac{\mathrm{Diameter}}{2}) \end{gathered}$$

Total pressure due to the water column using equation (ii),$p=pg(L+H)$

Substituting the values of A and p in equation (i), thehydrostatic force on the bottom of the barrel is given by:

role="math" localid="1657194416544" $F_1=\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2\mathrm{pg}(L+H)..............\left(1\right)$

But, the gravitational forceon the water contained in the barrel using equation (i) and the above values is given by:

$F_2=\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^2\mathrm{pg}H..............\left(2\right)$

The ratio of the hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel is given by:

$$\begin{gathered} \frac{F_1}{F_2}=\frac{\mathrm{\pi }{\left({\displaystyle \frac{\mathrm{D}}{2}}\right)}^2(\mathrm{L}+\mathrm{H})}{\mathrm{\pi }{\left(\frac{\mathrm{D}}{2}\right)}^{2}H} \\ =\frac{(L+H)}{H} \\ =\frac{1.8+1.8}{1.8} \\ =2 \end{gathered}$$

Therefore, the ratio of the hydrostatic force on the bottom of the barrel to the gravitational force on the water contained in the barrel is 2 and is not equal to1 because of the additional pressure exerted by the water in the tube.