Chapter 6.6, Problem 31E

To determine

To Calculate: The hydrostatic force on the lateral face of a semicircular plate dipped in water

Answer to Problem 31E

The total hydrostatic force on one side of the plate is $3095800N$ .

Explanation of Solution

Given: The tank contains water with density $1000kg/m^3$ .

• The diameter of the plate is $6m$ .
• The height of the plate above water is $1m$ .

Concept Used:

• The hydrostatic pressure of a water of density $\rho$ at a depth $d$ from the surface of the water is $P=\rho \text{gd}$ .
• The force on a surface due to pressure $P$ exerted is $F=PA$ , where $A$ is the area of the surface being considered.

If $\text{θ}$ is the angle with the horizontal made by a radius passing through a point $p$ located on the circular edge of the plate, then

• The depth of the point is $rsin\text{θ}-1$ .
• The width of horizontal strip of the plate passing through the point is $2rcos\text{θ}$ .
• The length of an infinitesimal section of perimeter around the point $p$ making an angle $d\text{θ}$ with the center of the disc is $rd\text{θ}$ .

The height of the horizontal strip of disc that correspondsto this small section of perimeter is $rd\text{θ}\cdot cos\text{θ}$ Therefore, the area of the infinitesimally thin strip is

  $d{A}_{\text{θ}}=width\times height=2\text{rcosθ}\times \text{rdθcosθ}$

  $=2r^{2}co{s}^2\text{θ}d\text{θ}=18co{s}^2\text{θ}d\text{θ}$

and the hydrostatic pressure at the level of point $p$ is

  $P_{\text{θ}}=\text{ρ}g\left(rsin\text{θ}-1\right)=\text{ρ}g\left(3sin\text{θ}-1\right)$

Therefore, the force on this thin strip is

  $\begin{array}{l}dF=P_{\theta }d{A}_{\theta }=\rho g(r\sin \theta -1)\cdot 2r^{2}co{s}^2\theta d\theta \\ =18\rho g(3sin\theta -1)\cdot {\cos }^2\theta d\theta \\ =176400(3sin\theta -1)\cdot {\cos }^2\theta d\theta \end{array}$

Now the total force on this lateral surface is the integral of $dF$ from $\theta ={\tan }^{-1}\frac{1}{3}$ (surface of the water) to $\theta =\pi -{\tan }^{-1}\frac{1}{3}$ (surface of the water on the other side). So

  $F={{\displaystyle \int }}_{{\tan }^{-1}\frac{1}{3}}^{\pi -{\tan }^{-1}\frac{1}{3}}dF={{\displaystyle \int }}_{{\tan }^{-1}\frac{1}{3}}^{\pi -{\tan }^{-1}\frac{1}{3}}176400(3sin\theta -1)\cdot {\cos }^2\theta d\theta$

  $={176400(-{\cos }^3\theta -\frac{\theta }{2}-\frac{sin2\theta }{4})]}_{{\tan }^{-1}\frac{1}{3}}^{\frac{\pi }{2}}\times 2$

  $=352800\left[-\left(\frac{\pi }{2}\right)+\left({\left(\frac{2\sqrt{2}}{3}\right)}^3+\frac{ta{n}^{-1}\frac{1}{3}}{2}+\frac{\sqrt{2}}{9}\right)\right]$

  $=358200\times 8.64$

  $=3095800N$

Therefore, the total hydrostatic force on one side of the plate is $3095800N$ .