Chapter 6.6, Problem 35E

To determine

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

Answer to Problem 35E

The total hydrostatic force on one side of the plate is $23927lb$ .

Explanation of Solution

Given: The tank contains water.

• The diameter of the plate is $10ft$ .
• The height of the plate below water is $2ft$ .

Concept Used:

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

Assume density of water is $62.5lb/f{t}^3$

The radius of the plate is $\frac{10}{2}=5$

Therefore, the depth of the center of the plate from the surface of water is $2+5=7ft$

So, 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 $p$ is $7-r\sin \theta =7-5\sin \theta$ .
• The width of a horizontal strip of the plate passing through the point is $2r\cos \theta$ .
• 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 (thickness) of the horizontal strip of disc that corresponds to this small section of perimeter is $rd\theta \cdot \cos \theta$

Therefore, the area of this infinitesimally thin strip is

  $d{A}_{\theta }=width\times height=2r\cos \theta \times rd\theta \cos \theta$

  $=2r^2{\cos }^2\theta d\theta$

  $=50{\cos }^2\theta d\theta$

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

  $P_{\text{θ}}=\rho gh=\rho g\left(7-5\sin \theta \right)$

Therefore, the force on this thin strip is

  $\begin{array}{l}dF=P_{\theta }d{A}_{\theta }=\rho g(7-5\sin \theta )\cdot 2r^2{\cos }^2\theta d\theta \\ =3125(7-5\sin \theta )\cdot {\cos }^2\theta d\theta \end{array}$

Now the total force on this lateral surface is the integral of $dF$ from $\theta =0$ to $\theta =\pi$ . So

  $F={{\displaystyle \int }}_0^{\pi }dF={{\displaystyle \int }}_0^{\pi }3125(7-5\sin \theta )\cdot {\cos }^2\theta d\theta$

  $={3125\left(\frac{5{\cos }^3\theta }{3}+7\left(\frac{\theta }{2}+\frac{sin2\theta }{4}\right)\right)]}_0^{\pi }$

  $=3125\left(5\left(\frac{-2}{3}\right)+7\left(\frac{\pi }{2}+0\right)\right)$

  $=3125\times 7.65\overline{6}$

  $=23927$

Therefore, the total hydrostatic force on one side of the plate is $23927lb$ .