Chapter 6.6, Problem 40E

To determine

To Calculate: The hydrostatic force on a semicircular gate at the bottom of a dam with water

Answer to Problem 40E

The total hydrostatic force on the gate is $11263467N$ .

Explanation of Solution

Given: The tank contains water.

• The diameter of the gate is $4m$ .
• The height of the gate below water surface is $12-2=10m$ .

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.

The radius of the gate is $\frac{4}{2}=2m$

The depth of the center of the gate from the surface of water is $10m$

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 gate, then

• The depth of the point $p$ is $10-r\sin \theta =10-2\sin \theta$ .
• The width of a horizontal strip of the gate 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

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

  $=2r^2{\cos }^2\theta d\theta$

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

  $P_{\text{θ}}=\rho gh=\rho g\left(10-2\sin \theta \right)$

Therefore, the force on this thin strip is

  $\begin{array}{l}dF=P_{\theta }d{A}_{\theta }=\rho g(10-2\sin \theta )\cdot 2r^2{\cos }^2\theta d\theta \\ =784000(10-2\sin \theta )\cdot {\cos }^2\theta d\theta \end{array}$

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

  $F={{\displaystyle \int }}_0^{\pi }dF={{\displaystyle \int }}_0^{\pi }784000(10-2\sin \theta )\cdot {\cos }^2\theta d\theta$

  $={784000\left(\frac{2{\cos }^3\theta }{3}+10\left(\frac{\theta }{2}+\frac{sin2\theta }{4}\right)\right)]}_0^{\pi }$

  $=784000\left(2\left(\frac{-2}{3}\right)+10\left(\frac{\pi }{2}+0\right)\right)$

  $=784000\times 14.3\overline{6}$

  $=11263467$

Therefore, the total hydrostatic force on the gate is $11263467N$ .