Question

The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.

Answer 1

Answer:

$26852726.19\ \text{N}$

$57.52^{\circ}$

Explanation:

r = Radius of circle = 7 m

w = Width of dam = 60 m

h = Height of the dam will be half the radius = $\dfrac{r}{2}$

A = Area = $rw$

V = Volume = $w\dfrac{\pi r^2}{4}$

Horizontal force is given by

$F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times 7\times 60\\\Rightarrow F_x=14420700\ \text{N}$

Vertical force is given by

$F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}$

Resultant force is

$F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}$

The hydrostatic force on the dam is $26852726.19\ \text{N}$.

The direction is given by

$\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}$

The line of action is $57.52^{\circ}$.