Water stands upto height $h$ behind the dam as shown in the figure. The front view of the dam gate is also shown in the adjoining figure. Density of water is $\rho$ and acceleration due to gravity is $g$. If atmospheric pressure force is also considered, then the point of application of total force acting on the dam due to water above $O$ is

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Aicharya · Accepted Answer

Force on dam $=$ Pressure of water $	imes$ Area $=\{($ density of water $)$ (acceleration due to gravity) (height of water 2$\} 	imes\{$ (width of dam) (height of water)\} $ \begin{aligned} F & =\left(ho g \frac{h}{2}ight) 	imes[(b-h) \cdot h] \ & =\frac{ho g}{2}\left(b h^2-h^3ight) \end{aligned} $ At equilibrium, atmospheric pressure $=$ pressure on dam So, $\frac{d F}{d h}=0$ $ \begin{array}{rlrl} & \Rightarrow & \frac{\partial}{\partial h}\left\{\frac{ho g}{2}\left(b h^2-h^3ight)ight\} & =0 \ \Rightarrow & \frac{ho g}{2}\left(2 b h-3 h^2ight) & =0 \ \Rightarrow & 2 b h-3 h^2 & =0 \ \Rightarrow & & 2 b=3 h \quad \Rightarrow \quad b & =3 h / 2 \ & & b-h=\frac{3 h}{2}-h & =h / 2 \end{array} $ is the location where the total weight of water acts at a particular point. To find the point of action of total force, $ y_R=\frac{I_{x c}}{y_c A}+y_c $ where, $y_R=$ location where point of force acts, $A=$ area, $y_c=$ location where total weight acts $=h / 2$ and $I_{x c}=$ moment of inertia (here rectangular plate). $ =\frac{1}{12} A h^2=\frac{1}{12} b h^3 $ So, $ \begin{aligned} y_R & =\frac{(1 / 12) b h^3}{(1 / 2) h \cdot b h}+\frac{1}{2} h \ & =\frac{1}{6} h+\frac{1}{2} h=\frac{2}{3} h \end{aligned} $ (from the top) So, $y_R=\frac{h}{3}$ (above the base $)=\frac{2}{3} h$ (from the top $)$

Water stands upto height $h$ behind the dam as shown in the figure. The front view of the dam gate is also…

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