Water stands up to height h behind a dam as s | vedclass

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(B) The pressure at a depth $y$ from the free surface of water is $P = P_{atm} + \rho g y$. The force $dF$ on a small strip of height $dy$ at depth $y

Vedclass · Accepted Answer

$\frac{h}{3}$

Vedclass · Answer

(B) The pressure at a depth $y$ from the free surface of water is $P = P_{atm} + \rho g y$. The force $dF$ on a small strip of height $dy$ at depth $y$ is $dF = P \cdot dA = (P_{atm} + \rho g y) \cdot (a \cdot dy)$,where $a$ is the width of the dam. The total force $F$ is the integral of $dF$ from $y=0$ to $y=h$. The point of application $y_R$ (center of pressure) is given by $y_R = \frac{\int y dF}{\int dF}$. For a rectangular gate of height $h$ submerged in water,the center of pressure is at a depth of $\frac{2}{3}h$ from the free surface. Since the question asks for the point of application relative to the base $O$ (which is at depth $h$ from the surface),the distance from the base is $h - \frac{2}{3}h = \frac{h}{3}$.

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