A square gate of size 1,m times 1,m is hinged | vedclass

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(C) Let the gate be of height $H = 1\,m$ and width $W = 1\,m$. The hinge is at the mid-point,i.e.,at a depth of $0.5\,m$ from the top edge of the gate

Vedclass · Accepted Answer

$\frac{\rho g}{6}$

Vedclass · Answer

(C) Let the gate be of height $H = 1\,m$ and width $W = 1\,m$. The hinge is at the mid-point,i.e.,at a depth of $0.5\,m$ from the top edge of the gate. Let $y$ be the depth from the top of the gate.The pressure at depth $y$ is $P = ho g y$.The force on an elemental strip of height $dy$ at depth $y$ is $dF_p = P 	imes W 	imes dy = ho g y 	imes 1 	imes dy = ho g y dy$.The torque $d	au$ due to this force about the hinge (located at $y = 0.5\,m$) is $d	au = dF_p 	imes (0.5 - y) = ho g y (0.5 - y) dy$.To keep the gate stationary,the total torque about the hinge must be zero. The force $F$ is applied at the bottom edge $(y = 1\,m)$,so its lever arm relative to the hinge is $0.5\,m$.$\int_0^1 ho g y (0.5 - y) dy - F 	imes 0.5 = 0$$ho g \int_0^1 (0.5y - y^2) dy = 0.5 F$$ho g [0.5 \frac{y^2}{2} - \frac{y^3}{3}]_0^1 = 0.5 F$$ho g [0.25 - 0.333] = 0.5 F$$ho g [\frac{1}{4} - \frac{1}{3}] = 0.5 F$$ho g [-\frac{1}{12}] = 0.5 F$Taking the magnitude,$F = \frac{ho g}{12 	imes 0.5} = \frac{ho g}{6}$.

$A$ square gate of size $1\,m \times 1\,m$ is hinged at its mid-point. $A$ fluid of density $\rho$ fills the space to the left of the gate. The force $F$ required to hold the gate stationary is

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