Question

A square gate of size $1m\times 1m$ 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

• A. $\frac{\rho g}{3}$
• B. $\frac{1}{2}\rho g$
• C. $\frac{\rho g}{6}$
• D. None of these

Answer 1

Answer: (C)

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

$dF=\rho gy.dby=\rho gbydy$

$\Rightarrow F(0\to \frac{h}{2})=\rho gb{\int }_0^{\frac{h}{2}}ydy$

$\frac{1}{8}\rho gb{h}^2$

$\tau (0\to \frac{h}{2})=\int dF>(\frac{h}{2}-y)$

$={\int }_0^{\frac{h}{2}}(\frac{h}{2}-y).\rho gbydy$

$=\rho gb{\int }_0^{\frac{h}{2}}(\frac{h}{2}y-y^2)dy$

$=\rho gb{[\frac{h}{2}\frac{y^2}{2}-\frac{y^3}{3}]}_0^{\frac{h}{2}}$

$=\rho gb{h}^3(\frac{1}{16}-\frac{1}{24})$

$=\frac{1}{48}\rho gb{h}^3$

$\tau (\frac{h}{2}-h)={\int }_{\frac{h}{2}}^h(y-\frac{h}{2})\rho gbydy$

$=\rho gb{[\frac{y^3}{3}-\frac{h}{2}.\frac{y^2}{2}]}_{\frac{h}{2}}^h$

$=\rho gb[\frac{1}{3}(h^3-\frac{h^3}{8})-\frac{h}{4}(h^2-\frac{h^2}{4})]$

$=\rho gb{h}^3[\frac{1}{3}.\frac{7}{8}-\frac{3}{16}]$

$=\rho gb{h}^3\frac{14-9}{48}$

$=\frac{5\rho gb{h}^3}{48}$

$\therefore (\frac{5}{48}-1)\rho gb{h}^3=F\times \frac{h}{2}$

$\frac{1}{12}\rho gb{h}^3=\frac{1}{2}Fh$

$\Rightarrow F=\frac{1}{6}\rho gb{h}^2=\frac{1}{6}\rho g$