Question

A wooden plank of length $1m$ and uniform cross section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water up to a height of $0.5m$. The specific gravity of the plank is $0.5$ If the angle $\theta$ by the inclination of that the plank makes with the vertical in the equilibrium position (exclude the case $\theta =0)$. Find the value of $\frac{1}{{\cos }^2\theta }$.

Answer 1

Buoyancy force will act on the center of submerged part of plank ,

distance from O is $d=\frac{0.5}{2\cos \theta }$

distance of center of mass from is $a=1/2=0.5m$.

let the cross section be A and density of water be $\rho$

weight of plank is w=$1\times A\times 0.5\rho \times g$,}

writing torque about O

we get $Bd\cos \theta =W\times 0.5\times \cos \theta$

we get ${\cos }^2\theta =1/2$

so $\frac{1}{{\cos }^2\theta }=2$