Question

A wooden plank of length $1\text{m}$ 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 $0.5\text{m}$. The specific gravity of the plank is $0.5$. Then the angle $\theta$ that the plank makes with vertical in the equilibrium position is _______.$($in degree$)$

$($Exclude the case $\theta =0^{\circ }$)

Answer 1

$W=$ Weight
$F=$ Upthrust

$OA=\frac{OB}{2}=\frac{0.5}{2\cos \theta }=0.25\sec \theta$

About point $O$, clockwise moment of $W=$ anticlockwise moment of $F$

$\Rightarrow W\left(\frac{L}{2}\sin \theta \right)=F\left(OA\sin \theta \right)=F\left(0.25\sec \theta \sin \theta \right)$

$\Rightarrow \cos \theta =\frac{F}{2W}=\frac{0.5\sec \theta \left(A\right)\left(1\right)g}{2\left(1\right)\left(A\right)0.5g}$

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

$\Rightarrow \theta ={45}^{\circ }$