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 figure. The tank is filled with water up to a height of $0.5m$. The specific gravity of the plank is $0.5$. Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position ( exclude the case $\theta =0$)

Answer 1

Taking moment about O

$\frac{mgxl}{2sing\theta }=FT(l-\frac{x}{2})sin\theta ⟹\left(i\right)$

Also $FT=$wt. of fluid displaced $=(l-x)\ast \rho wg⟹\left(ii\right)$

And $m=\left(lA\right)\ast 0.5\rho w⟹\left(iii\right)$

Where A is the area of cross section of the rod

From (i), (ii) and (iii)

$\frac{\left(lA\right)\ast 0.5\rho wgxl}{2sin\theta }=\left[(l-x)A\right]\rho wgx\left(\frac{l-x}{2}\right)sin\theta$

Here, $l=1m$

$\therefore {(1–x)}^2=0.5$

$x=0.293m$

From the diagram

$cos\theta =\frac{0.5}{1-x}=\frac{0.5}{0.707}$

$\theta ={45}^o$