Question

A wooden plank of length $1$ m and uniform cross section is hinged at one end to the bottom of a tank as shown in figure (13-W5). The tank is filled with water up to a height of $0.5$ m. 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

$\frac{mgxl}{2\sin \theta }=FT(l-\frac{x}{2})\sin \theta$ …. (i)

Also $F_T=wt.offluiddisplaced=\left[\right(l-x\left)\right]x{\rho }_{w}g$ …(ii)

And $m=\left(lA\right)0.5{\rho }_w$… (iii)

Where A is the area of cross section of the rod

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

$\left(lA\right)\frac{0.5{\rho }_{w}gx\ell }{2\sin \theta }=\left[\right(\ell -x\left)A\right]{\rho }_{w}gx(\ell -\frac{x}{2})\sin \theta$

Here, $l=1m$

$\therefore (1–x{)}^2=0.5\Rightarrow x=0.293m$

From the diagram

$\cos \theta =\frac{0.5}{1–x}=\frac{0.5}{0.707}\Rightarrow \theta ={45}^{\circ }$