A wooden plank of length 1m and unform cross-section is hinged at one end to bottom of a tank as shown in figure. The tank is filled with water upto 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

`theta=45^@`

Explanation

Let `y` be the length of the plank inside water.
`y=0.5/(costheta)`
let `A` be the cross section area of the plank. Then buoyant force on it is `F_(b)=Vrho_(w)g=(Ay)rho_(w)g`

Since plank is in rotational equilibrium so `sumvectau_(0)=0`
or `mgxxl/2sintheta-F_(b)xxy/2sintheta=0` or `mgl-F_(b)xxy=0`
or `(Alxx0.5)gl-(Ay)d_(2)y=0` or `0.5^(2)=y^(2)`
or `0.5xx(1)^(2)=(0.5/(costheta))^(2)` or `cos^(2)theta=1/2`
`:. theta=45^@`