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 0 that the plank makes with the vertical
in the equilibrium position. (Exclude the case ( = 0).

Answer 1

For equilibrium Fnet. = 0 and τnet = 0

Taking moment about O

mg x ℓ / 2 sing θ = FT (ℓ - x / 2 ) sin θ …. (i)

Also FT = wt. of fluid displaced = [(ℓ - x )] x ρw g …(ii)

And m = (ℓ A) 0.5 ρw … (iii)

Where A is the area of cross section of the rod

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

(ℓ A) 0.5 ρwg x ℓ/ 2 sin θ = [(ℓ - x) A] ρw g x (ℓ - x / 2) sin θ

Here, ℓ = 1 m

∴ (1 – x )2 = 0.5 ⇒ x = 0.293 m

From the diagram

cos θ = 0.5 / 1 – x = 0.5 / 0.707 ⇒ θ = 45°