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Grade 11Mechanics

A wooden plank of length 1 m and uniform cross-section is hinged at one end to the botto of a tank as shown in fig The tank is filled with water upto a height 0.5 m. The specific gravity of the plank is 0.5. Find the angle θ that the plank makes with the in the equilibrium position. (Exclude the case θ = 0°)

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Profile image of Radhika Batra
12 Years agoGrade 11
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1 Answer

Profile image of Kevin Nash
12 Years ago
Hello Student,
Please find the answer to your question
(a) 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 ρw g 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°
Thanks
Kevin Nash
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