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Let two uniform rods each of mass m and length l joined at angle θ constitute the V shaped rigid body ABC. The rigid body is suspended freely from point A and its BC part becomes horizontal and AB part remains inclined at angle θ with the horizontal when the system comes in equilibrium.
Let the midpoint of AB be P and that of BC be Q.The vertical line drawn through the point of suspension A intersects BC rod at O.The weight of AB part mg will act vertically downward through P and its line of action intersects BC at R. Again weight (mg) of BC will act downward through its mid point Q .These two weights will lie in two opposite sides of O
The perpendicular distannce of the weight acting through P from O is OR=l2cosθ.Again the perpendicular distance of the weight acting through Q from O is
OQ=BQ−BO=l2−lcosθ
So at equlibrium of two weigts
mg×OR=mg×OQ
mg×l2cosθ=mg×(l2−lcosθ)
⇒32cosθ=12
⇒cosθ=13
⇒θ=cos−1(13)=70.5
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