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A “V” shaped rigid body has two identical uniform arms. What must be the angle between the two arms so that when the body is hung from one end the other arm is horizontal ?
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 OThe 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 isOQ=BQ−BO=l2−lcosθSo at equlibrium of two weigtsmg×OR=mg×OQmg×l2cosθ=mg×(l2−lcosθ)⇒32cosθ=12⇒cosθ=13⇒θ=cos−1(13)=70.5
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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