In the following figure, the pulley P1 is fixed and the pulley P2 is movable. If W1=W2=100N, what is the the value of ∠AP2P1? The pulleys are light and frictionless.
The correct option is D120∘ Since a single string has been used to connect W1,P1,P2 let the tension in the entire string be T. Applying the equilibrium condition on block with weight W1 in vertical direction, we get: T=W1=100N....(i)
Further let ∠AP2P1=2θ, since the vertical line passing through centre of pulley P2 acts as its angle bisector.
Resolving tension in horizontal and vertical direction and since pulley P2 is massless we get : ⇒2Tcosθ=T1, where T1 is the tension in string connecting block and pulley P2 Again applying equilibrium condition on block, T1=W2 ⇒2Tcosθ=W2...(ii) From Eq.(i) ⇒2W1cosθ=W2 ∵W1=W2=100N ⇒cosθ=12 ⇒θ=60∘ ∴∠AP2P1=2θ=120∘