Question

# In the figure pulley p1 is fixed and p2 is movable. If W1=W2=100 N, what is the angle (in degrees) AP2P1 when the system is in equilibrium? The pulleys are massless and frictionless

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Solution

## Considering P2 and W2 to be system and drawing its FBD Let the tension in the string AP2 and P2P1 be T. Accordingly tension in the entire string is T. Considering the force on weight W1, we get T=W1 Further, let ∠AP2P1=2θ Resolving tensions in horizontal and vertical directions and considering the forces on weight W2, we get 2T cos θ=W2 or 2W1 cosθ=W2 or cos θ=12 or θ=60∘ So ∠AP2P1=2θ=120∘

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