Question

In the figure pulley $p_1$ is fixed and $p_2$
is movable. If $W_1=W_2=100N,$ what is the angle (in degrees) $A{P}_2{P}_1$ when the system is in equilibrium?

The pulleys are massless and frictionless

Answer 1

Considering $P_2$ and $W_2$ to be system and drawing its FBD


Let the tension in the string $A{P}_2$ and $P_2{P}_1$ be T.
Accordingly tension in the entire string is T.

Considering the force on weight $W_1$, we get
$T=W_1$
Further, let $\angle A{P}_2{P}_1=2\theta$
Resolving tensions in horizontal and vertical directions and considering the forces on weight $W_2,$ we get
$2Tcos\theta =W_2$
or $2W_{1}cos\theta =W_{2}orcos\theta =\frac{1}{2}$
or $\theta ={60}^{\circ }$
So $\angle A{P}_2{P}_1=2\theta ={120}^{\circ }$