Question

In the following figure, the pulley $P_1$ is fixed and the pulley $P_2$ is movable. If $W_1=W_2=100\text{N}$, what is the the value of $\angle A{P}_2{P}_1$? The pulleys are light and frictionless.

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is D ${120}^{\circ }$

Since a single string has been used to connect $W_1,P_1,P_2$ let the tension in the entire string be $T$.
Applying the equilibrium condition on block with weight $W_1$ in vertical direction, we get:
$T=W_1=100\text{N}....\left(i\right)$

Further let $\angle A{P}_2{P}_1=2\theta$, since the vertical line passing through centre of pulley $P_2$ acts as its angle bisector.

F.B.D



Resolving tension in horizontal and vertical direction and since pulley $P_2$ is massless we get :
$\Rightarrow 2T\cos \theta =T_1$, where $T_1$ is the tension in string connecting block and pulley $P_2$
Again applying equilibrium condition on block,
$T_1=W_2$
$\Rightarrow 2T\cos \theta =W_2...\left(ii\right)$
From Eq.$\left(i\right)$
$\Rightarrow 2W_1\cos \theta =W_2$
$\because W_1=W_2=100\text{N}$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\Rightarrow \theta ={60}^{\circ }$
$\therefore \angle A{P}_2{P}_1=2\theta ={120}^{\circ }$