Question

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

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

Answer 1

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

Let the tension in the string $A{P}_2$ and $P_2{P}_1$ be $T$. Considering the forces on block $W_1$, we get
$T=W_1$
Let $\angle A{P}_2{P}_1=2\theta$


Resolving tensions in horizontal and vertical directions and considering the forces on pulley$P_2$, we get
$2T\cos \theta =W_2$
Since $W_1=W_2=T$
$\Rightarrow 2W_1\cos \theta =W_1$
$\Rightarrow \cos \theta =1/2\Rightarrow \theta ={60}^{\circ }$
Therefore $\angle A{P}_2{P}_1=2\theta ={120}^{\circ }$