Question

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

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

Answer 1

The correct option is D ${120}^o$

Let the angle $A{p}_2{p}_1=\theta$

By glancing force:

$T=W_1=100NT\cos \theta /2+T\cos \theta /2=w_{2}2T\cos \theta /2=100N2\ast 100\ast \cos \theta /2=100\cos \theta /2=\frac{1}{2}\frac{\theta }{2}=60\theta =120$

Hence option D is correct.