question_answer

In the figure given, the system is in equilibrium. What is the maximum value that W can have if the friction force on the 40 N block cannot exceed 12.0 N?

A)  3.45 N                                 

B)  6.92 N

C)  10.35N                                

D)  12.32 N

Correct Answer: B

Solution :

                For equilibrium at point P                 \[T=T\cos {{30}^{o}}\] and        \[T=\sin 30=W\]                 \[T=\mu R=12N\] \[\therefore \]  \[12=T\cos {{30}^{o}}\]                 \[T=\frac{12\times 2}{\sqrt{3}}\]                 \[T=\frac{24}{\sqrt{3}}\] \[\therefore \]  \[T\sin 30=W\]                                 \[W=\frac{24}{\sqrt{3}}\times \frac{1}{2}\]                                 \[W=6.92\,N\]