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 \[T=\sin 30=W\] \[T=\mu R=12N\] \[\therefore \] \[12=T\cos {{30}^{o}}\] \[T=\frac{12\times 2}{\sqrt{3}}\] \[T=\frac{24}{\sqrt{3}}\] \[\therefore \] \[T\sin 30=W\] \[W=\frac{24}{\sqrt{3}}\times \frac{1}{2}\] \[W=6.92\,N\]You need to login to perform this action.
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