A) \[\mu mg\]
B) \[\mu \left[ mg+\left( \frac{P}{2} \right) \right]\]
C) \[\mu \left[ mg-\left( \frac{P}{2} \right) \right]\]
D) \[\mu \left[ mg-\left( \frac{\sqrt{3}P}{2} \right) \right]\]
Correct Answer: C
Solution :
Key Idea: Sketch the free body diagrams. Resoles the force \[P\] is horizontal and vertical components. The free body diagram of the various forces acting on body is shown. \[R\] is the reaction of the surface on mass, \[{{f}_{s}}\] is frictional force. Taking the horizontal and vertical components, we have \[R=mg-P\,\sin \,\,{{30}^{\circ }}=mg-\frac{P}{2}\] Limiting frictional force is \[F=\mu R=\mu \left[ mg-\frac{P}{2} \right]\]You need to login to perform this action.
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