question_answer

An inclined plane has an inclination \[\theta \] with horizontal. A body of mass m rests on it. If the coefficient of friction between the body and the plane is \[\mu ,\] then the minimum force that needs to be applied parallel to the inclined plane is

A)  \[mg\,\,\sin \,\theta \]

B)  \[\mu mg\,\,\cos \,\theta \]

C)  \[\mu mg\,\,\cos \,\theta +mg\,\,\sin \,\theta \]

D)  \[\mu mg\,\,\cos \,\theta -mg\,\,\sin \,\theta \]

Correct Answer: D

Solution :

The free body diagram showing the various forces acting on the block are as follows: The frictional force \[({{F}_{k}})\] acts opposite to direction of motion of block, given by \[{{F}_{k}}=\mu R\] where R is reaction of the plane on block. \[R=mg\,\,\cos \theta \] \[\therefore \] \[{{F}_{k}}=\mu mg\,\,\cos \theta \] Also, downward force is \[F'=mg\,\,\sin \theta \] The resultant of F' and F provides the necessary acceleration to the block. \[\therefore \]   Resultant force upwards \[=\mu mg\,\cos \theta -mg\sin \theta \]