`mu` is coefficient of friction between all surfaces . Block A is kept over block B on inclined plane . The minimum force required such that 'A' block can accelerate along applied force is mg sin `theta + n mu g "cos" theta ` , calculate n .

mu is coefficient of friction between all surface. Block A is kept of over block B on inclined plane. The minimum force required such that 'A' block can accelerate along applied force is mg sin theta+ n mug cos theta calculate n .

In the arrangement shown in figure coefficient of friction between 5kg block and incline plane is mu=0.5 . Friction force acting on the 5kg block is

A block of weight 1 N rests on a inclined plane of inclination theta with the horizontal. The coefficient of friction between the block and the inclined plane is mu . The minimum force that has to be applied parallel to the inclined plane to make the body just move up the plane is

If coefficient of friction between both the block is mu- (1)/(2) . Find friction force acting between both block.

A block is stationary on an inclined plane If the coefficient of friction between the block and the plane is mu then .

The coefficient of friction between the block A of mass m & block B of mass 2m is mu . There is no friction between block B & the inclined plane. If the system of blocks A & B is released from rest & there is no slipping between A & B then :

Two blocks A and B are lying on the table as shown in the figure. The coefficient of friciton between block A and block B is mu . The coefficient of friction between block B and ground also mu . An external force F is applied in the direction shown. The maximum value of the force F that can be applied without moving the blocks.

A block of mass m is at rest on a rough inclined plane, whose friction coefficient is mu. Net force exerted by surface on block is :-