Consider a block kept on an inclined plane (inclined at 45 ° ) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane ( µ ) is equal to : .

Step-by-step solution

The forces acting on block is shown in above figure. Here, normal force, N = m g cos 45 ° and frictional force, f = μ N . Balancing the force along incline, we have F 1 = m g   sin 45 ° + f = m g   sin 45 ° + μ N ⇒ F 1 = m g 2 + μ m g   cos 45 ° ⇒ F 1 = m g 2 1 + μ Now, the forces acting on block to just push it up on inclined plane is shown below. Balancing the force along incline, we have F 2 = m g   sin 45 ° - f = m g   sin 45 &#17