Question

a block of mass m placed at rest on inclined plane of inclination theta to the horizontal. if the coefficient of friction between the block and the plane is mu then the total force the inclined plane exerts on the block is

Answer 1

We first list all forces on the block. See figure.

Weight of the block, Mg, in vertically downward direction. This force is not due to inclined plane , but it is due to gravitation.

Normal force ,N. This force is due to inclined plane. Notice that for the equilibrium in the direction perpendicular to the inclined plane, N= Mg cos (theta). Here Mg cos(theta) is component of Mg in the direction perpendicular to the inclined plane.

Now, the block is also in equilibrium in the direction parallel to inclined plane. Therefore , the frictional force exerted by inclined plane, (mu) N= (mu) Mg cos ( theta) balances the Mg sin( theta) component of Mg.

Thus, we see that there are two forces exerted on the block BY inclined plane. These are N and ( mu) Mg cos(theta). These forces are perpendicular to each other.

Therefore , the total force on the block due to inclined plane is [N^2 + ( mu)^2 N^2]^1/2 = N[1 +(mu)^2]^1/2=Mg cos(theta) [1+(mu)^2]^1/2.