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

If block is rest on inclined plane then ,it's weight is resolved in two componenants one mgcos(theta) and mgsin (theta)

downward force=mgsin (theta)

frictional force is =mu

then frictional force opposes the downward force

hence net force= mgsin (theta)-mu

Answer 2

Answer:

If block is rest on inclined plane then ,it's weight is resolved in two componenants one mgcos(theta) and mgsin (theta)

downward force=mgsin (theta)

frictional force is =mu

then frictional force opposes the downward force

hence net force= mgsin (theta)-mu