Question

A block of mass 10 kg is placed on an inclined plane. When the angle of inclination is , the block just begins to slide down the plane. The force of static friction is

Answer 1

Answer:

Ff = u N     where  Ff is the frictional force, u the coefficient of friction and N is                               the normal force

N = W cos theta = m g cos theta   the component of weight perpendicular to the plane

Fp = W sin theta = m g sin theta   the component of weight along the plane

Fp = Ff      when the block just begins to slide

Then     u W cos theta = W sin theta

u = sin theta / cos theta = tan theta = tan 30 = .57

Answer 2

Answer:

Ff = u N     where  Ff is the frictional force, u the coefficient of friction and N is                               the normal force

N = W cos theta = m g cos theta   the component of weight perpendicular to the plane

Fp = W sin theta = m g sin theta   the component of weight along the plane

Fp = Ff      when the block just begins to slide

Then     u W cos theta = W sin theta

u = sin theta / cos theta = tan theta = tan 30 = .57