The coefficients of friction between a block of mass 1.5 kg and a surface are μs = 0.70 and μk = 0.40. Assume the only other force acting on the block is that due to gravity. What is the magnitude of the frictional force (in newtons) on the block if the block is at rest and the surface is inclined at 18o

Question

The coefficients of friction between a block of mass 1.5 kg and a surface are μs = 0.70 and μk = 0.40.  Assume the only other force acting on the block is that due to gravity.  What is the magnitude of the frictional force (in newtons) on the block if the block is at rest and the surface is inclined at 18o

anonymous · Answer

|dw:1370937942756:dw|
The gravity acting along the slope =mgsin18 =1.5*10*0.6 = 9N
The maximum static friction that can act is μs*N = μs mgcos18
           =0.7*1.5*10*0.8 = 8.4N

Hence static friction will fail to balance gravity and the object will start sliding downwards.
When it starts sliding downwards ,the kinetic friction will act to retard the motion which is given by μk *N = 0.4 * 1.5*10*0.8 =4.8N