The coefficients of friction between a block of mass 2.0 kg and a surface are μs = 0.55 and μk = 0.2. 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 21o? Use g = 9.79 m/s2.

Question

The coefficients of friction between a block of mass 2.0 kg and a surface are μs = 0.55 and μk = 0.2.  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 21o?  Use g = 9.79 m/s2.

anonymous · Answer

since the magnitude of component of gravitational force along the inclined plane is less than the maximum value of static frictional force. Therefore to prevent sliding, static friction will be there and that equal to  mgsin(21).

anonymous · Answer

F(down forse)=m*g*sin(a) and f(friction force)=u*N where N=m*g*cos(a)
therefore f(fr)=u*m*g*cos(a) but here we dont know which coefficient it can be static or kinetic.That is why we have to first Find the value of  F than compare both to kinetic and static friction.For clarity f(kinetic friction)=uk*m*g*cos(a) and f(static friction)=us*m*g*cos(a).if F<f(s) then answer is f(s) otherwise if F>f(s) answer is f(k).
Summary:just calculate the value of F and f(s) by putting value(numbers) inside of mentioned formulas and to find the answer use previous sentence.