A wooden block of mass 9 kg is at rest on an inclined plane sloped at an angle theta from the horizontal. The block is located 5 m from the bottom of the plane.

Question

A wooden block of mass 9 kg is at rest on an inclined plane sloped at an angle theta from the horizontal. The block is located 5 m from the bottom of the plane. 
A) Find the frictional force for theta = 20
B)If the angle is increased slowly the block starts sliding at an angle theta = 30. What is the coefficient of static friction?
C) If the block slides to the bottom in 2 s, what is the coefficient of kinetic friction?

Henry · Answer

M*g = 9 * 9.8 = 88.2 N. = Wt. of block.

A. Fp = 88.2*sn20= 30.2 N. = Force
parallel to the in.cline.

Fn = 88.2*Cos20 = 82.9 N. = Normal or force perpendicular to the incline.

Fs = u*Fn = u*82.9.

Fp-Fs = M*a. 30.2-82.9u = M*0 = 0, 82.9u = 30.2, u = 0.364 = Static coefficient of friction.

Fs = u*Fn = 0.364 * 82.9 = 30.2 N. = Force of static friction.

B. Fp = 88.2*sin30 = 44.1 N.

Fn = 88.2*Cos30 = 76.4 N.

Fs = u*Fn = u*76.4.

Fp-Fs = M*a.
44.1 - 76 .4u = M*0 = 0, u = 0.577.