Question

(1) A block is placed on a wooden plank, which is initially horizontal. One end of the plank is slowly raised to make it more and more inclined, and for a while the block stays in place on the plank and doesn't slide due to static friction. Finally, when the plank reaches an incline of 56.3^o above horizontal, the block begins to slide. What is the  coefficient of static friction between the block and the plank?

(2) Continuing the story... after the block begins sliding down the incline at 56.3^o, you measure its acceleration down the plank as 7.42 m/s². What is the coefficient of kinetic friction between the block and the plank?

These will be clicker questions on Wednesday, and we will work together through some problems involving tension and pairs of objects.

Answer 1

Part A.

When the block just starts to slide at that moment net force on block will be zero along the incline, So

Using force balance:

F_net = W*sin - Fs

W = Weight of block = m*g

Fs = static friction force = *N = *m*g*cos

F_net = 0, So

W*sin - Fs = 0

m*g*sin - *m*g*cos = 0

sin = *cos

tan =

= tan 56.3 deg

= 1.49944 = 1.50 = Coefficient of static friction

Part B.

Now when box starts moving, there will be kinetic friction force applied, So Now using force balance on block

F_net = W*sin - Fk

W = Weight of block = m*g

Fk = kinetic friction force = *N = *m*g*cos

F_net = m*a (from Newton's 2nd law), So

m*a = W*sin - Fk

m*a = m*g*sin - *m*g*cos

a = g*sin - *g*cos

= (g*sin - a)/(g*cos )

Using given values:

= (9.81*sin 56.3 deg - 7.42)/(9.81*cos 56.3 deg)

= 0.136 = Coefficient of kinetic friction

Let me know if you've any query.