An object with a mass of 10 kg10kg is on a plane with an incline of - pi/4 π4. If it takes 12 N12N to start pushing the object down the plane and 8 N8N to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Oct 1, 2017

mu_k ~~ 0.113μk0.113
mu_s ~~ 0.169 μs0.169

Explanation:

The minimum force to start pushing will help to overcome the static friction, so we can calculate the static friction from there.

Self created image, ( using MS paint )Self created image, ( using MS paint )

The Gravitational force acting on the block :
F_g = m*g = 10 * 10= 100 NFg=mg=1010=100N

Taking components along the incline and perpendicular to the incline, F_g cos θ = F_g cos 45° = 50sqrt(2)
and, F_g sin θ = F_g sin 45° = 50sqrt(2)

Since there is no motion perpendicular to the incline, the net force in that direction will be zero.

F_g cos θ = F_n = 50sqrt(2) ( where F_n is normal force. )

Static friction, f_s = mu_s * F_n

According to the question, the value of static friction force happens to be 12N.

implies 12 = mu_s * ( 50 sqrt(2))

implies mu_s ~~ 0.169

Now the kinetic friction will be applicable when the body starts to move.

Kinetic friction, f_k = mu_k * F_n

According to the question, the value of static friction force happens to be 8N.

implies 8 = mu_k * ( 50 sqrt(2))

implies mu_k ~~ 0.113