Question

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

Answer 1

`mu_k ~~ 0.113`
`mu_s ~~ 0.169`

Explanation:

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

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

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`