Question

An object with a mass of `9` `kg` is on a plane with an incline of `- pi/4`. If it takes `15` `N` to start pushing the object down the plane and `7` `N` to keep pushing it, what are the coefficients of static and kinetic friction?

Answer 1

The coefficient of static friction is `mu_s=1.24` and the coefficient of kinetic friction is `mu_k=1.11`.

Explanation:

The negative sign on the angle is not relevant in this case: the angle of the inclined plane is `pi/4` `radians` (`45^o`). The negative sign just shows that the slope increases right-to-left rather than left-to-right.

The force perpendicular to the plane, the normal force, will be given by:

`F_N=mgsinTheta=9*9.8*sin(pi/4)=62.4` `N`

Based on this, the frictional force will be:

`F_"frict"=muF_N`

Rearranging,

`mu=(F_"frict")/(F_N)`

We will find two different values, the static friction, `mu_s`, and the kinetic friction, `mu_k`.

The force parallel to the plane due to gravity will be given by:

`F=mgcosTheta=9*9.8*cos(pi/4)=62.4` `N`

Overcoming the force of static friction requires this parallel force plus `15` `N`, a total frictional force of `77.4` `N`.

`mu_s=(F_"frict")/(F_N)=77.4/62.4=1.24`

Once it is moving, keeping it moving only requires `7` `N` in addition to the parallel force, a total of `69.4` `N`.

`mu_k=(F_"frict")/(F_N)=69.4/62.4=1.11`

Coefficients of friction do not have units.