Question

An object with a mass of `8 kg` is on a plane with an incline of `- pi/6`. If it takes `9 N` to start pushing the object down the plane and `2 N` to keep pushing it, what are the coefficients of static and kinetic friction?

Answer 1

The static coefficient of friction is `0.7099` (4dp)
The kinetic coefficient of friction is `0.6068` (4dp)

Explanation:

For our diagram, `m=8kg`, `theta=pi/6`

If we apply Newton's Second Law up perpendicular to the plane we get:

`R-mgcostheta=0`
`:. R=8gcos(pi/6) \ \ N`

Initially it takes `9N` to start the object moving, so `D=9`. If we Apply Newton's Second Law down parallel to the plane we get:

`D+mgsin theta -F = 0`
`:. F = 9+8gsin (pi/6) \ \ N`

And the friction is related to the Reaction (Normal) Force by

`F = mu R => 9+8gsin (pi/6) = mu (8gcos(pi/6))`
`:. mu = (9+8gsin (pi/6))/(8gcos(pi/6))`
`:. mu = 0.7099051 ...`

Once the object is moving the driving force is reduced from `9N` to `2N`. Now `D=2`, reapply Newton's Second Law down parallel to the plane and we get:

`D+mgsin theta -F = 0`
`:. F = 2+8gsin (pi/6) \ \ N`

And the friction is related to the Reaction (Normal) Force by

`F = mu R => 2+8gsin (pi/6) = mu (8gcos(pi/6))`
`:. mu = (2+8gsin (pi/6))/(8gcos(pi/6))`
`:. mu = 0.6068069 ...`

So the static coefficient of friction is `0.7099` (4dp)
the kinetic coefficient of friction is `0.6068` (4dp)