Question

An object with a mass of `8 kg` is on a plane with an incline of `- pi/8`. If it takes `4 N` to start pushing the object down the plane and `1 N` to keep pushing it, what are the coefficients of static and kinetic friction?

Answer 1

The static coefficient of friction is `0.4694` (4dp)
The kinetic coefficient of friction is `0.4280` (4dp)

Explanation:

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

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

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

Initially it takes `4N` to start the object moving, so `D=4`. If we Apply Newton's Second Law down parallel to the plane we get:

`D+mgsin theta -F = 0`
`:. F = 4+8gsin (pi/8) \ \ N`

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

`F = mu R => 4+8gsin (pi/8) = mu (8gcos(pi/8))`
`:. mu = (4+8gsin (pi/8))/(8gcos(pi/8))`
`:. mu = 0.469437 ...`

Once the object is moving the driving force is reduced from `4N` to `1N`. Now `D=1`, reapply Newton's Second Law down parallel to the plane and we get:

`D+mgsin theta -F = 0`
`:. F = 1+8gsin (pi/8) \ \ N`

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

`F = mu R => 1+8gsin (pi/8) = mu (8gcos(pi/8))`
`:. mu = (1+8gsin (pi/8))/(8gcos(pi/8))`
`:. mu = 0.428019 ...`

So the static coefficient of friction is `0.4694` (4dp)
the kinetic coefficient of friction is `0.4280` (4dp)