Question

An object with a mass of `12 kg` is on a plane with an incline of `-(3 pi)/8`. If it takes `4 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 `2.5031` (4dp)
The kinetic coefficient of friction is `2.4587` (4dp)

Explanation:

For our diagram, `m=12kg`, `theta=(3pi)/8`

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

`R-mgcostheta=0`
`:. R=12gcos((3pi)/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+12gsin ((3pi)/8) \ \ N`

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

`F = mu R => 4+12gsin ((3pi)/8) = mu (12gcos((3pi)/8))`
`:. mu = (4+12gsin ((3pi)/8))/(12gcos((3pi)/8))`
`:. mu = 2.5030953 ...`

Once the object is moving the driving force is reduced from `4N` to `2N`. Now `D=3`, reapply Newton's Second Law down parallel to the plane and we get:

`D+mgsin theta -F = 0`
`:. F = 2+12gsin ((3pi)/8) \ \ N`

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

`F = mu R => 2+12gsin ((3pi)/8) = mu (12gcos((3pi)/8)`
`:. mu = (2+12gsin ((3pi)/8))/(12gcos((3pi)/8))`
`:. mu = 2.4586544 ...`

So the static coefficient of friction is `2.5031` (4dp)
the kinetic coefficient of friction is `2.4587` (4dp)