Question

An object with a mass of `8 kg` is on a plane with an incline of `- pi/12`. If it takes `12 N` to start pushing the object down the plane and `5 N` to keep pushing it, what are the coefficients of static and kinetic friction?

Answer 1

the static coefficient of friction is `0.4264` (4dp)
the kinetic coefficient of friction is `0.3340` (4dp)

Explanation:

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

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

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

Initially it takes `12N` to start the object moving, so `D=12`. If we Apply Newton's Second Law down parallel to the plane we get:

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

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

`F = mu R => 12+8gsin (pi/12) = mu (8gcos(pi/12))`
`:. mu = (12+8gsin (pi/12))/(8gcos(pi/12))`
`:. mu = 0.426409 ...`

Once the object is moving the driving force is reduced from `12N` to `5N`. Now `D=5`, reapply Newton's Second Law down parallel to the plane and we get:

`D+mgsin theta -F = 0`
`:. F = 5+8gsin (pi/12) \ \ N`

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

`F = mu R => 5+8gsin (pi/12) = mu (8gcos(pi/12))`
`:. mu = (5+8gsin (pi/12))/(8gcos(pi/12))`
`:. mu = 0.333974 ...`

So the static coefficient of friction is `0.4264` (4dp)
the kinetic coefficient of friction is `0.3340` (4dp)