Question

An object with a mass of `2 kg` is on a plane with an incline of `- pi/12`. If it takes `1 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

If angle of inclination is `(5pi)/12`:

`mu_s = 3.93`

`mu_k = 4.72`

(see explanation regarding angle)

Explanation:

We're asked to find the coefficient of static friction `mu_s` and the coefficient of kinetic friction `mu_k`, with some given information.

We'll call the positive `x`-direction up the incline (the direction of `f` in the image), and the positive `y` direction perpendicular to the incline plane (in the direction of `N`).

There is no net vertical force, so we'll look at the horizontal forces (we WILL use the normal force magnitude `n`, which is denoted `N` in the above image).

We're given that the object's mass is `2` `"kg"`, and the incline is `-(pi)/12`.

Since the angle is `-(pi)/12` (`-15^"o"`), this would be the angle going down the incline (the topmost angle in the image above). Therefore, the actual angle of inclination is

`pi/2 - (pi)/12 = ul((5pi)/12`

The formula for the coefficient of static friction `mu_s` is

`f_s >= mu_sn`

Since the object in this problem "breaks loose" and the static friction eventually gives way, this equation is simply

`f_s = mu_sn`

Since the two vertical quantities `n` and `mgcostheta` are equal,

`n = mgcostheta = (2color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos((5pi)/12) = 5.08` `"N"`

Since `1` `"N"` is the "breaking point" force that causes it to move, it is this value plus `mgsintheta` that equals the upward static friction force `f_s`:

`f_s = mgsintheta + 1` `"N"`

`= (2color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin((5pi)/12) + 1color(white)(l)"N" = 20.0` `"N"`

The coefficient of static friction is thus

`mu_s = (f_s)/n = (20.0cancel("N"))/(5.08cancel("N")) = color(red)(ul(3.93`

The coefficient of kinetic friction `mu_k` is given by

`f_k = mu_kn`

It takes `5` `"N"` of applied downward force (on top of weight) to keep the object accelerating constantly downward, then we have

`f_k = mgsintheta + 5` `"N"`

`= 19.0color(white)(l)"N" + 5` `"N"` `= 24.0` `"N"`

The coefficient of kinetic friction is thus

`mu_k = (f_k)/n = (24.0cancel("N"))/(5.08cancel("N")) = color(blue)(ul(4.72`

These values are fairly high, and it's also interesting to see that the kinetic friction force is greater than the static friction force...but this is merely theoretical!