Question

An object with a mass of `16 kg` is on a plane with an incline of `-(3 pi)/8`. If it takes `24 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 `pi/8`:

`mu_s = 0.580`

`mu_k = 0.449`

(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 `16` `"kg"`, and the incline is `-(3pi)/8`.

Since the angle is `-(3pi)/8` (`-67.5^"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 - (3pi)/8 = ul((pi)/8`

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 = (16color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/8) = 145` `"N"`

Since `24` `"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 + 24` `"N"`

`= (16color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/8) + 24color(white)(l)"N" = 84.1` `"N"`

The coefficient of static friction is thus

`mu_s = (f_s)/n = (84.1cancel("N"))/(145cancel("N")) = color(red)(ul(0.580`

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"`

`= 60.1color(white)(l)"N" + 5` `"N"` `= 65.1` `"N"`

The coefficient of kinetic friction is thus

`mu_k = (f_k)/n = (65.1cancel("N"))/(145cancel("N")) = color(blue)(ul(0.449`