An object with a mass of 5 kg is on a plane with an incline of  -(5 pi)/12 . If it takes 12 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?

Jul 22, 2017

If angle of inclination is $\frac{\pi}{12}$:

${\mu}_{s} = 0.521$

${\mu}_{k} = 0.310$

(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 $5$ $\text{kg}$, and the incline is $- \frac{5 \pi}{12}$.

Since the angle is $- \frac{5 \pi}{12}$ ($- {75}^{\text{o}}$), this would be the angle going down the incline (the topmost angle in the image above). Therefore, the actual angle of inclination is

$\frac{\pi}{2} - \frac{5 \pi}{12} = \frac{\pi}{12}$

Which is much more realistic considering the object remains stationary (due to static friction) unless a force acts on it.

The formula for the coefficient of static friction ${\mu}_{s}$ is

${f}_{s} \ge {\mu}_{s} n$

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

${f}_{s} = {\mu}_{s} n$

Since the two vertical quantities $n$ and $m g \cos \theta$ are equal,

$n = m g \cos \theta = \left(5 \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s}}^{2}\right) \cos \left(\frac{\pi}{12}\right) = 47.4$ $\text{N}$

Since $12$ $\text{N}$ is the "breaking point" force that causes it to move, it is this value plus $m g \sin \theta$ that equals the upward static friction force ${f}_{s}$:

${f}_{s} = m g \sin \theta + 12$ $\text{N}$

= (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/12) + 12color(white)(l)"N" = 24.7 $\text{N}$

The coefficient of static friction is thus

mu_s = (f_s)/n = (24.7cancel("N"))/(47.4cancel("N")) = color(red)(0.521

The coefficient of kinetic friction ${\mu}_{k}$ is given by

${f}_{k} = {\mu}_{k} n$

It takes $2$ $\text{N}$ of applied downward force (on top of weight) to keep the object accelerating constantly downward, then we have

${f}_{k} = m g \sin \theta + 2$ $\text{N}$

$= 12.7 \textcolor{w h i t e}{l} \text{N} + 2$ $\text{N}$ $= 14.7$ $\text{N}$

The coefficient of kinetic friction is thus

mu_k = (f_k)/n = (14.7cancel("N"))/(47.4cancel("N")) = color(blue)(0.310