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?

Jul 31, 2017

If angle of inclination is $\frac{\pi}{8}$:

${\mu}_{s} = 0.580$

${\mu}_{k} = 0.449$

(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 $16$ $\text{kg}$, and the incline is $- \frac{3 \pi}{8}$.

Since the angle is $- \frac{3 \pi}{8}$ ($- {67.5}^{\text{o}}$), this would be the angle going down the incline (the topmost angle in the image above). Therefore, the actual angle of inclination is

pi/2 - (3pi)/8 = ul((pi)/8

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,

$n = m g \cos \theta = \left(16 \textcolor{w h i t e}{l} {\text{kg")(9.81color(white)(l)"m/s}}^{2}\right) \cos \left(\frac{\pi}{8}\right) = 145$ $\text{N}$

Since $24$ $\text{N}$ 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 ${f}_{s}$:

${f}_{s} = m g \sin \theta + 24$ $\text{N}$

= (16color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/8) + 24color(white)(l)"N" = 84.1 $\text{N}$

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

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

It takes $5$ $\text{N}$ of applied downward force (on top of weight) to keep the object accelerating constantly downward, then we have

${f}_{k} = m g \sin \theta + 5$ $\text{N}$

$= 60.1 \textcolor{w h i t e}{l} \text{N} + 5$ $\text{N}$ $= 65.1$ $\text{N}$

The coefficient of kinetic friction is thus

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