# An object with a mass of 6 kg is on a plane with an incline of  -(3 pi)/8 . If it takes  15 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?

Sep 2, 2017

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

${\mu}_{s} = 0.690$

${\mu}_{k} = 0.605$

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 inclined 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 $14$ $\text{kg}$, and the incline is $- \frac{3 \pi}{8}$.

Since the angle is $- \frac{3 \pi}{8}$, 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

color(green)(ul(f_s = mu_sn

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

n = mgcostheta = (6color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/8) = color(orange)(ul(54.4color(white)(l)"N"

Since $15$ $\text{N}$ 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 ${f}_{s}$:

$\textcolor{g r e e n}{{f}_{s}} = m g \sin \theta + 15$ $\text{N}$

= (6color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/8) + 15color(white)(l)"N" = color(green)(37.5color(white)(l)"N"

The coefficient of static friction is thus

mu_s = (f_s)/n = (color(green)(37.5)cancel(color(green)("N")))/(color(orange)(54.4)cancel(color(orange)("N"))) = color(red)(ulbar(|stackrel(" ")(" " 0.690" ")|

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

color(purple)(ul(f_k = mu_kn

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

$\textcolor{p u r p \le}{{f}_{k}} = m g \sin \theta + 5$ $\text{N}$

$= 22.5 \textcolor{w h i t e}{l} \text{N} + 5$ $\text{N}$ = color(purple)(27.5color(white)(l)"N"

The coefficient of kinetic friction is thus

mu_k = (f_k)/n = (color(purple)(27.5)cancel(color(purple)("N")))/(color(orange)(119)cancel(color(orange)("N"))) = color(blue)(ulbar(|stackrel(" ")(" " 0.506" ")|