An object with a mass of 5 kg5kg is on a plane with an incline of -(5 pi)/12 5π12. If it takes 12 N12N to start pushing the object down the plane and 2 N2N to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Jul 22, 2017

If angle of inclination is pi/12π12:

mu_s = 0.521μs=0.521

mu_k = 0.310μk=0.310

(see explanation regarding angle)

Explanation:

We're asked to find the coefficient of static friction mu_sμs and the coefficient of kinetic friction mu_kμk, with some given information.

![upload.wikimedia.org](useruploads.socratic.org)

We'll call the positive xx-direction up the incline (the direction of ff in the image), and the positive yy direction perpendicular to the incline plane (in the direction of NN).

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

We're given that the object's mass is 55 "kg"kg, and the incline is -(5pi)/125π12.

Since the angle is -(5pi)/125π12 (-75^"o"75o), 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 - (5pi)/12 = pi/12π25π12=π12

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_sμs is

f_s >= mu_snfsμsn

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

f_s = mu_snfs=μsn

Since the two vertical quantities nn and mgcosthetamgcosθ are equal,

n = mgcostheta = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/12) = 47.4n=mgcosθ=(5lkg)(9.81lm/s2)cos(π12)=47.4 "N"N

Since 1212 "N"N is the "breaking point" force that causes it to move, it is this value plus mgsinthetamgsinθ that equals the upward static friction force f_sfs:

f_s = mgsintheta + 12fs=mgsinθ+12 "N"N

= (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/12) + 12color(white)(l)"N" = 24.7=(5lkg)(9.81lm/s2)sin(π12)+12lN=24.7 "N"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_kn

It takes 2 "N" of applied downward force (on top of weight) to keep the object accelerating constantly downward, then we have

f_k = mgsintheta + 2 "N"

= 12.7color(white)(l)"N" + 2 "N" = 14.7 "N"

The coefficient of kinetic friction is thus

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