An object with a mass of 2 kg is on a plane with an incline of - pi/3 . If it takes 12 N to start pushing the object down the plane and 4 N to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Aug 13, 2017

If angle of inclination is pi/6:

mu_s = 1.28

mu_k = 0.813

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.

upload.wikimedia.orgupload.wikimedia.org

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 2 "kg", and the incline is -(pi)/3.

Since the angle is -(pi)/3, 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 - (pi)/3 = ul((pi)/6

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

color(green)(ul(f_s = mu_sn

Since the two vertical quantities n and mgcostheta are equal,

n = mgcostheta = (2color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/6) = color(orange)(ul(17.0color(white)(l)"N"

Since 12 "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:

color(green)(f_s) = mgsintheta + 12 "N"

= (2color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/6) + 12color(white)(l)"N" = color(green)(21.8color(white)(l)"N"

The coefficient of static friction is thus

mu_s = (f_s)/n = (color(green)(21.8)cancel(color(green)("N")))/(color(orange)(17.0)cancel(color(orange)("N"))) = color(red)(ulbar(|stackrel(" ")(" " 1.28" ")|

The coefficient of kinetic friction mu_k is given by

color(purple)(ul(f_k = mu_kn

It takes 4 "N" of applied downward force (on top of the object's weight) to keep the object accelerating constantly downward, then we have

color(purple)(f_k) = mgsintheta + 4 "N"

= 9.81color(white)(l)"N" + 4 "N" = color(purple)(13.8color(white)(l)"N"

The coefficient of kinetic friction is thus

mu_k = (f_k)/n = (color(purple)(13.8)cancel(color(purple)("N")))/(color(orange)(17.0)cancel(color(orange)("N"))) = color(blue)(ulbar(|stackrel(" ")(" " 0.813" ")|