An object with a mass of 8 kg8kg is on a plane with an incline of - pi/3 π3. If it takes 9 N9N to start pushing the object down the plane and 7 N7N to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Aug 7, 2017

If angle of inclination is pi/6π6:

mu_s = 0.710μs=0.710

mu_k = 0.680μk=0.680

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.orgupload.wikimedia.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 88 "kg"kg, and the incline is -(pi)/3π3.

Since the angle is -(pi)/3π3, 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 = (8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/6) = color(orange)(ul(68.0color(white)(l)"N"

Since 9 "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 + 9 "N"

= (8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/6) + 9color(white)(l)"N" = color(green)(48.2color(white)(l)"N"

The coefficient of static friction is thus

mu_s = (f_s)/n = (color(green)(48.2)cancel(color(green)("N")))/(color(orange)(68.0)cancel(color(orange)("N"))) = color(red)(ulbar(|stackrel(" ")(" " 0.710" ")|

The coefficient of kinetic friction mu_k is given by

color(purple)(ul(f_k = mu_kn

It takes 7 "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 + 7 "N"

= 39.2color(white)(l)"N" + 7 "N" = color(purple)(46.2color(white)(l)"N"

The coefficient of kinetic friction is thus

mu_k = (f_k)/n = (color(purple)(46.2)cancel(color(purple)("N")))/(color(orange)(68.0)cancel(color(orange)("N"))) = color(blue)(ulbar(|stackrel(" ")(" " 0.680" ")|