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

1 Answer
Jul 22, 2017

Answer:

If angle of inclination is #pi/12#:

#mu_s = 0.521#

#mu_k = 0.310#

(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.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 #5# #"kg"#, and the incline is #-(5pi)/12#.

Since the angle is #-(5pi)/12# (#-75^"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 - (5pi)/12 = pi/12#

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_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

#f_s = mu_sn#

Since the two vertical quantities #n# and #mgcostheta# are equal,

#n = mgcostheta = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/12) = 47.4# #"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#:

#f_s = mgsintheta + 12# #"N"#

#= (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/12) + 12color(white)(l)"N" = 24.7# #"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#