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?

1 Answer
Sep 2, 2017

Answer:

If angle of inclination is #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.

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 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# #"kg"#, and the incline is #-(3pi)/8#.

Since the angle is #-(3pi)/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 >= 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 = (6color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/8) = color(orange)(ul(54.4color(white)(l)"N"#

Since #15# #"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 + 15# #"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# #"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 + 5# #"N"#

#= 22.5color(white)(l)"N" + 5# #"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" ")|#