An object with a mass of 3 kg3kg is on a ramp at an incline of pi/12 π12. If the object is being pushed up the ramp with a force of 2 N2N, what is the minimum coefficient of static friction needed for the object to remain put?

1 Answer
Aug 2, 2017

mu_s >= 0.198μs0.198

Explanation:

We're asked to find the minimum coefficient of static friction mu_sμs.

For this situation, the net horizontal force sumF_xFx is given by

sumF_x = mgsintheta - f_s - F_"applied" = 0Fx=mgsinθfsFapplied=0

where

  • F_"applied"Fapplied is 22 "N"N (directed up the incline)

  • f_sfs is the maximum static friction force (directed up the incline. Even though it is being pushed upward, the net force points downward in the absence of friction), which is given by

f_s = mu_sn = mu_soverbrace(mgcostheta)^n

So

sumF_x = mgsintheta - mu_smgcostheta - 2 "N" = 0

Now we solve the expression for the coefficient of static friction mu_s:

mu_smgcostheta = mgsintheta - 2 "N"

ul(mu_s = (mgsintheta - 2color(white)(l)"N")/(mgcostheta)

We know:

  • m = 3 "kg"

  • g = 9.81 "m/s"^2

  • theta = pi/12

Plugging these in gives

mu_s = ((3color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/12) - 2color(white)(l)"N")/((3color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/12)) = color(blue)(ulbar|stackrel(" ")(" "0.198" ")|)

The minimum coefficient of static friction is thus color(blue)(0.198.