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

1 Answer
Aug 2, 2017

Answer:

#mu_s >= 0.198#

Explanation:

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

For this situation, the net horizontal force #sumF_x# is given by

#sumF_x = mgsintheta - f_s - F_"applied" = 0#

where

  • #F_"applied"# is #2# #"N"# (directed up the incline)

  • #f_s# 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#.