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
Answer:
Explanation:
We're asked to find the minimum coefficient of static friction
For this situation, the net horizontal force
#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_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)(ulbarstackrel(" ")(" "0.198" "))#
The minimum coefficient of static friction is thus