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
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∑Fx=mgsinθ−fs−Fapplied=0
where
-
F_"applied"Fapplied is22 "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_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