Question

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?

Answer 1

`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`.