# 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?

Aug 2, 2017

${\mu}_{s} \ge 0.198$

#### Explanation:

We're asked to find the minimum coefficient of static friction ${\mu}_{s}$.

For this situation, the net horizontal force $\sum {F}_{x}$ is given by

$\sum {F}_{x} = m g \sin \theta - {f}_{s} - {F}_{\text{applied}} = 0$

where

• ${F}_{\text{applied}}$ is $2$ $\text{N}$ (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

${f}_{s} = {\mu}_{s} n = {\mu}_{s} {\overbrace{m g \cos \theta}}^{n}$

So

$\sum {F}_{x} = m g \sin \theta - {\mu}_{s} m g \cos \theta - 2$ $\text{N}$ $= 0$

Now we solve the expression for the coefficient of static friction ${\mu}_{s}$:

${\mu}_{s} m g \cos \theta = m g \sin \theta - 2$ $\text{N}$

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

We know:

• $m = 3$ $\text{kg}$

• $g = 9.81$ ${\text{m/s}}^{2}$

• $\theta = \frac{\pi}{12}$

Plugging these in gives

${\mu}_{s} = \left(\left(3 \textcolor{w h i t e}{l} \text{kg")(9.81color(white)(l)"m/s"^2)sin(pi/12) - 2color(white)(l)"N")/((3color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)cos(pi/12)) = color(blue)(ulbar|stackrel(" ")(" "0.198" }\right) |\right)$

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