An object with a mass of 14  kg is on a plane with an incline of  - pi/3 . If it takes 12  N to start pushing the object down the plane and 7   N to keep pushing it, what are the coefficients of static and kinetic friction?

Jun 19, 2017

$\setminus {\mu}_{\text{static" = F_"frict}} / \left(F \bot\right) = \frac{138.8}{68.6} = 2.02$

$\setminus {\mu}_{\text{kinetic" = F_"frict}} / \left(F \bot\right) = \frac{125.8}{68.6} = 1.83$

Explanation:

We have to assume that the motion is at constant velocity once the motion starts: the question doesn't say so, exactly, but it's the only way it makes sense.

There is a component of the weight force of the $14$ $k g$ mass that acts parallel to the inclined plane, which we can call ${F}_{| \setminus |}$, and which tends to cause it to move down the slope. There is another component, $F \bot$ that acts perpendicular to the slope, and dictates the magnitude of the frictional force. The frictional force itself acts up the slope and opposes movement down the slope.

The forces will be given by:

${F}_{| \setminus |} = m g \sin \setminus \theta = 14 \times 9.8 \times \sin \left(- \setminus \frac{\pi}{3}\right) = - 118.8$ $N$

$F \bot = m g \cos \setminus \theta = 14 \times 9.8 \times \cos \left(- \setminus \frac{\pi}{3}\right) = 68.6$ $N$

In this case, the negative sign just means the slope rises right-to-left rather than left-to-right. We can ignore the sign for our purposes.

To calculate the coefficient of static friction, we know that there is a force of $118.8$ $N$ acting down the slope due to the weight of the object. It takes an additional $12$ $N$ to make it start to move, for a total of $138.8$ $N$. This must just overcame - and therefore be virtually equal to - the force of static friction.

We have the expression for the static friction:

${F}_{\text{frict"=\mu_"stat}} F \bot$

Rearranging:

$\setminus {\mu}_{\text{stat" = F_"frict}} / \left(F \bot\right) = \frac{138.8}{68.6} = 2.02$

A frictional coefficient of 2 is a little unusual, but not impossible.

The argument for the coefficient of kinetic friction is the same, but the force that is required to maintain constant velocity versus the frictional force is $7$ $N$. Combined with the component of the weight force acting in this direction, the total force is $125.8$ $N$.

$\setminus {\mu}_{\text{kin" = F_"frict}} / \left(F \bot\right) = \frac{125.8}{68.6} = 1.83$