An object with a mass of 9 kg is on a plane with an incline of - pi/4 . If it takes 15 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?

Apr 25, 2016

The coefficient of static friction is ${\mu}_{s} = 1.24$ and the coefficient of kinetic friction is ${\mu}_{k} = 1.11$.

Explanation:

The negative sign on the angle is not relevant in this case: the angle of the inclined plane is $\frac{\pi}{4}$ $r a \mathrm{di} a n s$ (${45}^{o}$). The negative sign just shows that the slope increases right-to-left rather than left-to-right.

The force perpendicular to the plane, the normal force, will be given by:

${F}_{N} = m g \sin \Theta = 9 \cdot 9.8 \cdot \sin \left(\frac{\pi}{4}\right) = 62.4$ $N$

Based on this, the frictional force will be:

${F}_{\text{frict}} = \mu {F}_{N}$

Rearranging,

$\mu = \frac{{F}_{\text{frict}}}{{F}_{N}}$

We will find two different values, the static friction, ${\mu}_{s}$, and the kinetic friction, ${\mu}_{k}$.

The force parallel to the plane due to gravity will be given by:

$F = m g \cos \Theta = 9 \cdot 9.8 \cdot \cos \left(\frac{\pi}{4}\right) = 62.4$ $N$

Overcoming the force of static friction requires this parallel force plus $15$ $N$, a total frictional force of $77.4$ $N$.

${\mu}_{s} = \frac{{F}_{\text{frict}}}{{F}_{N}} = \frac{77.4}{62.4} = 1.24$

Once it is moving, keeping it moving only requires $7$ $N$ in addition to the parallel force, a total of $69.4$ $N$.

${\mu}_{k} = \frac{{F}_{\text{frict}}}{{F}_{N}} = \frac{69.4}{62.4} = 1.11$

Coefficients of friction do not have units.