An object with a mass of 10 kg is on a plane with an incline of  - pi/4 . If it takes 12 N to start pushing the object down the plane and 4 N to keep pushing it, what are the coefficients of static and kinetic friction?

Feb 21, 2017

The static coefficient is 1.17; the kinetic coefficient is 1.06.

Explanation:

The object on an incline has one component of the force of gravity pulling it down the slope, usually referred to as ${F}_{| |}$. As the diagram below shows, it is equal to $m g \sin \theta$

![http://thecraftycanvas.com/library/online-learning-tools/physics-homework-helpers/incline-force-calculator-problem-solver/]!

In this problem, the applied force also acts down the incline, and only friction acts up the ramp to oppose these two forces.

The force of friction is given by ${F}_{f} = \mu {F}_{N} = \mu m g \cos \theta$

Therefore the free body equation is

${F}_{\text{net" = mumgcostheta - mgsintheta - F_"applied}}$

Since there is no acceleration, ${F}_{\text{net}} = 0$ and the equation becomes

$\mu m g \cos \theta - m g \sin \theta - {F}_{\text{applied}} = 0$

Inserting what we know, I will find ${\mu}_{s}$ first:

${\mu}_{s} \left(10\right) \left(9.8\right) \cos \left(\frac{\pi}{4}\right) - \left(10\right) \left(9.8\right) \sin \left(\frac{\pi}{4}\right) - 12 = 0$

Since $\cos \left(\frac{\pi}{4}\right) = \sin \left(\frac{\pi}{4}\right) = .707$

${\mu}_{s} \left(69.3\right) - \left(69.3\right) - 12 = 0$

${\mu}_{s} = 81.3 \div 69.3 = 1.17$

The calculation of ${\mu}_{k}$ is similar, using 4N in place of the 12 N applied force:

${\mu}_{k} \left(10\right) \left(9.8\right) \cos \left(\frac{\pi}{4}\right) - \left(10\right) \left(9.8\right) \sin \left(\frac{\pi}{4}\right) - 4 = 0$

${\mu}_{k} \left(69.3\right) - \left(69.3\right) - 4 = 0$

${\mu}_{k} = 73.3 \div 69.3 = 1.06$