An object with a mass of 5 kg is on a plane with an incline of  - pi/8 . If it takes 8 N to start pushing the object down the plane and 3 N to keep pushing it, what are the coefficients of static and kinetic friction?

Jan 13, 2017

The static coefficient of friction is $0.5909$ (4dp)
The kinetic coefficient of friction is $0.4804$ (4dp)

Explanation: For our diagram, $m = 5 k g$, $\theta = \frac{\pi}{8}$

If we apply Newton's Second Law up perpendicular to the plane we get:

$R - m g \cos \theta = 0$
$\therefore R = 5 g \cos \left(\frac{\pi}{8}\right) \setminus \setminus N$

Initially it takes $8 N$ to start the object moving, so $D = 8$. If we Apply Newton's Second Law down parallel to the plane we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 8 + 5 g \sin \left(\frac{\pi}{8}\right) \setminus \setminus N$

And the friction is related to the Reaction (Normal) Force by

$F = \mu R \implies 8 + 5 g \sin \left(\frac{\pi}{8}\right) = \mu \left(5 g \cos \left(\frac{\pi}{8}\right)\right)$
$\therefore \mu = \frac{8 + 5 g \sin \left(\frac{\pi}{8}\right)}{5 g \cos \left(\frac{\pi}{8}\right)}$
$\therefore \mu = 0.5909306 \ldots$

Once the object is moving the driving force is reduced from $8 N$ to $3 N$. Now $D = 3$, reapply Newton's Second Law down parallel to the plane and we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 3 + 5 g \sin \left(\frac{\pi}{8}\right) \setminus \setminus N$

And the friction is related to the Reaction (Normal) Force by

$F = \mu R \implies 3 + 5 g \sin \left(\frac{\pi}{8}\right) = \mu \left(5 g \cos \left(\frac{\pi}{8}\right)\right)$
$\therefore \mu = \frac{3 + 5 g \sin \left(\frac{\pi}{8}\right)}{5 g \cos \left(\frac{\pi}{8}\right)}$
$\therefore \mu = 0.4804482 \ldots$

So the static coefficient of friction is $0.5909$ (4dp)
the kinetic coefficient of friction is $0.4804$ (4dp)