An object with a mass of 8 kg is on a plane with an incline of  - pi/8 . If it takes 4 N to start pushing the object down the plane and 1 N to keep pushing it, what are the coefficients of static and kinetic friction?

Jan 16, 2017

The static coefficient of friction is $0.4694$ (4dp)
The kinetic coefficient of friction is $0.4280$ (4dp)

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

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

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

Initially it takes $4 N$ to start the object moving, so $D = 4$. If we Apply Newton's Second Law down parallel to the plane we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 4 + 8 g \sin \left(\frac{\pi}{8}\right) \setminus \setminus N$

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

$F = \mu R \implies 4 + 8 g \sin \left(\frac{\pi}{8}\right) = \mu \left(8 g \cos \left(\frac{\pi}{8}\right)\right)$
$\therefore \mu = \frac{4 + 8 g \sin \left(\frac{\pi}{8}\right)}{8 g \cos \left(\frac{\pi}{8}\right)}$
$\therefore \mu = 0.469437 \ldots$

Once the object is moving the driving force is reduced from $4 N$ to $1 N$. Now $D = 1$, reapply Newton's Second Law down parallel to the plane and we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 1 + 8 g \sin \left(\frac{\pi}{8}\right) \setminus \setminus N$

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

$F = \mu R \implies 1 + 8 g \sin \left(\frac{\pi}{8}\right) = \mu \left(8 g \cos \left(\frac{\pi}{8}\right)\right)$
$\therefore \mu = \frac{1 + 8 g \sin \left(\frac{\pi}{8}\right)}{8 g \cos \left(\frac{\pi}{8}\right)}$
$\therefore \mu = 0.428019 \ldots$

So the static coefficient of friction is $0.4694$ (4dp)
the kinetic coefficient of friction is $0.4280$ (4dp)