An object with a mass of 4 kg is on a plane with an incline of  - pi/8 . If it takes 18 N to start pushing the object down the plane and 15 N to keep pushing it, what are the coefficients of static and kinetic friction?

Jan 16, 2017

The static coefficient of friction is $0.9112$ (4dp)
The kinetic coefficient of friction is $0.8284$ (4dp)

Explanation:

For our diagram, $m = 4 k g$, $\theta = \frac{\pi}{8}$

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

$R - m g \cos \theta = 0$
$\therefore R = 4 g \cos \left(\frac{\pi}{8}\right) \setminus \setminus N$

Initially it takes $18 N$ to start the object moving, so $D = 18$. If we Apply Newton's Second Law down parallel to the plane we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 18 + 4 g \sin \left(\frac{\pi}{8}\right) \setminus \setminus N$

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

$F = \mu R \implies 18 + 4 g \sin \left(\frac{\pi}{8}\right) = \mu \left(4 g \cos \left(\frac{\pi}{8}\right)\right)$
$\therefore \mu = \frac{18 + 4 g \sin \left(\frac{\pi}{8}\right)}{4 g \cos \left(\frac{\pi}{8}\right)}$
$\therefore \mu = 0.911230 \ldots$

Once the object is moving the driving force is reduced from $18 N$ to $15 N$. Now $D = 15$, reapply Newton's Second Law down parallel to the plane and we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 15 + 4 g \sin \left(\frac{\pi}{8}\right) \setminus \setminus N$

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

$F = \mu R \implies 15 + 4 g \sin \left(\frac{\pi}{8}\right) = \mu \left(4 g \cos \left(\frac{\pi}{8}\right)\right)$
$\therefore \mu = \frac{15 + 4 g \sin \left(\frac{\pi}{8}\right)}{4 g \cos \left(\frac{\pi}{8}\right)}$
$\therefore \mu = 0.828394 \ldots$

So the static coefficient of friction is $0.9112$ (4dp)
the kinetic coefficient of friction is $0.8284$ (4dp)