# An object with a mass of 8 kg is on a plane with an incline of  - pi/6 . If it takes 9 N to start pushing the object down the plane and 2 N to keep pushing it, what are the coefficients of static and kinetic friction?

Jan 13, 2017

The static coefficient of friction is $0.7099$ (4dp)
The kinetic coefficient of friction is $0.6068$ (4dp)

#### Explanation: For our diagram, $m = 8 k g$, $\theta = \frac{\pi}{6}$

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

$R - m g \cos \theta = 0$
$\therefore R = 8 g \cos \left(\frac{\pi}{6}\right) \setminus \setminus N$

Initially it takes $9 N$ to start the object moving, so $D = 9$. If we Apply Newton's Second Law down parallel to the plane we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 9 + 8 g \sin \left(\frac{\pi}{6}\right) \setminus \setminus N$

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

$F = \mu R \implies 9 + 8 g \sin \left(\frac{\pi}{6}\right) = \mu \left(8 g \cos \left(\frac{\pi}{6}\right)\right)$
$\therefore \mu = \frac{9 + 8 g \sin \left(\frac{\pi}{6}\right)}{8 g \cos \left(\frac{\pi}{6}\right)}$
$\therefore \mu = 0.7099051 \ldots$

Once the object is moving the driving force is reduced from $9 N$ to $2 N$. Now $D = 2$, reapply Newton's Second Law down parallel to the plane and we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 2 + 8 g \sin \left(\frac{\pi}{6}\right) \setminus \setminus N$

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

$F = \mu R \implies 2 + 8 g \sin \left(\frac{\pi}{6}\right) = \mu \left(8 g \cos \left(\frac{\pi}{6}\right)\right)$
$\therefore \mu = \frac{2 + 8 g \sin \left(\frac{\pi}{6}\right)}{8 g \cos \left(\frac{\pi}{6}\right)}$
$\therefore \mu = 0.6068069 \ldots$

So the static coefficient of friction is $0.7099$ (4dp)
the kinetic coefficient of friction is $0.6068$ (4dp)