# An object with a mass of 8 kg is on a plane with an incline of  - pi/12 . If it takes 12 N to start pushing the object down the plane and 5 N to keep pushing it, what are the coefficients of static and kinetic friction?

Jan 23, 2017

the static coefficient of friction is $0.4264$ (4dp)
the kinetic coefficient of friction is $0.3340$ (4dp)

#### Explanation: For our diagram, $m = 8 k g$, $\theta = \frac{\pi}{12}$

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

$R - m g \cos \theta = 0$
$\therefore R = 8 g \cos \left(\frac{\pi}{12}\right) \setminus \setminus N$

Initially it takes $12 N$ to start the object moving, so $D = 12$. If we Apply Newton's Second Law down parallel to the plane we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 12 + 8 g \sin \left(\frac{\pi}{12}\right) \setminus \setminus N$

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

$F = \mu R \implies 12 + 8 g \sin \left(\frac{\pi}{12}\right) = \mu \left(8 g \cos \left(\frac{\pi}{12}\right)\right)$
$\therefore \mu = \frac{12 + 8 g \sin \left(\frac{\pi}{12}\right)}{8 g \cos \left(\frac{\pi}{12}\right)}$
$\therefore \mu = 0.426409 \ldots$

Once the object is moving the driving force is reduced from $12 N$ to $5 N$. Now $D = 5$, reapply Newton's Second Law down parallel to the plane and we get:

$D + m g \sin \theta - F = 0$
$\therefore F = 5 + 8 g \sin \left(\frac{\pi}{12}\right) \setminus \setminus N$

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

$F = \mu R \implies 5 + 8 g \sin \left(\frac{\pi}{12}\right) = \mu \left(8 g \cos \left(\frac{\pi}{12}\right)\right)$
$\therefore \mu = \frac{5 + 8 g \sin \left(\frac{\pi}{12}\right)}{8 g \cos \left(\frac{\pi}{12}\right)}$
$\therefore \mu = 0.333974 \ldots$

So the static coefficient of friction is $0.4264$ (4dp)
the kinetic coefficient of friction is $0.3340$ (4dp)