# An object with a mass of 7 kg is on a ramp at an incline of pi/8 . If the object is being pushed up the ramp with a force of  2 N, what is the minimum coefficient of static friction needed for the object to remain put?

Jan 15, 2017

minimum friction coefficient is $0.3827$ (4dp)

#### Explanation:

For our diagram, $m = 7 k g$, $\theta = \frac{\pi}{8}$

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

$R - m g \cos \theta = 0$
$\therefore R = 7 g \cos \left(\frac{\pi}{8}\right) \setminus \setminus N$

It takes $2 N$ to keep the object in equilibrium, so $D = 2$. If we Apply Newton's Second Law down parallel to the plane we get:

$m g \sin \theta - D - F = 0$
$\therefore F = 7 g \sin \left(\frac{\pi}{8}\right) - 2 \setminus \setminus N$

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

$F \le \mu R \implies 7 g \sin \left(\frac{\pi}{8}\right) - 2 \le \mu \left(7 g \cos \left(\frac{\pi}{8}\right)\right)$
$\therefore \mu \ge \frac{7 g \sin \left(\frac{\pi}{8}\right) - 2}{7 g \cos \left(\frac{\pi}{8}\right)}$
$\therefore \mu \ge 0.382656 \ldots$