# An object with a mass of  5 kg is on a ramp at an incline of pi/8 . If the object is being pushed up the ramp with a force of  4 N, what is the minimum coefficient of static friction needed for the object to remain put?

$0.326$

#### Explanation:

The force $F$ causing the object of mass $5$kg to slide down parallel to the ramp, having coefficient of friction $\setminus \mu$ & inclined with the horizontal at an angle $\setminus \theta = \setminus \frac{\pi}{8}$, is given by the condition of minimum force applied to keep the object at rest

$F = m g \setminus \sin \setminus \theta - \setminus \mu m g \setminus \cos \setminus \theta$

But a force of $4 \setminus N$ is applied up to keep the object stationary then in equilibrium we have

$m g \setminus \sin \setminus \theta - \setminus \mu m g \setminus \cos \setminus \theta = 4$

$5 \setminus \cdot 9.81 \setminus \sin \left(\setminus \frac{\pi}{8}\right) - \setminus \mu \left(5 \setminus \cdot 9.81\right) \setminus \cos \left(\setminus \frac{\pi}{8}\right) = 4$

$\setminus \mu = \setminus \frac{5 \setminus \cdot 9.81 \setminus \sin \left(\setminus \frac{\pi}{8}\right) - 4}{5 \setminus \cdot 9.81 \setminus \cos \left(\setminus \frac{\pi}{8}\right)}$

$= 0.326$