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?

1 Answer

0.326

Explanation:

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

F=mg\sin\theta-\mumg\cos\theta

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

mg\sin\theta-\mumg\cos\theta=4

5\cdot 9.81\sin(\pi/8)-\mu(5\cdot 9.81) \cos(\pi/8)=4

\mu=\frac{5\cdot 9.81\sin(\pi/8)-4}{5\cdot 9.81\cos(\pi/8)}

=0.326