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

Answer:

#0.326#

Explanation:

The force #F# causing the object of mass #5#kg 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#