Question

A wooden box with a mass of 10.0kg rests on a ramp that is inclined at an angle of 25° to the horizontal. A rope attatched to the box runs parallel to the ramp and then passes over a frictionless pulley. A bucket with a mass of m hangs fro the end of the rope. the coefficient of static friction between the ramp and the box is 0.50. The coefficient of kinetic friction between the ramp and the box is 0.35. 3a). What is the friction force exerted on the box by the ramp? 3b). Does the box remain at rest relative to the ramp?

Answer 1

Below

Step-by-step explanation:

Force downplane from the box on the incline = mg sin 25 = 41.4 N

Force of friction = normal force * coeff static friction (if not moving)

= mg cos 25 * .50 =44.5 N (max)

Force up the plane from the 2kg bucket tension= mg = 2 * 9.81 = 19.62 N

This is a net downplane force of 41.4 - 19.62 N = =21.8N which is not enough force to overcome the static friction max....so the box does not move

So in this case the friction force = 21.8 N UP the plane