Question

A half-full recycling bin has mass 3.0 kg and is pushed up a 40.0^\circ40.0 ​∘ ​​ incline with constant speed under the action of a 26-N force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at constant velocity?

Answer 1

Answer:

11.82 N upward and parallel to the incline.

Explanation:

Let g = 9.81 m/s2. Gravity that acts on the bin has a magnitude of

W = mg = 3*9.81 = 29.43 N

The gravity component that is parallel to the 40 degree incline is

$Wsin40^0 = 29.43*0.643 = 18.9 N$

If the bin is moving at constant speed, then the net force (parallel to the incline) is also 0. Gravity is balanced with the push force of 26 N and the friction force.

The magnitude of friction force is 26 - 18.9 = 7.08 N

The the bin is moving down the incline, then gravity is 18.9N , friction is 7.08 N, one must create an upward force of 18.9 - 7.08 = 11.82 N parallel with the incline to balance this.