Please solve this. A box (w = 100 ?) rests on an inclined plane, angled 30 degre

Question

Please solve this.

A box (w = 100 ?) rests on an inclined plane, angled 30 degree above the horizontal. The coefficient of static friction is mus = 0.50, and the coefficient of kinetic friction is muk = 0.40. You push on the box with a force F, parallel to the plane. If the box is initially at rest, what is the minimum force F that will keep it from sliding downhill? If the box is initially at rest, what is the maximum force that you can exert without moving the box? If the box is sliding, what force F will keep the box moving uphill at a constant velocity? If the box is sliding, what force F will keep the box moving downhill at a constant velocity?

Explanation / Answer

Part 1)

Balance the forces

The up the plane forces will be friction and the applied force. The down the plane force will be a weight component

F + umgcos(angle) = mgsin(angle)

F + (.5)(100)(cos 30) = (100)(sin 30)

F = 6.70 N

Part 2)

Without moving the box, the force of friction will change direction, so...

F = mgsin(angle) + umgcos(angle)

F = (.5)(100)(cos 30) + (100)(sin 30)

F = 93.3 N

Part 3)

Same as part 2, but using kinetic friction

F = (.4)(100)(cos 30) + (100)(sin 30)

F = 84.6 N

Part 4)

Same as part 1, but using kinetic friction

F + (.4)(100)(cos 30) = (100)(sin 30)

F = 15.4 N

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