Question

What force would you have to exert on a 326-N trunk up a 19.0° inclined plane so that it would slide down the plane with a constant velocity? What would be the direction of the force? (The coefficient of friction between the plane and the trunk is 0.328.)

The force of friction is muFn, which is muW. F applied = F parallel (+)(-) F of friction. The trunk is going down the plane so you would subtract?? Am I thinking in the correct direction??

Answer 1

First, figure out whether the trunk would slide down the hill on its own if no external force were applied. If so, then a force UP the hill would have to be applied to have it move at constant velocity.

The friction force when moving is UP the hill, and is
(0.328)(326 cos 18) = 101.1 N
The gravity force component DOWN the hill is 326 sin 19 = 106.1. The difference of 5.0 N must be applied UP the hill (along the slope direction) for contant-velocity sliding.

Answer 2

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Answer 3

Thank you for the help @drwls!!!