- #1

**tension - "pully" problem w/ monkey**

Hello! The diagram in my book has a rope that is swung over a tree branch. Attached on one side is a 15 kg box of bananas and it is at rest on the ground. On the other side of the rope is a 10 kg monkey climbing up the rope. To me, this looks like a pully problem. Am I correct?

Then the questions are:

a) What is the magnitude of least acceleration the monkey must have if it is to lift the box off the ground? If after the box has been lifted, the monkey stops its climb and holds onto the rope, what are b) the magnitude and c) the direction of the monkey's acceleration, and d) what is the tension in the rope?

a) The "least acceleration" part confuses me. The box is just resting on the ground, so at the moment it has a = 0 m/s^2. So, wouldn't any acceleration on the monkey's part move the box up as long as we have Fmonkey > Fbox? I was thinking:

Fmonkey = T - mg = ma --> T = 10(a + 9.8)

Fbox = T - mg = ma --> T = 15(0 + 9.8) = 147N

So 10a + 98 = 147 --> a = 4.9 m/s^2

b and c) Okay, the box has been lifted and so the box and the monkey are hanging on the rope and are in the air. But then since the box has a greater mass than the monkey, the box starts to go downward and the monkey is accelerating upward? So, I think that the direction of the monkey's acceleration is upward. I know the direction, but I am lost as to the magnitude. The box has a greater mass than the monkey and since it is great enough to make the monkey accelerate upward as it falls down, I am thinking that Fbox > Fmonkey. But then, that is all I have understood. I am unsure of how to approach getting the magnitude of the monkey's acceleration upward.

d) I don't think I can get the tension of the rope until I have figured out b) :(

Thanks for your attention.