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Tension - pully problem w/ monkey

  1. Oct 2, 2003 #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.
  2. jcsd
  3. Oct 2, 2003 #2
    I agree with your acceleration for (a), and of course you're correct that in (b) the box accelerates downward & the monkey upward.

    So now set up the same kind of equations for the situation where both masses are simply hanging on the two ends of the rope. Now T and a are unknown, so you have two equations and two unknowns. (Note that a is the same for both equations, and T is the same for both.) Solve & you get the answers for (b) and (d).
  4. Oct 2, 2003 #3
    equations where the masses are simply hanging from the ropes?

    I thought I had those two equations already?

    T = 10a + 98
    T = 15a + 147

    10a + 98 = 15a + 147
    a = -9.8

    The answer in the example in the book says that the acceleration is supposed to be 2.0 m/s^2 and the T of the rope is supposed to be 120N.

    With my two equations, I didn't get the same answer :(
  5. Oct 3, 2003 #4
    Not exactly. You're confusing the direction of the acceleration. Remember, since they're attached, if you say that a is positive for the box accelerating downward, then a is positive for the monkey accelerating upward. I find it easier to see if I set the equations up in the same form as Newton:
    ∑F = ma
    Now, for the box...
    147-T = 15a
    And for the monkey...
    T-98 = 10a

    You should get a=1.96 m/s^2 and T=117.6 N.

    (Slightly different from the answers in your book. You and I were using 9.8 m/s^2 for g. Your book is rounding it to 10 m/s^2.)

    By the way, it's PULLEY. :)
  6. Oct 26, 2003 #5

    That problem was on my quiz, I got the same answer. :smile:
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