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Tension in a pulley cord

  1. Jan 13, 2008 #1
    [SOLVED] Tension in a pulley cord

    I'm having problems trying to solve a tension problem. It's very simple... and I don't know what I'm doing wrong.

    A light, inextensible cord passes over a light,
    frictionless pulley with a radius of 15 cm. It
    has a(n) 22 kg mass on the left and a(n)
    2.8 kg mass on the right, both hanging freely.
    Initially their center of masses are a vertical
    distance 2.6 m apart. The 2.8 kg block is the lower one.
    The acceleration of gravity is 9.8 m/s2

    The first question was at what rate are the two masses accelerating when they pass each other? I figured this one out and it was 7.5891 m/s2 and I got it right. I didn't have trouble at all figuring that one out because acceleration is constant, but then...

    It asks... What is the tension in the cord when they pass each other? I don't think that it matters when the tension is calculated, because it's still going to have the same amount of force being applied to it. I tried the sum of their forces (due to gravity), the differences of their forces, zero, and the sum of their forces divided by two... but none of them were right. Can anyone help me out on this one?
  2. jcsd
  3. Jan 13, 2008 #2

    Doc Al

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    Staff: Mentor

    Just apply Newton's second law to either one of the masses. What forces act? You already know the acceleration.
  4. Jan 13, 2008 #3
    I don't really understand... I multiplied 9.8 times 22 kg to get the force on one side of the cord. Then I multiply 9.8 times 2.8 kg to get the force on the other side... why isn't it the sum of those two forces? They're both pulling on the cord... so why isn't the tension that amount?
  5. Jan 13, 2008 #4

    Doc Al

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    Staff: Mentor

    The weights pull on the masses, not directly onto the cord. The tension in the cord depends on the acceleration.

    Example: Imagine you had a rock on the end of a string. You hold the string with the rock hanging down and not moving. The tension in the string will equal the weight of the rock. But if you jerk the string upwards with some acceleration, the tension in the string will be greater--even though the weight of the rock hasn't changed.

    You apparently took a short cut (which works in simple cases) when you calculated the acceleration. A better way is to analyze the forces acting on each mass, apply Newton's 2nd law, then combine the resulting equations.
  6. Jan 13, 2008 #5
    So if acceleration is 7.5891 m/s2... (I calculated that yesterday and I can't find my work for it), is the tension in the string going to be the mass of the smaller block times the acceleration? Or is it going to be the mass of the larger block times the acceleration.

    I'm thinking it would be the former because if the larger block were falling with the string attached to it and nothing else, there would be no tension in the string... so the large block isn't causing any tension. But since the small block is on the end, it's slowing down the over all acceleration and creating tension on the string.

    So would it be the smaller mass (2.8) times the acceleration of the system (7.5891)?
  7. Jan 13, 2008 #6
    Or would it be 2.8 times (9.8-7.5891)?
  8. Jan 13, 2008 #7
    I got it! Nevermind. I understand what you were saying. I have no idea how I was coming up with that acceleration... it must have been luck. But I found out how to. I know how to analyze the forces correctly now.
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