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Questions on Dynamics

  1. Jul 20, 2011 #1
     
  2. jcsd
  3. Jul 20, 2011 #2
    The logical place to start on the first question would be to write Newton's second law for every mass and the two pulleys. Did you try that?
     
  4. Jul 22, 2011 #3
    Just checking, since the 1kg pulleys are not massless, so the tension of the string on either side of the pulley is different?
     
  5. Jul 22, 2011 #4
    Yes, the tension would be different on both sides. You will have to write Newton's second law for both pulleys like normal.
     
  6. Jul 22, 2011 #5
    How do I proceed?

    d9Xit.jpg
     
  7. Jul 22, 2011 #6
    Since the pulleys are not massless, they will not have zero accelerations. Also, the tension in a thread is the same throughout, so the tensions T1 and T2 are equal, and T3 and T4 are equal. What I meant in the previous post was that the tension of the thread connecting the pulleys will not be equal to the tensions of the threads below them. Write Newton's laws for the pulleys in terms of the tensions. Then write Newton's laws for every mass, keeping in mind that the masses accelerate along with the pulleys they are connected to.
     
  8. Jul 22, 2011 #7
    Why is the tension in a thread the same throughout? If the pulley has a mass, a resultant torque is required to set it in rotational motion, thus the tension on one side has to be greater than the other side, no?
     
  9. Jul 23, 2011 #8
    That is correct, but you have no information of the radius of the pulleys, which is required to write a complete equation for torque. If you "ignore" the rotation of the pulley, you do end up with the answer you mentioned in your post. I'm not sure why you can do this, but I guess the two lower pulleys are not free to rotate about their central axes, so only the threads slide across them without friction (thus having the same tension)
     
  10. Jul 24, 2011 #9
    Is this a flaw in the question? Because when it says "pulley", it's only natural to assume that rotational motion is involved, rather than frictionless sliding.
     
  11. Jul 24, 2011 #10
    It would definitely appear so...where did you get the question from by the way?
     
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