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Homework Help: Building constraint equations

  1. Jun 30, 2013 #1
    1. The problem statement, all variables and given/known data
    In the system shown, the strings and the pulleys are ideal. There is no friction anywhere in the system. As the system is released, block m1 moves downward.

    If the magnitude of the acceleration of block m1 is a, what is the magnitude of the acceleration of block m2?

    2. Relevant equations

    3. The attempt at a solution

    If mass m1 moves down by x, the pulley attached to m2 moves towards left by 2x, Hence block moves by 2x. Hence, acceleration of block 2 is 2a. Am I right?

    I have a problem with constraint equations. If someone could clear it, please.

    Attached Files:

  2. jcsd
  3. Jun 30, 2013 #2
    No. Think of it this way : if mass m1 moves down by distance x, then all the points on the string attached to it move by distance x. Which means the center of the horizontal pulley also moves to the left by x, not 2x, since after all it is fixed to the string.
  4. Jun 30, 2013 #3
    Ya, I realized. How do I relate the acceleration of the two blocks?
  5. Jun 30, 2013 #4
    When you displace the horizontal pulley by a distance x to the left, how much excess string would you need for m2 to remain stationary?
  6. Jun 30, 2013 #5
    Also, the string it needs to move the distance x, is compensated by block 2 moving to the left by a distance x. Because the other side of the string that run through the pulley is attached to the wall. Therefore, block going down by x, would move the pulley to the left by x. Differentiating twice would give a1 = a2. But that's no the right answer.
  7. Jun 30, 2013 #6
    How can it remain stationary if the pulley is being displaced to the left? Won't the pulley pull the string, hence pulling the block 2?
  8. Jun 30, 2013 #7
    You didn't get my question - if, hypothetically, we were to make the block remain stationary, how much excess string would we need to add to the existing piece?
  9. Jun 30, 2013 #8
    EDIT : Double post
  10. Jun 30, 2013 #9
    Wouldn't it b x? Because the pulley is moving by x, and therefore it's going to try moving the block 2 by x. If block 2 had an excess x lenght of string, block 2 wouldn't have to move at all. Am I right?
  11. Jun 30, 2013 #10
    What's your approach to this problem? How do you relate a1 to a2?
  12. Jun 30, 2013 #11
    Nope. Think of it this way : the top of the pulley moves by a distance x, but the bottom also moves by a distance x. That means both the top portion of the string, and the bottom portion have increased by x, which can't happen as the length of the string is fixed. So the mass moves by 2x to the left.

    Basically, your answer is correct, but the method by which you reached it was not.
  13. Jun 30, 2013 #12
    Still haven't got it. You mean a1 = 2a2 is correct? Yup, it is. But I got it by manipulation. Haven't understood the problem. I always had a problem with constraints. By the very basics of it.
  14. Jun 30, 2013 #13
    If both the top and bottom strings have moved by x, shouldn't it be x1 = x2?
  15. Jun 30, 2013 #14


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    Can the total length of the string(s) change?
  16. Jun 30, 2013 #15
    If m2 is extremely large, m1 won't move at all. If m1 is zero, again, the pulley and m2 won't move at all.
  17. Jun 30, 2013 #16


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    In case it's not clear, dreamLord is pointing out that although your answer below is right, the statement above is wrong. To see this, ignore m2 and the string attached to it, and think of the top pulley as just a block. What you may have had in mind is that the top pulley both moves (by x) and rotates.
  18. Jun 30, 2013 #17
    In this drawing, the figure in red is when the pulley has moved by distance x to the left. Does this help?

    Attached Files:

  19. Jun 30, 2013 #18
    Okay, Thanks! Got it. Haruspex is right. I was considering the pulley to be rotating.
  20. Jun 30, 2013 #19
    Also, what would be the tension in the string attached to block 2? Would it be T? I know the tension on the string running from pulley to block 1 is T, because the tension is same in the string throughout. But what about the tension in the string attached to block 2?

    Would it be T11 = 2T2? therefore tension on string attached to block 2 would be T1/2. Correct?
  21. Jun 30, 2013 #20


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    What are the forces on the horizontal pulley? Can there be a net force on the pulley if its massless? Use that to find the relationship between the two tensions.
  22. Jun 30, 2013 #21
    Horizontal force on the pulley would be T11 = 2T2, so block 2 would be T11/2 = T2
  23. Jun 30, 2013 #22


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    I think you mean ##F_p = T_1 - 2T_2##. And yes if you then use the fact that ##F_p = 0## you will get what you originally stated (I just wanted to make sure you knew why the result followed).
  24. Jun 30, 2013 #23
    Yep, that relation is correct for the tensions.
  25. Jun 30, 2013 #24
    What I don't get is, given the setup in the original diagram, how can ##m_1## ever move? The right endpoint of the string attached to ##m_1## is fixed to the center of the horizontal pulley and the string around the horizontal pulley is itself anchored to the wall. If ##m_1## tries to accelerate down it will pull on the horizontal pulley but since the string around the horizontal pulley is anchored to the wall, won't the horizontal pulley refuse to budge thus keeping ##m_1## in place?

    The only movement I can see happening is that ##m_2## gets pulled forward due to the tension in the string on its end generated by the string attached to the wall resisting the pull of ##m_1## on the horizontal pulley. Thanks!
  26. Jun 30, 2013 #25
    The horizontal pulley is not anchored to the wall - I think that would happen if we tied a piece of string joining the center of the pulley to the wall.

    EDIT : Sorry, you're talking about the string, my bad.

    In that case, why should the pulley be stationary? If there is a net force on it, it will move. Where is the constraint?
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