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Conservation of the string length in pulleys

  1. May 27, 2017 #1
    1. The problem statement, all variables and given/known data
    84bf5fb6de3049c098b9b20e511f9d7f.png
    This is a common massless string and pulleys problem. I'd just like to understand why, according to the solution, [itex]l_2 + 2l_1 =[/itex] constant. It doesn't seem to me that two times the [itex]l_1[/itex] lenght is equivalent to [itex]l_2[/itex]. Can somebody explain?

    2. Relevant equations


    3. The attempt at a solution
     

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  3. May 27, 2017 #2

    haruspex

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    Indeed not, but that would lead to ##2l_1-l_2## is constant, which is not what is claimed.
    Besides, you need to think about changes in the lengths, not the absolute lengths. If l1 increases by x, what happens to l2?
     
  4. May 27, 2017 #3
    Hm. Will it decrease? Because if [itex]m_1[/itex] goes down, it means [itex]m_2[/itex] has to go up, right?
     
  5. May 27, 2017 #4
    Let x be the distance in elevation between the center of pulley 1 and the center of pulley 2, y be the distance between the elevation of the center of pulley 2 and the ceiling, z be the distance in elevation between mass 2 and the center of pulley 2, and w be the distance in elevation between mass 1 and the center of pulley 1. So the total length of string is L=2x+y+z. The distance l2 is given by ##l_2=z+y## and the distance l1 is ##l_1=x+y+w##. So, $$2l_1+l_2=(2x+z)+3y+2w=L+3y+2w$$ But, since L, y, and w are constant, ##2l_1+l_2## must be constant.
     
  6. May 27, 2017 #5
    I was able to understand everything before this point. Why do you take [itex]2l_1[/itex]?
     
  7. May 27, 2017 #6
    I do ##2l_1## because I need 2x.

    Incidentally, I would never have done this problem in terms of l1 and l2. I would have immediately written down the total length as ##L=2x+y+z##, and then differentiated to get $$\frac{dL}{dt}=0=2\frac{dx}{dt}+\frac{dz}{dt}$$ If m1 moves down at the rate of dx/dt, m2 moves up at the rate of -dz/dt.
     
  8. May 27, 2017 #7
    OK. I get it now, but I don't quite understand the differentiation.
     
  9. May 27, 2017 #8
    What part don't you understand?
     
  10. May 27, 2017 #9
    Where did the [itex]y[/itex] term go to?
     
  11. May 27, 2017 #10
    What is the derivative of y with respect to time?
     
  12. May 27, 2017 #11
    [itex]\frac{dy}{dt}[/itex]? But since it is constant, it equals 0? Is that so?
     
  13. May 27, 2017 #12
    Sure
     
  14. May 27, 2017 #13
    But isn't [itex]x[/itex] also constant? I mean, the distance in elevation between the center of pulley 1 and the center of pulley 2 will always be the same, won't it?
     
  15. May 27, 2017 #14

    haruspex

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    No, why?
     
  16. May 27, 2017 #15
    No
     
  17. May 27, 2017 #16
    Nevermind. I can see that now. Thank you!
     
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