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2 masses and 3 pulleys/accelaration

  1. Aug 9, 2007 #1
    beztytuuzj6.jpg

    We must count the accelaration of mass two(M2). Pulleys and ropes are massless. I've made the free body diagram and i've got two equations and three unknowns.

    Please help:)
     
  2. jcsd
  3. Aug 9, 2007 #2

    Doc Al

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    I assume that you've applied Newton's 2nd law to both masses and that gave you your two equations. But what about the constraint equation that relates the accelerations of M1 and M2? That will be your third equation.

    (To get additional help, show exactly what you've done so far.)
     
  4. Aug 9, 2007 #3
    bez tytu?u.JPG

    and two equations to those masses

    but how i can write the third?
     
  5. Aug 9, 2007 #4

    Doc Al

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    That's a good diagram showing the forces. What equations did you get?
     
  6. Aug 9, 2007 #5
    ok
    i understood it with those two tensions
    and of course we can write equation for the center( downward) pulley
    which is contradicted
     
  7. Aug 9, 2007 #6

    Doc Al

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    What are your two equations? What are your three unknowns?

    The constraint equation that I refer to relates the movement of M1 to the movement of M2.
     
  8. Aug 9, 2007 #7
    a1, a2, and tension



    but I think that topic is closed(write free body diagram for the upward center pulleys):rolleyes:
     
  9. Aug 9, 2007 #8

    Doc Al

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    So you've solved the problem?
     
  10. Aug 10, 2007 #9
    no, sorry for my previous posts, I made a mistake.
    bez tytu?u.JPG
    This scheme is good but I still don't know how to solve this.
     
  11. Aug 10, 2007 #10

    lightgrav

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    what does the attachment say?
     
  12. Aug 10, 2007 #11
    //the name?

    ok I wrote Newton's second law for the most left pulley:

    0 * a = 2T-T

    and for the rest massess if T=0 then accelaration of each mass is g, but they cannot move

    is it correct?^^
     
  13. Aug 10, 2007 #12

    learningphysics

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    Yes, that's what I get as the answer also. It was a kind of trick question. They can move... there's just no tension in the rope...
     
    Last edited: Aug 10, 2007
  14. Aug 10, 2007 #13

    lightgrav

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    No, the tension is the same all through the single rope ...

    If you try to apply Newton's 2nd to a "massless" object,
    you run into infinities.

    Always apply Newton#2 to NONzero mass! (like m1 or m2 !)
     
  15. Aug 10, 2007 #14

    learningphysics

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    Suppose we used a tiny nonzero mass for the pulleys... then the result would be that the accelerations of each of the two masses is approximately g... I think these zero mass situations are taken as "limit" behavior of small masses...
     
    Last edited: Aug 10, 2007
  16. Aug 10, 2007 #15
  17. Aug 10, 2007 #16

    Doc Al

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    Still waiting for you to show your two equations. (For m1 & m2.) Then we can worry about the third.
     
  18. Aug 10, 2007 #17
    m2a2=m2g-T
    m1a1=m1g-2T
     
  19. Aug 10, 2007 #18

    Doc Al

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    OK. Now you must connect the two accelerations. How are they related? Imagine the system in motion--if m2 drops 1 m, what happens to m1?

    Assume that the acceleration of m2 has magnitude "a" and is downward. Figure out the direction and magnitude of the acceleration of m1 in terms of "a".
     
  20. Aug 10, 2007 #19
    yes i've got problem with this
    I also tried to do this using conservation of energy (kinematic and
    potential) and then differentiate it and solve equation but I've got problems with what you wrote
     
  21. Aug 10, 2007 #20

    Doc Al

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    constraint equation

    No matter how you try to solve this, you'll still need to understand the constraint equation.

    Since m1 and m2 are connected by a string, their accelerations are not independent. If you are having difficulty seeing how they relate, get a piece of string or a strip of paper and figure it out. (I'm serious.)

    Here's an example of a constraint equation for the simple case of two masses hanging over a single pulley. If one mass goes up 1 m, then the other must go down 1 m. Thus, if one has acceleration +a, the other has acceleration -a. (Written as an equation, that would be: a1 = -a2.)
     
    Last edited: Aug 10, 2007
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