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Kleppner/Kolenkow Pulley Example

  1. Feb 5, 2016 #1
    1. The problem statement, all variables and given/known data

    Hey, all. I am working through Kleppner/Kolenkow and encountered a problem when trying to follow the example pictured below. My issues comes when they say "Equations (1)-(3) are easily solved..."

    As it turns out, they are not so easily solved for me! So, it appears I've found a hole in my mathematical training that needs to be filled (K&K have proved to be good at exposing these weak spots).
    e8e583d28b.png

    2. Relevant equations
    All pictured above.

    3. The attempt at a solution
    Frankly, I'm not sure where to begin. I've tried expanding the ##\ddot{y_1}## and ##\ddot{y_2}## terms using ##(3)##, but the equations quickly become quite long. I haven't had the time to really sit down and see if they eventually clean up nicely and give the desired result, but truth be told, I don't feel too confident going into this and would love some help.
     
  2. jcsd
  3. Feb 5, 2016 #2

    BvU

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    Build confidence by starting off with A = 0 !
    Take your time to really sit down and try another tack if things still become too complicated.
     
  4. Feb 5, 2016 #3

    SammyS

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    First of all, you may have noticed that the xp in the figure should have been yp .

    Solve Eq.(3) for either ##\ \ddot y_1\ ## or ##\ \ddot y_2\ .\ ## Plug the result into Eq (1) or (2).

    etc.
     
  5. Feb 5, 2016 #4
    Update: Got it! I can type up my solution if you guys would like to see it.
     
  6. Feb 5, 2016 #5

    SammyS

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    Sure. Why not?
     
  7. Feb 5, 2016 #6
    I actually solved it right before you posted your advice (I solved it in the same way):

    I solved Eq. (3) for ##\ddot {y_2}## and plugged that solution into Eq. (2). This gave [tex] T = W_2 + M_2 (2A - \ddot{y_1}) [/tex]
    I then set the right-hand sides of Eq. (1) and Eq. (2) together: [tex] W_1 + M_1 \ddot {y_1} = W_2 + M_2 \ddot {y_2} [/tex]
    From here, it was just a matter for breaking down the ##W## terms into their corresponding ##M_ig## terms and rearranging:

    [tex] M_1g+M_1 \ddot{y_1} = M_2g + 2A M_2 - M_2 \ddot {y_1} [/tex]
    [tex] \ddot{y_1} =(2A+g) \frac {M_2 - M_1g}{M_1 + M_2} [/tex]

    To solve for ##T##, I simply plugged this result into Eq. (1), and the desired result followed immediately.
     
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