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Find the acceleration of the hanging block

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    In terms of m1,m2, and g. find the acceleration of the hanging block in the two-block/ two-pulley setup shown in the diagram. The stands holding the string rests on the floor. All pulleys are massless, and there is no friction anywhere in the system.
    1418ljc.jpg

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    First I drew free body diagrams of both blocks and of the lower pulley. I numbered the tensions in the pictures. The equations that I got for m1:
    Fy= N=mg
    Fx= T1=m1a
    For m2 I got:
    Fy= T2-m2g=-a*m2
    For pulley I got:
    T1+T3=T2

    Then I substituted T2 Into the equation for m2 and I got:

    T1+T3-m2g=-a*m2
    m1*a+T3-m2g=-a*m2
    Now I am confused but I decided to call T3 just T because T is same in the rope.
    As a result my answer is a= (m2g-T)/(m2+m1), but it is probably wrong. Could you please check if I am on track here?
    Thanks for the help!
     
  2. jcsd
  3. Oct 18, 2014 #2
    It is sort of hard to tell if you are on the right track (imo) because the picture you posted is basically impossible to read
     
  4. Oct 18, 2014 #3
  5. Oct 18, 2014 #4
    Also ignore what it says on the picture it is not part of the question.
     
  6. Oct 18, 2014 #5
    Edit: nvm, I can't read XD
     
  7. Oct 18, 2014 #6
    No, all it says "the lab stand holding the string at the upper right rests on the floor. Thanks for help I appreciate it.
     
  8. Oct 18, 2014 #7

    NascentOxygen

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    Staff: Mentor

    Hi Emily. Did you take into consideration their different accelerations? Determine their ratio this way― if m1 slides 1cm to the right, m2 will drop by what distance?

    BTW, the pic in your first post is perfectly fine on my Android.
     
  9. Oct 18, 2014 #8
    Would it be 1 cm too?
     
  10. Oct 18, 2014 #9

    NascentOxygen

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    Staff: Mentor

    Do you have half a metre of string and a wheel or small bottle that can take the place of the pully? Arrange something over the edge of your table and you can try it and see!

    m1 can be a heavy book or something.

     
  11. Oct 18, 2014 #10
    The professor gave a hint to draw 3 body diagrams. One for block 1 second for block 2 and last one for the lower hanging pulley.
     
  12. Oct 18, 2014 #11
    I think m2 will fall faster than m1 will so m1 will still stay on the table…
     
  13. Oct 18, 2014 #12
    ok then disregard my comment, sorry
     
  14. Oct 18, 2014 #13
    Its ok ! I appreciate your help.
     
  15. Oct 18, 2014 #14

    NascentOxygen

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    Staff: Mentor

    I didn't mean falling. oo)

    As you slide the book 10cm along the table, take notice how far your improvised "pulley" is allowed to descend.

    No falling!
     
  16. Oct 18, 2014 #15
    All the top would get "down" and the pulley will hang but the block would stay on the table. Anyways this problem seems to be difficulties and I am not very good with pulleys so sorry if I do not get what your saying. This problem is very important for me it worths 22 pts. Thanks for your time.
     
  17. Oct 18, 2014 #16
    no problem
     
  18. Oct 18, 2014 #17
    Here are my new equations for the FBD:

    For the first block (m1):
    Fx T=m1a

    For the second block (m2):
    Fy T-m2g=-m2a

    For the pulley:
    2T-T-m2g=0 Not sure for this one if m2 is right and not sure that ma=0
     
  19. Oct 18, 2014 #18
    And for the pulley its for Fy cause no Fx
     
  20. Oct 18, 2014 #19
    Well remember that the pulley is without mass..
     
  21. Oct 18, 2014 #20
    For the first block (m1):
    Fx T=m1a

    For the second block (m2):
    Fy T-m2g=-m2a

    For the pulley:
    2T-T=0 Is it right now?
     
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