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Find tension between blocks 2 & 3

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data

    There are three blocks connected by strings as shown in the figure. Each block has the mass as following mass,
    m1=0.8kg
    m2=2.2kg
    m3=1.7kg
    Assuming there is no friction,


    2. Relevant equations

    F=MA

    3. The attempt at a solution

    In my FBD of the hanging box
    [itex]\uparrow[/itex] <--- Tension
    []
    [itex]\downarrow[/itex] <--- W=mg

    ^^ in y direction: T-W=0
    T=W

    The FBD of the box on the table
    [itex]\uparrow[/itex] <---Normal Force
    [] [itex]\rightarrow[/itex] <--- Tension
    [itex]\downarrow[/itex] <--- W=mg

    ^^ x direction: t=0
    y direction: Fn-W=O

    But im completely confused on how to find the tension of the boxes
     

    Attached Files:

  2. jcsd
  3. Feb 29, 2012 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Rey4312! Welcome to PF! :smile:

    Call the tensions "T1" and "T2", and the acceleration "a" (it'll be the same for all three boxes, won't it? :wink:)

    Then do F = ma three times, once for each box, and solve …

    what do you get? :smile:
     
  4. Feb 29, 2012 #3
    But if i do that the acceleration = 0 and so it makes all the forces 0... right??


    or am i missing something and it should be gravity?
     
  5. Feb 29, 2012 #4

    tiny-tim

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    Science Advisor
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    Yes, you need to include W (=mg) for the hanging box.
     
  6. Feb 29, 2012 #5
    W for the hanging box is 16.66

    and W for box 2(m2) is 21.56....

    Where do i go from there, because that doesn't give me acceleration....
     
  7. Feb 29, 2012 #6
    i figured it out... i ended up having to do m2a+m1a
     
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