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Tension between Blocks

  1. Feb 5, 2009 #1

    TG3

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    1. The problem statement, all variables and given/known data
    A block of mass 3 kg pulls a block of mass 2 kg across a floor with a force of 24 N. The coefficient of friction is mew = .19. What is the tension between the blocks?
    Note:
    Acceleration = 2.936. (Solved for this earlier and was correct.)

    2. Relevant equations
    T =F- m1a
    T = m2a
    T=(m2/(m1+m2)) x F


    3. The attempt at a solution
    T = 2.936 X 2 = 5.872 Wrong.
    T = (24-9.3195) - (3x2.936) = 5.8725 Wrong.
    T = (2 / (2+3)) x (24-9.315) = 5.8722 Wrong again.

    I've tried 3 different methods, come up with very similar answers that when rounded are equivalent, labeled units, and still gotten it wrong.... what's up?
     
  2. jcsd
  3. Feb 5, 2009 #2

    rl.bhat

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    Homework Helper

    T =F- m1a
    Try this!
     
  4. Feb 5, 2009 #3

    TG3

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    That was the second method I used:
    (24-9.3195) is the net force (since 9.3195 is the force of friction pushing back) and (3x2.936) is m1a.
    Still no luck...
     
  5. Feb 6, 2009 #4
    How did you find the acceleration for the first part? I know you got the correct answer, just want to see how you got it
     
  6. Feb 6, 2009 #5

    TG3

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    The Acceleration equals the Net force / (mass 1+ mass 2)
    The net force equals 24 - resistance due to friction. [(m1+m2) x 9.81 x .19]

    Still no luck with the tension though...
     
  7. Feb 6, 2009 #6
    Draw a free-body diagram for m1.
    Use SF=ma, as applied to the mass m1. Since we know the acceleration from part (a), we can find the net force. The net force is the sum of the forces. One of those forces is the tension T.
     
  8. Feb 6, 2009 #7

    TG3

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    Got it. Thanks!
     
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