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Kinetic friction and acceleration

  1. Dec 4, 2013 #1
    Hello everyone, I would appreciate some help with this problem:

    The coefficient of kinetic friction between the 1.2 kg block in the figure below and the table is 0.38. What is the acceleration of the 1.2 kg block?

    2. Relevant equations

    F=M*A
    Fk = μk * N

    3. The attempt at a solution

    I created freebody diagrams and came up with these

    Block A: +T1 - WAE = Ma * a
    Block B: + T2 - T1 - fk = Mb * a
    Block C: + WCE - T2 = Mc * a

    So I combined equations and came up with

    WCE - WAE - fk = (Ma * a)+(Mb * a)+(Mc * a)

    and fk= μk * N

    and I thought it would be .38*1.2*9.8

    but I know I am doing something wrong I just don't know what
     

    Attached Files:

  2. jcsd
  3. Dec 4, 2013 #2

    PhanthomJay

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    Looks ok to me....o wait I didn't check your combined equation..stand by....

    Yes, looks good! Solve for a .
     
    Last edited: Dec 4, 2013
  4. Dec 4, 2013 #3
    Alright, I redid my algebra and that is where the problem was.....I am rather embarrassed! Thank you for your help PhanthomJay.
     
  5. Dec 4, 2013 #4

    PhanthomJay

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    Great, good work, welcome to PF! :approve:
     
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