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Friction between two boxes

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  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Selection_017.png

    2. Relevant equations
    F=ma
    Ff = N u

    3. The attempt at a solution
    Pretty much, I did the same thing in the solution.

    For box A:
    N = mg
    Ffs = mg*uk = 9.81*20*.4=78.48N

    I note that Ffs > 60 N, so the boxes will move with the same acceleration

    Now there are two ways I can approach this problem, each gives me a different answer??

    1) Take two boxes as the same system, then Fx = 70*a = 60 ==> a = 0.857 m/s^2
    2) Look at the FBD of box 2, note that Fx = 50*a = 60 ==> a = 1.2 m/s^2

    However, both of these answers are different than the solution.

    Where did I mess up (and why do 1 &2 give different solutions?)
     
  2. jcsd
  3. Mar 1, 2015 #2

    PhanthomJay

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    Gold Member

    Solution messed up the math, although equation is correct. . When you look at the lower Block , you did not properly address the force acting on the lower block. It isn't 60. Draw an FBD.
     
  4. Mar 1, 2015 #3
    Check for errata. Looks like they did 60 = 70a --> a = 70/60, which is wrong.
     
  5. Mar 1, 2015 #4
    Isn't this the FBD? Ffs on the top block is -60N, and since its an action-reaction pair it should be +60N on the 50kg block


    EDIT: Actually, now that I think about it, the friction on the bottom/top block should be different, because its making the top block accelerate.

    So the I have more equations to solve:

    20a = -Ff + 60
    50a = Ff

    20a = -50a + 60
    70a = 60
    a = 60/70

    Thank you everyone for your help
     
    Last edited: Mar 1, 2015
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