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Friction and 3 blocks

  1. Dec 19, 2007 #1
    You are given that the coefficient of kinetic friction between the block and the table in the figure is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg.
    (a) What should m3 be if the system is to move with a constant speed?

    this is what i have:
    T2 - T1 - f = m3a

    m2g - m1g - ukg = m3
    2.45 - 9.65 - 5.488 = 4.508kg

    the answer is wrong, what is it that i am doing wrong?

    Attached Files:

    Last edited: Dec 19, 2007
  2. jcsd
  3. Dec 19, 2007 #2

    Doc Al

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    Staff: Mentor

    This is good.

    This is not good. For one thing, the tensions are not equal to the weights.

    Hint: Apply Newton's 2nd law to each mass separately. You took care of M3, now come up with equations for M1 & M2.

    You'll combine all three equations to solve for the acceleration and the tensions.
  4. Dec 19, 2007 #3
    i dont understand, i have for:
    T - m1g = m1a
    m2g - T = m2a
  5. Dec 19, 2007 #4

    Doc Al

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    Staff: Mentor

    Looks good, except that the tensions are different. Use T1 for m1 and T2 for m2.
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