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Homework Help: Acceleration of blocks on a slope

  1. Aug 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Two blocks are connected by a string that goes over an ideal pulley. Block A has a mass of 3.00 kg and can slide over a rough plane inclined at 30 degress to the horizontal. The coefficient of kinetic friction between block A and the plane is 0.400. block B has a mass of 2.77 kg. What is the acceleration of the blocks?

    2. Relevant equations
    fk = ukFn

    3. The attempt at a solution

    Fn - 3gcos(30)
    Fn = 25.4611

    fk = 0.4*25.4611
    = 10.1845

    Using force diagrams I identified the forces acting on the blocks to be:

    T is the tension in the rope

    For block A
    T - fk - 3gcos(30) = 3a

    For block B
    T - 2.77g = 2.77a

    Then I combine them

    fk - 3gsin(30) -3a = 2.77g - 2.77a
    a = 98.4

    I know the answer is 0.392m/s^2 but I don't know where I'm going wrong. I assumed that block B is being pulled up because block A has a greater weight. Is this a wrong assumption?
    Last edited: Aug 9, 2012
  2. jcsd
  3. Aug 9, 2012 #2


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    Hi Northbysouth! :smile:
    but you still need to find fk :confused:
    depends on the slope …

    obviously, if the slope was 0°, then B would win! :wink:
  4. Aug 9, 2012 #3
    I found fk = 10.1845

    Could you be more specific about the slope? How do I determine whether the friction is acting in the same direction as the tension or in the opposite direction?
  5. Aug 9, 2012 #4


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    to find the direction of friction, imagine that there's no friction, and see which way the blocks go :smile:

    (and i'm off to bed :zzz:)
  6. Aug 9, 2012 #5
    I ran the calculations again with friction acting opposite to the tension.

    T - 10.1845 - 3gsin(30) =3a for block A

    T =2.77a + 27.146 for block B

    I substituted T into block A's equation:

    2.77a+27.146 -10.1845 - 3gsin(30) = 3a
    a = 0.520145

    This still seems a ways off from 0.392. Am I missing something?
  7. Aug 10, 2012 #6


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    (just got up :zzz:)
    i] if you subtract equation B from A (so that the RHS is 3a), those first two terms on the LHS should be minus

    ii] but your B equation is wrong if you're measuring a downwards (ie, if B is going down) …

    then a = mg - T, not T - mg :wink:
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