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Acceleration of a three mass system

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data

    The system shown in the diagram is accelerating toward the right (clockwise). Find the acceleration in terms of m1, m2, m3, uk, and g.

    [​IMG]

    Note: There aren't any numbers used in this problem. It's just working with the variables to create an expression equal to the acceleration.

    2. Relevant equations

    Fg = mg
    Ff = ukFn
    SumofF= ma

    3. The attempt at a solution

    I began by creating free body diagrams of each mass, starting with m3.

    [​IMG]

    I used the forumla Fg = mg to get m3g. This is also equal to the tension in the string.

    T = m3g

    I moved on to creating a free body diagram of mass2.

    [​IMG]

    I used the forumla Fg=mg, where this downward force would be equal to the normal force. I then used the equation Ff = ukFn to get the friction force. Subtracting this friction force from the tension in the string would give me the net force for m2.

    T = m3g - m2guk

    This force would be equal to the tension in the string pulling m1 which is m3g - m2guk. Because the system is moving clockwise I subtract m1g to get the sum of all the sytem's forces moving in the clockwise direction.

    F = m3g - m2guk - m1g

    Since I need to find the acceleration of the system, I set the expression m3g - m2guk - m1g equal to m1, because m3g - m2guk - m1g is the force pulling m1 up.

    Thus.

    (m3g - m2guk - m1g) / (m1) = a

    Did I solve this problem correctly or was there some flaw to the logic I used?

    Thanks ahead of time. I apologize for the quality of the diagrams.
     
  2. jcsd
  3. Oct 28, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Vaati ! Welcome to PF! :smile:

    (have a mu: µ :smile:)

    Sorry … the flaw is that T = m3g is valid only if the acceleration is zero.

    But anyway, why work out the tensions when the question doesn't ask for them? :confused:

    Hint: call the distance moved "x", and use the work-energy theorem. :wink:
     
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