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Question about labeling forces and solving for acceleration of a system.

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a question about labeling forces for a problem that consists of combined motion taking rotational motion of the pulley into account.

    Set up: table with a pulley on the end. Two masses connected by a string. There is one mass on the table, the other hanging mass is hanging from the pulley. The table is rough.

    General question. no specific numbers. I'm supposed to solve for the acceleration of the hanging block. My question is, am I labeling my axes wrong?

    For example: Let m_1 be the mass on the table to the left. Let m_2 be the mass hanging and Let M be the pulley. T, T',S, S' be tensions in the string (T,T' between m1 and pulley; S,S' between pulley and hanging mass)

    I describe the forces for each as the following:

    T- force of friction= m_1*a for the mass on the table

    R*(T'-S)= I*alpha for the pulley

    S'- Fg=m_2*a

    When I work it out, I get the right symbols, but I'm off by one or two signs. When I plug in numbers, I don't get the right answer. When I changed my y-axis. Made y down positive and y up negative, then that gave me the right answer.

    Am I labeling my axes wrong? For example, The book says for rotational motion, clockwise is negative, counter-clockwise is positve, so I have the tensions on the pulley as T'-S. What should I change?

    Any help is appreciated.
     
    Last edited: Nov 14, 2012
  2. jcsd
  3. Nov 14, 2012 #2

    haruspex

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    In which direction is the acceleration 'a' defined?
     
  4. Nov 15, 2012 #3
    @haruspex,

    The acceleration of the two masses connected by the string is defined as negative in the y-direction, I guess. I don't know what I'm doing.
     
  5. Nov 15, 2012 #4

    haruspex

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    Well, it will be different for each mass. Since you are using the one variable for both (as you should) you need to define the directions in a consistent way. The obvious way is to say it's towards the pulley for the mass on the table and downwards for the other.
    That being so, check the signs in your equation S'- Fg=m_2*a.
     
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