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Quick forces conceptual question

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data

    Find an expression for the acceleration of m1.

    2. Relevant equations

    3. The attempt at a solution
    I did this earlier, and looking back on it I have a question:

    Sum of forces for m1:
    Fx= F2 on 1= m1a1

    Sum of forces for m2:
    Fy = F1 on 2 + T1 - m2g = m2a2

    I got the expression after substituting them into each other and saying a2 = -.5a1, I'm just wondering why you can say T1= F1 on 2 , because since there is acceleration, can you consider the tension throughout the rope to be the same?
  2. jcsd
  3. May 18, 2008 #2
    It's a double pulley problem with m1 sitting on a table, attached to a rope that goes around a pulley at the edge of the table. Then the string passes through another pulley attached to the top of a suspended m2. Then the rope goes up and is attached to a ceiling.
  4. May 18, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    The only horizontal force on m1 is the rope tension T.

    Careful. The tension force pulls up twice on m2 (via the pulley).

    As long as the rope is massless and the pulley is massless and frictionless, then the tension is the same throughout the rope.
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