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Pulley acceleration help

  1. Apr 2, 2010 #1
    1. The problem statement, all variables and given/known data

    hi in this picture ( http://img26.imageshack.us/img26/8458/53487724.png [Broken] ) the teacher says that even if m1 is greater than m2 m2 will accelerate down and m1 will accelerate to the right but by newtons third law shouldn't m1 have a leftward force to the rope which should pull m2 up?

    2. Relevant equations

    F(m2 to rope ) = -F(rope to m2)
    F(m1 to rope ) = -F(rope to m1)

    3. The attempt at a solution
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 2, 2010 #2
    Re: pulleys

    Newton's 3rd law says that the rope pulls m1 to the right, and m1 pulls the rope to the left (and this in turn pulls m2 in an upwards direction*). There is no force on m1 to the left.
    * This upwards force on m2 acts against the weight of m2 downwards.
    So as far as m2 is concerned, there are 2 forces, its weight downwards, and the tension in the rope upwards.
    As far as m1 is concerned, there is a force from the rope pulling to the right.
     
  4. Apr 2, 2010 #3
    Re: pulleys

    so why does m2 accelerate downwards? since tension is up and weight is down shouldn't Fnet = 0 ?
     
  5. Apr 2, 2010 #4
    Re: pulleys

    It accelerates downwards because
    a) its weight mg is greater than T, the tension in the string; and
    b) the mass m1 is pulled by a force (T) to the right, and the two masses are connected to the same string.
    Think about it; how could m1 accelerate to the right yet m2 not accelerate down?
    Because they are connected, both masses must have the same acceleration, a.
    The equation for m2 is M2g-T=M2a
    The equation for m1 is T=M1a
    Given g, m1 and m2 you have 2 unknowns, a and T, and two equations.
    Eliminate T and solve for a will give you the common acceleration of both masses.
     
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