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Find an expression for the acceleration

  1. Aug 9, 2005 #1
    In the figure, find an expression for the acceleration of m_1

    i dont understand why

    (-m_2*g)/(((-m_2)/2)-m_1) isnt right

    because the a of mass#2 is (-a_1)/2 right?

    Attached Files:

  2. jcsd
  3. Aug 9, 2005 #2
    So i am assuming that the pulley has no friction and its rotation does not need to be taken into account, also the pulley does has no mass and can be treated as a point particle.

    Just draw a FBD for each mass seperately.

    The first mass has friction f with the table (i assume this, if it is not the case than f =0 ofcourse). T is the tension in he rope
    [tex]m_1a_1 = T-f[/tex]

    For mass 2 you get

    hence [tex]m_1 = \frac{m_2(a_2+g)-2f}{2a_1}[/tex]

    This is the most general form. I do not know whether extra info was given on the masses standing still or not ? If so, some variables in the above formula will vanish

    Last edited: Aug 9, 2005
  4. Aug 9, 2005 #3
    oh thats right, it never did say anything about it being frictionless, maybe thats what i did wrong.

    and there was no other information about the masses standing still
  5. Aug 9, 2005 #4

    Doc Al

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    Staff: Mentor

    Show how you arrived at this answer.

    That's correct. (Be careful with signs.)
  6. Aug 9, 2005 #5
    x m_1: T=m_1*a_1

    y m_2: T-m_2*g=m_2*a_2

    a_2= a_1/2

    2 (m_1*a_1)*g=m_2*(a_1/2)
  7. Aug 9, 2005 #6

    Doc Al

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    Staff: Mentor

    Two problems here:
    (1) The rope pulls twice on the pulley (and thus on m_2)
    (2) Careful with your sign convention. If you want to use up for positive, then be consistent. (You chose a positive a_1 going to the right.)

    If m_1 moves to the right, then m_2 moves down: this means that if you call a_1 positive, then by your sign convention for m_2, its acceleration should be negative a_2.

    Regarding sign conventions, here's what I do: I always assume a_1 and a_2 to be positive numbers. I also visualize how the accelerations relate to each other. If a_1 is to the right, then a_2 is down. Then I write my equations accordingly.

    Try it again.
  8. Aug 11, 2005 #7
    oh ok.. so i redid it and i got

    m2 y: 2T-m_2*g=m_2*a_2

    since T= m_1*a_1 i subsituted that it and solved for a_1 and got

    a_1= (2*m_2*g)/(4m_1+m_2) is this right?
  9. Aug 11, 2005 #8

    Doc Al

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    Staff: Mentor

    Looks good to me.
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