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Dynamic problem with pulley

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  1. Mar 17, 2016 #1
    1. The problem statement, all variables and given/known data

    You have a a mass, m1, in a plane. This mass is connected to a pulley by a thread, to a mass m2. Prove that:

    \begin{equation}
    m_{1} = \dfrac{m_{2}(2g - a)}{4a}
    \end{equation}

    2. Relevant equations

    0e6a5d2340e77c6e9b97cafbb2f27f0d.png

    9e4631226608aba70dca3fc61ca115d1.png

    3. The attempt at a solution

    I don't know why, but I can get this:

    \begin{equation}
    m_{1} = \dfrac{m_{2}(g - a)}{a}
    \end{equation}

    It's a similar equation, but it's not the same. How can I get the right solution?

    Thank you!
     
  2. jcsd
  3. Mar 17, 2016 #2
    how you take a start-give free body diagram of the masses?
     
  4. Mar 17, 2016 #3
    m1 is in a plane (it's not inclinated) and m2 is is hanging from a pulley. They are connected by a thread.

    In m1 we only consider Tension force (because N = mg), and in m2 we consider Tension force and m_{2}g

    Thanks
     
  5. Mar 17, 2016 #4
    so , how you write the equation of motion? the pulley is massless or not? if m1 is sitting on a plane will there be motion?
     
  6. Mar 17, 2016 #5
    The pulley is massless. The plane has no motion.

    I use this:
    \begin{equation}
    m_{1}\vec{g} + \vec{N} + \vec{T_{1}} = m_{1}\vec{a_{1}}
    \ \\
    m_{2}\vec{g} + \vec{T_{2}} = m_{2}\vec{a_{2}}
    \end{equation}
    Therefore:
    \begin{equation}
    T_{1} = m_{1}a_{1} \ \\
    T_{2} - m_{2}g = m_{2}a_{2}
    \ \\
    T_{1} = T_{2} (massless)
    \ \\
    a_{1} = -a_{2}
    \end{equation}
    Then, I got the equation I put before... But it's not what I must get.
     
  7. Mar 17, 2016 #6
    i just feel that how the mass m1 can move sitting tight on the plane- if it is raised from the plane by the string then what happens to gravitational pull-sorry i can not visualize it ,pl make the dynamics more clear
     
  8. Mar 17, 2016 #7
    This is a pic of the situation:

    zxv1g2.jpg
     
  9. Mar 17, 2016 #8
    in the above case your equations seem to be correct but the accelerations are not related like a(1) = -a(2) as one is horizontal and the other is vertical
     
  10. Mar 17, 2016 #9
    Thanks, but, anyway, I can't get the solution if I suppose a1 = a2...
     
  11. Mar 17, 2016 #10
    Problem solved. The exercise was wrong, there were 2 pulleys, not only 1.
     
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