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Moving pulley question

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data
    http://prntscr.com/otut8
    Find acceleration of block 1, knowing that the mass of block 1 is 2kg and of block 2 is 4kg. You may ignore friction and pulley mass.

    2. Relevant equations

    Fr = ma

    3. The attempt at a solution

    I Drew free body diagrams to block 1, 2 and the pulley, then made the equations:
    [itex]
    \\
    \sum F_{r2} = m_2a_2\\
    P_2 - T_2 = m_2a_2\\
    \\
    \sum F_{rpulley} = m_{pulley}*a\\
    2T_2 = T_1\\
    \\
    \sum F_{1}=m_1a_1\\
    T_1=m_1a_1\rightarrow T_2 = \frac{m_1a_1}{2}

    \\\\a_1=2a_2 \rightarrow T_2=\frac{m_1(2a_2)}{2} = m_1a_2

    \\(Substituting\ in \ first\ equation)\\
    P_2-(m_1a_2)= m_2a_2\\
    a_2(m_2+m_1) = m_2*g\\\\
    a_2=\frac{m_2g}{m_2+m_1} = 6.53\frac{m}{s^2}\rightarrow a_1 = 13.06\frac{m}{s^2}
    [/itex]
    The answer, however, is supposed to be 4.4 m/s²


    Thanks in advance!
     
  2. jcsd
  3. Jan 11, 2013 #2

    gneill

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    Staff: Mentor

    Is ##a_1 = 2 a_2## true?
     
  4. Jan 11, 2013 #3
    Well, I think it is, isn't it?

    If 1 moves a distance "d", 2 will move only d/2 because the rope has to run on both sides of the pulley.
     
  5. Jan 11, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    On the other hand, if 1 moves a distance "d", the pulley "takes" a length "d" of rope from both sides of the pulley. This rope can only come from the free end holding m2.

    attachment.php?attachmentid=54631&stc=1&d=1357924275.gif
     

    Attached Files:

  6. Jan 11, 2013 #5
    Wow, that... that makes a lot of sense, so in fact a1 = 2a2

    Thanks a lot, gneill!
     
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