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Pulley system on a big block

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  1. May 23, 2017 #1
    1. The problem statement, all variables and given/known data

    Find the accelaration of ##M_1## in the given system if ##F = 0##.
    upload_2017-5-23_17-43-35.png


    2. Relevant equations


    3. The attempt at a solution
    upload_2017-5-23_17-49-23.png


    ##x_3 -x_1 = k \iff \ddot x_3 = \ddot x_1##

    and ##h - y_3 + x_3 - x_2 = l \iff \ddot y_3 + \ddot x_2 = \ddot x_3 \qquad (*)##

    h is the height of block ##M_1## and ## l ## is the length of string between ##M_2## and ##M_3##.

    Now from the Free body diagram,

    upload_2017-5-23_17-43-42.png

    ##-M\ddot x_1 = N^{\prime \prime \prime}##

    ##T = M_2 \ddot x_2##

    ##N^{\prime \prime \prime} = -M_3\ddot x_3##

    ##\therefore -M\ddot x_1 = M_2 \ddot x_2 - M_3\ddot x_3 \qquad (1)##

    Now from vertical force on ##M_3##,

    ##M_3 - T = -M_3 \ddot y_3##

    ##-M_3 g + M_2 \ddot x_2 = M_3 \ddot y_3##

    Substituting for ##\ddot y_3## in ##(*)##

    ##\ddot x_3 = x_2 + \dfrac{-M_3 g + M_2 \ddot x_2 }{M_3}##

    Solving for ##\ddot x_2##

    ##x_2 = \dfrac {M_3(\ddot x_3 + g) }{M_3 + M_2}##

    Substituting this in ##(1)##

    ##-M_1 \ddot x_1 = \dfrac {M_3M_2(\ddot x_3 + g) }{M_3 + M_2} - M_3\ddot x_3##

    Since ##\ddot x_1 = \ddot x_3##

    ##-M_1 \ddot x_1 = \dfrac {M_3M_2(\ddot x_1 + g) }{M_3 + M_2} - M_3\ddot x_1##

    Solving for ##\ddot x_1##

    ##\ddot x_1 = \dfrac{-g(M_2M_3)}{M_1M_2 + M_3M_1 - M_3^2}##

    Which is incorrect as the given answer is ##\ddot x_1 = \dfrac{-g(M_2M_3)}{M_1M_2 + M_3M_1 \color{red}{ + 2M_2M_3 +} M_3^2}##.

    What is the problem ?
     
  2. jcsd
  3. May 23, 2017 #2

    kuruman

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    This equation doesn't look right.
     
  4. May 23, 2017 #3
    Oh sorry that was a pure typo.
    That equation should be,
    ##M_3g - T = -M\ddot y_3##.
     
    Last edited: May 23, 2017
  5. May 23, 2017 #4

    kuruman

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    Still not right. On the left side (Fnet) "down" is positive, but on the right side (mass x acceleration) "down" is negative.

    Also, check equation (1). The center of mass does not accelerate because the only external force to the three mass system is a vertical force. So ##M_1 \ddot{x}_1+M_2 \ddot{x}_2+M_3 \ddot{x}_3=0##
     
  6. May 23, 2017 #5
    Yes I got the correct answer, but I think ##M_3g - T = -M\ddot y_3## is correct because ## y_3## is pointing upwards and so is ##\ddot y_3## and therefore I put a negative sign, since ##M_3g - T## is downwards as you said. Am I wrong ?
     
  7. May 23, 2017 #6

    kuruman

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    No, it's OK. I fooled myself because I didn't notice ##y_3## in the diagram.
     
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