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Absolute Dependent Motion

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    The cylinder C is being lifted using the cable and pulley system shown. If point A on the cable is being drawn toward the drum with a speed of 2 m/s, determine the speed of the cylinder.

    bfql2a.png

    2. Relevant equations

    [tex]2s_{A} + s_{b} = l[/tex]

    3. The attempt at a solution

    I set my points to this:

    ftk2dd.png

    I don't think its right because I am getting a negative number when it should be positive.

    [tex]2s_{A} + s_{b} = l[/tex]

    [tex]2v_{A} + v_{b} = 0[/tex]

    [tex]v_{b} = -2(2m/s) = -4 m/s[/tex] This would mean that the cylinder is going down not up.
     
  2. jcsd
  3. Oct 3, 2009 #2

    tiny-tim

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    Hi KillerZ! :wink:

    I think the cylinder does go down when the cable is drawn up.

    But I don't think it's 2:1.

    Try using sc instead of sb, where sc is the distance between the two lowest pulleys …

    and use the fact that the total length of the string is constant. :smile:
     
  4. Oct 3, 2009 #3
    So sc would be like this the difference between s1 and s2?

    2yoocxz.png
     
  5. Oct 3, 2009 #4

    tiny-tim

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  6. Oct 3, 2009 #5
    I got it:

    103ues4.png

    [tex]s_{B} + (s_{b} - h) + (s_{B} - h - s_{A}) = l[/tex]

    [tex]3s_{B} - s_{A} - 2h = l[/tex]

    [tex]3v_{B} - v_{A} - 0 = 0[/tex]

    [tex]v_{B} = -v_{A}/3 = -0.667m/s = 0.667m/s[/tex] up
     
  7. Oct 3, 2009 #6

    tiny-tim

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    Hi KillerZ! :smile:

    Yes, except it's +vA/3.

    (ignore what I said originally … I misread the diagram … the cylinder does go up! :redface:)
     
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