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Homework Help: 3D Equilibrium

  1. May 31, 2014 #1


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    1. The problem statement, all variables and given/known data

    I'm having a bit of trouble understanding something from this:


    2. Relevant equations

    3. The attempt at a solution

    I understand how to get the components of reaction at ##A## as well as the tension in segment ##BE##.

    What I'm having trouble with is the FBD of the pulley used to determine the tension in ##DEC##. I'm not quite sure how they have come up with a part of the force equation, namely:

    ##\sum F_z = 0 \Rightarrow 2(\frac{4}{\sqrt{96}})T - \frac{1}{\sqrt{5}}(1677.05) = 0##

    The term ##2(\frac{4}{\sqrt{96}})T## is what is confusing. I understand you need twice the tension since it's symmetric, but how on earth are they getting ##\frac{4}{\sqrt{96}}##?

    I figure they are applying ##\frac{T_z}{T} = \frac{4}{\sqrt{96}}## to a triangle, but I'm not seeing the triangle.
  2. jcsd
  3. May 31, 2014 #2


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    If you look at the diagram, write the coordinates of points C, D, and B.

    When the pulleys are under tension, one can see that CDE make a triangle and a plane. You can also assume that the line BE will also lie in that same plane. You can determine the position of the point E by drawing a projection of the plane CDE in the y-z coordinate plane, and extending it to the line AB on the y-axis. Once the coordinates of point E are known, you can calculate the length of line DE, which is Sqrt (96) feet.
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