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How to find forces in wires at equilibrium

  1. Aug 30, 2015 #1
    I need to find tensions in suspensions wires on the picture:
    fZruy.png

    I wrote 4 equilibrium equations for moments in A, B, C, D and equilibrium about y-axis, but i got matrix with rank two.

    [tex]\begin{pmatrix}
    3 & 2 & 1 & 0 \\
    0 & 1 & 2 & 3 \\
    -1 & 0 & 1 & 2 \\
    -2 & -1 & 0 & 1 \\
    1 & 1 & 1 & 1
    \end{pmatrix}
    \begin{pmatrix}
    F_A \\
    F_B \\
    F_C \\
    F_D
    \end{pmatrix}
    =
    \begin{pmatrix}
    1 \\
    2 \\
    1 \\
    0 \\
    1
    \end{pmatrix}[/tex]

    How to find the forces?
     
  2. jcsd
  3. Aug 30, 2015 #2

    haruspex

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    Yes, there ard only two equations to be had from the usual linear and rotational force balances. To get further, you will need to make assumptions about how the wires and beam deform under loads.
     
  4. Aug 30, 2015 #3

    haruspex

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    ... Specifically, I would suggest taking the beam to be completely straight and rigid, but allow the wires all the same modulus of elasticity. That is enough to get a solution.
     
  5. Aug 30, 2015 #4
    [tex]\delta_A = \delta_C - 2x[/tex][tex]\delta_B = \delta_C - x[/tex][tex]\delta_D = \delta_C + x[/tex]
    Thus,
    [tex]F_A = \frac{\delta_A SE}{L} = \frac{\delta_C SE}{L} - 2x\frac{SE}{L}[/tex]

    And further,
    [tex]F_B - F_A = F_C - F_B = F_D - F_C.[/tex]

    Now matrix looks like this:
    [tex]\begin{pmatrix}
    -1 & 2 & -1 & 0 \\
    0 & -1 & 2 & -1 \\
    3 & 2 & 1 & 0 \\
    -1 & 0 & 1 & 2
    \end{pmatrix}
    \begin{pmatrix}
    F_A \\
    F_B \\
    F_C \\
    F_D
    \end{pmatrix}
    =
    \begin{pmatrix}
    0 \\
    0 \\
    1 \\
    1
    \end{pmatrix}[/tex]

    It has solution 0.1, 0.2, 0.3, 0.4.
     

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    Last edited: Aug 31, 2015
  6. Aug 31, 2015 #5

    haruspex

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    That's the answer I get.
     
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