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Engineering Cable question

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A cable of length L is supported at different elevations and is used to suspend two masses m1=10kg and m2=25kg at B and C respectively, as shown in Fig 2. The vertical heights of B and C relative to the support at A are denoted by h1 and h2 respectively. For simplicity, use g=10m/s^2. Obtain two independent equations relating h1 and h2 and then use them to determine the cable length L if it is desired that h1=h2. If the cable length L=5m, computer h1, h2 and the maximum tension in the cable.


    t8n1ih.jpg

    I believe this question is about parabolic cable.

    Can give me hints how to start this question? Do i have to make an imaginary cut at part of cable?
     
  2. jcsd
  3. Oct 3, 2011 #2

    PhanthomJay

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    This is not about a parabolic curve or catenary curve. The cable is assumed massless and taut. I would try using the equilibrium equations for the system, and at the joints and supports. The cable length is the sum of the diagonals.
     
  4. Oct 3, 2011 #3
    But i cannot seem to get equations with h if i take moments about point A or D. And if i take at B or C, it seems to have many unknowns in the equations.
     
  5. Oct 3, 2011 #4

    PhanthomJay

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    Yeah, this seems to be a time consuming process with a lot of unknowns. When you sum moments about D, you get 2 unknowns, Ax and Ay. Ax and Ay are related by trig (since the resultant must line up with the cable), but that introduces another unknown, h1. But you still have more equilibrium equations (sum of Fx = 0 and sum of Fy = 0), which you can apply to the system and to each joint. You end up with a lot of equations but with an equal amount of unknowns. I don't know of any shortcuts offhand.
     
  6. Oct 3, 2011 #5
    Taking moments about D, -3Ay + 0.5Ax + 200 + 250 = 0

    Relating Ay with Ax, Ay = h1Ax

    Taking summation of upward forces of the system, Ay + Dy = 350

    Taking summation of all horizontal forces of the system, Ax = -Dx

    Relating Dy and Dx, Dy = (h+0.5)Dx

    I'm stucked now. If i were to apply moments or summation of vertical/horizontal forces at point B or C as you mentioned, do i have to make an imaginary cut?
     
  7. Oct 4, 2011 #6

    gneill

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    Why not start with the first requested item, the length of the cable if h1 = h2? Call h1 and h2 just "h" in this case. We want to find h and L.

    If you step back and look at the diagram in this case you'll see two suspended masses hanging from cables that are being pulled away from vertical by a horizontally applied force. Further, these horizontal forces are the same magnitude because they're being delivered by tension in the same section of cable that stretches between the mass suspension points.

    Here's the scenario for one of the masses:

    attachment.php?attachmentid=39601&stc=1&d=1317736775.gif

    The diagram for the other mass is almost a mirror image of this one, the difference being the added length (0.5m) added to h.

    This should be enough hints to be getting on with :wink:
     

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  8. Oct 5, 2011 #7
    I have solved this question. Thanks everyone for the hints and help assisted. Good day.
     
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