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Static Equilibrium and tension of cables

  1. May 27, 2014 #1
    1. The problem statement, all variables and given/known data
    A 30Kg neon sign is suspended by two cables, as shown. Three neighbourhood cats (5.0 Kg each) find the sign a comfortable place. Calculate the tension in each cable with the cats are in the positions shown.


    2. Relevant equations

    xCM = m1x1 + m2x2 ...
    mtotal

    yCM = m1y1 + m2y2 ...
    mtotal

    F = mg

    3. The attempt at a solution

    I found the centre of mass (CM) by assigning each cat a coordinate using the bottom left corner of the sign as the origin with (x,y). So:

    Cat1 = (0.2,0.5)
    Cat2 = (1.8,0.5)
    Cat3 = (2,0)
    Sign = (1,0.25) --> the centre of the sign

    Using the above two equations I found the centre of mass of the cats and sign to be (1.11m,0.278m)

    There should be no forces in the x axis and the forces in the y axis should all cancel out so:

    T1cable + T2cable = mg

    I know the position of each cable from the picture so: (using only y axis position)

    T1(0.5m) + T2(0.5m) = (45kg)(9.81)(0.278m)

    The tension in each cable should be equal? Or, because one cable is farther from the CM it should have less tension? Does this mean I should be taking into account the x axis? Please help!
     

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    Last edited: May 27, 2014
  2. jcsd
  3. May 27, 2014 #2

    PhanthomJay

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    Looks like you y'd when you should have x'd. Torque is force times perpendicular distance from line of action of force to any point chosen wisely.
     
  4. May 27, 2014 #3
    Hm ok. So:

    T1(0.2m) + T2(1.8m) = (45kg)(9.81m/s2)(1.11m)

    T = 245N

    So that means each cord has a tension of 245N?
     
  5. May 27, 2014 #4

    PhanthomJay

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    No, your equation was ok but then you assumed that T1 and T2 were equal and got an erroneous result. To simplify matters, try summing torques not about your assumed origin at (0,0), but rather, about the point at the base of the sign at coordinate (0.2,0). You can choose any point, but keep it simple. In this manner, you can solve for T2 directly, because there will be no torque from T1, then continue......
     
  6. May 27, 2014 #5
    Ok. So coordinates for each point would be in relation to (0.2,0) If I do it how you suggest:

    T1(0m) + T2(1.6m) = (45kg)(9.81m/s2)(0.91m)

    T2 = 250.8N

    Then can I use my earlier equation to solve for T1?

    T1(0.2m) + T2(1.8m) = (45kg)(9.81m/s2)(1.11m)
    0.2T1 + (250.8N)(1.8m) = (45kg)(9.81m/s2)(1.11m)
    T1 = 190.4N
     
  7. May 27, 2014 #6
    You look to be correct. I worked it out separately and got this result, so I hope you're right :p.
     
  8. May 27, 2014 #7

    PhanthomJay

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    yes, looks good!
     
  9. May 28, 2014 #8
    Thanks!!
     
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