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Tension and Equilibrium: Hanging sign

  1. Jul 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A 30 kg 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 when the cats are in the positions shown.

    Screen Shot 2015-07-17 at 11.28.03 PM.png
    Ft1 = tension in left cable
    Ft2 = tension in right cable
    T=torque

    2. Relevant equations
    Xcm=(M1*x1+M2*x2+...)/Mtotal
    ƩF(y) = 0
    Ʃτ = 0

    3. The attempt at a solution
    Xcm = (5.0kg)(0.2m) + (30kg)(1m) + (5.0kg)(1.8m) + (5.0kg)(2.0m) / 45kg = 1.11 m from the left edge of the sign.

    With left hand cable as reference point:
    ΣT = Ft1*(0) + Ft2(1.6 m) - (45kg)(9.80 m/s^2)(1.11m-0.2m)=0
    Ft2 = (45kg)(9.80)(0.91m) / 1.6 m = 250.82 N , 2.5 x10^2 N

    ∑Fy = Ft1 + Ft2 = mg
    Ft1 = (45kg)(9.80m/s^2) - 250.82 N = 190.1 N 1.9 x10^2 N

    I feel as though though I should somehow be taking into about the vertical distribution of the weight on the sign using Ycm but I'm not sure if I need to or how to go about doing that.

    Thanks!
     
  2. jcsd
  3. Jul 18, 2015 #2
    You should find Ycm too .

    τ= r × F = r F sin(θ) where θ is angle between r and F .
    Com is not at the top surface of the sign and so the torque of mg is is not horizontal distance multiplied by mg .
     
  4. Jul 18, 2015 #3

    haruspex

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    No need. In fact, there's no benefit in finding Xcm either. Just sum the torques from the different weights.

    Yes it is. The horizontal distance is the perpendicular distance from the reference axis to the line of action of the force.
     
  5. Jul 18, 2015 #4
    Haruspex , with regard to your two points , obviously calculating resultant of all initially is not required .

    However , to your second , no it is not . The force of gravity acts at a point below the point from which torque is balanced . So only one component of M(resultant)g produces a torque .
     
  6. Jul 18, 2015 #5

    haruspex

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    The point of action is not crucial. What matters for torque is the line of action.
    See https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/
     
  7. Jul 18, 2015 #6
    Ok I think I just made a mistake .

    τ = r⊥ × F
    or r × F ⊥ .

    You were saying r⊥×F and I r × F⊥ .


    Please excuse my mistake . I wasn't thinking clearly .
     
    Last edited: Jul 18, 2015
  8. Jul 18, 2015 #7
    Thanks for your reply! Does this mean that where I used (1.11m-0.2m) as the distance in the torque equation for the sign, I should have just used (1.0m-0.2m)?

    And Qwertywerty thanks for adding your thoughts as well!
     
  9. Jul 18, 2015 #8
    .
    Nevermind, I gave this way a shot and then the tensions come out as equal. Which wouldn't make sense.
     
  10. Jul 18, 2015 #9

    haruspex

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    I mean, for each force, find its torque about the reference axis (vertical force times horizontal displacement) and add these up. Be careful with signs. Equate the result to zero.
     
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