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Tension and Equilibrium: Hanging sign

  • Thread starter cassie123
  • Start date
15
0
1. Homework Statement
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. Homework 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!
 
501
65
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 .
 

haruspex

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I feel as though though I should somehow be taking into about the vertical distribution of the weight on the sign using Ycm
No need. In fact, there's no benefit in finding Xcm either. Just sum the torques from the different weights.

the torque of mg is is not horizontal distance multiplied by mg .
Yes it is. The horizontal distance is the perpendicular distance from the reference axis to the line of action of the force.
 
501
65
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 .
 

haruspex

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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 .
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/
 
501
65
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:
15
0
No need. In fact, there's no benefit in finding Xcm either. Just sum the torques from the different weights..
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!
 
15
0
.
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)?.
Nevermind, I gave this way a shot and then the tensions come out as equal. Which wouldn't make sense.
 

haruspex

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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)?
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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