Statics problem: cats on a beam, tension of cables calculation

In summary, the tension in the right cable is approximately 250 N and the tension in the left cable is approximately 190 N when three 5.0 kg cats are sitting on a 30 kg neon sign suspended by two cables, with a height of 0.5 m, and a length of 2.0 m. The calculations were found using the equations for net force and net torque, and taking into account the center of mass of the sign and the perpendicular distances from the pivot point to the forces.
  • #1
Persimmon
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Homework Statement



A 30 kg neon sign is suspended by two cables, as shown. Three
neighborhood 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.

[URL=http://s1152.photobucket.com/user/rusalka4/media/cats_zpse2fe500d.png.html][PLAIN]http://i1152.photobucket.com/albums/p498/rusalka4/cats_zpse2fe500d.png[/URL][/PLAIN]

M = 30 kg
m = 5.0 kg
T1 = tension in right cable
T2 = tension in left cable

Homework Equations



ƩF(y) = 0
Ʃτ = 0


The Attempt at a Solution



ƩF(y) = 0 = T1 + T2 - g(3m + M)
T1 + T2 = 9.8(3*5 +30)
T1+T2 = 441
T1 = 441 - T2

For net torque, I chose the axis at the point where the third cat is sitting, the one that's hanging off the side. Mainly because I'm a bit confused as to what forces that cat is exerting on the beam and in what directions.

Ʃτ = 0 = 0.2*mg - 0.2*T1 + 1*Mg + 1.8*mg - 1.8*T2
0 = 9.8 - 0.2*T1 + 294 + 88.2 -1.8*T2
0.2*T1 + 1.8*T2 = 392
0.2(441 - T2) + 1.8*T2 = 392
88.2 + 1.6*T2 = 392
T2 = 189.875 ≈ 190 N
T2 = 441 - T2 = 251.125 ≈ 250

Is this correct?? That third cat is making me nervous, as is the fact that the height of the beam is given as 0.5m and I didn't use that anywhere. Did I correctly calculate the distance to the pivot point for each of the forces contributing to the net torque? I just took them as the perpendicular distance the pivot, ie the length away from the third cat.

Thanks in advance!
 
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  • #2
I'm guessing the height of the beam is given to make you think about the "perpendicular" distance from each line of force to the pivot point you have chosen.

Looks like you did the math right. A good check is to see if your tension values make sense:
(1) Do the tensions add up to the right total upward force?
(2) From the picture, which cable would you expect to have a larger tension? Does your answer agree with this?
 
  • #3
how did u get 0.2 and 1.8?
 
  • #4
robbyrandhawa said:
how did u get 0.2 and 1.8?

Look carefully at the dimensions of the sign and its supports.
 
  • #5
In this case in order to take the cats out of the equation, I calculated the new cm for the sign.

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.

The i simply used that in the torque equation.

With T1 as the pivot point:
T2 = (45kg)(9.80)(0.91m) / 1.6 m = 250.82 N , 2.5 x10^2 N

∑F = T1 + T2 = mg
T1 = (45kg)(9.80m/s^2) - 250.82 N = 190.1 N 1.9 x10^2 N

I realize i probably should have calculated a ycm but it seems to match the answers you got from the other approach.
 
  • #6
SteveS said:
In this case in order to take the cats out of the equation, I calculated the new cm for the sign.

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.

The i simply used that in the torque equation.

With T1 as the pivot point:
T2 = (45kg)(9.80)(0.91m) / 1.6 m = 250.82 N , 2.5 x10^2 N

∑F = T1 + T2 = mg
T1 = (45kg)(9.80m/s^2) - 250.82 N = 190.1 N 1.9 x10^2 N

I realize i probably should have calculated a ycm but it seems to match the answers you got from the other approach.

Hey steve,

how did you decide to use the 1.6m when dividing by the torque force
 
  • #7
Hey Brendan, lucky I was just checking something else out on here when you asked.

I used 1.6m for the torque because T1 is the pivot point so from T1 to T2 is 1.6m.
 
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  • #8
SteveS said:
Hey Brendan, lucky I was just checking something else out on here when you asked.

I used 1.6m for the torque because T1 is the pivot point so from T1 to T2 is 1.6m.
ahh yes, I was having a hard time understanding. Thanks!
 
  • #9
I didn't put it in the post I made before but the ∑T equation is:

ΣT = T1*(0) + T2(1.6 m) - (45kg)(9.80 m/s^2)(0.91m)

distances are all from T1 since it's the pivot point. This let's you zero out T1 and find for the unknown T2.
 

1. How do you calculate the tension in the cables supporting the cats on the beam?

To calculate the tension in the cables, we need to use the principle of static equilibrium. This means that the sum of all the forces acting on the beam and the cats must equal zero. By analyzing the forces acting on the beam and using the equations of equilibrium, we can find the tension in the cables.

2. What is the significance of determining the tension in the cables?

The tension in the cables is crucial in ensuring the safety and stability of the cats on the beam. If the tension is too low, the cables may not be able to support the weight of the cats, causing them to fall. On the other hand, if the tension is too high, it may put unnecessary stress on the cables and potentially lead to their failure.

3. How does the weight and position of the cats affect the calculation of tension in the cables?

The weight and position of the cats directly influence the distribution of weight on the beam and, therefore, the forces acting on it. The further away the cats are from the support point, the higher the tension in the cables will be. Additionally, the weight of the cats must be taken into account when calculating the total weight on the beam.

4. Are there any assumptions made in the calculation of tension in the cables?

Yes, there are some assumptions made in this calculation. One key assumption is that the beam and cables are in perfect condition and can handle the calculated tension. Additionally, the cats are assumed to be stationary and not moving around on the beam, which would affect the distribution of weight and forces.

5. Can the same principles be applied to other statics problems involving beams and cables?

Yes, the principles of static equilibrium and calculating tension in cables can be applied to other similar problems. However, the specific calculations and assumptions may vary depending on the specific scenario. It is important to carefully analyze the forces and conditions in each problem to accurately determine the tension in the cables.

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