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

1. Jun 15, 2013

### Persimmon

1. The problem statement, all variables and given/known data

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

2. Relevant equations

ƩF(y) = 0
Ʃτ = 0

3. 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.

2. Jun 16, 2013

### Redbelly98

Staff Emeritus
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. Dec 16, 2013

### robbyrandhawa

how did u get 0.2 and 1.8?

4. Dec 16, 2013

### SteamKing

Staff Emeritus
Look carefully at the dimensions of the sign and its supports.

5. Apr 14, 2015

### SteveS

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. May 16, 2015

### Brendan Webb

Hey steve,

how did you decide to use the 1.6m when dividing by the torque force

7. May 16, 2015

### SteveS

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.

8. May 16, 2015

### Brendan Webb

ahh yes, I was having a hard time understanding. Thanks!

9. May 16, 2015

### SteveS

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 lets you zero out T1 and find for the unknown T2.