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1. Sep 6, 2015

### whitejac

1. The problem statement, all variables and given/known data
The Tensions in four cables are equal:

|T1| = |T2| = |T3| = |T4| = T

Determine the value of T so that the four cables exert a totl force of 12,500-lb magnitude on the support.
(Inser a picture of 4 connected cables angled 9, 29, 40, 51 degrees respectively)

My question is what exactly does T stand for? I can do the vector algebra so I will not attempt that until I understand what exactly I'm looking for. If T is the resultant vector then I can find that using vector addition, but if it's just the magnitude then I divide the desired magnitude evenly amongst the 4 vectors.

I'm fairly certain T is not a vector, thought, because in the chapter the vectors were bolded. Is there something I'm not seeing here?
2. Relevant equations

3. The attempt at a solution

2. Sep 6, 2015

### C. Lee

The problem states "The tensions in four cables are equal".

T is the magnitude those 4 vectors share.

Although those 4 vectors have the same magnitude, their components that are really 'supporting' should be all different(for example, if 4 cables are keeping an object from falling to the groud, these components should be the components vertical to the ground).

Your job is to calculate the magnitude T that makes the sum of these components 12,500lb.

3. Sep 7, 2015

### whitejac

Then this confuses me because if they have to equal 12,500-lb and each cable shares an equal load then the magnitude will be divided equally.

I agree that the components will not be the same, but it seems a little trivial in the scope of this question. Am I right?

4. Sep 7, 2015

### C. Lee

They do not share an equal load. How could they while the components are all different?

5. Sep 7, 2015

### whitejac

I believe they could because the magnitude of each cable is the same. So where the Fx diminishes the Fy expands in magnitude.

Otherwise I do not understand how |T1| can equal |T2|

6. Sep 7, 2015

### C. Lee

Yeah but I think a load on a cable is not the magnitude of tension of the cable, but a certain component of it.

For example, think about two cables are keeping an object from falling to the ground at the ceiling, one making 30° and another 120° with the x-axis(horizontal to the ground).

Suppose that they have the same tension T.

Their x components cancel out, and only their y components are doing their job against gravitational force.

The load on the first cable is its y component, which is T*sin(30°) = 0.5*T.

Similarly, the load on the second one is T*sin(120°) = (√3/2)*T.

They share the same magnitude for their tension, but have different load on them.

This is what I think. Are you convinced, or do you think there is any flaw in this? Please do not hesitate to point out if you think so.

Sorry for my deficient English.

7. Sep 7, 2015

### whitejac

Okay, I agree wth your explanation.
This is why I think I'm not quite grasping the notation.

So the T stands for tension? This would drastically change things. The loads would all be different but in order to maintain equilibrium the tension would have to be uniform. This is correct, yes?

8. Sep 7, 2015

### C. Lee

T stands for "the magnitude" of tension.

T1, T2, T3, and T4 in bold text stand for tension for each cable.

To maintain equilibrium, the sum of all, say, y-components of tensions should balance out the gravitational force on the object.

9. Sep 7, 2015

### vela

Staff Emeritus
If by "tensions would have to be uniform," you mean $\|\vec{T}_1\| = \|\vec{T}_2\| =\|\vec{T}_3\| =\|\vec{T}_4\|$, then I'd say the answer generally is no. That's a given condition for this problem; the condition doesn't follow just because the object is in equilibrium.

10. Sep 7, 2015

### whitejac

That makes sense. I guess if we just taped a rope to the support and the bridge then technically there'd be a fifth 'cable' but by no means would it share equal tension. Then I am a little confused about the intentions of this problem again. It seems a little bit trivial.

If the 'magnitude' of the tensions are equal to each other, given by this problem, then finding |T1| would be the given load divided equally amongst them; in this case 3125-lb, I believe. To find the components and make up the vectors T1 would be sin(9)*|T1| and then cos(9)*|T1| for the y and x components, respectively. However, the tension would be the magnitude of these two components otherwise we'd have a system of too many unknowns (Fy and |F|). This implies to me that I should use the magnitude I said, 3125-lb for |T1| and given the set up of this problem I have coincidentally found T as well.

11. Sep 7, 2015

### C. Lee

Again, |T1| is the magnitude of the tension on Cable 1, not the load on it. Furthermore, each cable does not have to share an equal load with others; those 4 cables could have all different loads on them. For example, if they are set appropriately, Cable 1 can have 500lb for its y-component, 2000lb for Cable 2, 4000lb for Cable 3, and the rest for Cable 4: to sum up, they are exerting total 12500lb to the object, upward in y-direction.

Load on each cable depends on tension of the cable, but since tension is the same for all cables they only differ by factor of sin(θ), where θ is the angle that a cable makes with the x-axis.

Therefore, the sum over Tsin(θ) for all cables should be equal 12,500-lb. This is how I'm understanding the situation.

Sorry if I'm not getting the point of your question.

12. Sep 7, 2015

### whitejac

Okay, so I'm gathering then that the load and tension are indeed not the same thing. Tension is not a force, while load can produce one. (This is justified by the fact I am equating load with Fy in this scenario).

So, using your understanding of the situation,
Fy = sin(9)T + sin(29)T + sin(40)T + sin(51)T = 12,500-lb
Fy = (2.061)T = 12,500-lb
=> T = (12,500 / 2.061) - lb

I guess the only thing i'm curious about now would be why do you choose to equate T with Fy/Sin(x) instead of cos(x)?

Maybe for better understanding of the problem, the graphic has 4 cables attached to a post in the ground angled up and to the right. So, in my free body diagram, I have the tension resisting the upward force of the bridge but also the horizontal force of the bridge. For example, if we had two kids pulling on a rope then the sin(x) would be zero and cos(x) would be 1. Are we primarily interested in the scaling theta does to this situation and therefore it woud be redundant to also seek out cosine? I'm neglecting gravity at the moment.

13. Sep 7, 2015

### vela

Staff Emeritus
C. Lee's interpretation isn't quite correct. Posting the picture right at the start would have helped a lot.

You can't simply divide 12500 lb four ways to find T because the tensions are vector quantities and act in different directions. You're looking for $T$ so that $\| \vec{T}_1 + \vec{T}_2 + \vec{T}_3 + \vec{T}_4 \| = 12500\text{ lb}$.

14. Sep 7, 2015

### whitejac

Okay, that makes a little more sense.
Then, with that in mind, i find the Sin(x)T and the Cos(x)T and this gives me Fx and Fy.
Fortunately, since T is equal everywhere, this means I can find the magnitude of F by (Fy^2 + Fx^2)^0.5
Then I divide the load 12,500 by the coefficient of T.