# B Tension in a spring

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1. Mar 11, 2016

### Benoit

Imagine a spring that has its two ends attached to a wall at the same height. The spring will adopt a kind of a ''U'' shape (maybe a parabollic shape would be more precise). Why are the coils near the wall more distant from each others than the coils at the middle of the spring ? I think it is because the tension is higher, but the tension isn't suppose to be same everywhere in a spring ? How could I find the difference between the tension near the wall and the tension in the middle of the spring ?

2. Mar 11, 2016

### .Scott

How much does your spring weigh?

3. Mar 11, 2016

### Benoit

It was a conceptual question, just consider the mass of the spring as non-nigligible.

4. Mar 11, 2016

### .Scott

Actually, a little more detail is needed. Measure the length of the spring when it is relaxed, that is compressed with no load. We'll call that the "compressed length". Now consider the bent spring. It will have an inside edge where all the coils tend to be compacted and an outside edge where they are the most loose.

When you attach the sprint to the wall, measure the distance between where the two inside edges are attached to the wall. We'll call that the inside distance.

We'll also assume that we are doing this in free fall - no gravity.

Then, if pi times the inside distance is equals the compressed length, then the spring will form an arc and each full coil along it length will be bent in the same fashion.

If pi times the inside distance is greater than the compressed length, then the spring is being stretched and the ends closest to the wall will take up most of the bend.

If pi times the inside distance is less that the compressed length, then the spring is loose and can easily flop around. In gravity, it would flop down. Without gravity, it's far more complicated - but it will probably end up spreading away from the attachment points before eventually closing back in along an arc.

5. Mar 11, 2016

### Benoit

Can you explain why is this happening and quantitatively describe the difference between the tensions ?

6. Mar 11, 2016

### Qwertywerty

Could you explain to me, why the spring forms a U-shape (Under the action of which forces?)

7. Mar 11, 2016

### Benoit

Gravitational force of course, so the U-shape can be explained by the center of mass. The problem I have is that I don't get why the tension doesn't seem constant everywhere.

8. Mar 11, 2016

### Qwertywerty

Tension being constant in a spring/ string is subject to a constraint. What is it?

9. Mar 12, 2016

### Leong

I think if you draw a free body diagram for a coil in the middle and another one for the coil near the wall, you may find that the sum of the tensions (the pulling force that acts on the coil) for the coil near the wall is greater than the sum of tensions that pulls the coil in the middle. Perhaps, that is why the separation distance is greater at the wall than at the middle because the coil near the wall is pulled harder than the one in the middle.

10. Mar 12, 2016

### Benoit

We consider the mass as negligible ? Or equally distributed across the lenght ?

I agree with this. Is the sum of the tensions that pulls the coil in the middle is lower because the gravitational force we have to consider here is only the weight of a single coil, instead of the case where we consider a coil near the wall, where we have to consider the weight of the coils below it?

11. Mar 12, 2016

### Qwertywerty

We assume negligible mass.
What does 'single coil' mean? You may want to rephrase that. But yes, as we consider a larger mass of the spring (symmetrically, about the center) as our system, a greater spring force is required to balance the said system's weight. And from the equation $F = kx$, we get the result you desire.

12. Mar 12, 2016

### Benoit

It was really unclear, pardon me. When I draw a free body diagram, I draw one that considers only a ring of the coil in the middle (a single loop). The forces applied on this ring are weaker than the forces applied on a ring near the wall, because in the last situation, the weight of the rings below the upper one are pulling it down. Is this right ?

13. Mar 12, 2016

### Qwertywerty

That's correct. You finally got it. Congratulations

14. Mar 12, 2016

### Benoit

Many thanks for your time

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