# Problem regardin an elastic wire.

1. Jul 8, 2013

### DorelXD

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

An elastic wire has the folowing charcteristics:initial length $l_0=60cm$,section $S=6mm^2$,Young's Modulus $E=\frac{12MN}{m^-2}$.Initially the wire has both ends fixed. Its initial position is horizontal, untensioned. At the middle of the wire it's hanged a body that produces an elongation equal to 8 cm. Find the potential energy of deformation stored in the wire, when the body reaches its equilibrium position.

2. Relevant equations
$$\frac{F}{S}=E\frac{Δl}{l_0}$$
$$F_e=kx$$
$$E_e=\frac{k*Δl^2}{2}$$

3. The attempt at a solution

Well what bugs me is that I don't know what to do with the elongation in the problem. It confuses me that we pull from the middle of the wire, and not from one of its ends. Can somebody help me to understand this problem, please? Also I don't know what the spring constant is.

2. Jul 8, 2013

### SteamKing

Staff Emeritus
When you hang something in the middle of the wire, you are going to pull the wire downward from its original position. Think what happens to a clothesline (if you know what that is). It would be helpful to draw a free body diagram of the wire suspending the weight. The vertical components of the tension in the wire must balance the weight suspended.

3. Jul 8, 2013

### DorelXD

First of all, thank you for your answer!
Yes, I know what a clothesline is. I understand that the wire it's pulled downward, this is easy.

This is easy too. I had drawn it before I posted it, but it was on a piece of paper. Here is a picture of the free body-diagram (I know it's not complete, I should have decomposed the tensions(is this term correct? I'm not a native speaker) in the vertical and horizontal components. I drew it in paint because I don't know any better way to do it. If you know, please tell me.

I get it. This is pretty straight forward from the inital supposition of the problem that the body remains in equilibrium.

So we now have:

$$mg=2Tcos\Theta$$

4. Jul 8, 2013

### haruspex

So do you understand what to do with the elongation now?

5. Jul 8, 2013

### Staff: Mentor

From the geometry that you have drawn, how much longer are each of the two halves of the wire now than they were before you hung the weight?

6. Jul 9, 2013

### DorelXD

Well, not yet.
Well, the initial length of one half is: $l_i=\frac{l_0}{2}$.

The final length is: $l_f=\frac{li}{sin\Theta}=\frac{l_0}{2sin\Theta}$, isn't it?

So $\Delta l=l_f-l_i=\frac{l_0}{2sin\Theta}-\frac{l_0}{2}=\frac{l_0(1-sin\Theta)}{2sin\Theta}$,right?

Thank you both for your answers!

Last edited: Jul 9, 2013
7. Jul 9, 2013

### voko

Should that not be $\Delta l = 2(l_f - l_i)$? And, $l_f - l_i = \frac{l_0}{2sin\Theta} - \frac{l_0}{2}$.

8. Jul 9, 2013

### DorelXD

Thank you for your time! But why is it like this? Have I made some algebra mistake or have I missed something?

9. Jul 9, 2013

### voko

All of that follows from your definitions in #6.

10. Jul 9, 2013

### DorelXD

Excuse me, but I don't undersatnd. What definition? And what is #6?

11. Jul 9, 2013

### voko

#6 is message number 6 in this thread:

In it, you defined that $l_i$ and $l_f$ are the initial and the final lengths of one half of the wire.

12. Jul 9, 2013

### DorelXD

Yes, it's true. But $\Delta l$ is the elongation of one half, not the elongation of the whole wire. The elongation of the whole wire would be $2\Delta l$, isn't it? This confusion is beacuse of me, I get it. When I first created this theard and I posted the main equations I used $\Delta l$ as the elongation of the whole wire. So, given the fact that $\Delta l$ is the elongation of only one half (besides in #1 where $\Delta l$ is the elongation of the whole wire) is my judgement correct?

13. Jul 9, 2013

### voko

Correct, but mind the order of $l_i$ and $l_f$ in the difference.

14. Jul 9, 2013

### Staff: Mentor

So now that you know the change in length of half the wire and the original length of half the wire, what is the strain?

15. Jul 9, 2013

### DorelXD

Yes, I've corrected. I was very uncareful. Thank you for pointing that out!

Well it'll be:

$2\Delta l=2(l_f-l_i)=2(\frac{l_0}{2sin\Theta}-\frac{l_0}{2})=\frac{l_0(1-sin\Theta)}{sin\Theta}$,right?

16. Jul 9, 2013

### haruspex

Yes.

17. Jul 16, 2013

### DorelXD

I've managed to solve the problem, much more simpler than the initial approach. If everyone is intersted please let me know, and I'll post it here. Thank you again for your answers!

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