1. Oct 23, 2008

### Jaklynn429

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

A 76 cm long, 1.0 mm diameter steel guitar string must be tightened to a tension of 2200 N by turning the tuning screws. By how much is the string stretched?

2. Relevant equations

F=kx
K=YA/L

3. The attempt at a solution
I dont know how to approach this problem because I feel like I dont have everything i need! I know that F=2200N. but How do i get K to find delta x?

2. Oct 23, 2008

### CompuChip

What is Y? What does it depend on?

3. Oct 23, 2008

### Jaklynn429

Y is Young's Modulus, which in my notes is just referred to as a measure of the substances inherent stiffness, and i dont know where to get this.....

4. Oct 23, 2008

### LowlyPion

5. Oct 23, 2008

### Jaklynn429

I understand that, but how is this going to help me in my problem? I'm sorry if I sound stupid....

6. Oct 23, 2008

### LowlyPion

Look at the formula for Young's modulus

$$Y = \frac{Stress}{Strain} = \frac{F/A}{\Delta L/L}$$

7. Oct 23, 2008

### Jaklynn429

Would the stress be 2200 then?

8. Oct 23, 2008

### LowlyPion

No. But the Force 2200 N divided by the cross sectional area of a 1 mm diameter string would be though.

9. Oct 23, 2008

### HallsofIvy

10. Oct 23, 2008

### Jaklynn429

Okay. So now i have y=200. yay! So now i need area A which is [im assuming?] l*w? So .76*.001=A?

11. Oct 23, 2008

### LowlyPion

A is cross section = π*R2 = π*(.5*10-3)2

12. Oct 23, 2008

### LowlyPion

By the way Y is 200 * 109N/m2

13. Oct 23, 2008

### Jaklynn429

I got my final answer to be .0106. Is this in meters, then? Do I need to convert to centimeters?

14. Oct 23, 2008

### Jaklynn429

I figured it out, converted to CM and got my final answer as 1.06 cm, and it was right! Thank you so much for all your help!

15. Oct 23, 2008

### LowlyPion

Hope you didn't get too stressed or that it was too much of a strain.

Cheers

16. Oct 24, 2008

### CompuChip

* grins at Pions puns *

Jaklynn, try to keep the units throughout your calculation, you can just consider them as ordinary variables. For example, if you try to do (0.76 m) * (0.001 m^2) you will get 0.00076 m^3. But m^3 is a unit of volume, not of area.
Similarly, if you forget converting something, you will end up with something like: cm * m, instead of m * m = m^2. This makes it easier to spot your errors beforehand.