# Tensil Stress

1. Nov 22, 2008

### anubis01

1. The problem statement, all variables and given/known data
A string or rope will break apart if it is placed under too much tensile stress. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density 7890 kg/m^3 and will break if the tensile stress exceeds 7.0x10^8 N/m^2. You want to make a guitar string from a mass of 4.4g of this type of steel. In use, the guitar string must be able to withstand a tension of 900 N without breaking. Your job is the following. Ysteel=20x10^10

a)Determine the maximum length the string can have.
b)Determine the minimum radius the string can have.
c)Determine the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.

2. Relevant equations
u=m/L
v=sqrt(F/u)
v=sqrt(Y/p)
p=density

3. The attempt at a solution

a)
okay so I equated the tension =force and combined the v equations to form
sqrt(F/u)=sqrt(Y/P)
FL/m=Y/P
mY/FP=L
(4.4x10^-3)(20x10^10)/(900X7890)=123.92m

Now considering the asinine length of the string The answer is wrong, Can anyone help me figure out whats wrong with my answer. As always any help is appreciated.

2. Nov 23, 2008

### Staff: Mentor

Take a much simpler approach. Start by using the given tension and breaking stress to figure out the smallest allowable cross-sectional area.

3. Nov 23, 2008

### anubis01

Okay I found out my Area to be 1.286x10-6m2, I just a little bit confused on how to proceed. The rope is a cylinder so its area is 2pirh. I replaced r with h/2 to give the equation
1.286x10-6m2=2pi(h/2)h
(1.286x10-6m2/2pi)=(h/2)h
(1.286x10-6m2x2/2pi)=h^2
sqrt(1.286x10-6m2x2/2pi)=h

I'm I correct in my reasoning?

4. Nov 23, 2008

### Staff: Mentor

Still more complicated than necessary. What's the volume of a cylinder with cross-section A and length L? What's the volume of material you have to work with?

5. Nov 23, 2008

### anubis01

Okay I think I got it. To determine the amount of volume we have we divide the mass by density=4.4x10^-3/7890=5.07x10^-7 m^3

V=AXL
V/A=L
5.07x10^-7/1.286x10^-6=0.394m is the length. Is my line of thinking correct or am I still missing something.

6. Nov 23, 2008

### Staff: Mentor

7. Nov 23, 2008

### anubis01

Thanks a bunch I got part a) and I figured out the rest of the problem. Your a life saver.

8. Nov 30, 2008

### carlee172

So, I had a similar question, except the length of my string came out to be 0.3575 m. I don't know what I'm doing wrong, but I got the radius to be 2.05*10^-7 m. Apparently this is wrong.

Here's what I did:
A=2pi*r
r=1.286*10^-6/2pi=2.05*10-7

Can someone try to explain what I'm missing?

9. Nov 30, 2008

### anubis01

It would help if you show the work you did, then I could help you determine if you made a mistake. and when writing out your work please try to make it as readable as possible.

10. Dec 1, 2008

### carlee172

A string or rope will break apart if it is placed under too much tensile stress. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density 7830 kg*m^-3 and will break if the tensile stress exceeds 7*10^8 N*m^-2. You want to make a guitar string from a mass of 3.6 g of this type of steel. In use, the guitar string must be able to withstand a tension of 900 N without breaking. Your job is the following.

a) Determine the maximum length the string can have.
b) Determine the minimum radius the string can have.
c) Determine the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.

I did a) as I said above and I got 0.36 m, and this is the correct answer according to the program, and I used the same method as you described at the beginning.

11. Dec 1, 2008

### anubis01

b)You did not take into account the length of the string.so whats the formula for the volume of a cylinder and what did you find in part a. use that information to solve for r.
c)since this equation is not clearly given in the book I'll help because the assignment's due at 3:00pm(I'm guessing your in my class). you use the formula for fundemental frequency with the formula u=m/L use those two equations and its easy to solve.

Best of luck on the assignment.

Last edited: Dec 1, 2008
12. Dec 1, 2008

### carlee172

Thanks. :)

13. Dec 1, 2008

### carlee172

How did you do the interference problem?

14. Dec 1, 2008