# Homework Help: Mass hanging from a flexible wire

1. Dec 11, 2012

### ASmc2

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

2. Relevant equations

The definition for Young's modulus. Newtons second law. Pythagoras' formula.

3. The attempt at a solution

I know from physics 1 that I have to apply F=ma to the point where two "triangles" meet. The phrase "the tension" implies that the tension is the same everywhere on the string. If this is true, though, my answer for tension does not fit with the book's answer. I do not know how the tension would be related to the force when the force is perpendicular to object that is extended. Please provide me with hints, i am clueless. I attempted a solution assuming that extension of the string was solely vertical, but I got L_0/50 for y. The answer is L_0/20. The tension is supposedly 5mg. If I plug that into my equation for T, it spits out L_0/20=y. Otherwise, I have not a clue how to get L_0/20=y. Thanks!

#### Attached Files:

• ###### PROB1.png
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2. Dec 11, 2012

### haruspex

Please post your working, in particular your analysis of the forces acting on the mass and any equations you derive. Create unknowns as necessary. Include nan equation for the modulus based on the initial info.

3. Dec 12, 2012

### CWatters

Ah but it is not exactly perpendicular in the final position is it.

What haruspex said, draw a diagram showing the forces acting on the weight when it's in it's final position.

Note that their diagram is wrong. The problem says the mass is hung in the midpoint.

4. Dec 12, 2012

### ASmc2

Yes. Their diagram IS wrong. If we call the tension on the right leg of the triangle T1 and the other leg T2. Then T1cos(theta)=T2cos(beta), right? Theta and beta are angles between the respective legs and the horizontal. If the mass IS at midpoint, THEN beta=theta. If that is true, my calculation for y gives about L0/50! (See attachment) The book says the answer is L0/20. That answer would be possible with sig figs, but even then y^2 has to be missing a factor of 4. I checked a billion times, the author must have made a mistake. Again, see the attached analysis. When you do the numerical calculation, you will see what I am saying.

#### Attached Files:

• ###### prob1-2.png
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5. Dec 12, 2012

### TSny

It appears to me that in your analysis you are assuming that the fractional increase in length will still be .001 when hanging the mass on the sagging horizontal wire. But when the mass is hung on the horizontal wire, the tension will be greater than if you just hang the mass vertically from one end of the wire. Will the amount of stretch be the same?

6. Dec 12, 2012

### CWatters

What Tshy said.

Vertically you have mg = 2 * T * Sin(θ)

so

T = mg/2Sin(θ)

When θ approaches zero, Sin(θ) approaches zero and T approaches ∞.

With the weight hanging vertically Tv = mg so the tension and extension scales by 1/2Sin(θ).

7. Dec 12, 2012

### ASmc2

OK. TShy is right. Be assured that my formula says the same thing.

8. Dec 12, 2012

### ASmc2

Can somebody tell me if my equation for y^2 is correct?

9. Dec 12, 2012

### TSny

Not if you're letting a = .001.

10. Dec 12, 2012

### ASmc2

Thanks. I realize that now. I am bilingual, so I will blame that for my misinterpretation of the problem.

11. Dec 12, 2012

### ASmc2

Ok!!! The problem was solved after my last reply. Thank you very much everyone!