# Homework Help: Stretching Wires Problems

1. Jul 14, 2012

### Catchfire

Hello first post here. These problems aren't for a specific course just working through an old textbook specifically Vibrations and Waves by A.P. French. I hope it's alright that I include two problems in one thread.

1. The problem statement, all variables and given/known data
Q 3-7) A wire of unstretched length L0 is extended by 10^-3L0 when a mass m is hung from it's end. If this same wire is connected between two points A,B horizontally and then the same mass is hung from the middle of the wire, how large is the depression y of the midpoint?

2. Relevant equations
x^2 + y^2 = h^2

3. The attempt at a solution
I assume the total length of the wire will still be (1 + 0.001)L0. So dividing the length in two and the unstretched length in two you get a right angle triangle.

(L0/2)^2 + y^2 = (L0(1+0.001)/2)^2

y = (L0(1+0.001)/2)^2 - (L0/2)^2))^0.5
= ((0.5005^2 - 0.25)L0^2)^0.5
= 0.022 L0

1. The problem statement, all variables and given/known data
Q 3-8b) A 0.5kg mass is hung from a 2m long steel wire with diameter 0.5mm. What is the maximum height h the mass can be dropped from if the wire is not to break?

Young's modulus of steel: 2x10^11 N/m^2
Ultimate strength of steel: 1.1x10^9 N/m^2

2. Relevant equations
Y = (F/A) / (ΔL/L)
US = F/A
F = ma
v0^2 = v - 2aΔx
KE = 0.5mv^2
GPE = mgh

3. The attempt at a solution
F = US x A = 1.1x10^9 x ∏(0.0005/2)^2 = 216 N

The thing that's throwing me off is that if I have a falling mass I have a energy and not a force. So I'm thinking maybe look at the acceleration over the distance the wire stretches.

a = 216/0.5 = 432 m/s^2

ΔL = FL/YA = (216 x 2) / ((∏(0.0005/2)^2)(2x10^11)) = 0.011m

v0^2 = 0 - 2(9.8)(0.011) = 9.5 m/s

0.5mv^2 = mgh
0.5(9.5) = 9.8h
h = 0.48m

Any help with this problems is much appreciated, thanks.

Last edited: Jul 14, 2012
2. Jul 14, 2012

### TSny

Catchfire, this is not a correct assumption. The wire stretches by .001Lo if the wire is suspended vertically and then mg is hung on the end. In this case the tension in the wire is mg. However, when the wire is suspended horizontally and mg is hung in the middle, the tension in the wire will be much greater than mg. So, the wire will stretch much more than .001Lo.

So, you might start by trying to find an expression for the tension in the wire in terms of mg and the angle θ, where θ is the angle the wire makes to the horizontal when the mass is hung in the middle. Hint: Draw a free body diagram for the hanging mass.

3. Jul 14, 2012

### Catchfire

Ok so I got this far:

2Tsinθ = mg

T = mg/2sinθ

I know T is supposed to be 5mg so sinθ = 0.1 but I'm lost as to how that works out.

4. Jul 14, 2012

### TSny

Very good. Note that you have shown that the tension has increased by a factor of 1/(2sinθ) compared to just hanging the mass on the vertical suspended wire. So, if the wire stretched by .001Lo when it was vertical, how much will it stretch when horizontal? Express this in terms of Lo and sinθ.

Try to construct a right triangle with the angle θ and with the sides related to the unstretched and stretched lengths. You can use trig on this triangle to get a relation between θ, Lo, and L (where L is the stretched length).

Then try to put it all together so that you can solve for θ.

5. Jul 14, 2012

### Catchfire

For the horizontal it should be:
Tcosθ = mg cotθ/2
L_x = cotθ/2 L0

for the triangle:

(1/2)^2 + (x)^2 = (1/2sinθ)^2
1 + x^2 = 1/sin^2θ
x = (1/sin^2θ -1)^0.5

so
L_x = (1/sin^2θ -1)^0.5 L0

and

cotθ/2 = (1/sin^2θ -1)^0.5
cot^2θ/4 = 1/sin^2θ -1
cos^2θ/4 + sin^2θ = 1
cosθ = 0
so θ = π/2

...was it my trig?

6. Jul 15, 2012

### TSny

Sorry, I didn't understand much of your last post. This is a tricky problem.

You know from your triangle: (Lo/2)2 + y2 = (L/2)2

Try substituting L = Lo+ΔL in the right hand side of the above equation. Square out the right hand side and see if you can simplify to

y2 = (LoΔL)/2 + (ΔL)2/4

Now we know that ΔL will be small compared to the overall length Lo. So, the last term may be neglected:

y2 ≈ (LoΔL)/2

We need to find an expression for ΔL. From the statement of the problem, we know that ΔL = .001Lo when the tension in the wire is mg. But when the the mass is hung on the horizontal wire, you found earlier that the tension is mg/(2sinθ). In other words the tension has increased by a factor of 1/(2sinθ).

Hooke's law tells you that the amount of stretch ΔL is proportional to the tension. So, see if you can write down an expression for ΔL when the mass is hung on the horizontal wire. [You should get that ΔL is a numerical factor times Lo/sinθ]

We can write this expression for ΔL in terms of our unknown y by writing sinθ in terms of y. Another approximation you can use is that sinθ≈tanθ (since θ should be a small angle). Use your right triangle to express tanθ in terms of y and Lo. So, you will be able to express ΔL in terms of Lo and y. If you substitute this expression for ΔL into the right hand side of the above equation for y2 you can solve for y.

7. Jul 15, 2012

### Catchfire

Ok hopefully this is a bit clearer. :)

ΔL = 0.001 L0/2sinθ
= 0.001 L0/2tanθ
= 0.001 L0/2(y/(L0/2))
= 0.001 L0$^{2}$/4y

y$^{2}$ = L0ΔL/2
= L0/2 0.001 L0$^{2}$/4y

y$^{3}$ = 0.001/8 L0$^{3}$

y = 0.05 L0

Which is the answer. *sigh of relief*

Thanks for your assistance and patience, mind lending a hand on my other question?

8. Jul 15, 2012

### TSny

Nice work on the first problem!

For the second problem, your calculation of the tension in the wire at the breaking point (216 N) looks good to me. Also the maximum stretch of the wire before breaking (0.011 m) looks good.

However, note that the acceleration of the mass is not constant during the stretching of the wire because the tension force of the wire varies as the amount of stretch varies. So, constant-acceleration kinematic formulas are not applicable.

To use energy concepts, try ΔKE + ΔPEgrav + ΔPEwire = 0. You will need to choose appropriate initial and final points and you will need to construct a potential energy function for the stretching wire. But this will just be the usual type of potential energy function for Hooke's law (like a spring).

9. Jul 15, 2012

### Catchfire

Ok here it goes:

ΔKE + ΔPEgrav + ΔPEwire = 0

ΔKE should = 0 if we're looking at the limit

so we have

mgh - 0.5kΔx2 = 0

h = 0.5kΔx2/mg

Δx = 0.011m
mg = 4.9 N

k = AY/L0

L0 = 2m
Y = 2x1011 N/m2 N/m2
A = 0.000252π m2

k = 1960 N/m

so h = 0.5(1960)(0.011)2/4.9 = 0.242m

but that h is really 0.011m below x0

so the answer is h = 0.242 - 0.011 ≈ 0.23m.

Thanks, I'm glad that one was a bit more painless.
I really appreciate your time and effort.

10. Jul 15, 2012

### TSny

Great! You even saw that you needed to subtract .011 m off your initial result for h.