Correct statement about size of wire to produce larger extension

[songoku]

Homework Statement

Please see below

Relevant Equations

E = stress / strain , E = Young Modulus

stress = F/A

strain = x / L , x = extension and L = original length

The answer is (B) but I don't really understand why.

Based on formula of Young Modulus:
$$x=\frac{FL}{AE}$$

The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of $x$ so to get larger value of $x$ we can increase $F$ and $L$ and decrease $A$

I am not sure whether there is change in $F$ for first and second wire so I will just assume $F$ does not change. It leaves (B) and (C) as possible options so why is (C) wrong?

Thanks

 

[haruspex]

just assume F does not change

Not a good assumption. You have not considered what determines when a wire of a given material will break.

 

[wrobel]

Hooke's law, by the way, does not work under large stress when the material is about to break.

 

[songoku]

Not a good assumption. You have not considered what determines when a wire of a given material will break.

I see. I need to consider breaking stress

Hooke's law, by the way, does not work under large stress when the material is about to break.

What about the formula: Young Modulus = stress / strain? Does it work under large stress because I am thinking the breaking stress of both wire is the same so the strain on the second wire must be constant and can be achieved by using longer wire?

Thanks

 

[wrobel]

https://en.wikipedia.org/wiki/Stress–strain_curve

 

[haruspex]

I see. I need to consider breaking stress


What about the formula: Young Modulus = stress / strain? Does it work under large stress because I am thinking the breaking stress of both wire is the same so the strain on the second wire must be constant and can be achieved by using longer wire?

Thanks

Yes, same strain with longer wire means greater elongation.
Take note of two things about your first attempt:

1. you made an assumption which you realised might not be valid
2. you failed to use all the information provided

So next time you are tempted to make an assumption, what should you do?

 

[256bits]

Homework Statement: Please see below
Relevant Equations: E = stress / strain , E = Young Modulus

stress = F/A

strain = x / L , x = extension and L = original length

The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of x so to get larger value of x we can increase F and L and decrease A

Sometimes one has to read the whole problem, start to finish.

That is the correct way to initially address the problem.
So the answer could be B) or C) if one assumes a constant F.
ie a 1 gram load will produce double the extension in the second wire if it is longer or of smaller diameter, as long as the wire(s) can support the loading.

Then the problem gives the criteria:
Second wire of the same material.
The second wire wire has a larger extension before breaking.

At breaking, both wires will have had the same stress or strain applied.
Meaning at breaking, the ratio F/A stress is the same, and the ratio ΔL/L strain is the same.

Then from formula,

Homework Statement: Please see below
Relevant Equations: E = stress / strain , E = Young Modulus

stress = F/A

strain = x / L , x = extension and L = original length

Based on formula of Young Modulus:
x=FL/AE

E = stress / strain = ( F/A ) / ( ΔL/L )
one can substitute, to arrive at the answer.

the breaking stress of both wire is the same so the strain on the second wire must be constant and can be achieved by using longer wire

 

[Herman Trivilino]

What about the formula: Young Modulus = stress / strain?

That relation only holds if, when the force is removed, the wire returns to it's former length. If you apply enough force the wire will not return to its former length. It will have deformed permanently. We call this exceeding the elastic limit. But it happens when the force is not large enough to break the wire.

So in this scenario the student could add enough weight to exceed the elastic limit. Then add some more weight until the wire breaks.

 

[256bits]

If you apply enough force the wire will not return to its former length. It will have deformed permanently.

Is that true for a brittle material such as glass?
For all we know the wire material in the problem may show elastic only deformation only before breaking.

 

[Herman Trivilino]

Is that true for a brittle material such as glass?

No, I don't think so. Glass has some elasticity. You could witness this, for example, by reflecting a beam of light off the surface of a widowe pane at a glancing angle so that the reflected light makes a spot on a distant wall. Push on the other side of the glass pane with your finger and watch the spot on the wall move. Proof that the glass is deforming without breaking.

I think the spot would return to its previous position when you stop pushing, demonstrating the elasticity of the glass. The question is, can you push hard enough so the spot doesn't return? In other words, could you exceed the elastic limit without breaking?

I keep going back and forth on my answer?

And more over, if you could get sheet of plexiglass to exhibit the same behavior would you still attribute the behavior to brittleness?

For all we know the wire material in the problem may show elastic only deformation only before breaking.

But it's wire! Part of the manufacturing process requires the material to be ductile.

 

[256bits]

could you exceed the elastic limit without breaking

With a prolonged load, creep would become a factor, with the glass fibre then continuously having its strain increase. I suppose one could then say that the glass fibre is plasticly deformed.

But it's wire! Part of the manufacturing process requires the material to be ductile.

A thin piano wire of steel should break while experiencing little or no plastic deformation.
The manufacturing process of drawing the wire leads to work hardening, giving a less ductile material with more strength, sometimes up to the high nineties of the UTS.

Given that the problem did not say of what material the wire is made up of - ductile, brittle, elastomer, plastic - and the the only equation is that relating to Young's Modulus, the problem might be one of analyze with the information given and pick the best qualitative answer in the allocated 2 minutes per question.

Nevertheless, the answer should be the same for all common materials.

 

[haruspex]

That relation only holds if, when the force is removed, the wire returns to it's former length. If you apply enough force the wire will not return to its former length. It will have deformed permanently. We call this exceeding the elastic limit. But it happens when the force is not large enough to break the wire.

So in this scenario the student could add enough weight to exceed the elastic limit. Then add some more weight until the wire breaks.

View attachment 368119

The given answer does not require exact linearity.

 

[Steve4Physics]

Here's one way to look at it...

For a given material, there is a unique maximum value of strain, $\epsilon_{max}$, at which fracture occurs.* This correponds to the unique end-point on a stress-strain graph.

For a uniform wire of unstretched length $L$ and maximum possible extension $\Delta L_{max}$, we have $\Delta L_{max} = \epsilon_{max} ~ L$.

$\Delta L_{max}$, and therefore the total length at the moment of fracture, can only be increased by increasing $L$. Hence answer B.

The shape of the stress-strain curve up to $\epsilon_{max}$ is irrelevant in this question; the Young modulus, linearity, elastic limit, etc., play no part.
____________

*That’s a bit of an oversimplification because it assumes the material is isotropic. Also $\epsilon_{max}$ will depend on the sample’s history, the temperature and no doubt other factors. But at an introductory level, $\epsilon_{max}$ can be thought of as a simple intrinsic property of the material.

 

[Herman Trivilino]

But $\epsilon_{max}$ depends on the wire's diameter, no?

 

[jbriggs444]

But $\epsilon_{max}$ depends on the wire's diameter, no?

Why should it? Are there any specific confounding concerns that you have in mind?

 

[Steve4Physics]

But $\epsilon_{max}$ depends on the wire's diameter, no?

No. The shape of a stress-strain graph doesn't depnd on the wire's diameter. So the breaking stress and strain are 'constants' for the material (in the same way that a material's density doesn't depend on its size/shape).

E.g. wire-1 has cross-sectional area A and wire-2 has cross-sectional area 2A. For the stretching process, wire-2 behaves the same as two wire-1s in parallel; wire-2 fractures when each of its 'constituent' wire-1's fractures. So $\epsilon_{max}$ is the same for wire-1 and wire-2.

 

[Herman Trivilino]

Why should it? Are there any specific confounding concerns that you have in mind?

No. It was my misunderstanding.

 

[songoku]

Yes, same strain with longer wire means greater elongation.
Take note of two things about your first attempt:

1. you made an assumption which you realised might not be valid
2. you failed to use all the information provided

So next time you are tempted to make an assumption, what should you do?

avoid making the assumption and use all the information provided.

For this question, consideration about breaking stress was not in my head until I read your reply