Change in length of wire after adding a load

[coconut62]

Homework Statement

Please refer to the image attached.

I don't understand why can they just take 1.8m as the original length, since there is already a load hanging there.

My calculations yield the value of strain to be e/L, but in the answer it is e/1.8.

Please explain this to me.

Homework Equations

E=stress/strain

The Attempt at a Solution

My working is together with the question.

 

[Chestermiller]

The problem statement is a little ambiguous. It isn't clear whether there was a previous weight hanging from the wire, or whether the 25 N force represents the weight. Either way, it doesn't really matter much. If there was already a small weight hanging on the wire, the change in the length of the wire would have been insignificant compared to the 1.8 m. So, even in the incremental problem, the length of the wire could still be taken as 1.8 m (with virtually no loss in accuracy of the answer).

 

[coconut62]

If there was already a small weight hanging on the wire, the change in the length of the wire would have been insignificant compared to the 1.8 m.

Why is this so? If there was initially no weight hanging, would the change in length be more significant?

 

[Chestermiller]

Why is this so? If there was initially no weight hanging, would the change in length be more significant?

Consider this: If, with the small weight, the length of the wire had increased to 1.8001, would this really have materially changed the answer to your problem (for the incremental change in length). Try it out in your problem and see.

When we use Hooke's law, we assume that all strains are small and linearly superimposible, so that the changes in length are always small compared to the original length. If the change in length were larger, and comparable to the original length, we would have to use large strain theory.

 

[coconut62]

What is "linearly superimposible" ?

 

[Chestermiller]

What is "linearly superimposible" ?

It means that the strains from each of the two individual loads (in isolation) can be added together to get the strain from the combined load.