Elasticity limit of copper Problem

[loudgrrl4_ever]

Okay, here is the problem:
If the elastic limit of copper is 1.50E8 N/m2, determine the minimum diameter a copper wire can have under a load of 10.5 kg if its elastic limit is not to be exceeded.
Obviously I am looking for d in this problem. I am given the downward force (102.9 N = ma); A = (pi * d2)/4.
I am having trouble identifying the correct symbol for the elastic limit. My professor has a habit of hading out worksheets full of formulas with no explanation on them. For some reason I know that when given modulus, it will be called modulus, so this is the TSmax? I'm quite confused.
Trying it with 1.50E8 = stress, I get stress = F/A; A = F/stress; (pi/4)d*d = F/stress; d = (sqrt)[(4*102.9)/(pi * 1.5E8)]; d=.000934m; d=0.93400mm
If I knew the unit strain, or the length and change in length this problem would be much easier. I haven't done any with so little information before.
I haven't tried that answer yet (I just checked and I answered 0.093400 earlier when I thought I used this) but I would just like to know that I have done it correctly. My textbook spends all of a page and a half talking about this section, and it makes little sense to me unless the problems are simple.
Any help you can give towards understanding this section of Physics would be very helpful and appreciated (especially links/explanations).
Thanks :)

 

[Doc Al]

Trying it with 1.50E8 = stress, I get stress = F/A; A = F/stress; (pi/4)d*d = F/stress; d = (sqrt)[(4*102.9)/(pi * 1.5E8)]; d=.000934m; d=0.93400mm

The elastic limit is the maximum stress (just check the units) that can be applied while staying within the parameters of Hooke's law. So your thinking here is just fine. (I did not check your arithmetic.)

If I knew the unit strain, or the length and change in length this problem would be much easier.

All that is irrelevant.

I haven't done any with so little information before.

This problem is probably easier than you are used to.

 

[PhanthomJay]

Following up on this, do not confuse the elastic stress limit of a material, often called yield stress, and quite often denoted as $$\sigma_y$$, which has units of $$force/length^2$$, with the modulus of elasticity of a material, often called $$E$$, which also has units of $$force/length^2$$. Both are properties of the material that are independent of each other. The yield stress denotes the stress level above which point the material will no longer obey Hookes law. The modulus of elasticity is the slope of the stress - strain curve within the elastic range, and is a measure of how much the material will strain (e) or elongate/compress under a given stress, per $$\sigma = eE$$. As Doc Al noted, you have the correct equation for determining the required diameter; the strain in the material is irrelevant in this problem.