What is the elastic modulus of the wire?

AI Thread Summary
To calculate the elastic modulus of a wire with a diameter of 0.2mm that stretches by 0.20% under a 6.28-N force, the correct formula is Elastic Modulus = (F/A) / (ΔL/L). The area should be calculated using A = πr², where r is 0.1mm or 1x10^-4m. The correct value for ΔL/L is 0.002, not 0.2. The initial calculation mistakenly included an extra factor of two, leading to an incorrect modulus value of 5x10^9 N/m² instead of the correct 1.0x10^11 N/m². Understanding the proper application of the formula and calculations is crucial for finding the elastic modulus accurately.
joseg707
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Homework Statement


A wire of diameter 0.2mm stretches by 0.20% when a 6.28-N force is applied. What is the elastic modulus of the wire?


Homework Equations


Elastic Modulus=F/A/L/\DeltaL


The Attempt at a Solution


r=.1mm=1x10-4m

6.28N/(2\pi1x10-8/.02=5x109N/m2

The right answer is 1.0x1011.

What am I doing wrong?
 
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joseg707 said:

Homework Equations


Elastic Modulus=F/A/L/\DeltaL
That should be (F/A)/(ΔL/L).
 
Oh, yeah. Well how do I calculate that? I know that the change is .2% if I use .002 or .2 as a value for \DeltaL/L I don't get the right answer. What is wrong with my calculation?
 
joseg707 said:
6.28N/(2\pi1x10-8/.02=5x109N/m2
Area = pi*r². Looks like you have an extra factor of two in there. And ΔL/L = 0.002.
 
Ah! Thank you so much! I don't know why I was thinking are was 2*pi*r^2. Thanks a lot!
 

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