Calculate Tensile Stress of Nylon Fishing Wire for 20N Force

[nerofiend]

Im stuck on this one and its driving me a bit mad

Here is what I have so far...

Calculate the tensile stress in a nylon fishing wire of diamater 0.36mm which a fish is pulling with a force of 20 N.

What I get:
Stress = Force / Area

Stress = 20 / π*(0.18x10^-3)^2

Stress = 20 / 1.02x10^-7

Stress = 1.96x10^8 Pa


OR

Area = π*(0.18)^2 mm
Area = 1.02x10^-1 mm
Area = 1.02x10^-4 m

Stress = 20 / 1.02x10^-4

Stress = 1.96x10^8 Pa

Stress = 1.96x10^5 Pa


WHEREAS it should be:

Stress = 1.96x10^6 Pa


Could somebody tell me what the answer is and PLEASE tell me how it was supposed to be worked out. This isn't really for homework per say, as its just a question I seen while working far ahead of everyone and it struck my curiosity.

 

[dextercioby]

Stress=Force/Area
Force=20N
$$A=\frac{\pi d^{2}}{4}\sim\frac{3.14\cdot(0.36mm)^{2}}{4}\sim 1.01\cdot 10^{-1}mm^{2}=1.01\cdot 10^{-7}m^{2}$$
$$Stress\sim\frac{20N}{1.01\cdot 10^{-7}m^{2}}\sim 1.97\cdot 10^{8}Pa$$

Daniel.

 

[learningphysics]

Hmmm I get 1.96x10^8 Pa.

Remember when converting areas that
$$1mm=10^{-3}m$$ but
$$(1mm)^2=(10^{-3})^2m^2$$
so
$$1mm^2=10^{-6}m^2$$

So your second answer should also come out to 1.96x10^8 Pa.