Calc Min Steel Wire Diameter for 323kg Load

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To determine the minimum diameter of a steel wire that can support a 323 kg load without stretching more than 7.38 mm, the Young's modulus of steel (200.0 GPa) is used in the calculation. The formula involves the relationship between force, area, and elongation. Participants in the discussion confirm that using g = 9.81 m/s² yields a radius of approximately 0.0069 m, leading to a diameter of about 0.0138 m. Some confusion arises over significant figures and rounding, but the calculations align closely with expected results. The correct diameter for the wire is confirmed to be around 6.86 mm.
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Homework Statement


Find the minimum diameter of an l = 17.2 m long steel wire that will stretch no more than 7.38 mm when a mass of 323 kg is hung on the lower end. (Hint: The Young's modulus of steel is 200.0 GPa.)


Homework Equations



Youngs = (Force / Area) / (change L / L)
= (mg / pi*r^2) / (change L / L)

The Attempt at a Solution


Simple enough... just plug and chug right?
just to clarify, 200.0GPa = 2x10^11 N/m^2 correct?
and 7.38mm = 0.00738m
If so,

Then after solving for r... i got .006853m, which is wrong. And i did multiply by 2 for the diameter. 0.0685 is also wrong, so is .685m

What am i doing wrong?? or am i just way off??

Thanks
 
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What value of g did you use?
 
I used the same formula as you did and I get 0.0069 m. Do you know the correct answer?
 
Sorry for not replying earlier... fell asleep.
I used 9.81 as g, and idk what the right answer is... =(
 
You have the right answer. Maybe you have the wrong number of significant figures, or are rounding incorrectly.

g=9.81 gives 6.86 mm, and g=9.80 gives 6.85 mm for the diameter.
 
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