Young's modulus

  • Thread starter naeblis
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  • #1
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hi i was wondering if i did this correctly.

Q: A wire 2.5m long has a cross-sectional area of 2.5 mm^2. It is hung vertically and a 4.5 kg mass is hungfrom it. Byhow much does the wire stretch if Young's modulus for that material is 2.0 x 10^11 N/m^2 ?

so i did Y = ( F / A ) / ( del L / L )

2.0 x 10 ^ 11 N/m^2 = (( 4.5 kg * 9.81 m / s^2 ) / (0.0025 m^2)) / ( del L / 2.5 )

delta L = ((4.5 kg)(9.81 m/s^2)(2.5m))/((.0025m^2)(2x10^11 N / m^2))
= 2.21 x 10 ^ -7 meters
 

Answers and Replies

  • #2
Dale
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You set it up correctly. Assuming your arithemetic is right you got it.

-Dale
 
  • #3
Fermat
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1 mm = 10^(-3) m
1 mm² = 10^(-6) m²
2.5 mm² = 2.5*10^(-6) m²

Apart from that, your working is fine. Just multiply your answer by 1000.
 
  • #4
Dale
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Oops, I missed that. Quite correct Fermat.

-Dale
 

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