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Homework Help: Young's modulus

  1. Dec 10, 2005 #1
    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
     
  2. jcsd
  3. Dec 10, 2005 #2

    Dale

    Staff: Mentor

    You set it up correctly. Assuming your arithemetic is right you got it.

    -Dale
     
  4. Dec 10, 2005 #3

    Fermat

    User Avatar
    Homework Helper

    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.
     
  5. Dec 10, 2005 #4

    Dale

    Staff: Mentor

    Oops, I missed that. Quite correct Fermat.

    -Dale
     
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