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Young's Modulus and Interatomic Bond Length PLEASE HELP

  1. Feb 10, 2013 #1
    A hanging wire made of an alloy of titanium with diameter 0.14 cm is initially 2.9 m long. When a 97 kg mass is hung from it, the wire stretches an amount 1.49 cm. A mole of titanium has a mass of 48 grams, and its density is 4.54 g/cm3.

    Based on these experimental measurements, what is Young's modulus for this alloy of titanium?
    Y = CANNOT FIGURE OUT

    As you've done before, from the mass of one mole and the density you can find the length of the interatomic bond (diameter of one atom). This is 2.60 10-10 m for titanium. As shown in the textbook, the micro quantity ks,i (the stiffness of one interatomic bond) can be related to the macro property Y:
    ks,i = CANNOT FIGURE OUT
     
  2. jcsd
  3. Feb 10, 2013 #2

    haruspex

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    Pls quote the equations you are aware of that relate to the definition of Young's modulus and (from your textbook) the relationship between interatomic bond stiffness and the modulus.
     
  4. Feb 10, 2013 #3
    young's modulus is stress/ strain which is (F*Lo)/(Ao*delta_L) I do not know how to get the numbers and I don't understand how they relate to interatomic stiffness
     
  5. Feb 10, 2013 #4

    haruspex

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  6. Feb 10, 2013 #5
    okay so i know that Lo= 2.9m and delta_L= 0.0149m but I'm not sure how to get F and Ao
     
  7. Feb 10, 2013 #6

    haruspex

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    You're given the wire diameter, so what do you think the cross-sectional area will be?
    What force will be exerted by a suspended 97kg mass?
     
  8. Feb 10, 2013 #7
    Ao= 0.9079

    I know force= m*a and I know that m is 97 but i don't know a
     
  9. Feb 10, 2013 #8

    haruspex

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    What units? But I don't think that's right whatever units you chose. Pls show how you calculated it.
    What does it mean to say something is hanging? What is the nature of the force that makes the wire stretch?
     
  10. Feb 10, 2013 #9
    To find Ao i converted cm to m and then divided it by 2 to get the radius then used the area of a circle formula

    and the force would be gravity
     
  11. Feb 10, 2013 #10

    haruspex

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    Correct procedure, but the answer was way off. Please post all that working.
    Yes, so what force will the 97kg mass exert?
     
  12. Feb 10, 2013 #11
    So i redid it and got Ao= .01539

    and the force should be -950.6
     
  13. Feb 10, 2013 #12
    it didn't work with those numbers idk what to do anymore but thank you
     
  14. Feb 10, 2013 #13

    haruspex

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    That's in sq cm, right?
    Newtons. Always state the units.
    So, plug those numbers in, be careful with the units, and post your working.
     
  15. Feb 10, 2013 #14
    no its in meters because its supposed to be in meters.

    i got 12021839.34 N/m and it was wrong
     
  16. Feb 10, 2013 #15
  17. Feb 10, 2013 #16

    haruspex

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    Then you calculated it wrongly. The diameter is 0.14cm, not 0.14m.
     
  18. Feb 10, 2013 #17
    i converted it to meters because everything we do is always in meters
     
  19. Feb 10, 2013 #18
    ok i redid my calculations again and got -264308724.8 N/m^2
     
  20. Feb 10, 2013 #19
    would -264308724.8 be correct or should it be positive
     
  21. Feb 10, 2013 #20

    haruspex

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    It should be positive. If you are consistent, ΔL has the same sign as the force, which you've taken to be negative.
    But I don't think that number is right yet. Radius = 0.07 cm, so area ≈ 0.015 sq cm = 1.5E-6 sq m. L = 2.9 m.; F = 97g ≈ 950N; ΔL ≈ 1.5 cm = 1.5E-2m.
    950*2.9/(1.5E-6 * 1.5E-2) ≈ 1.2E11. That's roughly 500 times your number.
     
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