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Properties of matter

  1. Mar 11, 2006 #1
    Sir,
    Please help me with this problem.
    # A wire of cross sectional area a, length L and young’s modulus Y is extended by an external force F. What is the total energy stored in the wire?
    I solved it in the following way:

    Energy stored = ½ x stress x strain
    = (½) x (F/a) x (dL/L)
    = (½) x (F/a) x (F/aY)
    = (1/2) x (F^2/a^2Y)

    Is it right? But the answer given in my book is ½(YF^2L/a).
    Here the symbol ^ represents power and x represents multiplication.
     
  2. jcsd
  3. Mar 11, 2006 #2

    Hootenanny

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    Hooke's law relates tension (T) to extension (x) in a wire in relative to it's origional length(L) and young's modulus ([itex]\lambda[/itex]);
    [tex]T = \frac{\lambda x}{L}[/tex]
    Energy stored is simply work done by stretching the wire, which is force multiplied by distance moved, which is given by integrating Hooke's law between the limits of zero and maximum extension (e);
    [tex]E_p = \int_{0}^{e} \frac{\lambda x}{L} \;\; dx = \frac{\lambda e^2}{2L}[/tex]
     
  4. Mar 11, 2006 #3

    Doc Al

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    You're missing the area. If [itex]\lambda[/itex] is Young's modulus, then:
    [tex]T = A \frac{\lambda x}{L}[/tex]
     
  5. Mar 11, 2006 #4

    Hootenanny

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    Indeed, I stand corrected, I have just work through the derivation using stress and strain. However, glancing through my textbooks it appears that they make no mention of area and states the [itex]\lambda[/itex] is simply the modulus of elasticity, which is the same as young's modulus. I am now rather confused and worried with regards to my upcomming exam :confused: Could you enlighten my Doc Al?
     
  6. Mar 11, 2006 #5

    Doc Al

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    Not sure about bringing enlightenment before having coffee, but the standard definition of Young's modulus is Stress (F/A) over Strain ([itex]\Delta L / L[/itex]). How does your text define it?

    See here: http://hyperphysics.phy-astr.gsu.edu/HBASE/permot3.html#c2

    Except for leaving out the area, your analysis is perfectly correct.

    Note to Amith2006:
    Check the units of that answer; the correct answer must have units of energy.
     
  7. Mar 11, 2006 #6

    Hootenanny

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    I've just had a big cup :tongue2:

    Yeah, I've just been reading through that and agree with it totally.

    Just reading through my text(applied mathematics textbook) it says that;
    However, I know from my physics that youngs modulus is defined as;
    [tex]\lambda = \frac{FL}{Ax}[/tex]
    which should leave units as [itex]N\cdot m^{-3}[/itex]. Perhaps the applied mathematics textbook is using a different constant and incorrectly naming it the 'modulus of elasticity'?
     
  8. Mar 11, 2006 #7

    Doc Al

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    That must be it. No problem as long as you use the definition consistently. (But it looks like the OP is using the standard definition--so be careful!)

    Funny, I just saw another problem where this same issue came up (https://www.physicsforums.com/showthread.php?t=113574); wonder if that fellow is using the same text.
     
  9. Mar 11, 2006 #8

    Hootenanny

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    Looks like he is, it's a pretty standard text for A-Level Mathematics, funny I haven't noticed it before. I'll just have to remember that my physics exam uses the 'proper' youngs modulus! Thanks for you help.
     
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