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Homework Help: Young's modulus for bone

  1. Mar 26, 2005 #1
    Young's modulus for bone is 1.5x10^10 N/m^2 and that bone will fracture if more than 1.5X10^8 N/m^2 is exerted.

    What is the max force that can be exerted on the femur if the effective diameter is 2.5cm?

    [tex]Y=F/A[/tex]
    [tex]1.50x10^8N/m^2 =\frac{F}{\pi (0.0125m^2)^2}[/tex]
    [tex]F=73631N[/tex]

    is that correct?

    if this force is applied compressively, by how much does the 25cm long bone shorten?

    [tex]\Delta L=\frac{FL}{AY}[/tex]
    [tex]\Delta L=\frac{73631N*0.25m}{\pi (0.0125m^2)^2 * 1.5x10^{10} N/m^2}[/tex]
    [tex]\Delta L=0.0025m[/tex]

    I think I did some of this wrong but I'm not sure how to approch these problems. Any guidance?
     
    Last edited: Mar 26, 2005
  2. jcsd
  3. Mar 26, 2005 #2

    Doc Al

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    Staff: Mentor

    Yes. But realize that 1.5x10^8 N/m^2 is not Y (which stands for Young's modulus) but is the maximum stress the bone can support. (Stress = F/A). Also: round off to a sensible number of significant figures.

    Looks good. But realize you could have saved a bit of arithmetic by starting with the maximum stress (F/A) instead of the force.
     
  4. Mar 26, 2005 #3
    so what was the point in giving me [tex]1.5X10^8 N/m^2[/tex]? Was that to throw me off?
     
  5. Mar 27, 2005 #4

    Doc Al

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    Staff: Mentor

    I don't understand your question. You were given two numbers: max stress and Young's modulus. You used them both.
     
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