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Homework Help: Another Diffcult Bone Compression problem.

  1. Nov 5, 2006 #1
    A gymnast does a one-arm handstand. The humerus, which is the upper arm bone between the elbow and the shoulder joint, may be approximated as a 0.26-m-long cylinder with an outer radius of 1.12 x 10-2 m and a hollow inner core with a radius of 3.7 x 10-3 m. Excluding the arm, the mass of the gymnast is 58 kg. Bone has a compressional Young's modulus of 9.4 x 109 N/m2. (a) What is the compressional strain of the humerus? (b) By how much is the humerus compressed?

    Ok here is what I have so far on this problem I know that I'm suppose to find the Cross sectional area (A) in this problem. But I'm not sure how I would go about doing that with the 2 radius. Do I find the area of each and add them together? Or perhaps subtracting them?

    I know that I'm suppose to use the formula F = Y (Delta L/ L0) (A) . Y = Youngs Modulus, Delta L = Change in Length/ L0 is initial Length, A = crossectional area.

    The force applied here would be (58 kg)(9.8 m/s) right? The F value
    young's Modulus is given as well as the initial length.

    I know that to find the strain it is the ratio between Delta L and L0 right? And to find the change in how much it is compressed by solving for the Delta L.

    So my main difficulty is the cross sectional area of this problem and perhaps not working right to find the total force?
  2. jcsd
  3. Nov 5, 2006 #2
    If you want the right cross sectional area you actually have to use both radii. Since the inside of the bone is hallow, you wouldn't want to include that would you? You are right for the force though.
  4. Nov 5, 2006 #3
    do you mean for me to subtract the second radius area from the 1st since it is a hollow object?

    I tried that and plugged into my formula of F = Y (D L/L0) A and I could not get the right answer for the 1st part.

    Just to make sure that my logic is right the Delta L/ L0 would give me the compression strain "answer to the first part of the question" correct?
  5. Nov 5, 2006 #4


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    Yes, it would. Regarding the area, check your calculation again.
  6. Nov 5, 2006 #5
    oh crap, is this suppose to be an crossectional area of a cylinder? I think I was thinking of it interms of a circle.....

  7. Nov 5, 2006 #6


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    It is the area you get when you subtract the area of the hollow cylinder's cross section from the area of the 'outer' cylinder's cross section.
  8. Nov 5, 2006 #7
    ok well I'm just wondering how do u find a cross sectional area of a cylinder? I could not find it anywhere in the book.

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