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Elastic Deformation through compression (stress)

  1. Nov 18, 2008 #1
    1. The total cross-sectional area of the load-bearing calcified portion of the two forearms bones (radius and ulna) is approximately 2.16cm^2. During a car crash, the forearm is slammed against the dashboard. Tha arm comes to rest from an initial speed of 81.9km/h in 7.72 s. If the arm has an effective mass of 1.88kg what is the compressional stress on the arm? Answer in units of Pa.



    2. Stress=F/A ,
    additional
    F=ma
    Vf=Vo+at



    3.
    Conversions
    A=2.16cm^2*1m^2/10^4cm
    =2.16e-4m^2
    Vo=81.9km/h*(1hr/60mins)(1min/60s)(1000m/1km)
    =22.75m/s

    F=ma
    a=(Vf-Vo)t=-2.94689.

    I assumed I could make it positive since the problem's context is in forward motion not retardation.

    F=1.88kg*2.94689
    =5.540kg*m/s^2

    P=F/A
    P=5.540kg*m/s^2/2.16e-4
    =25.648.8677N/m^2= same in Pa.


    I feel like I put in the appropriate conversions and formulas but when I plug in the answer, I'm told I am wrong. Does it have to do with the fact that it mentions "two forearm bones" in the beginning? If it is, where do I half the cross sectional area?

    I hope I'm not too far off.
     
  2. jcsd
  3. Nov 18, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Your method looks fine. Units are OK.

    Check the significant digits is about all I can suggest.

    Your concern about 2 cross sections should be allayed by the statement of total cross section.
     
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