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Force Required To Prevent Breaking an Ankle

  1. Jun 23, 2009 #1
    1. The problem statement, all variables and given/known data

    The compressive force per area necessary to break the tibia in the lower leg, is about F/A =1.6 ×10 N m^2. The smallest cross sectional area of the tibia, about 3.2 cm^2, is slightly above the ankle. Suppose a person of mass m =6.0 ×10 kg jumps to the ground from a height h0 =2.0 m and absorbs the shock of hitting the ground by bending the knees. Assume that there is constant deceleration during the collision. During the collision, the person lowers his center of mass by an amount d=1 cm.

    a) What is the collision time, t?
    b) Find the average force of the ground on the person during the collision.
    c) What is the average impulse of the ground on the person?
    d) Will the person break his ankle? How much would you need to lower your center of mass so you do not break your ankle?


    2. Relevant equations

    p = mv
    F = p/t
    J = pf - pi

    3. The attempt at a solution

    a) I first started by finding the final velocity.

    vf = sqrt(2ad) = sqrt(2(9.8)(2)) = 6.26 m/s

    I then divided the distance he lowers his mass by this amount.

    t = d/v = 0.01/6.26 = 1.6 x 10^-3 s

    b) The average is force is simply the change in position over the time.

    F = m(vf-vi)/t = 60(0-6.26)/(1.6 x 10^-3) = -234750 N

    c) The impulse is the change in momentum.

    J = pf - pi = 0 - 60(6.26) = -375.6

    d) P = F/A = -234750/(3.2x10^-4) = 7.3 x 10^8

    This pressure is greater, therefore, he will break his ankle.


    I'm not sure if I did anything correctly. Can someone show me what I'm supposed to do?
     
    Last edited: Jun 24, 2009
  2. jcsd
  3. Jun 24, 2009 #2
    Don't worry. I do not have a calculator with me to definitely confirm your results, but you did it the right way.
     
  4. Jun 24, 2009 #3
    Are you certain? Because the pressure I got was MUCH larger than the required pressure.
     
  5. Jun 24, 2009 #4
    Is he landing on BOTH feet?
    Well it makes sense that the pressure is going to be very large.

    Wait, t=d/v will not work since velocity is NOT constant. t is supposed to be twice as large.
     
    Last edited: Jun 24, 2009
  6. Jun 24, 2009 #5
    Yes, he's landing on both feet.

    So I'm supposed to use average velocity?
     
  7. Jun 24, 2009 #6
    MyNewPony, I think you have your question wrong. 1.6x10^1 is a bit small, you think? If that were true, I calculated my tibia would break every time I stand up.

    Reread the question, I am sure the minimum pressure is several orders of magnitude larger. :)
     
  8. Jun 24, 2009 #7
    I checked numerous times. That is indeed the minimum pressure it gives in the question.
     
  9. Jun 24, 2009 #8
    The reason I asked is because I found this: http://ocw.mit.edu/NR/rdonlyres/hs/physics/e/8_01t_fall_2004_ps09.pdf [Broken]

    Question #1 looks the same as yours but his tibia is ten million times stronger!
     
    Last edited by a moderator: May 4, 2017
  10. Jun 24, 2009 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Note that in your problem statement the exponent of the 10 got left out. Don't assume that it's 101. (It's more like 108.)

    Edit: Pinu7 beat me to it. Check out the link he posted.
     
  11. Jun 25, 2009 #10
    Ugh, that's not fair.

    I had my exam yesterday, and this was the exact question that was on it. I figure I should complain.
     
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