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Finding the height of a fall given force on impact

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data
    This is the second part of a two part question. The first part asked for the maximum force that can be applied to a bone with minimum cross-sectional area is 3.0cm2, Young's modulus 1.4 × 1010 N/m2, and can withstand a 1.0% change in length before fracturing. I calculated it to be 4.2x108 N.

    This is the part I'm struggling with:

    (b) Estimate the maximum height from which a 50 kg student could jump and not fracture her tibia. Take the time between when she first touches the floor and when she has stopped to be 0.03 seconds.

    2. Relevant equations
    change in momentum = net force * change in time
    p = mv
    v = at (or possibly a different kinematics equation?)

    3. The attempt at a solution
    Just after the student hits the ground, the ground exerts and equal and opposite force of -mg = 490N on the student:

    p = 490 * .03 = 14.7

    I am confused about how to relate the maximum force I calculated earlier to the force of the ground on the student when she lands. The force on the student seems like it should be 490 just after she impacts no matter how far she fell, but that obviously isn't the case.

    I asked my professor for help and he said that it's the small time interval that results in the force getting so large, which I sort of understand, so I tried this (with the ultimate goal of getting the total time it took the student to reach the ground so I could solve for the height of the drop):

    14.7 = 4.2 x 108 * t

    But that comes out even smaller than the impact t. So just in case I had done things backwards, I tried:

    50 * v = 4.2 x 108 * .03

    But that resulted in a time interval that was several hours long, which is also unreasonable.

    I am generally very confused about momentum and the force can change on impact. I understand that exerting the force over a longer time interval will make the force at a given time smaller and vice versa but I still can't figure out how to solve this problem. Thank you.
     
  2. jcsd
  3. Oct 16, 2015 #2

    Nathanael

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    The force "just after the student hits the ground" is not relevant. The bone would fracture while the student is hitting the ground, not after. So what is the average force while the student "is hitting" the ground? Hint, this equation:
    (It should really be, "change in momentum = time-averaged net force * change in time.")
     
  4. Oct 16, 2015 #3
    So if the force is 4.2x108N at one instant and 490N at another instant, is the time-averaged net force just the average of those two forces (2.1x108)? Or am I misunderstanding the meaning of net force in this context?
     
  5. Oct 16, 2015 #4

    Nathanael

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    We have a certain time interval (0.03 seconds) over which the normal force acts. In reality, this normal force will not be constant over this interval. (It will not jump from 4.2x108N to 490N in a single instant.) But since there is no way for us to know how the normal force changes over this time of impact, we treat it as a constant over time (the 'time-average of the force').

    You can calculate the momentum of the person just before impact, and you know just after impact the momentum will be ______, so you know the change in momentum over this 0.03 second time interval. You have to use this to find the (average) net force over that time interval, and then use that to find the (average) normal force. This normal force is what causes the fracture.
     
  6. Oct 16, 2015 #5

    haruspex

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    I read your 'just after hitting the ground' as meaning 'on making contact with the ground'. At that point there would, in practice be effectively no force. Forces won't exceed zero until there is at least some compression of the skin of the foot. The eventual force, when all has come to rest, will be mg, but that is of no interest here.
    You cannot take the average of two forces at different instants, it wouldn't mean anything. By time average Nathanael means ##\frac{\int F.dt}{\Delta t}##.

    However, that won't tell you what you really want to know, which is the maximum force during impact. That is quite hard to assess, and maybe beyond the intended scope of the problem. You could model the leg as a spring (but one which gives you back very little energy on being allowed to expand). In this model, the force rises linearly with the degree of compression, so reaches a maximum when her speed is zero. To find the maximum force, you would use the usual equations for spring compression.
     
  7. Oct 16, 2015 #6
    In order to calculate the momentum before and after impact would I need to know the velocities (p=mv)?

    Ultimately the goal isn't to find the maximum force since I already know that; I need to find the maximum height the student can fall without breaking their bone.
     
  8. Oct 16, 2015 #7

    JBA

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    Since the acceleration due to gravity is constant the time of the fall is proportional to her initial distance above the ground. The farther she falls, the greater the time, and the greater the velocity and therefore her momentum when she touches the ground.

    The .03 sec between the time she touches the ground and the break occurs indicates that some of the momentum and the resulting force are being absorbed by her leg before the bone breaks.
     
  9. Oct 16, 2015 #8

    Nathanael

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    Can you calculate the velocity based off the height of the building?

    In reality, you would want to find the maximum of the normal force over this 0.03 second interval (in terms of the height of the building) and then equate this to the force which causes the bone to fracture.
    I think, as haruspex said, that doing this is beyond the scope of the problem. I think they want you to find the time average force and pretend that it's constant (so that the average equals the maximum). This wouldn't be accurate in reality, (it would fracture from shorter buildings than will be calculated) but I think it is what the problem expects.

    "Proportional" has a specific meaning, namely that a relationship is linear (y=kx). The time does not vary linearly with the height above the ground (it varies by the square root) so this statement is not true. You have the right idea, when one increases so does the other, but that is not what "proportional" means.
     
  10. Oct 16, 2015 #9

    JBA

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    Clearly, I should have used "increases with" rather than "is proportional to".

    However, in a general statement, there is a difference between "proportional" and "directly proportional".
     
  11. Oct 16, 2015 #10
    Now I have that v = √2gh, p before impact = 50*√2gh, and p after impact = 0, making Δp = -50* √2gh = Fnet * 0.03. But I'm not sure what is appropriate to plug in to find h. I thought that Fnet should be 4.2x108 but that doesn't give me a reasonable answer.
     
  12. Oct 17, 2015 #11

    JBA

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    Previously you determined the p at the time her foot hit the ground and p at the time her leg broke. Now you have the p at which her leg will not break and you know how to by using t = .03 sec to calculate the difference between those two values of p, doesn't that give you a method to determine the p and then the v at first contact for this new case. Once you have that new v and using the equation for h as a function of v & g, you can calculate the h for this question.
     
  13. Oct 17, 2015 #12
    Ok, I figured out something that was throwing off my calculations quite a bit. The force I calculated was actually off by a factor of 104, but I believed it was correct because when I showed it to my professor and asked if it was too high, he glanced at it and said it was ok. The real Young's modulus is 4.2x104.

    The solution is:
    Δpmax = FmaxΔt = 1260N
    So 50v = 1260 N and v = 25.2
    v = √2gh --> h is 32.4m.

    Thank you all for the help.
     
  14. Oct 17, 2015 #13

    JBA

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    Congratulations, Good work!
     
  15. Oct 17, 2015 #14

    haruspex

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    Yes, I'm sure that is the intended answer, but please be aware that it is an overestimate for the reasons I gave in post #5. The force is not constant during the impact, so the maximum force will be more than ##\Delta p/\Delta t##. If you model it as a spring, I think you will get a multiplier on the force of something like ##\pi/2##.
     
  16. Oct 17, 2015 #15

    JBA

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    That is clearly a part of a rigorous solution but is beyond the scope of this lesson as presented.
    This is a preliminary physics course where this student started from the point of not understanding the relationship between the height of a fall and the resulting force to, with a little guidance, being able to understand the problem basics as presented and calculate a solution based upon the information given; and, to my mind that is quite an accomplishment all on its own, regardless of the level of sophistication of calculations.
     
  17. Oct 17, 2015 #16

    haruspex

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    It is never right to teach a lie just because it is simpler. It is usually quite easy to keep the question simple without lying about the significance of the answer. In the present case, it could have asked for the average force during impact, instead of for the maximum force, or asked for an upper bound on the maximum height, or whether it would be safe to fall from a certain height, etc.
     
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