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Force exerted: on a foot run over by a car VS a car staying on a foot for longer

  1. Jan 18, 2012 #1
    i am having a hard time figuring this one out "Does it hurt more/cause more damage when a car runs over your foot? or when it STAYS on your foot for a longer amount of time?"

    Mathematically the FORCE exerted on your foot is the same --- (but, yes the MOMENTUM is different).. I had a car run over my foot when I was younger and I didnt feel much at all. but I think if that same car STAYED on my foot for 15 seconds it would hurt a lot and physically damage my foot more.

    I cannot exactly figure out why the difference in pain sensation & damage done to foot. time does not really change how much force is exerted. Can somebody help explain in an elaborate (mathematical) way??

    is it the kinetic energy or momentum that causes "more damage" or is it the force? Shear force? pls help :)
     
  2. jcsd
  3. Jan 18, 2012 #2
    I'm not sure about defining 'pain' and 'damage' but here are a couple of physics relationships that come to mind.

    FT = mv, so while the car is going real fast (lots of 'v') it might seem it would hurt more, but that 'v' is lateral (horizontal speed) so it doesn't count. The 'v' that 'hurts' would be while the car is lifting ever so slightly off the ground as it rolls over your foot. That would be the force the spring of the shock absorber [strut] exerts...every so slightly more than the stationary weight of the car it would seem, then likely and briefly a bit less as some momentum carries the car up every so slightly....so maybe there is a touch more damage to the side of your foot towards the front of the car, a smidgen less towards the rear, assuming forward motion. Depending on how the tire deforms, and how fast, the pressure on different parts of your foot seems like it would also vary. A really,really soft tire would spread the pressure (force per unit area) and a really soft tube say, would keep some pressure around your foot but still on the ground.

    If the car 'sits' immobile on your foot, seems like about 1/4 the weight of the car, or 1/2 the front or rear weight, if you prefer, would be the force you feel....F = ma = mg. Obviously you'll have more damage the longer it sits...say blood flow, for example, would be partially cutoff.....so I assume that's part of why emergency workers try to extricate injured people from crushing situations.

    I've seen measures of the forces mixed martial arts fighters exert and 500 lbs is not unusual, over a thousand pounds possible...and other guys seem to withstand most of that, I guess sustain some bruises, and continue to fight but maybe with subsequent pain the next few days.
     
  4. Jan 19, 2012 #3

    sophiecentaur

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    When a high force is applied for a short time on your foot, the bits inside don't have time to move as far as if the force is applied for ever. I guess that's why it doesn't hurt /damage as much.
     
  5. Jan 19, 2012 #4
    not considering the horizontal impact by the car, the vertical force is the same no matter how long the car is on your foot 1 sec or 10.

    Or is it? Another confusing thing is looking at the formula
    F=Ma=Mg when g=9.8m/s^2

    the longer s is (or longer the car stays on your foot) F gets smaller, since s is in the denominator. "The car's velocity downwards to your foot decreases with longer time" so it should be less and less force every second on your foot, eventually 0 Newtons.

    I know I am missing something BIG here but not sure what. Is it not the force that causes the "deformation/damage" ? Is it impulse?..momentum?

    an analogy would be a concrete column inside an old building, which will crumble over time because of it cannot carry the weight "anymore" - but was fine the first few years. shear force?
     
  6. Jan 19, 2012 #5

    Drakkith

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    The s in 9.8 m/s^2, is simply the units that acceleration is measured in. An object with a force equal to Earth's gravity at it's surface will increase in velocity by 9.8 meters per second every second. You would not put in a number for m/s^2. The equation linked doesn't have a spot to input time, as it isn't for that. Instead it is simply a measurement of force downward from an object due to Earth's gravity. Input the mass of the object as M and 9.8 m/s^2 in G, and that gives you the force downward from whatever object.
     
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