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I just thought of something I have no good way of ruling out in my mind. Say someone is in free-fall, and is meant to, unfortunately, splat on the ground. Due to it accelerating, its speed will be increasing constantly with time. However, by considering two different definitions of Newton's Second Law, $\vec F = m \vec a$, and $\vec F = \dot p$, I seem to find a contradiction with my knowledge. According to the first definition, the force on the person in free-fall is constant (which I would agree with). However, when the person hits the ground, the force that which it imparts on the ground will have to do with its final velocity, I would think. One would say if the person is falling with a force of $50 \ N$ it will apply $50 \ N$ to the ground on impact, but this seems to be the case without considering how long the person was experiencing $50 \ N$ before hitting the ground I've found, or, at least when I've done exercises about these sorts of things. The final speed of the object, its mass, and the time in which the mass comes to a stop should determine the force felt by the ground, not just the force on the person falling. Is my confusion getting through? I feel like when I've considered the force a falling object applies on impact, it's merely the force experienced on the falling object applied to the ground, but I feel like this is misleading - whether the object falls for $1$ second or $10$, the force on it is the same, but its momentum would not. Perhaps this is just an oversimplification of my high school. Am I correct in thinking that the things I've listed are actually the important factors and not just the theoretical force experienced by the free-falling object?


marked as duplicate by sammy gerbil, John Rennie newtonian-mechanics Dec 20 '17 at 8:00

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  • $\begingroup$ Why are you saying the force acting on the object is the same as the impact force? $\endgroup$ – JMac Dec 19 '17 at 11:57
  • $\begingroup$ Oh this is a good point I forgot to address. I am lead to assume this due to energy conservation. How can there be more impact force than the force on the object? Where would the extra force come from? $\endgroup$ – sangstar Dec 19 '17 at 11:58
  • $\begingroup$ Energy for this is force times displacement. The object falling has been experiencing that $50N $ force a lot longer than the ground experiences the impact. $\endgroup$ – JMac Dec 19 '17 at 12:01
  • $\begingroup$ Gah! How did I not think about kinetic energy?? Okay, I'm on board with you on that then, since even though energy is conserved for the falling object, its potential energy has been mostly converted to kinetic on impact. What then relates the kinetic energy to impact force, out of curiosity? $\endgroup$ – sangstar Dec 19 '17 at 12:10
  • $\begingroup$ Velocity at the time of impact. With a constant force, that means constant acceleration. The higher velocity at impact; the greater acceleration the ground has to provide to stop it. You can also think that it has more kinetic energy that gets released. $\endgroup$ – JMac Dec 19 '17 at 12:52

You seem to be trying to equal forces in one situation to forces in another, even though there is no law requiring them to be equal.

  • During the fall, the gravitational force acts.
  • During the impact the gravitational force as well as the normal force act.

In terms of force, one case has got nothing to do with the other.

The gravitational force is constant, while the normal force is a variable force; a "holding back" force. The ground uses it to "hold back" against the impacting object. "Hold back" in the sense that, if it didn't, then it would break and the object would pass right through. And the normal force can adjust to balance out impacting force (until it's limit).

If something impacts with an impact speed, then the ground has to slow down the object from this speed to zero. It can either do that by

  • flexing / deforming (trampoline) which gives a certain force over longer time (smaller acceleration over longer time), or by
  • rigidly pushing back (asphalt) which gives a much larger force over much shorter time (must larger acceleration over shorter time).

Both methods give the same momentum change. Because momentum change is equivalent to speed change, and the ground wants to slow then object down.

The force exerted by the ground to stop an object depends on many things. The impact speed, the object's mass, the flexibility and rigidity of the ground, the hardness and strength of the ground, how much time it takes, etc...

Acceleration and momentum-change-over-time are equivalent ways to talk about how objects are stopped by forces. What causes their impact speed (their initial momentum) is not important. It could be from a fall or a throw or a push... Don't equate whatever force caused this impact speed to the forces that must dampen it. There is no law combining those two directly. The force during impact depends in many more things.

  • $\begingroup$ I see.. so, in itself, the force felt by a falling object will have no indication as to the force the ground will feel on impact? $\endgroup$ – sangstar Dec 19 '17 at 14:03
  • $\begingroup$ @sangstar Correct. $\endgroup$ – Steeven Dec 19 '17 at 20:02

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