Force acted on a falling object.

[JZC]

I was having a discussion with a friend and neither of us could come to a conclusion.
The question is, what is the force acting on an object, by the ground, if it is released from a heigh?

This is how we related it:
If the object is on the floor, there is a weight and normal force acting on it, and since it isn't accelerating, Weight is equal to normal force by Newtons second law.

If it falls for a distance:
I am sort of stuck with a FBD and KD. When the object is free falling, ma = mg. However, if I draw a FBD and KD at the instant it is at impact, I get a net force of 0.

positive in the up direction
Normal Force - Weight = -ma
normal force = weight - ma
________
this would show that the normal force applied on the object is less than if the object was at rest. It does not make sense to me.

My friend suggested that the force will be the same for both, but I am reluctant about that.

 

[dextercioby]

Okay.So what?If it's on the earth,there are 2 forces acting on it.If it's in the air (okay,vacuum,so we'd have no friction with air),only one.What's so surprizing about that?


Daniel.

 

[shyboy]

An impact is not instant, what happens if you drop you object on a spring or on a car tire?
At the moment when they touch, the object will experience no normal force. Then, when the object stops for a moment, the normal force will be equal to the object weight. After that, the object will strike the curious experimentator (joke, no offence, please)

 

[FredGarvin]

Impact equations are not an easy thing to figure out. The defining factor is how long does it take the objecto to come to a complete rest once it hits the stationary object. In theory you can trat it as a perfectly elastic collision and use momentum and go about your merry way. Real life does not involve perfectly elastic collisions. There is a finite, however usually quite large deceleration of the object when it hits the ground. It will all depend on how much of a distance the object is going to decelerate. The longer the distance, the less the deceleration and thus the less the force that is a result.

If you were to drop something on to a hard surface that was instrumented with an accelerometer, you would see that the peak acceleration will be much more that of gravity. That is the real force experienced in the drop. The curve would show you the impule the object sees. Impulse is what you need to look into.

 

[MusePhysick]

i was in class the other day and i was asking Sir about how would i find the decceleration of a person if they fell from a height and broke their leg. I wanted to find out if a 6m drop would break bone to prove a point. The thing was i diddnt know the time it took to deccelerate. Sir said that "assume the time it takes to deccelerate is a small value such as 0.2s". If that helps