force of a falling object

[casrip1@gmx.com]

ok, so i get that net force is calculated as m*a, but what about the actual force of the object would exert and not just the net force?

 

for example, a 50kg mass is dropped from a 45m height with v1 = 0. how much force would it exert on the ground when it hits? surely not 450N (rounding gravity to 10 m/s^2 for simplifying calculations) because that would mean doesn't matter whether the object is dropped from 1m 10m 50m or 100m it would hit with the force of 450N, which doesn't make sense (if you drop a hammer on your foot from a couple centimeters it doesn't hurt nearly as much as if you dropped it from say, 1 meter). so how would you calculate the amount of force an object would exert on impact given mass and acceleration, or velocity right before crashing

Edited April 2, 2013 by casrip1@gmx.com

[elfmotat]

It depends mainly on the material of the object being dropped. From 45 meters, the object will have a final velocity of about 30 m/s before it hits the ground. The ground will apply a net impulse of 1500 Ns, but the actual force as a function of time F(t) will depend on what the object is made of. A big rock, for example, will have a much faster impulse than a 50 kg bundle of feathers.

[swansont]

To put what elfmotat said a different way, the acceleration you have to worry about for the force of impact is not g, it's the acceleration that beings the object to rest.

[daniton]

And the the hammer thing you said, hurts because as height increase the velocity of the hammer increase since the 2 event are performed by same material aka hammer would have same contact time as the hammer goes to rest, which indicates great acceleration -> great force ,so it hurts you very badly.

[Nehushtan]

for example, a 50kg mass is dropped from a 45m height with v1 = 0. how much force would it exert on the ground when it hits? surely not 450N (rounding gravity to 10 m/s^2 for simplifying calculations) because that would mean doesn't matter whether the object is dropped from 1m 10m 50m or 100m it would hit with the force of 450N, which doesn't make sense (if you drop a hammer on your foot from a couple centimeters it doesn't hurt nearly as much as if you dropped it from say, 1 meter).

Where did you get [latex]\mathrm{450\ N}[/latex] from? The weight of the object (force exerted on it due to gravity) is [latex]\mathrm{500\ N}[/latex] taking [latex]g=\mathrm{10\ m\,s^{-2}}[/latex]; this is NOT the force the object would exert on the ground on impact. The latter depends (for a given object) on the object’s final velocity, which in turn depends on the height from which it is dropped. Hope this is clear.

Edited April 4, 2013 by Nehushtan

[ewmon]

Collision mechanics can be very complex. You're looking to decelerate a falling object to zero velocity. It requires knowing the nature of the object and the "ground", and to what extent the collision is elastic or inelastic, The distance through which the object decelerates is a function of these factors.

 

various factors → distance → time → deceleration → force

 

So far, you've recognized the last calculation, deceleration → force, which is F=ma, assuming that the deceleration is constant.

[John Cuthber]

It can be very complicated, but there's sometimes a useful simplification.

If an object of weight w accelerates under gravity for a distance x and is then brought to a halt over a distance y the mean f deceleration is g times x/y and the mean force is w times x/y.

 

The peak forces involved an be a lot higher or a bit higher depending on the nature of the colliding materials.

[casrip1@gmx.com]

thanks everyone for their response, i think i understand the concept a little better now. so to sum it up (and make sure that i didn't misunderstand), the "poor man's solution" (i'm only in high school) would be that the final velocity when it just reaches the collision surface becomes the new v1 and v2 = 0 (since the object will crash to a stop) and then you need either how much the ground gave in OR the time it took for it to go from contact to full stop (the impulse time? if i'm not wrong?) and then you can use acceleration equations to find the deceleration rate and the collision force = m*a to find the collision force? it would make sense because the v1 (aka the velocity at which the object contacted the ground) would be greater for greater heights and the impulse would probably be same/similar (totally guessing that part) and so therefore the deceleration would be greater, producing a greater force

 

 

Where did you get [latex]\mathrm{450\ N}[/latex] from? The weight of the object (force exerted on it due to gravity) is [latex]\mathrm{500\ N}[/latex] taking [latex]g=\mathrm{10\ m\,s^{-2}}[/latex]; this is NOT the force the object would exert on the ground on impact. The latter depends (for a given object) on the object’s final velocity, which in turn depends on the height from which it is dropped. Hope this is clear.

sorry about that i made a mistake, i did m*d instead of m*a by accident but that wasn't the point, the point was how would one calculate the force of an object crashing to a stop given the height and mass. but thanks for pointing it out

 

EDIT: and i assume that if one was to find a force at which an object strikes another object (i.e a bullet vs different objects) it would be the same principal? you take the velocity right before contact, then the time it took for the bullet to come to a rest, and then find deceleration of the bullet and then plug it into f = ma

Edited April 6, 2013 by casrip1@gmx.com