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Freefalling force

by trekkie1701c
Tags: force, freefalling
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trekkie1701c
#1
Feb17-07, 03:05 PM
P: 7
Stupid question here - what's the equation for determining the impact force of an object in freefall?
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arunbg
#2
Feb17-07, 03:18 PM
P: 600
The question is very vaguely put. What sort of object ? what sort of impact in freefall ?!
Kindly give you question in a more precise form.
trekkie1701c
#3
Feb17-07, 03:25 PM
P: 7
Alright. I dropped an approximately 40 pound rock from a 30 foot height onto a sheet of ice to determine whether the ice could widthstand enough force to hold my weight (trying to prove it without actually taking the risks associated with being wrong had I done it traditionally).

The only equations I've been able to find dealing with force pretty much boil down to getting the weight of an object, not force if an object has been falling for several seconds.

rdx
#4
Feb17-07, 03:29 PM
P: 50
Freefalling force

Quote Quote by trekkie1701c View Post
Stupid question here - what's the equation for determining the impact force of an object in freefall?
A falling object accelerates according to gravity, gaining velocity and thus kinetic energy. Force really isn't directly part of the problem. So let us rephrase it to say what is the impact energy. That is (mv^2)/2. In the absence of air, the kinetic energy of impact will equal the potential energy the object had when dropped (assuming it simple falls from a high place) but air resistance would slow the object and reduce the impact in real world conditions.
rdx
#5
Feb17-07, 03:32 PM
P: 50
Quote Quote by trekkie1701c View Post
Alright. I dropped an approximately 40 pound rock from a 30 foot height onto a sheet of ice to determine whether the ice could widthstand enough force to hold my weight (trying to prove it without actually taking the risks associated with being wrong had I done it traditionally).

The only equations I've been able to find dealing with force pretty much boil down to getting the weight of an object, not force if an object has been falling for several seconds.
40 pound rock from 30 feet! That should have bashed it good. Did the ice break? If not I bet it held you. If you slid a 40 pound rock onto the ice and listened for cracking, that would tell you more.
CPL.Luke
#6
Feb17-07, 03:52 PM
P: 444
yeah impact force has to be directly measured as different objects will behave differently on impact.

for instance if you dropped a buckets worth of water onto the ice, the force would be very small as it would take a while for all of the molecules to come to rest.

If you drop a brick than the molecules of the brick will come to a halt very quickly (on the order of ns I'd guess) and as such there will be alot of force.

you get a similar situation when you take out the material its falling on, if a rock falls onto a truck filed with pillows, the force will be smaller than if it fell onto cement
arunbg
#7
Feb17-07, 04:09 PM
P: 600
I'd say the best eqn to use would be F=dp/dt. Assuming that the rock comes to complete rest on impact, the force acting would be the mass of the rock times its velocity as it hits the ground (approximately).
cesiumfrog
#8
Feb17-07, 04:38 PM
P: 2,056
I'd say the rock is more rigid than you are. Therefore, the impulse (the duration of time over which the rock and ice were in contact, each deforming like a spring) should be shorter than if you jump onto the ice (with your highly elastic knees and all).

You can calculate (in the old empire's crazy units) how much momentum must be exchanged to stop the rock (from continuing through the ice), and you can then find how high your own weight would need fall to accumulate the same momentum. But since the ice has a longer time to oppose your momentum, it exerts a lower force.

So as long as that height for you was about a meter, you can trust you can't be putting nearly as much force on the ice as what the rock did (in it's instant), though of course in the messy real world you can't account for every factor influencing the ice's behaviour, so you should keep someone along with you anyway.
CPL.Luke
#9
Feb17-07, 04:47 PM
P: 444
I'd say the best eqn to use would be F=dp/dt. Assuming that the rock comes to complete rest on impact, the force acting would be the mass of the rock times its velocity as it hits the ground (approximately).
no, mv would be the momentum the rock had when it hit the ice, the force is the rate at which that quantity changed with respect to time over the course of the impact, hence dp/dt=ma
trekkie1701c
#10
Feb17-07, 06:45 PM
P: 7
Quote Quote by rdx View Post
40 pound rock from 30 feet! That should have bashed it good. Did the ice break? If not I bet it held you. If you slid a 40 pound rock onto the ice and listened for cracking, that would tell you more.
Just barely; the ice was indented downwards a bit and some water leaked up from below.

As for the other replies:

I guess I misused the word "force". I'm trying to get an equation that would allow me to get how much weight/mass another object would need to do the same amount of damage to the ice as the rock (so I can figure out the breaking point of it).

I did try an equation for kinetic energy (KE=1/2mv^2), and came up with an answer of 5334 joules, which I then converted to kilograms (I found the equation 1 joule = 1 kg * (m/s)^2, which gave me an answer of around 390 kg.

I also looked at F=dp/dt, and, unless I'm getting something wrong somewhere, this is just another variant of F=ma, which doesn't appear to give me any relavent information to the problem (it gives me what kind of force Earth's gravity has on the rock, from what I know). Am I missing something or did I just misstate my problem? (I will admit that I'm still pretty much a noob when it comes down to Physics, although I do want to learn more).
arunma
#11
Feb17-07, 07:39 PM
P: 906
Quote Quote by arunbg View Post
I'd say the best eqn to use would be F=dp/dt. Assuming that the rock comes to complete rest on impact, the force acting would be the mass of the rock times its velocity as it hits the ground (approximately).
Hey, another Arun!

Anyway, certainly this equation is what we need, but not in the calculus form. We can use [tex]F = \dfrac{\Delta p}{\Delta t}[/tex]. The change in momentum is easy (it is equal to the object's mass times the final velocity immediately before impact). The [tex]\Delta t[/tex] is a bit more tricky. It's difficult (perhaps impossible) to theoretically calculate the time length over which the impact occurs, and usually this is done experimentally. But we can qualitatively deduce that the longer period of time for which an impact occurs, the less the force. This is why many cars have "crunch zones," which compress during a crash to increase the time of impact. It's also why boxers pull their heads backwards when they are punched, and why humans tend to bend their legs when they fall on their feet.
CPL.Luke
#12
Feb17-07, 10:47 PM
P: 444
you cannot convert joule to kilograms except through certain nuclear reactions which I can certainly tell you is not occuring here.

you won't get much information on the breaking point of ice from the energy perspective either as the ice breaks from experiencing to much force upon impact (when you're neglecting shockwaves) so the energy is meaningless to you.


what you need to find is the critical amount of pressure required for the ice to break. I would recommend staking weights on top of eachother with a footprint equivalent to that of your foot.

keep stacking the weights until you reach something above 400 lbs. or so, that way you know that you'll be safe.
arunbg
#13
Feb19-07, 01:56 AM
P: 600
I'd say the best eqn to use would be F=dp/dt. Assuming that the rock comes to complete rest on impact, the force acting would be the mass of the rock times its velocity as it hits the ground (approximately).
My bad, i forgot to add about the impact time. Anyway I don't think this approach is practical for the case at hand. You should much rather try Luke's approach since the impact parameters are difficult to determine.


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