Solving a Quick Falling Object Problem in Physics - Simplified Solution

[minger]

Hi, I need help with a problem here at work. Unfortunately, I completely forget physics and dynamics. Here's the problem:

I have a 1000 lbf object falling 32 feet. What is the impact force on the ground assuming that it stops instantaneously. I'm pretty sure I can get the number if I'm given a time or distance the object takes to stop to v=0, but I am at a loss trying to find force given that it just stops instantaneously.

I would really appreciate some help. I'd like to get this problem done before I go home today, haha.

 

[dextercioby]

How do you compute the force...?

What's its relation wrt momentum/impulse...?

Daniel.

 

[minger]

OK, here is my plan, please tell me if its flawed, or if there is a better way.

I can easily calculate potential and kinetic energy. Now, I am going to assume that all of the kinetic energy is converted into work done on the "ground." (1/2)mv^2 = Fd My problem before was I did not know the distance, or time it took to stop the object from falling. What I will do to get around this is assume a large time for the total impulse, like 0.5s. Obviously I know that it will take much less time, but this is where I'm starting. Based on this, I can now solve for force. From a given force, and properties of the plate that it is falling onto, I can calculate how much the plate deflects. I can also check at this point to see if the pressure exceeds the yield of the metal. Now, I will keep decreasing the time (and thus decreasing d) until either one of two things happen. Either I calculate that my deflection is correct, at which point I will be done, or the pressure exceeds the yield of the metal, in which case we will need a thicker plate.

Please let there be an easier way.

 

[minger]

How do you compute the force...?

What's its relation wrt momentum/impulse...?

Daniel.

The maximim force is the points on the impulse curve. The problem is that I do not have that curve (this is not a homework problem, this is a real life problem, but resembles a homework problem).

 

[dextercioby]

Do you agree that
$$F=\frac{\Delta p}{\Delta t}$$ ?
What do these words "it stops instantaneously" tell u about \Delta t...?

Daniel.

 

[minger]

yes yes. I realize this. I know the equations. I mean, typically, problems out of a book will give you a delta t, or deflection. What do you propose that I do to solve this problem?

 

[dextercioby]

Plugging in the 'Delta t' in this case...?

Daniel.

 

[Doc Al]

I have a 1000 lbf object falling 32 feet. What is the impact force on the ground assuming that it stops instantaneously. I'm pretty sure I can get the number if I'm given a time or distance the object takes to stop to v=0, but I am at a loss trying to find force given that it just stops instantaneously.

If you can find the average impact force if the object is stopped within a finite time of $\Delta t$, then imagine how that force would need to change if $\Delta t$ went to zero. Bottom line: Stopping "instantly" is highly unrealistic.

(Edit: Just realized that Daniel already made the same point!)

 

[minger]

If you can find the average impact force if the object is stopped within a finite time of $\Delta t$, then imagine how that force would need to change if $\Delta t$ went to zero. Bottom line: Stopping "instantly" is highly unrealistic.

Yes, I now realize that instantaneous stopping is unreasonable, that is where my other proposal came from. I just wasn't sure if there was a good formula for approximating the force, given just momentum.

and Daniel, I do not have delta t or deflection, THAT IS THE PROBLEM! I'm going crazy. I'm actually trying to get some equations from the civil guys from deflection in plates so I can start working my method out.

Again, if anyone has another method for doing this, please let me know.

 

[dextercioby]

In the problem you initially formulated,'Delta t' is zero and that's why the force of impact is infinite.Of course,it's never the case in 'the real world'.

Daniel.

 

[minger]

In the problem you initially formulated,'Delta t' is zero and that's why the force of impact is infinite.Of course,it's never the case in 'the real world'.

Daniel.

Yes, I realize this. Do you have a solution to this problem without knowing any deflections or delta times?

 

[dextercioby]

No.That is the only solution the problem may have in the form it was given.You'd have to change several details to obtain a (this time reasonable) answer.

Daniel.

 

[minger]

Well, therein lies the problem. I can't change any details, and I can't get anymore information. This was unfortunately all I was given, and there is no way to get anythign esle. I think I can manage following the procedure I stated above. I am just waiting to get the formulas for deflection in plates (even to view them on the internet, it looks like a $150.00 subscription).

 

[dextercioby]

You haven't told us what about that "deflection in plates"...?Is it relevant to this problem in any way...?

Daniel.

 

[minger]

Yes, it's the road block right now. In my process of solving this, I'm basically assuming a value, then calculating it. I'm not sure if you've ever done any calculations for engines, but to do that, you assume a residual mass fraction, and assume an exhuast temperature. Then, at the end of your calculations, you can find your residual mass fraction and exhaust temp based on what you've found. If they aren't equal, then you assumed incorrectly.

That is basically the principle which I'm working this problem. Since I do not have a displacement or delta time, I am going to assume one. Work is force times distance, so given an assumed value for distance (distance of course can also be found from time), I can calculate the force I'm looking for. Now obviously I will need a way to check if I assume correctly. Based on force and the given geometry of the plate, I can calculate deflection in that plate. I will need to keep iterating until my deflections are equal. Then, I will take my force from that iteration and check to see if the pressure it creates exceeds the yield strength for the metal. The problem right now is that I don't have access to those equations. Apparently, even to look at them on the internet costs like $150 for a subscription.

 

[dextercioby]

Ouch,that's really bad...
As for the first part,assume realistic (nonzero) values for all your parameters including time of impact.

Daniel.

 

[FredGarvin]

I replied in the engineering forum.