Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Effective Force Exerted by a Moving Mass


Date: 08/23/2001 at 23:50:42
From: david
Subject: Effective force exerted by a moving mass

I have to determine how much force a mass of 1000 lbs. moving at 50 
inches/sec will exert on an object in its path about 0.1 inches from 
where it started (assuming it was moving at the specified velocity 
throughout the 0.1" distance).

I.e., at the instant of stopping at the 0.1" mark, what did the
1000 lb. weight effectively weigh?


Date: 08/24/2001 at 09:08:18
From: Doctor Rick
Subject: Re: Effective force exerted by a moving mass

Hi, David, thanks for writing to Ask Dr. Math.

There is not enough information here to calculate an effective force. 
Force is related to acceleration, not velocity. What matters is not 
how fast the mass is moving when it hits, or how far it has moved, but 
how quickly it is slowed down by the impact. The force is the rate of 
change of momentum. The momentum is the velocity times the mass, which 
is initially 4.17 foot-pounds per second in your example. If the 
impact stops the object, then the momentum goes to zero. What we don't 
know is how long it took for the object to stop. If it takes 1 second, 
then the average force is 4.17 foot-pounds per second divided by 1 
second, or 4.17 foot-pounds per second squared. If it takes half as 
long, the average force is twice as much.

I will copy below a discussion that I had nearly three years ago 
concerning the effective weight of a dropped object. It's lengthy, but 
it may be of interest to you.
----------------------------------

TimeStamp: 11/11/98 at 23:44:27
From: Lou
To: Doctor Rick

Hello,

I know that this is a physics problem but it came up in a math
discussion. If a weight of, say, 40 lbs is dropped from a distance of
a given height (let's say 4 feet), what is the impact force when it
hits the ground? I calculated it using f = ma and s = 1/2 at^2. No 
matter what distance I use, I always come up with 40 pounds. 
Intuitively, it seems that the higher the object is dropped, the 
greater the force on impact. Also, for very small heights, it seems 
that the impact in these cases would always be 40 pounds. I'm 
confused!  Thanks for your help.

TimeStamp: 11/12/98 at 13:23:24
From: Doctor Rick
To: Lou

Hi, Louis. I think you are using f = ma with a = g, the acceleration 
due to gravity. Obviously the height does not enter into the equation 
for f. But this is not the relevant acceleration. Did you ever hear 
the joke, "Falling doesn't hurt me a bit - it's the stop at the end 
that hurts"?

While the weight is falling, its acceleration is g. When it hits the
ground, its acceleration is higher, because it stops in a much shorter 
time than it took to get up to its final speed. It's the higher 
acceleration that hurts!

How fast does the weight stop? That's the hard question. Just suppose 
that it stops in a distance of 1/4 inch - that's how deep a dent it 
makes. You can work out how fast the weight is moving when it hits 
(v0), and then use

  0 = v0 - at
  1/4 inch = (1/2) at^2

to solve for a. I don't think this is a very good model for how the 
weight actually stops (surely the dent will get deeper with height), 
but at least you will see that the stopping force increases with 
height.

If you have learned about spring forces, you might suppose that the 
weight lands on a spring (a trampoline, perhaps) with some spring 
constant, and work out a better model for the impact. You might also 
think about why air bags save lives.

- Doctor Rick, The Math Forum

TimeStamp: 11/12/98 at 23:05:16
From: Lou
To: Doctor Rick

Dr. Rick,

I appreciate you getting back to me on my question on free-falling 
objects. If you could answer a couple of follow-up questions, it would 
become clearer to me.

I think what you are saying is that the greater the "stopping 
distance" (time of impact to final dead stop), the less it "hurts." 
Does this mean that the energy is spread over a greater distance? Is 
it correct to say that dropping a 40-pound weight from a 4-foot drop 
is still 40 pounds? Would it still be 40 pounds from a 16-foot, or 
even a 400-foot drop? I dropped a weight on a bathroom scale from 
different heights and I see massive differences at small changes in 
heights. Again, I thank you very much for your time.

TimeStamp: 11/13/98 at 09:03:24
From: Doctor Rick
To: Lou

Hi again, Lou. Yes, when the kinetic energy of the object is 
dissipated over a longer distance, it takes less force to stop it. 
Remember that work (which is energy) = force * distance; for the same 
energy, if the distance is greater then the force is less.

You must distinguish the free fall from the impact. This is the point 
of the joke about the fall versus the stop. While the object is in 
free fall, the only force acting on it (neglecting air resistance) is 
gravity. This force is 40 pounds regardless of how far the object has 
fallen.

At the moment of impact, a new force begins to act on the object. This 
is a force upward from the ground. As I said before, this force must 
be much greater than the force of gravity if the object is going to 
stop in much less time than the time that it has been falling. It is 
this force that the bathroom scale measures.

The scale reading will bounce up and down a bit before settling down 
at 40 pounds, or whatever the weight of the object you dropped. This 
is called damped harmonic motion, as the energy of the dropped object 
is converted into compression of a spring and then back into upward 
motion. Gradually friction absorbs all the kinetic energy and you are 
left with just the force of gravity acting on the object.

- Doctor Rick, The Math Forum


TimeStamp: 11/13/98 at 17:50:58
From: Lou
To: Doctor Rick

Dear Dr. Rick,
One more time, please. So, if the object is dropped from, say, 400 
feet, then the upward force (acting against the object) is much higher 
than 4 feet? How much force would it be? at 4 feet? at 400 feet?
Thanks again.

TimeStamp: 11/14/98 at 15:01:32
From: Doctor Rick
To: Lou

Hi, Lou. The only way I can answer this question is (once again) if I 
make assumptions about what the object hits. Since the last thing you 
and I both mentioned was a spring scale, let's continue along this 
line.

The force depends on the spring constant of the scale, that is, how 
fast the force increases as the spring is compressed. The lower the 
spring constant, the greater the distance over which the object is 
decelerated, and the lower the force.

I can work out the maximum force by thinking about energy. The kinetic
energy of the object as it hits the scale is 

  E = mgh

where h is the height from which it fell. (When the object was at that
height, it had this much potential energy, and at the bottom this has 
all been converted to kinetic energy.)

By the time the spring reaches its greatest compression, the kinetic 
energy has all been converted to potential energy in the spring. This 
energy is

  E = (1/2)kx^2

where k is the spring constant and x is the distance the spring is
compressed. The total energy is the same as before if we ignore the 
damping (friction) in the scale. Since this is the maximum compression 
of the spring, the force exerted by the spring on the object is at its 
greatest. So what is this maximum force? Let's work it out:

  mgh = (1/2)kx^2

  x = sqrt(2mgh/k)

  f = kx = sqrt(2mghk)

So we see that (1) the greater the spring constant k, the greater the 
force (as I said); and (2) the force is proportional to the square 
root of the height from which the object was dropped.

Therefore if the object is dropped from 400 feet, the maximum force it
exerts on the scale is sqrt(400/4) = 10 times the force exerted if it 
is dropped from 4 feet.

If you look carefully, you might wonder why my equations say that the 
force is 0 if the object is just placed on the scale. The object 
should weigh something! The problem is that I left out the decrease in 
potential energy as the scale goes down by the distance x, on the 
assumption that x is much less than h. If h = 0, the correct equation 
is

  0 = (1/2)kx^2 - mgx
  mg = (1/2)kx
  x = 2mg/k
  f = kx = 2mg

Now you wonder why the force is TWICE the force of gravity on the 
object. The answer is that the scale overshoots; when its oscillation 
has been damped and the object is stationary, x will be mg/k and f 
will be mg. You actually observe this effect every time you step on a 
scale. No matter how careful you are to step on the scale from zero 
height, the weight reading will overshoot to well over your weight and 
take a little time to settle down (unless you have a well damped 
scale).

You have asked some very good questions. Thanks for writing.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Physics/Chemistry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/