Effective Force Exerted by a Moving MassDate: 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/ |
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