DrRocket Posted February 18, 2012 Share Posted February 18, 2012 OK, so among the things I doubt is that you actually considered the momentum of the system. It's easy enough to calculate the mass of the car from the velocity and the energy using 1/2 MV^2. But everyone agrees that there is not actually enough information to calculate the force asked for, even if you do calculate the momentum. So, the momentum does not actually help. If you still think I'm wrong then calculate them momentum and solve the problem. To a good approximation, at voltages high enough to kill, the body acts as a simple resistive load. It might not be true in general, but it's true in the particular instance that was defined. Nice straw man. You again miss the points being made with respect to the fundamental physics. 1. As has been repeatedly stated, there is not enough information to solve the problem. Nevertheless the fundamental quantity of interest is momentum. That is because the question was phrased in terms of time and momentum is naturally related to force acting over time, and is a vector, while energy is naturally related to force acting over a distance, and is a scalar. This does not mean that energy is not a consideration as well. You have a complex problem involving non-linear material response, including buckling, fracture and ordinary material failure. There are a great many mechanisms through which the initial kinetic energy of the car can be dissipated and ultimately realized as heat. That is because energy is a scalar. Thus one has no means, outside of a sophisticated computer model, to keep track of all of these mechanisms. Momentum, a vector quantity, is somewhat easier to handle in this situation. While one cannot calculate the peak force, or the precise force-time curve, the resisting force integrated over time will equal the initial momentum (mass x speed) of the car. That much can be said. Using energy considerations alone you can say nothing. But once again, your call to "solve the problem" demonstrates nothing more than a lack of understanding of the underlying physics. As an observation I would note that this problem is somewhat analogous to that of calculating recoil forces when a gun is fired. Without knowing details about how the gun is supported one cannot calculate peak recoil force. Nevertheless, momentum conservation demands that the time integral of the recoil force be equal to the momentum of the projectile and expelled gas. 2. Electricity commonly kills by means of disrupting the rhythm of the heart, and is not simply a matter of either volts or amps. Frequency is also a very significant factor, as is the current path through the body. Frequency near the pulse rate can be particularly problematic and 60 Hz is sufficiently close. Frequency is also a factor in determining the primary current path, and the load characteristics are not just simple dc resistivity. Link to comment Share on other sites More sharing options...
John Cuthber Posted February 18, 2012 Share Posted February 18, 2012 What straw man? You said "Conservation of momentum will be a helpful principle in this problem.". I say it is not . If it helps then give an answer. If you can't do that then momentum didn't really help did it? Exactly what help did it provide? And I understand the physics well enough to not take kindly to your attitude. However I also think that you are mistaken in saying " Nevertheless the fundamental quantity of interest is momentum. " for the simple reason that the momentum is not known. It may have the advantage of being strictly conserved, unlike KE, but you don't know what it is. It is also possible (if you know the distance over which the force acts) to calculate the force directly by appeal to the law of conservation of energy. It won't be absolutely accurate because of losses -that's the effect of a "crumple zone" but it will be close. On the other hand, the point of the crumple is to alter the forces received by the occupants of the car. Force times distance = energy. If you know the energy and the distance, you can calculate the (mean wrt distance) force. Someone already pointed that out. BTW, you can kill someone perfectly well with DC. In any particular case- say a 400 Hz power system on board a 'plane, the impedance is defined (and very largely resistive). The point is that you can not independently change the voltage or the current. There is only one V vs I curve (and it's pretty close to a straight line) Defining one, defines the other. You can't say it was one that killed someone, but not the other. Incidentally, it's horribly complicated, people survive very large power dissipations, provided that the energy is low and they survive relatively large energy shocks provided that the power is low. The point remains that you cannot have volts without amps and vice versa. Link to comment Share on other sites More sharing options...
InigoMontoya Posted February 18, 2012 Share Posted February 18, 2012 I am still trying to persuade my audience that the points made are in fact inappropriate. I don't believe that they were inappropriate. Given the nature of the initial request it is pretty clear that we're almost certainly dealing with an entry-level physics student either at the HS senior or college freshman level. I'd wager that a sizable chunk if not an outright majority of such students get the ol' "what is the force of the car wreck" problem. I know I did in HS *and* in college. Assuming (yes, there's that word) that this is what we're dealing with, there is absolutely noting inappropriate about the time averaged approach. It is the only approach available given their educational experience and exactly the approach that the instructor is anticipating. 1 Link to comment Share on other sites More sharing options...
Xittenn Posted February 18, 2012 Share Posted February 18, 2012 I don't believe that they were inappropriate. Given the nature of the initial request it is pretty clear that we're almost certainly dealing with an entry-level physics student either at the HS senior or college freshman level. I'd wager that a sizable chunk if not an outright majority of such students get the ol' "what is the force of the car wreck" problem. I know I did in HS *and* in college. Assuming (yes, there's that word) that this is what we're dealing with, there is absolutely noting inappropriate about the time averaged approach. It is the only approach available given their educational experience and exactly the approach that the instructor is anticipating. We also get questions that are incomplete that test how well we understand first principles. I don't know is often a viable answer. If the student made a mistake then they should correct themselves. As stated the most appropriate answer is I don't know. If someone had felt the need to address this point, there was plenty of room to have done so when the question was asked. Link to comment Share on other sites More sharing options...
DrRocket Posted February 18, 2012 Share Posted February 18, 2012 What straw man? You said "Conservation of momentum will be a helpful principle in this problem.". I say it is not . If it helps then give an answer. If you can't do that then momentum didn't really help did it? Exactly what help did it provide? Yep you say things. Sometimes they are correct. This is not one of those times. As noted above it gives you the integral of the restraining force over time, from which one can trivially obtain the time-average of the force. And I understand the physics well enough to not take kindly to your attitude. But apparently not well enough to see the essence of the problem. I don't think much of your attitude or your claimed "expertise" either. However I also think that you are mistaken in saying " Nevertheless the fundamental quantity of interest is momentum. " for the simple reason that the momentum is not known. It may have the advantage of being strictly conserved, unlike KE, but you don't know what it is. If this is truly an engineering problem, and not some foolish exercise, then one knows the mass and velocity of the car, hence the momentum. That informatioin is also necessary in order to have determined the kinetic energy. It is also possible (if you know the distance over which the force acts) to calculate the force directly by appeal to the law of conservation of energy. It won't be absolutely accurate because of losses -that's the effect of a "crumple zone" but it will be close. On the other hand, the point of the crumple is to alter the forces received by the occupants of the car. The whole point of a collision analysis is that you DON'T know the distance over which the force acts. Here nothing has been stated regarding the second obect in the collision. It might be another car or it might be a large tree. The distances involved could vary quite a bit. Both are reasonable scenarios. The point is that energy considerations don't tell you very much, that detailed analysis takes a lot more information than is given, but that one can obtain a time-average of the force if one knows the initial momentum. A more important issue is why one is particularly interested in the force in the first place. That is not at all clear. There are more important parameters for most applications than either the peak or time-average force. Moreover, conservation of energy does NOT get you the restraining force. The initial kinetic energy of the vehicle is NOT the line integral of the restraining force over the applied path. Energy does not work like that. Energy is NOT a scalar and is not countered by the restraining force. There are lots of mechanisms for the dissipation of the initial kinetic energy of the vehicle -- plastic deformation and tearing of the material of the car, deformation of that with which it has collided for instance -- but the restraining force does not absorb energy. In particular it is NOT true that the restraining force x distance over which it acts is equal to the initial kinetic energy of the vehicle. Let's take an extreme example. Suppose the car hits an indeal (immovable) brick wall. And let's look at the problem in the reference frame of the wall. Then the restraining force (contact forct) exerted by the wall on the car acts through no distance at all, and therefore does no work. Energy tells very little. But the time average of that force is still determined by the momentum of the vehicle. The distance-average of the restraining force is meaningless in this idealization, as the distance is 0. Force times distance = energy.If you know the energy and the distance, you can calculate the (mean wrt distance) force. Someone already pointed that out. Newton But see above. That energy has nothing to do with the kinetic energy of the vehicle, and that mean with respect to distance could be nearly infinite (in the idealization above it would be infinite). BTW, you can kill someone perfectly well with DC. Yep. But not as easily as with AC at the proper frequency with a current path that disrupts the heart. With enough potential and current capacity you can vaporize someone. The Los Angeles Department of Water and Power has a very large DC transmission line that goes from hydroelectric generating stations in Oregon to an inverter station at Saugus, and I would advise against inserting oneself in that circuit. Do you have a point ? In any particular case- say a 400 Hz power system on board a 'plane, the impedance is defined (and very largely resistive). That is absolutely absurd. The power system does not define the impedance of the load. The load defines the impedance of the load. 400 Hz is a typical frequency in aircraft applications because of the widespread use of servomechanisms. Servo motors have a significant reactive component to the impedance. But there are certainly other devices that are also served by the power supply, and they will impedances that are characteristic of their configuration. TDefining one, defines the other.You can't say it was one that killed someone, but not the other. Incidentally, it's horribly complicated, people survive very large power dissipations, provided that the energy is low and they survive relatively large energy shocks provided that the power is low. The point remains that you cannot have volts without amps and vice versa. This is getting silly. People survive all sorts of things. People also die from all sorts of things. People survive lightning stikes, and people die from lightning strikes. Damn few people are electrocuted by DC current, but it is not impossible for some idiot to find a way to do it. Yes you can survive long duration current with an associated high energy. You just have to have low enough power that there is no serious immediate effect. People who work with car batteries don't worry a lot about 12 v DC, unless they get a piece of metal across the terminals. Ditto for very low voltage AC power. So what ? You can also survive high power shocks if the duration is so short that no appreciable energy is transferred. Power system electricians have been known to test large busses by tapping them with the back of their hand -- not recommended but it has been done. So what ? You can most certainly have volts without amps. Ever hear of an open circuit ? There is a potential difference between the terminals at each and every wall switch in your house, until you turn that switch to the"on' position and energize an appliance in your home. If this were not the case none of your appliances would work. And you can have amps without volta as well. Ask the Cap'n about the experiment that employed him that deals with superconductors. I was recently asked if I had reason to believe that designated experts might not be quite so expert. Thanks for a sterling example. -1 Link to comment Share on other sites More sharing options...
baxtrom Posted February 19, 2012 Share Posted February 19, 2012 1) Energy is NOT a scalar 2) Let's take an extreme example. Suppose the car hits an indeal (immovable) brick wall. And let's look at the problem in the reference frame of the wall. Then the restraining force (contact forct) exerted by the wall on the car acts through no distance at all, and therefore does no work. Energy tells very little. But the time average of that force is still determined by the momentum of the vehicle. 3) I was recently asked if I had reason to believe that designated experts might not be quite so expert. Thanks for a sterling example. My numbering above. 1) Please consult an introductory textbook on physics, page 1. Energy is a scalar quantity and is obtained by means of the dot product if working with force and distance vectors. I'm sure you are aware of the properties of the dot product, otherwise, consult an introductory textbook on linear algebra, page 1. 2) We are of course referring to the energy dissipated by predominantly plastic work on the car during impact, which btw is assumed not to be made of your "indeal" rigid bricks. 3) You are yourself quite an example. Really, you are so full of yourself you need to get your hoop stress checked. Yeah, you saved "milliions of dollars, and likley a life or two", that doesn't sound like a load of manure at all. Myself I saved thousands of orphans and got Nelson Mandela out of prison. Top that if you can. Funny this, how my two line reply on friday could start this chain reaction of indignation. It's like that small pertubation of a thread seemingly in equilibrium caused DrRocket's ego to suddenly reach critical mass. Wonder what the man is like IRL. Link to comment Share on other sites More sharing options...
John Cuthber Posted February 19, 2012 Share Posted February 19, 2012 Dr Rocket, either answer the original question using the concept of momentum, or accept that you were wrong in saying that it would help. Then we can get back to your failure to understand the point about current through, and voltage across, a body not being independent; and to your ability to straw-man on the issue such as pretending that humans might be superconductors. Link to comment Share on other sites More sharing options...
DrRocket Posted February 19, 2012 Share Posted February 19, 2012 My numbering above. 1) Please consult an introductory textbook on physics, page 1. Energy is a scalar quantity and is obtained by means of the dot product if working with force and distance vectors. I'm sure you are aware of the properties of the dot product, otherwise, consult an introductory textbook on linear algebra, page 1. You are correct. Mistyped. Meant to say energy is not a vector. However, my logic basec on the corrected text is quite correct. Thanks for catching that typo. 3) You are yourself quite an example. Really, you are so full of yourself you need to get your hoop stress checked. Yeah, you saved "milliions of dollars, and likley a life or two", that doesn't sound like a load of manure at all. Myself I saved thousands of orphans and got Nelson Mandela out of prison. Top that if you can. Please feel free to go do something anatomically unlikely. Link to comment Share on other sites More sharing options...
Xittenn Posted February 19, 2012 Share Posted February 19, 2012 3) You are yourself quite an example. Really, you are so full of yourself you need to get your hoop stress checked. Yeah, you saved "milliions of dollars, and likley a life or two", that doesn't sound like a load of manure at all. Myself I saved thousands of orphans and got Nelson Mandela out of prison. Top that if you can. Realistically speaking this isn't a very difficult thing to do when machines can cost upwards of half a million dollars and there are are a dozen or so of them on the floor. In addition to this the costs of contracts, again often upwards of half a million for a small inconspicuous facility that produces trivial goods, and multiplied by any number over a year. Then there is the consideration of injury and the costs of increasing insurance and health care and so on . . . . there is absolutely nothing incredulous about this and your response says a lot about your experience. In fact I had to nod when I read it because it is in fact so true. You are still arguing a moot point that was irrelevant in the first place. Is there room for thinking like this in a class room? Absolutely, and had the OP taken the care to properly post their question they would have been given the proper treatment of the matter. Does it apply to the automobile engineering industry? Not in the slightest. Try to relax Mr. you are really emanating a great deal of needless anger, and DrRocket's answer to the OP was appropriate given the improper posing of the question in the first place. Students are best when they are pushed to do things properly, and throwing things out like avg. empirical observation which doesn't apply is not at all helpful. Please ruin someone else's brain, I see really unusual replies to homework questions all the time that tout all sorts of inappropriate matter! note: it is unfortunate that DrRocket made a typo and seems to do so fairly often, mostly small letter mistakes; sucks don't it. -1 Link to comment Share on other sites More sharing options...
John Cuthber Posted February 19, 2012 Share Posted February 19, 2012 OK Xitten, if you think Dr Rocket's post saying that momentum helps here is correct then could you use momentum to answer the question? It seems Dr Rocket himself is ignoring the issue in the hope it will go away. Link to comment Share on other sites More sharing options...
Klaynos Posted February 19, 2012 Share Posted February 19, 2012 ! Moderator Note Wowsers...Can we all remember to be civil.I think there might have been a couple of misunderstandings and assumptions made along the way here that have spiralled a little. Taking a step back and a breather is probably to be advised.Please do not reply to this modnote. Link to comment Share on other sites More sharing options...
Xittenn Posted February 19, 2012 Share Posted February 19, 2012 (edited) The time average of the force (presumably we are talking about the contact force between the car and whatever it hits) is tied to momentum, not energy. Better use a little engineering judgment and go read a physics book. Conservation of momentum will be a helpful principle in this problem. Conservation of energy is not so helpful as there are a lot mechanisms for dissipation of energy -- including material failure in the vehicle, which is a significant design objective in modern vehicle design -- that one cannot estimate without detailed knowledge of the car design and the pramaters of the collision. To make use of energy considerations you are right back to those detailed non-linear mechanics models. OK Xitten, if you think Dr Rocket's post saying that momentum helps here is correct then could you use momentum to answer the question? It seems Dr Rocket himself is ignoring the issue in the hope it will go away. All I see is DrRocket stating that trying to conserve energy in one form is futile while momentum is conserved indefinitely. So . . . if there are any presuppositions made it would be in favour of calculations that involve momentum and not energy, where energy calculations would in fact not be helpful because energy could not easily be accounted for. Absorbing x joules is an indication of impulse and which isn't an accurate indication of what energy the initial system had. I'm not sure why you are pressing the issue of putting this in context of the OP and solving the equation. The relevant equations could be written out in full form demonstrating proper knowledge of first principles and would not require any algebra as is also the case of an academic situation (the algebra is irrelevant). Could you point directly at the problem that you are seeing? I suggest you unnegaffi the man, but this is only my suggestion :/ Edited February 19, 2012 by Xittenn Link to comment Share on other sites More sharing options...
John Cuthber Posted February 19, 2012 Share Posted February 19, 2012 "I'm not sure why you are pressing the issue of putting this in context of the OP" Because I have this weird idea that the OP asked the question because he wanted an answer. Dr Rockets's saying "Conservation of momentum will be a helpful principle in this problem." is, at best unhelpful, because the momentum is unknown and also is, as far as I can see, flat out wrong. I'm happy to see anyone actually show that the statement is true by using momentum as a helpful principle in this problem. But until then, the point remains that, since the problem is not open to solution, neither momentum nor energy nor slaughtering a goat will actually help. Link to comment Share on other sites More sharing options...
Xittenn Posted February 19, 2012 Share Posted February 19, 2012 "I'm not sure why you are pressing the issue of putting this in context of the OP" Because I have this weird idea that the OP asked the question because he wanted an answer. Dr Rockets's saying "Conservation of momentum will be a helpful principle in this problem." is, at best unhelpful, because the momentum is unknown and also is, as far as I can see, flat out wrong. I'm happy to see anyone actually show that the statement is true by using momentum as a helpful principle in this problem. But until then, the point remains that, since the problem is not open to solution, neither momentum nor energy nor slaughtering a goat will actually help. Thanks for modifying my statement for me, it is much appreciated! Are you a reporter? x Joules Energy Absorbed Just to be more clear you can evaluate the problem without introducing fictitious numbers, you simply will not be able to solve it. Link to comment Share on other sites More sharing options...
John Cuthber Posted February 19, 2012 Share Posted February 19, 2012 I didn't modify it. I cropped it but, if it makes you feel better... "I'm not sure why you are pressing the issue of putting this in context of the OP and solving the equation." Because I still have this weird idea that the OP asked the question because he wanted an answer. And Dr Rocket said that considering momentum would help and as far as I can tell, it won't. Anyway, to answer you question. No. reporters don't just paraphrase, they actually change the meaning. I'm pressing this because I think the OP wanted an equation and a solution. Incidentally re "Just to be more clear you can evaluate the problem without introducing fictitious numbers, you simply will not be able to solve it. " That seems odd given the meaning of the word evaluate "Mathematics . to ascertain the numerical value of (a function, relation, etc.)." from http://dictionary.reference.com/browse/evaluate Really this doesn't seem to be getting us anywhere. Is there any evidence to support DrRocket's assertion that "Conservation of momentum will be a helpful principle in this problem."? Unless there is, I don't see any point doing anything but accepting that he was simply wrong and carrying on. Link to comment Share on other sites More sharing options...
Xittenn Posted February 19, 2012 Share Posted February 19, 2012 (edited) That seems odd given the meaning of the word evaluate "Mathematics . to ascertain the numerical value of (a function, relation, etc.)." from http://dictionary.reference.com/browse/evaluate e·val·u·ate/iˈvalyo͞oˌāt/ Verb: 1) Form an idea of the amount, number, or value of; assess. 2) Find a numerical expression or equivalent for (an equation, formula, or function). well I bolded what I feel is appropriate to my usage, which I can because of the 'or'. I feel his evaluation of the problem was quite appropriate, if you wish to ignore the obvious, or feel that it is somehow a disservice to the op I understand and I have no quarrel or need to press the issue further. I feel it is obvious and am really not willing to explain what I consider obvious especially when the end result will be a complete waste of my time. There are obviously two sides if you feel the OP absolutely necessitates a solution--which is really weird because there wasn't even one single value given--so I don't understand the need to force a poster to comply with general silliness. I'm done with the thread and moving on, I have some 20 hours of biology studying to do. I'm sure if you ask DrRocket nicely and without condescension he will provide a well thought out derivation of how to evaluate a problem like this. Edited February 19, 2012 by Xittenn Link to comment Share on other sites More sharing options...
John Cuthber Posted February 19, 2012 Share Posted February 19, 2012 Enjoy your studies. I wonder if Dr Rocket will drop by and explain how momentum can help solve something that's not solvable. Link to comment Share on other sites More sharing options...
Xittenn Posted February 19, 2012 Share Posted February 19, 2012 Enjoy your studies. I wonder if Dr Rocket will drop by and explain how momentum can help solve something that's not solvable. I'm sorry I would like to modify my statement slightly because I wondered off and got caught up in the mess of thoughts. This is now a circular argument and evaluating it simply brings us back to the original arguments which baxtrom was making, which are still arguably incorrect. I think the point I was trying to make before I fell apart was, it is not what you say but how you say it and why you say it, and DrRocket was--as far as I can tell--trying his best to say things in the most appropriate manner, as opposed to simply spewing any old non-sense. Formalization makes this profession what it is, the ability to articulate the problem is of utmost concern. I really need to go do my work now! **whether or not both objects are moving is, and correct me if I am wrong here, irrelevant if the frame of reference is appropriately adjusted. Link to comment Share on other sites More sharing options...
DrRocket Posted February 19, 2012 Share Posted February 19, 2012 Enjoy your studies. I wonder if Dr Rocket will drop by and explain how momentum can help solve something that's not solvable. You continue with the straw man arguments. I stated rather clearly at the outset that more information is required in order to solve such a collision problem. I then stated that conservation of energy would be a helpful principle to apply in analyzing a collision problem. I further demonstrated how that prinicple applies, though you seem to not understand that point and conventiently ignore it as well. The time-average restraining force is trivially determined from momentum considerations, and that is about as close as one can come to "solving" the given problem. Momentum conservation is useful in such problems precisely because momentum is a vector quantity and there are relatively few mechanisms for change in momentum. An external force is required. Energy conservation, because energy is a scalar, is much less useful as there are a myriad of mechanisms available for energy dissipation, and non-conservative forces are involved in these mechanisms. Link to comment Share on other sites More sharing options...
swansont Posted February 19, 2012 Share Posted February 19, 2012 The problem is not solvable as-is. You would need to apply conservation of momentum to do so, were the appropriate information made available. Work-energy could be applied (especially if the interaction is elastic), again only if enough information was available. Link to comment Share on other sites More sharing options...
Klaynos Posted February 19, 2012 Share Posted February 19, 2012 ! Moderator Note I think we're done here. Op, if you find a more complete version of the question please start the thread again. Link to comment Share on other sites More sharing options...
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