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From: Remy McSwain on 25 Mar 2010 07:57 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4bab162f$0$16520$afc38c87(a)news.optusnet.com.au... > > "Remy McSwain" <Paradis70080(a)gmail.com> wrote in message > news:TvadnZCU-_yf9TrWnZ2dnUVZ_judnZ2d(a)giganews.com... >>> The "cause" of the collapse was the failure of the steel beams >>> supporting one floor (the floor of the crash). These were >>> weakened by fire to the extent that they could not support the >>> upper structure (as they were designed to do) and a catastrophic >>> pancake collapse ensued. >> >> And when it ensued, the upper block came crashing down upon the >> lower support structure. And then that support structure, which >> was designed to support the weight of that upper block, failed. >> So how did it come to be subjected to a greater force than mg? > > > The upper part of the building - the pancake - crashed into it. LOL! And what were the relevant forces that caused the structure to fail, Shake&Bake? Give us the most relevant couple of equations. >> Is it your claim that the catastrophic crash never imposed a >> force upon the lower support structure of greater than mg? Yes >> or no? >> > > The catastrophic crash caused force. Obviously. Really? This, by you, just yesterday..... "The forces that occur *during* a collision cannot be determined, and do not form part of the equations of motion for an inelastic collision. http://en.wikipedia.org/wiki/Inelastic_collision See any mention of forces? No? So why do you think the forces are an issue during an inelastic collision?" So now, let me ask you why YOU now think that forces are an issue DURING AN INELASTIC COLLISION? > In fact, in classical mechanics the force is infinite but last for > zero time. Een in an inelastic collision? Hmmmmm. (see above) > This is easily verified. > > The equations here: > http://en.wikipedia.org/wiki/Inelastic_collision give velocity as > a function of time. If you differentiate this wrt to time and > multiply by m you can re-formulate this as an equation for Force > as a function of time (as F = m (d/dt)V, and the equation for V is > given in terms of t). > > You may like to try this at home. > > You get an infinite force for zero time. Now its not really an > infinite force, and its not for zero time, as the materials deform > and hence the concept of V is not uniquely defined. The actual > function F(t) will depend upon the materials used, their shapes, > etc. However, you don't need to know any of this to solve the > equations of motion (ie produce the equations on the web page); it > simply does not use the function F(t) so the fact that it cannot > be determined doesn't matter. Again, you may note the absence of > force terms in the equations of motion. > > I could elaborate on this in several ways, Incorrectly, I'm sure. as it used often in physics. For > example, the model tacitly assumes a dirac delta function for the > force, and another interesting line of argument is looking at the > integral of the Force over time to measure momentum and over > distance to measure kinetic energy. But quite frankly, as you do > not seem to understand the basic physics of an inelastic > collision, trying to explain any of the more advanced physics is a > waste of my time. But if you do want to learn more, google these > words. > > >
From: Peter Webb on 25 Mar 2010 08:00 "Remy McSwain" <Paradis70080(a)gmail.com> wrote in message news:04adnQiXG-dz1TbWnZ2dnUVZ_h-dnZ2d(a)giganews.com... > > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message > news:4baaf062$0$19545$afc38c87(a)news.optusnet.com.au... >> >> "Remy McSwain" <Paradis70080(a)gmail.com> wrote in message >> news:8bSdndtHMvatFzfWnZ2dnUVZ_u2dnZ2d(a)giganews.com... >>> >>> "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message >>> news:4baa82e1$0$28464$afc38c87(a)news.optusnet.com.au... >>>> >>>> "Remy McSwain" <Paradis70080(a)gmail.com> wrote in message >>>> news:tKOdnbob24BtiDfWnZ2dnUVZ_oydnZ2d(a)giganews.com... >>> >>> [snipped a bunch of lies about not having previously told you where >>> you're wrong] >>> >>> >>>> I think there is an "upward, normal force deriving fom its interaction >>>> with the lower part of the building". >>>> >>>> I have already told you about 200 times this statement is "absolutely >>>> correct". >>> >>> LOL! Well, as subtle as that lie is, it's a clear indication that you >>> know you're wrong. THAT isn't the statement you keep claiming is >>> correct. >>> >>> THIS, AGAIN, is the statement which you're said, about 200 times, is >>> "absolutely correct": >>> >>> "The only two relevant forces acting on the falling block are >>> gravity (mg) and an upward normal force (N) due to its interaction >>> with the lower section of the building." >>> >>> Clearly, the N to which he refers is the mg of the upper section. >> >> >> Clearly not. He says there are two forces; if they were the same force >> there would only be one force. Furtermore, gravity (mg in your >> terminology) is a downward force, and he says that N is an upward force. >> He could not possibly be talking about gravitational force (mg). >> >> >> >>> And just as clearly, he's claiming that it's DURING THE COLLISION. >> >> >> No, he actually says when it is "falling", and doesn't even mention >> collision in the quote you gave. > > Go look at it again, KOOK. He says DUE TO ITS INTERACTION WITH THE LOWER > SECTION OF THE BUILDING. That interaction, and the normal upward force of > the lower section that you went to great lengths to describe, could not > possibly happen except DURING THE COLLISION. I would disagree with that. Whilst the major "reaction" is from the floors slamming together, collapsing the outer curtain wall of the building would offer a (minor) reactive force during the fall between floors. And if you are going to put something in caps like that, or even if you are not, its only one sentence, you could actually quote it verbatim. > For you to keep claiming that he's talking about some time period before > the collision would mean that the lower section was supplying NO upward > force to it, and so it would be wrong on that basis as well. > I'm not really sure exactly which time period they mean, but it doesn't really matter; as I said it is true under either interpretation. > Keep trying, KOOK, but either way you look at it, the statement is WRONG, > and so are you. > How is it wrong? What is the additional force? (BTW momentum is not a force) > >> Not that it matters, the statement is still absolutely correct. > > So it doesn't matter if it's the time period DURING THE COLLISION? Then > he's wrong on that basis because he ignored the force of the crash DURING > THE COLLISION. > Hoe is the crash not "an upward normal force (N) due to its interaction with the lower section of the building" to use their exact words? > So now you're proven that you're doubly wrong. > >>> Therefore, taken as a whole, that statement is wrong. DURING THE >>> COLLISION, the relevant force is F=(d/dt)mv (up to the point of failure >>> of the support structure). >> >> >> That is always true, and not just for falling buildings. It is the >> definition of force. > > And, whereas you JUST said that it doesn't matter if his claim means > DURING THE COLLISION, then you've just admitted he's wrong to have ignored > the forces that were at play DURING THE COLLISION! > > Huh?
From: Peter Webb on 25 Mar 2010 08:11 >> >> The formulas show the changes of speed that result from inelastic >> collisions. Which is what you need to work out how long the building will >> take to fall. > > Sakin' & Bakin again! LOL! Go waaaaay up to the top of this post. Did > you see it? Did you see what the subject of this particular exchange is? > AY challenges you to give YOUR opinion of what YOU think were the forces > that were actiing upon the the upper block and the lower support > structure, DURING THE COLLISION. > > Did you catch that. Since then, all you've done is refer to the equations > that were developed in order to determin the velocities of the masses > AFTER THE COLLISION. > The equations completely specify the position of the pancake at every time, including the exact instant of collision. > I don't know if it's ADD, or excessive pride, or just plain stupidity that > causes you to be all over the place, and I really don't care. But the > fact remains that for whatever reason, most of your responses have nothing > to do with whatever issue is at hand. > Gee, mostly I just say "If you think that I have said something which is wrong, please quote me exactly" and try and explain the equations of motion for an inelastic collision that appear on about 1,000 websites. > The question, this time, was your opinion of the forces YOU think were > acting upon the upper block, DURING THE COLLISION. My opinion? They were very large. And impossible to calculate. ASnd completely irrelevant to the equations of motion of an inelastic collision, which is why no web pages on this subject mention them. > As you try to either answer, or deflect from answering in your response, > keep in mind that just today, you said that the statement by the other > KOOK that the only two forces that he said were relevant to that upper > block would be correct even it he was referring to the time DURING THE > COLLSISION. > Yes, it is correct. Unless you can identify a third, relevant force. (Momentum is not a force). >>> As I've been explaining to you all along, your confusion stems from the >>> fact that because this was an inelastic collision, and because you found >>> those equations on a web site entitled "inelastic collisions", you've >>> mistakenly concluded that those equations apply to a consideration of >>> the dynamics of what happens DURING A COLLISION. >> >> No, I derived my equations from first principles using conservation of >> momentum. I didn't and don't need to refer to a web page for this stuff, >> its really basic physics. > > OK, then. I accept your response that you picked out the wrong equations > all by yourself! LOL! > The same ones as on the web pages. Which completely determine the position of the objects at all times. Which is what you need if you want to calculate the time to the bootom. >>> But they do not. As that site clearly indicates, those equations are >>> only for the purposes of determining the velocities of the masses >>> involved in a collision, but only AFTER the collision, and not DURING >>> THE COLLISION, when accelerations and forces are most definitely at >>> work. >> >> It gives before and after velocities. Exactly what you need. > > For what? Please go there. PLEASE, PLEASE, PLEASE go there. "you need" > for what DURING THE COLLISION? > You don't need to know anything at all about what happens during the collision, other than it is inelastic, it doesn't affect the equations, as you can see by looking at them. What they do - and I did - was to consider this using conservation of momentum, and not force, which lietarlly doesn't enter the equations. > > >>> Just admit your mistake, and be done with it. >>> >> >> If you think I have said something incorrect, you should quote my exact >> words. > > All of the above by you as it pertains to what forces you think were > relevant to what was going on DURING THE COLLISION. And the third force is .... > Quiet in here!
From: Remy McSwain on 25 Mar 2010 08:13 "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message news:4bab20e8$0$16520$afc38c87(a)news.optusnet.com.au... > > "Remy McSwain" <Paradis70080(a)gmail.com> wrote in message > news:trWdnVaV0eV1-zrWnZ2dnUVZ_gGdnZ2d(a)giganews.com... >> >> "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in >> message news:4ba759ff$0$6090$afc38c87(a)news.optusnet.com.au... >>>>> But what do you believe to be true? What is your theory? >>>> >>>> F=ma. F=mg (weight) >>>> What is YOUR theory for the increase in the value of 'a' beyond >>>> 'g' DURING THE CRASH? >>>> >>> >>> Let me get this straight. You think that the buildings fell >>> faster than gravity? >> >> For you to come to that conclusion from what I said shows your >> complete lack of understanding of physics. In fact, it's exactly >> why you're so confused about how to analyze what it was that >> imposed more of a force in the support structure than it was >> designed to withstand. >> > > But isn't "g" the acceleration due to gravity in your equation? > > And if a>g, doesn't that by definition mean that it is falling > faster than gravity? I'm not at all surprised that someone like yiou who knows nothing about physics would come to that conclusion. But, as usual, you're looking at it all wrong. Yes, of course a>g. But, in the case of a collision, the a is the (d/dt)v of the mass. IOW, it's the deceleration of the mass which is crashing upon the lower structure. And so yes, I'm sure that there were slices of time where a was much, much greater than g. I've tried to teach you this very many times now, and you still don't get it. You seem like you should be smart enough to do so, but yet, you just can't. <sigh> >> So answer the question.... Given that F=ma (which, in the case >> of weight, F=mg because a=g), and that the mass of the upper >> block didn't change appreciably, and that the structure was >> designed to support mg, then that would mean that for the >> structure to fail, the F being imposed upon the support structure >> had to be greater than mg. So that means that the 'a' DURING THE >> CRASH, had to be greater than 'a', right? > > If your question is whether the acceleration changes "during the > crash", yes it does. In fact at the moment of collision (ideally, > instantaneous) the acceleration is upwards and infinite. In > practice it is upwards and huge. Yes, upwards. In the opposite > direction to the motion itself. So now that you've agreed with me on that, how would you express it in mathematical terms? Keep in mind that just a couple of posts ago, you said that infamous statement that the only forces acting on the upper body as it fell onto the support structure would be mg, and the support structure's reaction to it, even if that meant (which it must) DURING THE CRASH! >> If not, then tell us of F increased beyond mg. If so, then how >> did 'a' come to be greater than 'g' DURING THE CRASH? LOL! Silence? Go back and look. I just gave you the answer. No wonder you're so confused. You really don't know what the hell you're talking about.
From: Peter Webb on 25 Mar 2010 08:26
"Remy McSwain" <Paradis70080(a)gmail.com> wrote in message news:RcGdnZOtmJD-zjbWnZ2dnUVZ_oCdnZ2d(a)giganews.com... > > "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message > news:4bab20e8$0$16520$afc38c87(a)news.optusnet.com.au... >> >> "Remy McSwain" <Paradis70080(a)gmail.com> wrote in message >> news:trWdnVaV0eV1-zrWnZ2dnUVZ_gGdnZ2d(a)giganews.com... >>> >>> "Peter Webb" <webbfamily(a)DIESPAMDIEoptusnet.com.au> wrote in message >>> news:4ba759ff$0$6090$afc38c87(a)news.optusnet.com.au... >>>>>> But what do you believe to be true? What is your theory? >>>>> >>>>> F=ma. F=mg (weight) >>>>> What is YOUR theory for the increase in the value of 'a' beyond 'g' >>>>> DURING THE CRASH? >>>>> >>>> >>>> Let me get this straight. You think that the buildings fell faster than >>>> gravity? >>> >>> For you to come to that conclusion from what I said shows your complete >>> lack of understanding of physics. In fact, it's exactly why you're so >>> confused about how to analyze what it was that imposed more of a force >>> in the support structure than it was designed to withstand. >>> >> >> But isn't "g" the acceleration due to gravity in your equation? >> >> And if a>g, doesn't that by definition mean that it is falling faster >> than gravity? > > I'm not at all surprised that someone like yiou who knows nothing about > physics would come to that conclusion. But, as usual, you're looking at > it all wrong. Yes, of course a>g. But, in the case of a collision, the a > is the (d/dt)v of the mass. IOW, it's the deceleration of the mass which > is crashing upon the lower structure. And so yes, I'm sure that there were > slices of time where a was much, much greater than g. > So you are agreeing I am correct? > I've tried to teach you this very many times now, and you still don't get > it. You seem like you should be smart enough to do so, but yet, you just > can't. <sigh> > No teaching required. Learned all this stuff in Physics 1. Known it for 30 years. Never denied it was true. >>> So answer the question.... Given that F=ma (which, in the case of >>> weight, F=mg because a=g), and that the mass of the upper block didn't >>> change appreciably, and that the structure was designed to support mg, >>> then that would mean that for the structure to fail, the F being imposed >>> upon the support structure had to be greater than mg. So that means >>> that the 'a' DURING THE CRASH, had to be greater than 'a', right? >> >> If your question is whether the acceleration changes "during the crash", >> yes it does. In fact at the moment of collision (ideally, instantaneous) >> the acceleration is upwards and infinite. In practice it is upwards and >> huge. Yes, upwards. In the opposite direction to the motion itself. > > So now that you've agreed with me on that, how would you express it in > mathematical terms? As a dirac delta function: http://en.wikipedia.org/wiki/Dirac_delta_function (or, more accurately, as a constant times the delta function) > Keep in mind that just a couple of posts ago, you said that infamous > statement that the only forces acting on the upper body as it fell onto > the support structure would be mg, and the support structure's reaction to > it, even if that meant (which it must) DURING THE CRASH! > > If you think I said something which is incorrect, you should quote it exactly. I don't think I have. >>> If not, then tell us of F increased beyond mg. If so, then how did 'a' >>> come to be greater than 'g' DURING THE CRASH? > > LOL! Silence? Go back and look. I just gave you the answer. No wonder > you're so confused. You really don't know what the hell you're talking > about. > I don't think there could possibly be any dispute about the answer, and hence it was a rhetorical question. There was a massive acceleration of the pancake upwards on every collision. |