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Natural Science Forum / Physics / General Physics / May 2007"Line of action of a force"
One of the problems from the Feynman Lectures asks us to show that three forces in static equilibrium must all be coplanar, and their lines of action must pass through a single point. If someone were to ask me what is meant by "line of action of a force", I would probably say the line parallel to the force and passing through the point of application. By that definition, the proposition of the problem is wrong. Here's someone else's definition: http://www.arch.virginia.edu/~km6e/references/glossary/struc-glossary.html#L_glossary "Line of Action: The line of action of a force is the infinite line defined by extending along the direction of the force from the point where the force acts." Take a long narrow board and apply three forces to it. let the forces be in the same plane, and let two of the forces act in the same direction perpendicular to the long axis of the board. Apply a third force in the center of the board in the opposite direction. That will achieve static equilibrium, and not all of the lines of action pass through the same point. Is there another definition of "line of action of a force"?  Signaturehttp://www.vho.org/GB/c/DC/gcgvcole.html http://www.vho.org/GB/Books/dth/ http://www.germarrudolf.com/ http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm >One of the problems from the Feynman Lectures asks us to show that three >forces in static equilibrium must all be coplanar, and their lines of [quoted text clipped - 18 lines] >equilibrium, and not all of the lines of action pass through the same >point. What a nonsense. You have no clue. (How to find the result of two parallel forces). >Is there another definition of "line of action of a force"? > >http://www.vho.org/GB/c/DC/gcgvcole.html >http://www.vho.org/GB/Books/dth/ >http://www.germarrudolf.com/ >http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm NAZI sh.t. hanson will like it, though :-) w. >>One of the problems from the Feynman Lectures asks us to show that three >>forces in static equilibrium must all be coplanar, and their lines of [quoted text clipped - 22 lines] > You have no clue. > (How to find the result of two parallel forces). Well, since I neglected to specify the magnitude of the third force, you might try to argue that I did not provide sufficient information. But, considering that the geometry of the problem determines the necessary magnitude, it was clearly implied. The board is of length L aligned along the X axis with its center at the origin. The force F1 acts in the direction of the negative Y axis, and is of magnitude f. It is applied at the end of the board located at {-L/2,0}. Another force F2 described by the same vector, F1=F2={0,-f} acts at {L/2,0}. A third force F3 is applied at the origin and has components F3={0 , 2f}. We therefore have the vector equation: F1+F2+F3=0. Since the forces do not lie along the same line of action, we must also take into account the torque, which is F1 (-L/2) + F2 L/2 + 0 F3 = f (L/2-L/2) = 0. It follows that the system is in static equilibrium. >>Is there another definition of "line of action of a force"? >> [quoted text clipped - 5 lines] > NAZI sh.t. > hanson will like it, though :-) No. Germar Rudolf is not a Nazi, nor is David Cole. Of course that might be a bit difficult to grasp for a person who can't figure out that 1+1-2=0. Signaturehttp://www.vho.org/GB/c/DC/gcgvcole.html http://www.vho.org/GB/Books/dth/ http://www.germarrudolf.com/ http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm >>>One of the problems from the Feynman Lectures asks us to show that three >>>forces in static equilibrium must all be coplanar, and their lines of [quoted text clipped - 37 lines] >account the torque, which is F1 (-L/2) + F2 L/2 + 0 F3 = f (L/2-L/2) = 0. >It follows that the system is in static equilibrium. Sounds a little better than your first posting. But what makes you think you discovered something exciting? It is the most basic stuff, from the very first minutes in any static-mechanics lesson. What is your sphere of competence? Raising sheep? >>>Is there another definition of "line of action of a force"? >>> [quoted text clipped - 8 lines] >No. Germar Rudolf is not a Nazi, nor is David Cole. Of course that might >be a bit difficult to grasp for a person who can't figure out that 1+1-2=0. Are you sure? Holocaust deniers are not NAZI idiots? Show me one, who is not -at least - an idiot, Of course that might >be a bit difficult to grasp for a person who can't figure out that 1+1-2=0. figure out NAZI + Idiot - brains = NAZI Idiot, with ZERO brains. w. ahahaha.. Helmut, are you still cranking yourself because of this: http://groups.google.com/group/sci.physics/msg/8414caa0fc349965 wherein you posted to the world that you can "live with being an idiot" [hetware] >>>>Is there another definition of "line of action of a force"? >>>>http://www.vho.org/GB/c/DC/gcgvcole.html >>>>http://www.vho.org/GB/Books/dth/ >>>>http://www.germarrudolf.com/ >>>>http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm [Wabnigger to hetware] >>> NAZI sh.t. >>> hanson will like it, though :-) [hetware to wabigger] >>No. Germar Rudolf is not a Nazi, nor is David Cole. Of course >>that might be a bit difficult to grasp for a person who can't figure >> out that 1+1-2=0. [Wabnigger to hetware] > Are you sure? Holocaust deniers are not NAZI idiots? > Show me one, who is not -at least - an idiot. Of course that might >>be a bit difficult to grasp for a person who can't figure out that >>1+1-2=0. > figure out NAZI + Idiot - brains = NAZI Idiot, with ZERO brains. > w. [hanson] Here are the precise explanations for your dilemma, Helmut: == Austria's Wabnig was fingering Jews to the Nazis, in here: http://groups.google.com/group/sci.physics/msg/9d6540b93d4f7a54 == Wabnig says he knew Nazi Generals, in here: http://groups.google.com/group/sci.physics/msg/5c71e19727e0ec9a == Helmut Wabnig's deepest obsession:his hate on Jews, in here: http://groups.google.com/group/sci.physics/msg/be8452e683364a49 Now then Wabi, can you still learn to live WITHOUT being an idiot? Thanks for the laughs > One of the problems from the Feynman Lectures asks us to show that three > forces in static equilibrium must all be coplanar, and their lines of [quoted text clipped - 20 lines] > > Is there another definition of "line of action of a force"? The farther away the point in common, the more parallel the lines are. How far away is the point in common of parallel lines? So, you don't need another definition for "line of action of a force". You need to use Feynman's implicit definition for "passing through a single point", namely including parallel lines, passing through a point at infinity. Dirk Vdm >> One of the problems from the Feynman Lectures asks us to show that three >> forces in static equilibrium must all be coplanar, and their lines of [quoted text clipped - 6 lines] >> >> Here's someone else's definition: http://www.arch.virginia.edu/~km6e/references/glossary/struc-glossary.html#L_glossary >> "Line of Action: The line of action of a force is the infinite line >> defined by extending along the direction of the force from the point [quoted text clipped - 18 lines] > > Dirk Vdm I don't know if you are having me on or not. That is certainly bending the rules of Euclidian geometry. OTOH, these guys assigned the Kepler Conjecture as a problem in the first lecture's homework - without bothering to explain that this problem had remained unsolved for 350 years. I would not put it past them to have expected that answer. BTW, I don't believe Feynman actually created the exercises. The names listed are Leighton and Vogt. Signaturehttp://www.vho.org/GB/c/DC/gcgvcole.html http://www.vho.org/GB/Books/dth/ http://www.germarrudolf.com/ http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm >>> One of the problems from the Feynman Lectures asks us to show that three >>> forces in static equilibrium must all be coplanar, and their lines of [quoted text clipped - 33 lines] > > I don't know if you are having me on or not. Perhaps you are the one having everyone on. > That is certainly bending the > rules of Euclidian geometry. In that context, projective geometry was created to save time. > OTOH, these guys assigned the Kepler > Conjecture as a problem in the first lecture's homework - without bothering [quoted text clipped - 3 lines] > BTW, I don't believe Feynman actually created the exercises. The names > listed are Leighton and Vogt. In that case you need to use their implicit definition for "passing through a single point", namely including parallel lines, passing through a point at infinity. It's what we call "shorthand". In Dutch we have a saying that goes like "For a good understander half a word is sufficient." BTW. Physics not part of mathematics. Dirk Vdm >>> The farther away the point in common, the more parallel >>> the lines are. How far away is the point in common of [quoted text clipped - 9 lines] > > Perhaps you are the one having everyone on. No. I'm pretty much what I present myself to be, and I try to be honest in what I post. When I posted the question, I was having second thoughts because I thought that it was simply a trivial error in the text. I _may_ have been exposed to the explanation of it being a degenerate case years ago, but it certainly was not clear when I saw it yesterday. Now I'm glad I asked. I learned something from it, and appreciate the feedback. Thank you. >> That is certainly bending the >> rules of Euclidian geometry. > > In that context, projective geometry was created to save time. Unless it keeps equations from null dividing when crunching numbers, or something like that, I don't see where it's more than a minor philosophical debate. >> OTOH, these guys assigned the Kepler >> Conjecture as a problem in the first lecture's homework - without [quoted text clipped - 10 lines] > we have a saying that goes like "For a good understander half > a word is sufficient." There is something to be said for having a reason to even want such an answer. > BTW. Physics not part of mathematics. If you mean to suggest that doing physics and doing "pure" mathematics are distinct endeavors, I agree. OTOH, not understanding the assumptions underlying the math one uses in physics can be as source of misunderstanding and error. Often, I find myself examining math _more_ closely when applying it to physics than when I'm just going through some formal development. Getting some final result from a sequence of steps without stopping to ponder what each step means has its merits in some circumstances, but so does understanding what all the intermediate stages really mean. Signaturehttp://www.vho.org/GB/c/DC/gcgvcole.html http://www.vho.org/GB/Books/dth/ http://www.germarrudolf.com/ http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm >>>> The farther away the point in common, the more parallel >>>> the lines are. How far away is the point in common of [quoted text clipped - 27 lines] > something like that, I don't see where it's more than a minor philosophical > debate. I assumed you were familiar with the primary concept of projective geometry, and its difference with Euclidean geometry. >>> OTOH, these guys assigned the Kepler >>> Conjecture as a problem in the first lecture's homework - without [quoted text clipped - 25 lines] > circumstances, but so does understanding what all the intermediate stages > really mean. ... and I still don't know whether you are satisfied with the answers given to you. Do you still think that "the proposition of the problem is wrong" ? Can you answer this with a yes or no? Dirk Vdm >>>>> The farther away the point in common, the more parallel >>>>> the lines are. How far away is the point in common of [quoted text clipped - 32 lines] > I assumed you were familiar with the primary concept of projective > geometry, and its difference with Euclidean geometry. Not in any formal sense. I cannot speak authoritatively of what is taught in secondary school in the US, but my impression is that projective geometry is not typically covered. When I think of projective geometry, I think of projection transformations used to render a 3D model onto the 2D screen in computer graphics. >> If you mean to suggest that doing physics and doing "pure" mathematics >> are [quoted text clipped - 11 lines] > Do you still think that "the proposition of the problem is wrong" ? > Can you answer this with a yes or no? No, regarding the second question. Regarding the first, and more interesting question, I'm not sure whether the idea that the lines of action of these parallel forces meet at infinity is useful, and it may actually be misleading. I'm thinking in terms of angular momentum. I have long puzzled over the distinction of angular momentum vs. linear momentum. Suppose I take the front forks of two bicycles and weld them top-to-top so that both wheels rotate in the same plane. That is, the axles are parallel. Assume the bearings are frictionless, so that the entire assembly could be suspended on a shaft which is parallel to the axles and rotated about the shaft without the wheels rotating about their axles. The experiment is to be conducted in a Lorentz frame. Also neglect the mass of the forks. If the wheels are suddenly and simultaneously released from the forks, they will head off in opposite directions tangent to the arc to which they were previously constrained. If we examine either of the wheels from the perspective if its center of mass coordinates, it is not rotating, and thus has no angular momentum. Neglecting the axles, there is no rigid body in this system which is rotating. Nonetheless, the original system clearly had angular momentum. What we see is that momentum is not fully specified by a vector. We must also specify its "line of action". In order that the angular momentum of a system be correctly represented, the perpendicular distance between these lines of action must remain constant. Perceiving them to somehow merge at infinity seems to undermine the very notion of angular momentum. Since force is the time derivative of linear momentum, and torque is the time derivative of angular momentum, it seems that the preceeding development applies to force as well. Signaturehttp://www.vho.org/GB/c/DC/gcgvcole.html http://www.vho.org/GB/Books/dth/ http://www.germarrudolf.com/ http://www.ice.gov/pi/news/newsreleases/articles/051115chicago.htm >>>>>> The farther away the point in common, the more parallel >>>>>> the lines are. How far away is the point in common of [quoted text clipped - 38 lines] > think of projection transformations used to render a 3D model onto the 2D > screen in computer graphics. No wonder that my remark didn't ring a bell. Projective geometry has no metric. Every pair of lines in a plane have 1 point in common. Parallel lines don't exist but have point in common "at infinity". Directions are in a sense "points at infinity". So in this context projective geometry was created to save time. >>> If you mean to suggest that doing physics and doing "pure" mathematics >>> are [quoted text clipped - 17 lines] > idea that the lines of action of these parallel forces meet at infinity is > useful, Of course it is useful. It saves a bit of time - but only for good understanders. [snipping part for which I lack the patience to look at] Dirk Vdm > One of the problems from the Feynman Lectures asks us to show that three > forces in static equilibrium must all be coplanar, and their lines of [quoted text clipped - 18 lines] > equilibrium, and not all of the lines of action pass through the same > point. It's the limiting case as the point of intersection of the lines of action (PILA) recedes to infinity. Try it without the forces being parallel. Keep it symmetric, so you have F1x = F2x; F1y = -F2y; F3x = -2F1x; F3y = 0; F3 applied at (0,0). Write the formula for the PILA, take the limit as F1y -> 0. Does the PILA require a new definition to take this into account? This may depend on whether one is inclined towards being a mathematician, or being a physicist. SignatureTimo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html >Take a long narrow board and apply three forces to it. let the forces be in >the same plane, and let two of the forces act in the same direction >perpendicular to the long axis of the board. Apply a third force in the >center of the board in the opposite direction. That will achieve static >equilibrium, and not all of the lines of action pass through the same >point. The case where the three forces are parallel is a degenerate one. You can either take it as an exception or consider that the three forces act through a common point at infinity. The latter may seem like cheating, but considered as a limiting case it is reasonable. -- Richard Signature"Consideration shall be given to the need for as many as 32 characters in some alphabets" - X3.4, 1963. |
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