hobz Posted July 24, 2008 Share Posted July 24, 2008 Assuming equally symmetric riders of the same density the heavy guy wins. The drag coefficient is probably the same or very close and the volume of the riders is proportional to their weight. The frontal area of the heavier guy is, however, proportionally less. Advantage heavy guy! Normally I would agree, but your statement incorporates that the area of the heavier guy is proportionally less with respect to his weight. However, the weight was "canceled out" as the math explains. Link to comment Share on other sites More sharing options...
DrP Posted July 24, 2008 Share Posted July 24, 2008 Normally I would agree, but your statement incorporates that the area of the heavier guy is proportionally less with respect to his weight. However, the weight was "canceled out" as the math explains. The weight was cancelled out when working out the accelaration due to gravity, NOT when working out the force he exerts against friction and drag. Link to comment Share on other sites More sharing options...
Pete Posted July 24, 2008 Share Posted July 24, 2008 The concept of drag has been introduced and that makes the problem more difficult. Two people can have the exact same drag coefficient and yet have more mass. Anyone who has learned about health and fitness knows that one can loose fat and yet not loose weight since its feasable to work out in a way such that you become stronger in the process and loose fat but gain muscle. It is therefore completely feasible to have two people with the same drag coefficient and yet have different masses. Tell your friend that while the force on a heavier person is greater than a lkight person one mustn't forget that the heavier person will be harder to accelerate (i.e. has more inertia) that the lighter person. As someone mentioned above, these two effects cancel. This is a fact and has been experimentally confirmed. Tell your friend that the laws of physics disagree with him. Does your friend have a reason for his assumption?? Consider a body in free fall in a uniform gravitational field (the gravitational field near the surface of the earth can be considered highly uniform if one confines the height to reaonable values (such as a few miles). The gravitational force on his body is given by F = mg where g is the acceleration due to gravity. The force on a body is given by F = dp/dt where p = mv. If the mass, m, is constant then F = ma where a = dv/dt = acceleration. Therefore F = mg = ma The mass cancels to give a = g Therefore the gravitational acceleration is independant of the mass of a body. The drag force is also indepentant of the mass of the body so that a person's weight does not contribute to the time it takes to roll down a hill. If your friend is really interested in this then he can pick up an introductoryu book on physics. However if you'd like I can create a web page to explain these facts. Just let me know. I hope this helped. It probably didn't contribute much because of the very good explainations given above. Let me say Well done! for your efforts here folks. Pete Link to comment Share on other sites More sharing options...
noob Posted July 24, 2008 Share Posted July 24, 2008 Hi, the small guy with 50 kg would win the race! If we talk about such exercises it is very important to look on the undergound. In an ideal case the mass is completly unimportant. The mass fall out and so it don't cares. 100 kg, 150 kg, 2000kg, no matter. But it is not ideal. We have friction and becaus of that the friction force is directed against the motion. If you look now to wintersports, the mass could be useful, because it melts the ice and snow and the friction coefficient becomes better, but on a street is more mass not good. [math]\vec{F}_{r} = \mu \vec{F}_{n} = \mu m \vec{g} \cos(\alpha)[/math] This trem slows the fatter guy down and the surface of the street wouldn't melt for a better friction coeffizient. greetings Link to comment Share on other sites More sharing options...
Pete Posted July 25, 2008 Share Posted July 25, 2008 Hi, the small guy with 50 kg would win the race! I disagree. If we talk about such exercises it is very important to look on the undergound. In an ideal case the mass is completly unimportant. Not if the person asking the question wishes to ignore things like friction and drag. The mass fall out and so it don't cares. 100 kg, 150 kg, 2000kg, no matter. But it is not ideal. We have friction and becaus of that the friction force is directed against the motion. If you look now to wintersports, the mass could be useful, because it melts the ice and snow and the friction coefficient becomes better, but on a street is more mass not good. No. The acceleration of a body is dependant on the coeffiction of kinetic friction but is independant of mass. [math]\vec{F}_{r} = \mu \vec{F}_{n} = \mu m \vec{g} \cos(\alpha)[/math] What is [math]\vec{F}_{r}[/math]? I.e. what does the "r" subscript denote? If this is simply the force on the body then once you replace it with ma you will see that the mass drops out yet once again! This trem .. What is a "trem"? ..slows the fatter guy down .. I disagree. Consider a body sliding on a surface under friction for which the coefficient of kinetic friction is [math]\mu_k[/math]. Then [math]F = ma = \mu_k F_n = \mu_k (mg) [/math] Cancel out m to yield [math]a = \mu_k g [/math] Thus the acceleration of a body sliding on a rough surface is a function of [math]\mu_k[/math] but not a function of m. ..and the surface of the street wouldn't melt for a better friction coeffizient. Huh? What does that mean? Pete Link to comment Share on other sites More sharing options...
noob Posted July 25, 2008 Share Posted July 25, 2008 Hi Pete, Not if the person asking the question wishes to ignore things like friction and drag. If the person wants to ignore the friction, the mass cancels out! This follwos directly of the conservation of energy. It is correct because the system is conservative. The dissipative friction force is zero. Without friction the two riders are simultaniously in the finish. If you have friction the structure of the surface is important to say something about the situation. No. The acceleration of a body is dependant on the coeffiction of kinetic friction but is independant of mass. think about this sentence. It is completely nonsense What is [math]\vec{F}_{r}[/math]? I.e. what does the "r" subscript denote? If this is simply the force on the body then once you replace it with ma you will see that the mass drops out yet once again! Oh, sorry. It thought in German. We have not the same indecis in German. The "r" stands for "Reibung" which is the German word for friction. When we talk with Newtons arguments, this is the friction force. Such things like energydissipation are only useful, if we want to build Euler-Lagrange or Hamilton functions what the creator of the thread probably doesn't want. What is a "trem"? This is a typo I meant "term" I disagree. Consider a body sliding on a surface under friction for which the coefficient of kinetic friction is [math]\mu_k[/math]. Then [math]F = ma = \mu_k F_n = \mu_k (mg) [/math] Cancel out m to yield [math]a = \mu_k g [/math] This is also completely nonsense! What you wrote is a coloumb assumption, but we are talking about sliding. Huh? What does that mean? It means that sliding in a bobsled run is better than sliding on a street. This is the fundamental argument of dissipative motion on a real surface like the street or the bobsled run. greetings Link to comment Share on other sites More sharing options...
Mr Skeptic Posted July 25, 2008 Share Posted July 25, 2008 If the person wants to ignore the friction, the mass cancels out! This follwos directly of the conservation of energy. It is correct because the system is conservative. The dissipative friction force is zero. Without friction the two riders are simultaniously in the finish. If you have friction the structure of the surface is important to say something about the situation. Even if you ignore friction, the heavier guy will win the bike race. Unless you replace the bike with a sled, the bike wheel has to rotate as well as translate. Then a lesser portion of the energy of the heavy guy will be wasted in rotating the bike wheel, so that the heavier guy will win anyways. That effect will be miniscule compared to the lower deceleration due to drag that the heavier guy will have. Link to comment Share on other sites More sharing options...
Pete Posted July 25, 2008 Share Posted July 25, 2008 If the person wants to ignore the friction, the mass cancels out! Even when a person does not ignore friction the mass still cancels out. think about this sentence. It is completely nonsense On the contrary. This is basic physics (physics 101) and what I said is 100% accurate. One merely has to read a basic physics text to learn this fact. This is also completely nonsense! What you wrote is a coloumb assumption, but we are talking about sliding. If I were you I'd restrain from using the term "nonsense". You have used it here to reject the most basic facts of physics so far. What I wrote has absolutely nothing to do with the Coulomb force. The force on a particle is defined as the time rate of change of the particle's momentum, i.e. F = dp/dt where p = mv. If the mass is constant then the force equals F = ma. The force of friction [math]F_f[/math] on a moving body is given ny the relation [math]F_f = \mu_k F_n[/math] where [math]F_n[/math] is the normal force acting on the body. In the case of the gravitational force [math]F_n = w(weight) = mg[/math]. Plug these into the above expression to obtain [math]F = ma = F_f = \mu_k F_n = \mu_k mg[/math] which is identical to what I showed you above and as I also said, the mass cancels out. Since you claim this is wrong then please provide a derivation for explicit expression for the acceleration of a body which is sliding on a rough surface. I am stating that the expression so derived will be independant of the mass of the body. Prove me wrong. That I'm correct can easily be seen from the fact that the force of kinetic friction is dependant only of two materials (the body which is sliding and the material on which it is sliding) which results in [math]\mu_k[/math]. One can see this by considering two identical bodies sliding side by side. Each body will accelerate at the same rate. Now tape/glue them together. The fact that they are glued together will not change the rate at which they accelerate. But now we have a body whose mass has doubled. Since the force of friction is not dependant on the area of the surface of the body which is in contact with the surface on which it is sliding then the acceleration will not change. Pete Link to comment Share on other sites More sharing options...
noob Posted July 28, 2008 Share Posted July 28, 2008 Even if you ignore friction, the heavier guy will win the bike race. Unless you replace the bike with a sled, the bike wheel has to rotate as well as translate. Then a lesser portion of the energy of the heavy guy will be wasted in rotating the bike wheel, so that the heavier guy will win anyways. No, think about the dynamic of a tire. You haven't got a rotation and no raotaion energy because of no friction. Only potential energy which transforms to kinetic energy and the mass cancels out. @Pete: I guess you missunderstood me. Tell me what is the coloumb force in your oppinion? greetings Link to comment Share on other sites More sharing options...
CaptainPanic Posted July 28, 2008 Share Posted July 28, 2008 Air resistance force goes up with a second order of the size of someone. (length x width) - frontal surface area is expressed in m2. Mass goes up with a third order of the size (it's volume) and is expressed in m3. The gravity force will accelerate both equally. The drag force however will be relatively smaller for the big guy. I don't know how the friction in wheels and bearings will go up with extra weight, but I do know that on a bike you feel more resistance from air than from any other friction... so that can all be ignored. Link to comment Share on other sites More sharing options...
Sisyphus Posted July 28, 2008 Share Posted July 28, 2008 Indeed. Surface friction will, of course, affect the heavier rider more, but surface friction in this case is just in the axles, which all experience and common sense say will be far less of a factor than air resistance, which in any likely scenario (i.e., we're assuming they're both physically normal humans and neither one is dragging a parachute, etc.) will clearly favor the heavier rider. Link to comment Share on other sites More sharing options...
Pete Posted July 28, 2008 Share Posted July 28, 2008 @Pete: I guess you missunderstood me. Tell me what is the coloumb force in your oppinion? My opinion? The definition of the term Coulomb force has rarely, if ever, be questioned. Its found in all textbooks on electrodynamics. The mathematical expression for it is given in Eq. (1) at http://scienceworld.wolfram.com/physics/CoulombForce.html Pete Link to comment Share on other sites More sharing options...
Mr Skeptic Posted July 29, 2008 Share Posted July 29, 2008 No, think about the dynamic of a tire. You haven't got a rotation and no raotaion energy because of no friction. No, I did address that in my post. If you don't have the tire spinning, then you have a sled with wheels rather than a bicycle, as I said. And this problem is about bicycles, not sleds. The bicycle will have some of its energy transferred into rotational kinetic energy, in addition to translational kinetic energy for the entire bicycle plus rider. This will have a lower effect on the heavy rider than on the light one, so that without friction the heavier rider will still win. And no, I don't care what makes the tires spin, so long as they do spin at the appropriate rate. Link to comment Share on other sites More sharing options...
noob Posted July 29, 2008 Share Posted July 29, 2008 My opinion? The definition of the term Coulomb force has rarely, if ever, be questioned. Its found in all textbooks on electrodynamics. The mathematical expression for it is given in Eq. (1) at http://scienceworld.wolfram.com/physics/CoulombForce.html Pete This is exactly the problem. You don't know what a coloumb force is. What you mean and what you post is the law of coloumb, not the comoumb force of friction! No, I did address that in my post. If you don't have the tire spinning, then you have a sled with wheels rather than a bicycle, as I said. And this problem is about bicycles, not sleds. The bicycle will have some of its energy transferred into rotational kinetic energy, in addition to translational kinetic energy for the entire bicycle plus rider. This will have a lower effect on the heavy rider than on the light one, so that without friction the heavier rider will still win. And no, I don't care what makes the tires spin, so long as they do spin at the appropriate rate. This is wrong! You will never have a rotation if there is no friction and you don't have rotation energy! Rotation energy coul only exist here if you have friction. Link to comment Share on other sites More sharing options...
doG Posted July 29, 2008 Share Posted July 29, 2008 Indeed. Surface friction will, of course, affect the heavier rider more, but surface friction in this case is just in the axles... Not true. Rolling resistance is that friction between the tire and the road. My copy of Marks' Standard Handbook for Mechanical Engineers has a table of coefficients that vary with tire type, size, width, road type and road condition for calculating rolling resistance in tractive effort calculations. Two riders on identical tires with identical inflation will compress those tires proportionately to the mass they carry and the heavier mass will have a greater rolling resistance because the tire will be flatter from the greater compression. Link to comment Share on other sites More sharing options...
Sisyphus Posted July 29, 2008 Share Posted July 29, 2008 Not true. Rolling resistance is that friction between the tire and the road. My copy of Marks' Standard Handbook for Mechanical Engineers has a table of coefficients that vary with tire type, size, width, road type and road condition for calculating rolling resistance in tractive effort calculations. Two riders on identical tires with identical inflation will compress those tires proportionately to the mass they carry and the heavier mass will have a greater rolling resistance because the tire will be flatter from the greater compression. Fair enough. I'll concede I hadn't thought of rolling resistance, but to be fair, that's not the same thing as friction. It's the energy used up in the continuous deformation of the wheel and/or surface. The surface friction is just in the axles. Link to comment Share on other sites More sharing options...
Mr Skeptic Posted July 29, 2008 Share Posted July 29, 2008 This is wrong! You will never have a rotation if there is no friction and you don't have rotation energy! Rotation energy coul only exist here if you have friction. And I maintain that if the wheel isn't spinning, then what you have is a funny looking sled rather than a bicycle. And instead of rolling down the hill as par the original post, it would be sliding down the hill. These are two distinct problems; I am answering the one that was asked. Link to comment Share on other sites More sharing options...
swansont Posted July 29, 2008 Share Posted July 29, 2008 This is exactly the problem. You don't know what a coloumb force is. What you mean and what you post is the law of coloumb, not the comoumb force of friction! However, the Coulomb friction is precisely what Pete used earlier, in post #33. So might I suggest a change in tactic, and you present information that helps the discussion along, instead of simply declaring that people are wrong or things are complete nonsense. If you feel that [math]F_{fr} = \mu F_n[/math] is incorrect or inappropriate, explain why you think this is so. Link to comment Share on other sites More sharing options...
doG Posted July 29, 2008 Share Posted July 29, 2008 Fair enough. I'll concede I hadn't thought of rolling resistance, but to be fair, that's not the same thing as friction. It's the energy used up in the continuous deformation of the wheel and/or surface. The surface friction is just in the axles. Well, you can look at it that way but I wonder why my handbook lists a coefficient of 0.003 for steel wheels on steel rails? Me thinks there is not any deformation going on there, especially with work hardened wheels rolling on work hardened rails. Link to comment Share on other sites More sharing options...
Sisyphus Posted July 29, 2008 Share Posted July 29, 2008 Well, you can look at it that way but I wonder why my handbook lists a coefficient of 0.003 for steel wheels on steel rails? Me thinks there is not any deformation going on there, especially with work hardened wheels rolling on work hardened rails. On the contrary, there most certainly is. Everything has some elasticity, and steel actually has quite a bit compared with, say, ceramics or crystals. Link to comment Share on other sites More sharing options...
doG Posted July 29, 2008 Share Posted July 29, 2008 No more than the deformation causing the surface friction at the axle bearing. You implied that one is different from the other but they seem the same to me. Aren't the hardened ball bearings riding on their races the same as a steel wheel rolling and a rail? Link to comment Share on other sites More sharing options...
Pete Posted August 1, 2008 Share Posted August 1, 2008 (edited) You don't know what a coloumb force is. On the contrary. I know it better than you do from what you've posted so far. In the first place if one were to look that relationship I referred you to up in a physics text then it would be referred to as Coulomb's Law. The term Coulomb force is not used that often' date=' but when it [i']is[/i] used it always refers to the electric force (in the absense of a magnetic force). Its also defined in these references http://scienceworld.wolfram.com/physics/CoulombForce.html http://www.britannica.com/EBchecked/topic/140084/Coulomb-force# Are you saying that the Enclyclopedia Britannica has the wrong definition? However if you wish to prove this wrong then find an instance in a physics text/book/article and post the quote here along with the reference. What you mean and what you post is the law of coloumb, not the comoumb force of friction! You don't need to tell me what I posted and what I mean. I know it all too well. Frankly you're wasting our time and when you complain about sematics when nobody here really cares about it. All you needed to do is say "force of friction" . Regarless of what you call it you have used it incorrectly and have refused to address the proof that was given to you. Let me remind you of where this came up. In response to an erroneous assertion made by you I posted the following explaination. I disagree. Consider a body sliding on a surface under friction for which the coefficient of kinetic friction is [math]\mu_k[/math]. Then [math]F = ma = \mu_k F_n = \mu_k (mg) [/math] Cancel out m to yield [math]a = \mu_k g [/math] Thus the acceleration of a body sliding on a rough surface is a function of [math]\mu_k[/math] but not a function of m. In response to that you responded with This is also completely nonsense! What you wrote is a coloumb assumption, but we are talking about sliding. While I was writing this post I did a search on the internet on the term Coulomb assumption and the only thing I could find was in regards to a relationship regarding stress. I recommend that you pay more attention to your assumptions rather than criticizing what you think other people did wrong. What I explained is all about frictional forces which act when a body is sliding. Hence the term coefficient of kinetic friction. The term "kinetic" here refers to the fact that the formula pertains to sliding bodies. Something which your response indicated that you did not understand. What I explained to you is, in all instances in the physics literature, defined as the Coulomb Force. It never means Coulomb friction force or whatever. Please look these terms up in a physics text before you make any further attempts at "correcting" anybody. In any case I never heard/read the term "Coulomb friction force" before. Perhaps because no physics teacher I ever had used that term and no physics text/book/article I ever read used it. For that reason and that reason only I'm glad you mentioned it. Thanks. This is wrong! You will never have a rotation if there is no friction and you don't have rotation energy! Rotation energy coul only exist here if you have friction. That is clearly wrong. There is absolutely no reason why, in principle, something can't rotate without frictional forces acting on it. Only in practice does one have to account for fruiction and then only when the frictional force are large enough to affect the results of a computation/prediction. For example; superconductors can be levitated in a magnetic field in a hard vacuum and set rotating. The frictional forces are neglegible (i.e. they are virtually absent for all practical purposes) in such cases and the rotational energy is very well defined. Rotational energy is not dependant on the presence of frictional force. There is nothing in the theory of mechanics that would imply that. Consider the rotational energy of the astronomical bodies such as moons or asteroids! There are virtually no frictional forces there either but one can't say that there is no rotational kinetic energy either. Your comments above regarding rotational energy indicate that you have a lack of understanding of basic physics so I'd like to ask you exactly how far in physics did you say you went? Did you finish the PHY 101-103 series? If so then what physics text did you use in that sequence? At least post a reference to one text that you learned/are learning from. I'd like to as the original poster what the purpose of this question was? Was the question posted to inquire about the dependance of the rider's mass on acceleration? Or was it to ask about all these frictional forces? If its the later then there is no answer until we know what frictional forces you wish to take into account, what accuracy are you considering, whether the bicycles are identical etc. I mention accuracy because if you don't have instruments which can distinguish 1 microsecond of travel time and the frictional forces turn out to effect times of less than 1 microsecond then it'd be worth knowing that. Pete Edited August 1, 2008 by Pete Link to comment Share on other sites More sharing options...
Pete Posted September 5, 2008 Share Posted September 5, 2008 This is exactly the problem. You don't know what a coloumb force is. What you mean and what you post is the law of coloumb, not the comoumb force of friction! I came to learn how the frictional force became associated with the Coulomb force. The reason being is that the Coulomb force (as in the force between two charged particles etc.) is the cause of the frictional force at the microscopic level. I.e. friction is a result of charges interacting at the surface of the two materials in contact. Pete Link to comment Share on other sites More sharing options...
npts2020 Posted September 12, 2008 Share Posted September 12, 2008 The question is nearly entirely dependant on whether the wind resistance is the same for both riders. At between 17-18 mph about 70% of the energy required to maintain speed is overcoming wind resistance, that increases to about 90% at around 25mph. Surely the riders under conjecture would go even faster, making wind resistance an even bigger factor. It seems to me that a heavier rider has more potential and will have a higher terminal speed than a lighter rider, therefore going faster, if all other things are equal. Link to comment Share on other sites More sharing options...
swansont Posted September 12, 2008 Share Posted September 12, 2008 I think that conclusion was reached ~30 posts ago. Link to comment Share on other sites More sharing options...
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