12
$\begingroup$

I'm a computer programmer that never studied physics in school and now it's coming back to bite me a bit in some of the stuff I'm being asked to program. I'm trying to self study some physics and I've got a few open source intro physics books and understanding it for the most part but I'm a bit confused on this statement I've stumbled upon in a section about static friction.

It's already gone over the formula for static friction and so forth. It gets into a section explaining that the weight of a train increases static friction between the wheels and the tracks. Alright, makes sense. But then it says this:

The reason locomotives are built to be so heavy is for traction. The upward normal force of the rails on the wheels, FN, cancels the downward force of gravity, FW, so ignoring plus and minus signs, these two forces are equal in absolute value, FN = FW. Given this amount of normal force, the maximum force of static friction is Fs = sFN = sFW. This static frictional force, of the rails pushing forward on the wheels, is the only force that can accelerate the train, pull it uphill, or cancel out the force of air resistance while cruising at constant speed. The coefficient of static friction for steel on steel is about 1/4, so no locomotive can pull with a force greater than about 1/4 of its own weight. If the engine is capable of supplying more than that amount of force, the result will be simply to break static friction and spin the wheels.

- "Newtonian Physics", Light and Matter - Book 1, p158 B. Crowell http://www.lightandmatter.com/bk1.pdf

I'm confused as to how static friction is the only thing that can move the train forward. I thought static friction was what kept it in place in the first place. There's another force - that I can't think of the name of, but I've heard of somewhere - that I thought was more what they're describing here, where the weight of the wheels pushing down and forward slightly on the tracks causes the tracks to push up and forward (from the opposite side).

Can someone explain to me what this is saying?

$\endgroup$

4 Answers 4

10
$\begingroup$

This is all a complicated (and confusing, or just plain confused) way to say that, if you want the locomotive to pull the train, you don't want its wheels to slip. It's friction that prevents the wheels from slipping.

I suggest you simply delete this sentence:

This static frictional force, of the rails pushing forward on the wheels, is the only force that can accelerate the train, pull it uphill, or cancel out the force of air resistance while cruising at constant speed.

The paragraph makes a lot more sense without it. The author is trying to get at Newton's third law (equal and opposite reaction) but this way of putting it provides more confusion than insight.

$\endgroup$
5
  • $\begingroup$ So.. really what I was thinking (and didn't think to include) is that it really isn't static friction that accelerates the train, or does anything else like cancelling out air friction, or whatever, but just that it keeps the wheels from slipping and the engine provides the force that actually moves it? $\endgroup$
    – trycatch
    Mar 27, 2011 at 18:04
  • $\begingroup$ Yes, that's right. $\endgroup$
    – nibot
    Mar 27, 2011 at 18:07
  • $\begingroup$ A more intuitive way to think about it is to ask, where does the energy exerted by the locomotive go? If the wheels are not slipping, it all goes into kinetic and potential energy of the train. If the wheels are slipping, then energy also goes into heat where the wheels are rubbing the tracks. The power being lost to this friction is the normal force * coefficient of friction * the rate of rotation of the wheel * their circumference. Static friction, however, consumes no power, since there is no displacement (by definition). $\endgroup$
    – nibot
    Mar 27, 2011 at 18:12
  • $\begingroup$ Got it, thanks to both of you for your answers. :) You both got me to the "understanding" point so.. I'm tossing the "accepted" check to nibot since he got in one minute before and I can't share it with both of you :( (And I don't have enough rep to upvote yet. Arg.) $\endgroup$
    – trycatch
    Mar 27, 2011 at 18:27
  • 1
    $\begingroup$ Nibots answer does not adress Your question! $\endgroup$
    – Georg
    Mar 27, 2011 at 18:58
8
$\begingroup$

The thing you need to think about here is that the wheels roll.

If there was no friction between the wheels and the track, firing up the locomotive would just cause its drive wheel to spin.

Friction acts to prevent or resist relative motion between the two surfaces. So, if there is a torque on the wheels and the point of contact can not move relative the rail (just where it touches) because of static friction, the only way the wheel can turn is if the train moves relative the wheel.

$\endgroup$
6
  • $\begingroup$ So.. rather than say exactly what I said to nibot.. is my analysis of his answer relatively accurate? $\endgroup$
    – trycatch
    Mar 27, 2011 at 18:05
  • 1
    $\begingroup$ @trycatch: Well, yes, and then no. For the train to move relative the rails, it must push on the rails (so that the rails push back---equal and opposite reaction, right?). No friction means no going forward no matter how powerful the locomotive. The engine provides the energy, but friction is part of the mechanism that provides the impulse. // Goes off imagining some steampunk fantasy of a fan propelled rail engine. $\endgroup$ Mar 27, 2011 at 18:10
  • $\begingroup$ Nobody doubts friction of wheels as far as braking is debated. Why then for the acceleration? Its exactly symmetric. $\endgroup$
    – Georg
    Mar 28, 2011 at 10:09
  • $\begingroup$ @trycatch technovelgy.com/ct/Content-Comments.asp?Bnum=459 $\endgroup$
    – mplungjan
    Mar 28, 2011 at 17:01
  • $\begingroup$ @mplungjan - I think you mean @dmckee :P $\endgroup$
    – trycatch
    Mar 28, 2011 at 20:05
5
$\begingroup$

There is a question of what you are trying to learn about friction from this convoluted example. What may be slightly confusing is that there are two types of friction static friction and kinetic friction. Kinetic friction is the friction associated with two substances sliding against one another, which can only happen when one is moving relative to another. Kinetic friction is therefore associated with motion. Static friction is the friction associated with keeping an object stationary. However when one considers rolling wheel motion the rolling wheel friction is static friction, despite the motion of the wheel, because the wheel never actually slips on the track (or at least it is not meant to).

The force name your are looking for in the last paragraph is called the Normal Force. Look at the Wikipedia article on Friction if that helps as it gives more equations and diagrams.

EDIT (after a comment below) From the original question and comments we can see that an additional component of this question was to have a better understanding of a more complex case of elastic deformation. To understand this consider a massive object on a springy mattress: then there will be a U or V shaped deformation in the material. Modelling this would require some mathematics and Hooke's Law (F=-kx) would seem the most appropriate as an approximation. Now if this object were moving forward at some velocity v then the corresponding deformation will travel at this velocity too. It would also be necessary to model the surface tension of the material (which could rip if the strain was too large). Together all this will cause leading and lagging deformations (and possibly oscillations) in the material.

As this is for a computer model I am not clear whether the physics of this material oscillation has to be exactly correct according to some specific parameters and equations, or whether it is just an effect for demonstration on some computer game. In either case some further modelling is required to determine the equation for such motion, based on the principles above. If we are dealing with motion of cars on roads or trains on tracks then this extra modelling should be unnecessary and such elastic deformation can be ignored.

$\endgroup$
6
  • $\begingroup$ This definitely helps too. And I'm not sure the Normal force was what I was looking for.. it was something to the effect where if an object is rolling across a surface, as it rolls forward, it's pressing into the surface in front of it, deforming it slightly, and the surface behind it is pushing back up... I came across this somewhere while I was trying to understand this example.. $\endgroup$
    – trycatch
    Mar 27, 2011 at 18:47
  • $\begingroup$ @Georg - nibot and dmckee had answered at essentially the same time, and both of them together got me to the point of understanding what was going on.. I gave the "accepted" to nibots since he got in a few seconds earlier and it was basically a tie to me between who got me understanding what was going on. I'd already given out the accept before Roy answered. $\endgroup$
    – trycatch
    Mar 27, 2011 at 19:51
  • $\begingroup$ @George, @trycatch I had noticed that the earlier answers did not completely answer the questions, but they helped. I will add a little to this answer about the latter question. $\endgroup$ Mar 27, 2011 at 21:13
  • $\begingroup$ I think this bit about "[the wheel] pressing into the surface in front of it, deforming it slightly, and the surface behind it is pushing back up" is also irrelevant. The track may be deformed by the weight of the train, but that is not essential to the train's locomotion. $\endgroup$
    – nibot
    Mar 27, 2011 at 21:43
  • 1
    $\begingroup$ ""The track may be deformed by the weight of the train, but that is not essential to the train's locomotion"" Nonsense! The deformation of the rail and wheel are essntial to get an area of contact (and friction) What is not important is the thoughts of work done in this process. The work is very small for steel on steel, because there is almost no hysteresis in the process. (As long as the speed is much smaller than sound velocity in steel.) $\endgroup$
    – Georg
    Mar 28, 2011 at 10:05
1
$\begingroup$

As the wheels try to roll they are prevented from rolling by the frictional force acting in the opp. direction.As the traction force exceeds the limiting frictional force the wheel starts rolling forward w.r.t rails. The force tries to induce relative motion between the wheels and rails.As the rails cannot move backwards (due to friction) the wheels have to roll forward to induce relative motion.The friction opposes wheel rotation up to limiting friction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.