Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How do the wheels of a train have sufficient grip on a metal track? I mean both of the surfaces are smooth (and not flexible) and it is okay if there is no inclination, but how about on an inclined track?

share|improve this question
1  
""(& not flexible)"" is simply wrong. No real material is "not flexible". –  Georg Mar 2 '11 at 10:15
    
what's the right term then? i meant "not flexible" relative to a tire. –  sterz Mar 2 '11 at 10:18
    
several : "high modulus of elasticity", "very hard", "brittle". (and some more) In general: if You mean some thing with a certain expression, then say it. –  Georg Mar 2 '11 at 10:42
2  
Greece where I live is full of mountains. There are mountain trains that have "wheels with teeth" so as not to slide at the inclines. It is called the "toothy train" :) . Here is a picture of the teeth odontotos.com . Built 113 years ago. More in en.wikipedia.org/wiki/Rack_railway . –  anna v Mar 2 '11 at 15:18
add comment

2 Answers

Sliding is prevented by friction and the friction force is equal to the product of the weight - the perpendicular force - and the dimensionless coefficient of static friction.

The coefficient of static friction between steel and steel can be as high as 0.78 so the angle would have to be hugely non-horizontal for the train to slide. And a lot of acceleration may be added, too.

The lowest coefficient of static friction in wet and greasy conditions may be 0.05 which is approximately the angle in radians where one could start to get worried. It is just 3 degrees and if there's lot of oil everywhere on the tracks, the train may get unsafe already for these small angles. However, in reality, the coefficient never drops this low and 15 degrees is usually a safe angle.

See also:

http://en.wikipedia.org/wiki/Rail_adhesion

Note that the coefficient of static friction is higher than the coefficient of kinetic friction so the hardest thing is to start the sliding. Once the train starts to slide, it is more likely that it will continue to do so.

All the text above was about the sliding - the stability in the front-rear direction. The stability in the left-right direction is guaranteed by the shape of the wheels: enter image description here

share|improve this answer
    
so, to sum it up: it is because of the weight of the whole train provides sufficient friction and steel-to-steel friction coefficient is also actually sufficient (not 'slippery' as i thought). did i get that correctly? –  sterz Mar 2 '11 at 15:12
4  
if you look at the link I gave for a cog train, in a comment to your question, you will see that the flat inclinations is about 3% and the inclination climbing with cogs is 17%. Sliding will happen over 3% with a margin of error for rain and freeze I guess, that is why they need the cogs for the mountain pass. –  anna v Mar 2 '11 at 15:33
    
@anna v: thank you. actually both of your comments are part of the answer. –  sterz Mar 3 '11 at 7:04
add comment

The reason trains stay on track is because the wheels are not cylindrical, but conical. See Feynman's explanation here: http://www.youtube.com/watch?v=y7h4OtFDnYE

share|improve this answer
1  
Thanks for the Feynman link, that was fun and informative. –  anna v Mar 2 '11 at 15:06
    
p.s. the conical does not address the sliding part along the track, though. This has to be friction according to how the force from gravity decomposes perpendicular and vertical to the inclination, and the coefficient of friction, as others have answered. –  anna v Mar 2 '11 at 15:28
    
this...doesn't really answer the real question at all. on an inclined track, cylindrical and conical wheels face the same problems. –  Justin L. Mar 2 '11 at 18:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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