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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?

sterz
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    ""(& not flexible)"" is simply wrong. No real material is "not flexible". – Georg Mar 02 '11 at 10:15
  • what's the right term then? i meant "not flexible" relative to a tire. – sterz Mar 02 '11 at 10:18
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    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 02 '11 at 10:42
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    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 http://www.odontotos.com/ . Built 113 years ago. More in http://en.wikipedia.org/wiki/Rack_railway . – anna v Mar 02 '11 at 15:18

2 Answers2

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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

Luboš Motl
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  • 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 02 '11 at 15:12
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    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 02 '11 at 15:33
  • @anna v: thank you. actually both of your comments are part of the answer. – sterz Mar 03 '11 at 07:04
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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

TROLLHUNTER
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    Thanks for the Feynman link, that was fun and informative. – anna v Mar 02 '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 02 '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 02 '11 at 18:30