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?
[Physics] Grip of the train wheels
friction
Related Solutions
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.
It's a surprisingly complicated question. Given your mention of friction, probably the main point is that for a car tyre the friction is not linearly dependant on load. Wikipedia has some information about this here.
If you had perfectly smooth surfaces the friction is actually proportional to the area of contact and independant of the load. This is because friction is an adhesive effect between atoms/molecules on the surfaces that are in contact. However in the real world surfaces are not smooth. If you touch two metal surfaces together the contact is between high spots on the two surfaces so the area that is in contact is much less than than the apparent area of contact. If you increase the load you deform these high spots and broaden them, so the effect of load is to increase the real area of contact. The real area of contact is approximately proportional to the load, and the friction is proportional to the area of contact, so the friction ends up being approximately proportional to the load.
However a rubber type is a lot softer than metal, and a road is a lot rougher than a metal plate. Even at low loads the tyre deforms to key into the irregularities in the road, so increasing the load has a lesser effect. That's why you get the sub-linear dependance described in the Wikipedia article.
But this is only the start of the complexity. If you use a wider tyre the contact patch area isn't necessarily bigger. A wider tyre has a wider shorter contact patch while a narrow tyre has a narrower longer contact patch. The contact patch area depends on the tyre pressure, the deformation of the sidewalls and probably lots of other things I can't think of at the moment.
And anyway, if by "grip" you mean grip when cornering, the grip isn't just controlled by the contact patch area. When a car is cornering the contact patch is being twisted. This is known as the slip angle. The wider shorter contact patch on a wide tyre has a smaller slip angle and as a result grips better.
Best Answer
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:
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: