# Friction of a train

## Homework Statement

Consider a train that rounds a curve with a radius of 570 m at a speeed of 160 km/h (a) Calculate the friction force needed on a train passenger of mass 75 kg if the track is not banked and the train does not tilt. (b) Calculate the friction force on the passenger if the train tilts at an angle of 8.0 degrees toward the center of the curve.

Ac= mV^2/r, F=ma

## The Attempt at a Solution

Ok so for part (a) I can use the equation for Centripital Acc, Ac=mv^2/r, to find the acceleration of the system. Then using F=mAc I can find the force that the train is putting on the passenger. Now the equation for friction is F=uN. Unfortunatly I do not know the coffeciant of friction. Is this problem solveable with the data given? Am I missing an assumption that I may make?

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the only thing you are missing is that you already solved the problem. the force the train exerts on the passenger is the friction force.

tiny-tim
Homework Helper
hi notsam! … Then using F=mAc I can find the force that the train is putting on the passenger. Now the equation for friction is F=uN. Unfortunatly I do not know the coffeciant of friction.
the question only asks for the force, and you've found it so why do you need the coefficient of friction? Duh! Because if he doesn't move then he hasn't over taken the force of friction...Ok so for the second part if the train tilts 8 degrees, then I just use COS(8)*(9.8) to find my new normal force?

tiny-tim
Homework Helper
i don't think so use F = ma (what is a ?)​

I'm really confused now, you're asking me what "a" is. Does this mean that the tilting of the train gives the system a new centripital acceleration? And if so I have no idea how to find it.

tiny-tim
Homework Helper
… I have no idea how to find it.
uhh? centripetal acceleration is always v2/r​

tiny-tim is, i think, trying to say that things are not so simple. just adjusting your value of g will not help you solve the problem. i believe he is also emphasizing the fact that a is a vector, and so its direction must be considered carefully (along with the direction of all the forces, of course). i don't want to put too many words in tim's mouth, but if he did mean that, it would be good advice :)

Oh ok so the second part of the problem is just simply that my acceleration has changed due to the tilt. A= v^2/r...how does the tilt affect this?

actually, everything except the acceleration (and gravity of course) tilts. the train is still going in the same circle, so the acceleration is exactly the same as before. but, the friction force and the normal force are rotated by the angle of tilt. you might have trouble with this problem if you don't have a good drawing.

OK! Now it makes much more since. I must use the first part of the problem to find the cofficient of friction. Using F=uN on the first part of the problem I should beable to find "u" by finding the normal force of the passenger, and using my awnser from par (a) for F. Then to solve part (b) I would use COS(8)*9.81*u to find the new force due to friction

well, not quite. we are not going to be able to find the coefficient of friction. for static friction, uN is just an upper limit, the force varies depending on what it needs to do.

neither the friction nor the normal force will be in the direction of acceleration, but both will have a component in that direction, so they both contribute to the centripetal acceleration.

Ok I'm lost. How am I supposed to figure out part b. Please at least get pointed in the right direction.

have you drawn a force diagram for part b? that is almost a necessity to understand this problem. i would suggest looking in a physics text book; they almost all have very nice diagrams for this very situation.

tiny-tim
either take components perpendicular to a (to eliminate a and to find N), or parallel to the slope (to eliminate N and to find a) 