# Circular motion

1. Jun 8, 2005

### amose093

I have recently completed my mid-year chemistry exam and came accross a question that i cannot find an answer to. Me and my friends completely disagree and i was wondering if anyone could help me?
-The question showed a simplified diagram of a train around a circular section of a track, and the question asked-"draw an arrow representing the force exerted on the wheels of the train by the track while the train is decelerating".
Does the braking of the train have any effect on the forces present between the track and the train?
Thanx guys...
Andrew

2. Jun 8, 2005

### ZapperZ

Staff Emeritus

Zz.

3. Jun 8, 2005

### amose093

So you can't actually help me, i just thought that i may have posted in the wrong section!

4. Jun 8, 2005

### p53ud0 dr34m5

wouldnt the braking create a greater magnitude of friction or create friction if there wasnt any initially present?

5. Jun 8, 2005

### amose093

See thats my main problem. Does the braking affect the interection between the track and the wheels, the question seemed to be quite specific. Is the brake force applied to the wheels or the track?

6. Jun 8, 2005

### ramollari

There are actually two forces: the normal component F_cent of the reaction force of the track (providing the centripetal acceleration), and the backward pointing frictional force f_k that arises during braking. So the resultant force that the track applies on the wheel points somewhere between the center of the circular track and the backward direction.

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7. Jun 8, 2005

### amose093

Wow this forum is great, thanx for the speedy replies.
See thats the answer i would give if asked for the net force on the system, however if it singled out the force ON THE WHEELS EXERTED BY THE TRACKS, is the answer still the same?

8. Jun 8, 2005

### ramollari

It's the same thing. The weels have the same acceleration as the whole system. The difference is that the acceleration is multiplied with the mass of a single wheel, rather than the mass of all the train system.