# Are both my textbooks wrong about banked tracks.

1. Oct 11, 2006

### Damned charming :)

a car of mass m is travelling around a banked track.
if F is the friction force up the track and x is the angle of the track

I say
F= mg sin(x)- mv^2/r cos(x)
N = mg cos(x) + mv^2/r sin(x)

but both my books say

F= mg sin(x)- mv^2/r cos(x)
N = mg cos(x) - mv^2/r sin(x)

Last edited: Oct 11, 2006
2. Oct 11, 2006

### OlderDan

I agree with you.

3. Oct 11, 2006

### Chi Meson

Your books are correct. Make a drawing of the situation. The componants that are perpendicular to the slope will show the perp componant of centripetal force and the perp componant of weight. The normal force will be such that the net force in this direction will be zero.

So N + (-mgcos(x)) + mv^2/rsin(x) = 0

the (-) above indicates that the perp componant of weight points opposite to N.

4. Oct 11, 2006

### OlderDan

I really don't think so. The centripetal force is the resultant of all the actual forces that are acting: Normal, Friction, Gravity. You can treat mv^2/r as a centrifugal force and add the forces to get zero. Either way you get the same result. That attachment is short on labels, but the work is all there and I think it's obvious which force is which.

#### Attached Files:

• ###### BankedTrack.pdf
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Last edited: Oct 11, 2006
5. Oct 11, 2006

### civil_dude

This site has a good picture and explanation. The Normal force is the resultant of the weight and acceleration forces.

http://www.siena.edu/rfinn/phys110/circular-motion.pdf#search='banked%20acceleration' [Broken]

Last edited by a moderator: May 2, 2017
6. Oct 11, 2006

### Staff: Mentor

You made an error with the sign of the "centripetal force" term. If you insist on treating that as a separate force (shocking!), you'd better treat it as a centrifugal force (as Dan says), which means it has the opposite sign as the centripetal force.

Even better is just to stick to an inertial frame with "real" forces and apply Newton's 2nd law:
N -mgcos(x) = mv^2/r sin(x)

Where "v^2/r sin(x)" is just the component of the acceleration perpendicular to the surface.

7. Oct 11, 2006

### Chi Meson

Oh, sloppy me.

I am unaccustomed to treating "centrifugal" as a real force. In fact I always rail against it. And the one time I try to "use" it (what am I, an engineer? Come on!) I go and draw the "centriptal" arrow. So I am in agreement with Doc Al and Dan. Mostly in that it's an annoying way to analyze this problem.

8. Oct 11, 2006

### OlderDan

We agree about that. I had to force myself to do it with centrifugal force. I did it first with centripetal, and then had to do a bit of algebra to get it into the form that was given. So I figured I'd best do it the centrifugal way too.

9. Oct 12, 2006

### Damned charming :)

Thanks for this guys I thought I was going mad. Fancy 2 books being wrong.

10. Oct 12, 2006

### Andrew Mason

I am not sure they are wrong. If the reference xy axes are parallel and perpendicular to the track with + being above and to the right (an odd way to look at it, but this is apparently what you have) and analysing it in the reference frame of the accelerating car and using the fictitious centrifugal force as a 'real' force, then with no banking:

$$N + W = 0$$

where W = -mg

With 90 degree banking:

$$N + F_c = 0$$

where [itex]F_c = - \frac{mv^2}{r}[/tex]

In between:

$$N + W\cos{x} + F_c\sin{x} = 0$$

$$N = -W\cos{x} - F_c\sin{x}$$

$$N = -mg\cos{x} + \frac{mv^2}{r}\sin{x}$$

AM

Last edited: Oct 12, 2006
11. Oct 12, 2006

### Staff: Mentor

Your "centrifugal force" terms have the wrong sign!

12. Oct 12, 2006

### Andrew Mason

I noticed that and was editing as you posted.

I should have changed it further: W = mg and g is negative.

Note that with 90 degree banking:

$$F_f + W = 0$$

so:

$$F_f = -W = - F$$

(that is, the friction force is in the -F direction, where F is in the direction of the positive x axis ie up)

If we agree with the book authors that:

$$F = mg$$ then the authors are necessarily saying that g is negative.

So the text books are right. They are just using g as a negative.

AM

Last edited: Oct 12, 2006
13. Oct 12, 2006

### Staff: Mentor

You are way too generous. If they really took g as negative then when the book says
that translates (using the standard positive g) to be:
F= - mg sin(x)- mv^2/r cos(x)

I don't think so.

14. Oct 12, 2006

### OlderDan

If g is negative, then the authors are saying both F and N are also negative.
It seems to me the discrepency is the minus in both of these equations when one of them should be plus. Changing the sign of g does not fix that.

15. Oct 14, 2006

### Andrew Mason

Ok. I agree. (I have to learn not to do PF at 4:00 in the morning). The normal force never changes direction whereas the force of friction does (as x changes from 0 to 90).

I would also agree with Chi's comment that this is an awkward way to analyse it because it requires using the frame of the moving car so the frame of reference (axes) changes with the angle. If one resolves the centripetal acceleration into components along each of the axes, the component along the surface is never positive and the normal component is never negative. Gravity components are always negative. Normal force is always positive and the direction of friction can be either depending on the angle

$$F_f - mg\sin{\theta} = ma_x = -\frac{mv^2\cos{\theta}}{r}$$

$$N - mg\cos{\theta} = ma_y = \frac{mv^2\sin{\theta}}{r}$$

AM

16. Oct 14, 2006

### Chi Meson

I keep having the sneeking feeling that perhaps the books are right, but...

I've reanalyzed the problem from three points of view: an "equilibrium-centrifugal" reference frame, a "parallel-perpendicular" reference frame and a "horizontal-vertical" reference frame. The latter two gave me a fun little riminder of trig identities, and it was a good algebra work-out (and it was essentially in the link provided by OlderDan). Each time I ended up with OP being correct and books being incorrect.

I'm wondering now, are both books by the same author or publisher? Which books are they, and how old are they? It seems so bizarre.

Oh yeah, and... the "centrifugal" reference frame is actually the easiest method to produce a statement for finding the normal force in terms of mass, speed and angle. No wonder engineers like it.

Last edited: Oct 14, 2006