Are both my textbooks wrong about banked tracks.

[Damned charming :)]

a car of mass m is traveling 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)

 

[OlderDan]

I agree with you.

 

[Chi Meson]

Your books are correct. Make a drawing of the situation. The components 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.

 

[OlderDan]

Your books are correct. Make a drawing of the situation. The components 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.

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.

 

[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'

 

[Doc Al]

Your books are correct. Make a drawing of the situation. The components 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.

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.

 

[Chi Meson]

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.

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.

 

[OlderDan]

Oh, sloppy me.

I am unaccustomed to treating "centrifugal" as a real force. In fact I always rail against it.

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.

 

[Damned charming :)]

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

 

[Andrew Mason]

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

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

 

[Doc Al]

With 90 degree banking:

$$N + \frac{mv^2}{r} = 0$$


In between:

$$N + W\cos{x} + \frac{mv^2}{r}\sin{x} = 0$$ so:

Your "centrifugal force" terms have the wrong sign!

 

[Andrew Mason]

Your "centrifugal force" terms have the wrong sign!

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 textbooks are right. They are just using g as a negative.

AM

 

[Doc Al]

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

You are way too generous. If they really took g as negative then when the book says

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

that translates (using the standard positive g) to be:
F= - mg sin(x)- mv^2/r cos(x)

I don't think so.

 

[OlderDan]

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

AM

If g is negative, then the authors are saying both F and N are also negative.

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

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.

 

[Andrew Mason]

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.

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

 

[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.