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Re: L2-"Negotiating" a curve.

A banked road is conceptually easier to deal with than
the flat road because here the centripetal force is mostly
gravitational.

Well, no. Not if the circle is horizontal. The centripetal force is the
horizontal component of the normal reaction from the banked road. I was
warning my class about this just this morning! It's an fme (frequently made
error).


Mark Sylvester


Maybe someone on the list can help me with a problem I've always had with
the standard textbook force diagrams for a car on a banked road. I'm
currently using Wilson & Buffa's College Physics, but I've seen the same
diagrams in other textbooks.

As Mark says, the horizontal component of the normal reaction (labeled N)
from the banked force is show to be the centripetal force. In addition, on
the diagram the vertical component of N is shown and implied to be equal to
mg. This would result in the equation, N cos(angle) = mg or
N = mg/cos(angle) .

However, when textbooks discuss incline planes (which I would think a small
section of a banked road could be considered to be), components of N are
not used in the force diagram, components of mg are used, resulting in the
equation
N = mg cos(angle) . If this diagram were applied to a banked road, the
mg sin(angle) component would equal some component of friction or the car
would slide toward the center of the curve.

I can't reconcile that N is mg TIMES cos(angle) and N is mg DIVIDED BY
cos(angle) depending on which diagram you use.

So..am I missing something? Or is friction the answer to this apparent
discrepancy? And if friction is the answer, why isn't it in the diagrams?

Lynn Aldrich


******************************************************
Lynn K. Aldrich Phone: 570-674-6376
Assoc Prof Physics email: laldrich@miseri.edu
College Misericordia FAX (8:30-4:30)
301 Lake St 570-675-4028
Dallas, PA 18612-1098 FAX 570-675-2441
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