Re: L2-"Negotiating" a curve.

[Bob Sciamanda <trebor@VELOCITY.NET>]

Now this is a problem that I have found commonly arises if the student is
merely given the solutions to the two problems (the inclined plane vs the
banked curve) with no explicit motivation for each line of reasoning.

The salient point is that each is an application of F=mA in which one is
GIVEN the DIRECTION of the resultant force, by the statement of the
problem.  That is the motivation for resolving the same pair of forces (mg
and N) along two different (x,y) coordinate systems.

It is instructive to solve each of these problems the hard (and dumb) way,
ie., by beginning with a resolution of forces along the "wrong" set of
(x,y) directions and then applying what is "given".

Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor

----- Original Message -----
From: Lynn Aldrich <laldrich@MISERI.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Thursday, November 04, 1999 1:54 PM
Subject: Re: L2-"Negotiating" a curve.


> 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