Re: L2-"Negotiating" a curve.

[Mark Sylvester <msylvest@SPIN.IT>]

One might add that in one case the thing is (potentially) sliding down the
slope, so that's where we look for the unbalanced force. In the other case
it's going round in a horizontal circle, so we assume that the unbalanced
force is horizontal. Different accelerations >> different forces >>
different diagrams.

Mark


At 14:17 04-11-99 -0500, you wrote:
> Two different vector diagrams.  In the case of the inclined plane, you are
> breaking up mg into components along and normal to the plane.  mg ends up to
> be the hypotenuse of the vector triangle and N one of those components.  In
> the banked turn you are breaking up the normal into vertical and horizontal
> components.  In that vector triangle it is N that is the hypotenuse with mg
> one of the components.  I think that explains the differences.
>
> Rick
>
> ----- 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.
>
>
> > > > 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
> > ******************************************************

Mark Sylvester
United World College of the Adriatic
34013 Duino TS
Italy
msylvest@spin.it