Re: L2-"Negotiating" a curve.

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

As Bob wrote last night, the centripetal force is the normal
component of the net force. The tangential component may
or may not be present. Just something to think about while
dealing with the three forces with which the road acts on the
wheels of a tricycle.

Arlyn DeBruyckere wrote:

> Lynn Aldrich wrote:
>
> > 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?
>
> I wish I could draw on e-mail - it would make it so much easier.  On the
> inclined plane, mg is straight down.  The normal force (N  the definition of
> "normal" is perpendicular to the surface) is perpendicular to the plane and is
> the balancing force for the part of mg that is perpendicular to the plane so N
> is part of mg or N=-mgcos(angle).  I think that many people "neglect" the
> negative part in that the normal force is acting in the direction opposite of
> the gravitational force (again it would be nice to be able to show vector
> notation in e-mail).
>
> On the banked curve the vertical component of N is NOT equal to mg.  If it
> implies that on the diagram then the diagram is not to scale or it suggests
> something that is not correct.  The vertical component of N on a banked curve
> will be less than -mg because "some of mg" is used as the centripetal force.  In
> other words, just like mg causes a horizontal force on an inclined plane
> (=mgsin(angle) for the force *along the plane*, part of this is horizontal {too
> many angles and forces to describe without sketches}) it will also cause a
> horizontal force in the circular motion.  Remember centripetal force in this
> case *IS* the net force.  The net force causes uniform circular motion and is
> directed to the center of the circle (if we are using the "standard" earth as
> the frame of reference).
> --
> Arlyn DeBruyckere
> Hutchinson High School