Re: L2-"Negotiating" a curve.

[Bob Sciamanda <trebor@VELOCITY.NET>]

No; if I understand your notation, you want a force component in the -x
direction; ie., at right angles to the present velocity; and that's just
what the road's reaction to the turned rudder/front-wheel will give you.

Bob

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

----- Original Message -----
From: Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, November 03, 1999 11:44 PM
Subject: Re: L2-"Negotiating" a curve.


> I agree. But how to explain this Ffx component? Why is
> it directed to the left? The wheel moves along the y axis
> and sliding friction force must be along the -y. The front
> wheel orientation should have not affect this, unless that
> wheel is already on the left side of the y axis. Then we
> have a net torque.
>
> Bob Sciamanda wrote:
>
> > > . . . How come
> > > that by turning the front wheel to the left the vector sum of
> > > Fr+Fl+Ff becomes the centripetal force directed to the left
> > > (and equal to m*v^2/R )?
> >
> > You don't mean that; there is still a tangential frictional force
> > component (affecting the speed).
> > It only takes a COMPONENT of the total force in the "centripetal
> > direction" to effect a change of direction; the "other" (tangential)
> > component will affect the speed.