Re: Why is it "static friction?"

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

Ludwik,

When you start walking from a stationary position, what force is acting
in the direction of your acceleration?

____________________________________________________
Robert Cohen; 570-422-3428; www.esu.edu/~bbq
East Stroudsburg University; E. Stroudsburg, PA 18301


> -----Original Message-----
> From: Forum for Physics Educators 
> [mailto:PHYS-L@lists.nau.edu] On Behalf Of Ludwik Kowalski
> Sent: Thursday, November 13, 2003 4:41 PM
>
> I have no doubt that friction has something to do
> with turning. Without friction a car would continue
> to move along the straight line, no matter how its
> wheels are turned.  But this alone does not justify
> the statement that  m*v^2/r=mu*m*g.
>
> On Thursday, November 13, 2003, at 04:02 PM, Ludwik Kowalski wrote:
>
> > Why do we say that the m*v^2/r (acting on a car) is
> > the force of static friction?
> >
> > I think of static friction as a "responding" force. For example, a 
> > crate pulled to the right (by a rope) will experience a responding 
> > force (from the floor) directed to the left. If the pulling force 
> > (cause) is small then the responding force (effect) is also 
> small. The 
> > net force acting on the crate remains zero when the causing
> > force increases (up to a limit). Kinetic frictional force
> > acts in the direction opposite to motion, static frictional
> > force acts in the direction opposite to the direction of
> > a causing force.
> >
> > Referring to the centripetal force acting on a turning
> > car a textbook states "the force in the radial direction 
> acting on the 
> > car is the force of static friction directed toward the 
> center of the 
> > circular path." If the static friction force is the effect 
> then where 
> > is the causing force? I am not convinced that the centripetal force,
> > acting on a turning car, is the static friction force (as
> > it was introduced in the first paragraph above).