Re: Why is it "static friction?"

[Charles Bell <charbell@BELLSOUTH.NET>]

There are several other forces acting on a car.
There is drag from the resistance to the airflow around it.
There is a positive or negative lifting force from the reaction forces from
the airflow around it.
There is a sidewards force or torque about it center of gravity tending to
tip it over as it changes direction.
The frictional force caused by the interaction of the tires with the road
surface has components resisting the tangential motion and a sidewards
force pushing the tires towards the center of the turn it is making or
radially inward equivalent to the mv^2 /r.

Without the inward radial force, the car would stop turning and move in the
direction of the last tangent, like an object being wirled in a circle on a
string when the string breaks.

There is a coriolis force from the earth's rotation.
The force of gravity from the moon and sun.
There is acceleration as the car traverses hills and dips in the road.
Gusts of wind....

At 08:27 AM 11/14/2003, you wrote:
> A friend of mine has this to offer:
>
> "There is an alternative way of regarding the problem that  obviates
> Kowalski's problem.  In reality there are but two forces acting on
> the car as it goes in a circular path - its weight and a force from
> the road, call it R.  This force R acts at an angle theta to the
> road.  Adding R and mg gives the net force - R  + mg = Fnet.  The net
> force equals, of course, mv^2/r. This force must increase as v
> increases.  The components of R are the normal force N and the
> "static friction" force.  We tend to forget that they are components
> of a single force.  After all, if we pull a wagon at an angle theta,
> we can resolve the pull into  horizontal and  vertical components.
> There is, however, just one force acting.
>
> When we reach the maximum speed possible before spin-out, we
> interpret the horizontal component in terms of the coefficient of
> static friction.
>
> Returning to Kowalski's last sentence, I would say that the
> centripetal force acting on a turning car is the resultant of the
> weight and the road force."
>
> -- Wolfgang