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Re: Why is it "static friction?"



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