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Herb Schulz wrote:
I'm not quite sure why everyone is assuming that sliding down the
the inclined plane is static? The difference is that in sliding
along a frictionless inclined plane the net force (& therefore the
acceleration) is paralle to the surface of the plane while in going
around (an assumed frictionless) banked turn at the design speed the
net force is horizontal, towards the center of the horizontal circle
that is the path of the object.
Acceleration along an inclined plane does not change N but
the circular motion does. In both cases N is a reaction
force. In the first case it the reaction to the perpendicular
component of mg, in the second case N also has a component
which is the reaction to the CENTRIFUGAL force. Is this an
acceptable interpretation?
Consider an extreme case, the road is the inner wall of
a vertical cylinder. In that case N is nothing else but the
reaction to the centrifugal force. The car acts on the road
and the road reacts on in the opposite direction. In other
words, the CENTRIPETAL force is N. Both forces, action and
reaction, are real to an outside observer. (The mass of the
cylinder is enormous and that is why the cylinder does not
turn when "pushed backward" by tires. The outside observer
would see the "free cylinder's" motion if the mass of the
road were not much larger than the mass of the car. I am
thinking about a "free cylinder" located somewhere between
galaxies.)
Ludwik Kowalski