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Thanks to all who responded. I was puzzled butdirection
now all is clear. Unfortunately, most authors do
not discuss this "quazi-paradoxical" situation.
Ludwik Kowalski
Bob Sciamanda wrote:
Both cases involve an object accelerating under the influence of two
forces : mg(down) and N(normal to the plane).
These problems are solvable only because it is assumed that the
Theof the acceleration (and therefore the net force) is known from the
statement of the problem (the ocurring motion is stated-or implied).
theacceleration is along the plane in one case, and toward the center of
knowncircular trajectory in the other case.
The equations you quote for N simply state that the net force (the
resultant of mg and N) must have zero component perpendicular to the
althoughdirection of the acceleration.
In the static case, a third (frictional) force is involved. Here
thethe acceleration is zero the rigidity of the plane still demands that
thisnet force into the plane be zero. The Normal force adjusts to make
so (so long as the structure is capable!).
Bob Sciamanda
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: Sunday, November 04, 2001 11:48 AM
Subject: Banked road
The normal force with which an inclined plane is acting on an
object is N= m*g*cosA, where A is the angle of inclination.
Students use this approach to calculate accelerations (with or
without friction) or to solve equilibrium problems.
But in dealing with banked roads they are suddenly asked to
accept that N=mg/cosA. How can this be explained?
Ludwik Kowalski