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Re: Banked road



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 direction
of the acceleration (and therefore the net force) is known from the
statement of the problem (the ocurring motion is stated-or implied). The
acceleration is along the plane in one case, and toward the center of the
circular 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 known
direction of the acceleration.

In the static case, a third (frictional) force is involved. Here although
the acceleration is zero the rigidity of the plane still demands that the
net force into the plane be zero. The Normal force adjusts to make this
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