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Here's my take. Look at the second situation. N = mg/cosA is only trueif
the object is in equilibrium. That is not generally true if the object isacceleration.
at rest--take the surface to be frictionless for example. That is, the
object (usually a car) has to be moving around the banked road so that the
horizontal component of N will produce the needed centripetal
In effect the car is trying to drive in a straight line but the road isnormal
pushing it towards the center point of the curve. This INCREASES the
force to be greater than the weight of the car. This is one I think you'see'
have to visualize in 3 dimensions to understand. That is, you have to
the straight line path versus the actual path of the car in 3D.
Rick
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Richard W. Tarara
Professor of Physics
Saint Mary's College
Notre Dame, IN 46556
219-284-4664
rtarara@saintmarys.edu
FREE PHYSICS INSTRUCTIONAL SOFTWARE
www.saintmarys.edu/~rtarara/software.html
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----- 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