Re: Banked road

[Rick Tarara <rbtarara@SPRYNET.COM>]

Here's my take.  Look at the second situation.  N = mg/cosA is only true if
the object is in equilibrium.  That is not generally true if the object is
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 acceleration.
In effect the car is trying to drive in a straight line but the road is
pushing it towards the center point of the curve.  This INCREASES the normal
force to be greater than the weight of the car.  This is one I think you
have to visualize in 3 dimensions to understand.  That is, you have to 'see'
the straight line path versus the actual path of the car in 3D.

Rick

*************************************************
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
NEW:  Photo Realistic Laboratory Simulations
**********************************************************


----- 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