Re: L2-"Negotiating" a curve.

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Thu, 4 Nov 1999, Richard Tarara wrote:

> Two different vector diagrams.  In the case of the inclined plane, you are
> breaking up mg into components along and normal to the plane.  mg ends up to
> be the hypotenuse of the vector triangle and N one of those components.  In
> the banked turn you are breaking up the normal into vertical and horizontal
> components.  In that vector triangle it is N that is the hypotenuse with mg
> one of the components.  I think that explains the differences.

This explanation is essentially true, but it might seem to imply
(improperly) that the difference is graphical rather than physical.  The
answers to physical problems can't depend on the coordinate system you
use--i.e., which forces you end up breaking into components.

In the case of the block sliding down or simply resting on an inclined
plane (frictionless or not but with no other forces acting on it), the
magnitude of the normal force normal force *is* mg cos(theta).

The case of a car rounding a banked curve with angle theta and radius R at
speed v is more complicated because the acceleration is not parallel to
the incline.  As long as the car is not slipping the normal force is given
by

  N = m[ (v^2/R) sin(theta) + g cos(theta)]

If, in addition, there is no friction, then mv^2/R = N sin(theta) which,
in combination with the general result above gives

  N = mg/cos(theta)

The bottom line is that they are different problems and one should not
expect them to have the same answers.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm