expurgated version: "Negotiating" a curve

[John Denker <jsd@MONMOUTH.COM>]

At 10:25 PM 11/7/99 -0500, Ludwik Kowalski wrote:

> 1) There were no external northwest push on the tricycle, the
>      only external forces are coming from the road. That is why
>      I did not try to follow the problem you wanted me to analyze
>      John. It was a good problem, but more complicated. Let us
>      deal with three forces first, then add another force. OK?

Not really OK.  If you had trusted me a little more, we'd be done by
now.  The homework I posed is not more complicated.  It is actually less
complicated, because it forces the key concept to the foreground.

> The words such as tensor, eigenvalue, etc., are not acceptable
> in the introductory physics course. (Even a teacher, like me,
> only vaguely remembers what stands behind such words.

OK, here's a version that doesn't use unacceptable terms like t*ns*r,
m*tr*x, or **g*nv*l**.

It is OK to multiply a vector by a scalar.  But make sure it's a
scalar.  If you ever find yourself trying to multiply a vector by two
different scalars at the same time, STOP!  You've got a problem.  You don't
have two scalars;  whatever you've got isn't a scalar at all.

This problem is fairly common.  Suppose that for vectors in a special
direction AAA you know you should multiply by a value aaa, while for
vectors in another direction CCC you know you should multiply by a
different value ccc. The key point is that for some intermediate direction
BBB the right answer is **not** to multiply the vector by some intermediate
value bbb.  Multiplication by scalars can only change a vector's magnitude,
but here we are faced with a situation where the physically-correct answer
involves a change in direction as well as magnitude.

You know intuitively that if you have a wheeled cart pointing north and
apply a push in the northeast direction, it will move north not
northeast.  So clearly a correct description of the motion cannot involve
simply multiplying the force vector by a scalar.

The elementary way to deal with a situation is as follows.  The first step
is to resolve the vector BBB into components aligned with the two
directions where you actually know what's going on, namely the AAA and CCC
directions.  Deal with these two components separately, then add the
contributions back together as a final step if necessary.

===============

Interesting tangential riddle:  You will find that for "most" problems you
encounter, the two directions (AAA and CCC) in which the behavior is simple
(almost scalar-like) will turn out to be perpendicular.  Is this a
coincidence?  Is it because the problem-posers are artificially selecting
simplified problems?  Or is there a law of nature at work here?  Exhibit a
proof or a counterexample.