Re: Newton's Second Law of Motion

["John S. Denker" <jsd@MONMOUTH.COM>]

Abul Kalam wrote:
> Introductory textbooks have routinely stated Newton's Second Law as: a =
> Sigma(F)/m, or Sigma(F) = ma.

That's odd.

Normally people start with plain old
        F = m a.                [1]

Writing it as, say
        Sigma(F) = m a          [2]
introduces extra complexity with disproportionately-small
improvement in correctness, accuracy, or generality.

Quite independent of Newton's laws, we know that we can
add forces and represent the sum by a single "resultant"
force.  So we could equally well write
  F_total = m a
but if we're going to be so fussy about the LHS we ought
to pay a comparable amount of attention to the RHS,
perhaps something like
  F_total = m_total a_center-of-mass

But normally we just write
  F = m a
and explain elsewhere what the symbols mean.


> This is a special case and is true only when the
> mass is constant.

OK.

> There should also be an equal emphasis on the fact
> that it is possible to generate a force at a constant velocity when the
> mass changes as well, such as in rocket motion or in water jetting out
> of a hose.

Those aren't really counterexamples to the F=ma law.
F=ma is valid moment-by-moment even for a rocket where
the mass (in a certain region) is changing.

And by the way, the recommended procedure (especially in
introductory courses) is not to worry about regions but
rather to keep track of individual particles.  A complex
object (such as a rocket) can be decomposed into particles
and the F=ma law applies to each particle separately (as
well as applying in the overall center-of-mass sense
mentioned above).

If you want a counterexample to F=ma, look to relativity.

> Even Newton does not state his Second Law the same way,
> instead indicating that force is proportional to the rate of change in
> momentum. I think that should be the more general and the correct way to
> state the second law.

You can't go wrong with that.  The flow of momentum is
sometimes easier to visualize than the corresponding
forces.

And F=dp/dt remains valid in relativistic situations.

In any non-relativistic situation, F=dp/dt is
formally equivalent to F=ma, so you can't think of
it as a "solution" to the alleged "problem" above.
But since I don't think the alleged problem is a
real problem, no solution is required.

If you still think there is a problem with F=ma
in situations such as (non-relativistic) rocketry,
please ask a more specific question.