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Re: The drag force -- a correction squared



On Sat, 7 Oct 2000, brian whatcott wrote:

Two comments in response.
1) A high school teacher expressed his unease with a text book
that purported to show a relation between drag force and the
product of mass and velocity.

To be precise Justin's question was:

... an example I pulled from Sears, Zemansky, and Young had a
resistive force proportional to mv. The Drag force given by
Halliday and Resnick, however, does not depend on the mass of the
object in question at all. My question is: is the resistive
force in the Sears et al. problem an accurate picture of reality
or simply made up to simplify the math? It doesn't seem
reasonable to me that the resistive force would depend on the
mass...otherwise no one would think that heavy objects fall
faster!

Justin later wrote with the correction:

I was mistaken ... The example to which I was refering is from
Classical Dynamics (3rd ed., Marion and Thornton)

a well-known intermediate mechanics text.

brian went on to write:

There was one contribution to his question which was responsive.
It was mine.

I respectfully disagree. In more formal terms, the original
questioner said he had seen it written that

F_drag = c*m*v

and wondered specifically about the presence of the "m" in that
formula. After responding to some confusion that I thought had
been generated by posts from others including brian, I wrote:

Notwithstanding all of the above, there is nothing "wrong" with
*expressing* a drag coefficient as the product of the object's
mass with another positive constant. This is often done in
intermediate and advanced mechanics texts in order to eliminate
the explicit appearance of mass in the differential equation of
motion for objects subject only to gravity and drag.

Unless I really have misunderstood the original question, which I
sincerely doubt, I think this *is*, in fact, *the* explanation of
the observation which motivated that question.

Perhaps I need to be more concrete ...

Consider an object subject to uniform gravitation and a velocity
dependent drag force, F_drag = -b * v^n * v_hat where v is the
magnitude of the velocity, v_hat is a unit vector in the direction
of the velocity, b is a positive constant (the "drag coefficient")
and n is generally taken to be 1 or 2 in reasonable (but far from
perfect) models. In this case Newton's Second Law becomes

a_vec + b/m * v^n * v_hat - g_vec = 0

and one can see at a glance that the general solution of this
equation depends only on b/m, the ratio of the drag coefficient to
the mass. We can make that feature of the solution more readily
apparent (and many intermediate textbooks do) by *expressing* the
drag *coefficient* (as I said and meant) as c * m which yields
F_drag = c * m * v (i.e., "a resistive force proportional to mv".)
Then Newton's second law becomes the somewhat simpler *looking*

a_vec + c * v^n * v_hat - g_vec = 0

brian goes on to write:

The remaining efforts leave me in no doubt at all why HS teacher
discussions particularly from women are relatively scarce here -
the contributions do not respond to the original question.

I'm disappointed to hear this. I certainly hadn't intended for
*my* contribution to muddy the waters. I *was* trying to be
helpful and I thought that I had *directly* addressed the original
question. Under the circumstances, I would be interested to hear
from Justin as to whether or not he has yet had his question
answered with sufficient directness and, if so, by whom.

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