Re: The drag force -- a correction squared

[Jack Uretsky <jlu@HEP.ANL.GOV>]

Am I being particularly dense this Saturday?  My free-body diagram of an
object in free fall at terminal velocity shows two force vectors; they are
air drag up and mg down.  The two are equal, so air drag =mg.  Have I
missed something?  Or is Brian nit-picking on my use of the word
"general"?
                                        Regards,
                                                Jack

Adam was by constitution and proclivity a scientist; I was the same, and
we loved to call ourselves by that great name...Our first memorable
scientific discovery was the law that water and like fluids run downhill,
not up.
                          Mark Twain, <Extract from Eve's Autobiography>

On Sat, 7 Oct 2000, brian whatcott wrote:

> At 11:35 10/7/00 -0500, you wrote:
> > Dear Justin-
> >        In the absence of a verbatim statement of the problem (which you
> > should have included), I can only give you a general statement as an
> > answer.  At terminal velocity, the drag force of a falling object is
> > proportinal to the mass; the up and down forces are in balance, the up
> > force is the drag force and the down force is mg.
> >        I would not want to critique a paraphrase of someone else's
> > statement.
> >
> >                                Regards,
> >                                        Jack
>
> Jack will undoubtedly want to reconsider the general statement
> that the air drag on an object is proportional to its mass at
> terminal velocity.
>
> It is not.
>
>
> brian whatcott <inet@intellisys.net>  Altus OK
>                 Eureka!