Re: FBD

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Fri, 22 Oct 1999, Ludwik Kowalski wrote:

> This is not different from the free body diagram of an accelerating
> Atwood machine block. We show only two forces, W and T; the
> m*a is not shown as a vector on the FBD diagram. We find the
> net force (W-T) and we equate it to m*a. Right?
>
> If I transferred the m*a to the left side (with the negative sign), and
> drew the -m*a vector on FBD, then  I would be saying "the particle
> is in equilibrium, the sum of all forces is zero".  Is this permissible
> for an accelerating object?

It works, of course, and is equivalent to going into the frame of the
accelerating object and properly treating the inertial force that arises
as a result.

As a matter of personal preference, I *do* instruct students to include
the acceleration vector in their FBD's because any knowledge about its
direction or magnitude is *just* as important as it is for each force.
However, because students often tend to take any vector they see
(including, for instance, velocity) and throw it on the "left side" of
NII, I have them draw the acceleration vector differently from the force
vectors to remind them that it is *not* a force.  My preference is to draw
the acceleration vector with two parallel tails, like an "implies"
symbol.  When we get to the dynamics of circular motion, I wear a button
that says "mv^2/r is *not* a force."

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