Re: Re; a mechanical problem

[Herb Schulz <herbs@interaccess.com>]

>   While the analysis of Ludwik Kowalski's given by Herb Schulz is certainly
> correct, it misses the problem.
>  Ludwik's confusion is due to the inconsistant usage of "reaction force"
> in many texts and in different situations.
>  When the block resting on the plate are both at rest, the Normal Force
> and the Reaction Force are identical. If the plate and block have a common
> acceleration, due to an external force on the plate, then the plate exerts
> BOTH a Normal Force AND a Friction Force.
>  The resultant of these TWO forces CAN be considered the "Reaction Force"
> of the plate on the block, and is therefor oblique. The Free-Body Diagram
> of the block would then show ONLY TWO forces, the weight of the block
> and the oblique reaction force.
>  It is cleaner, IMHO, to stick with the THREE forces, as Schulz did,
> the weight of the block, Normal Force due to the plate, and Friction Force
> due to the plate.

Howdy,

When I introduce the Normal force in class I start by talking about the
oblique force the surface applies on the block.  I then resolve that force
into two components parallel and perpendicular to the surface and state
that the component of the force perpendicular to the surface is called the
Normal force and that parallel to the surface is called the friction force.
then the students understand what I mean when I talk about frictionless
surfaces (even if they don't really exist).

There is always confusion over "action-reaction" pairs.  I try to emphasize
that  action-reaction pairs can NEVER act on the same body: A applies a
force on B and therefore B MUST apply an equal and opposite force on A.  I
also emphasize that only the forces acting ON A are needed to discuss the
motion of A with a similar statement for B.

Good Luck,

Herb Schulz
(herbs@interaccess.com)