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I've always taught my students that your weight is what the scale readsthat
you're standing on. This works whether you're accelerating in an elevator,but
in orbit around the earth, standing still on the earth's surface, or
anyplace else I can think of.
Paul O. Johnson
.
----- Original Message -----
From: "Larry Smith" <larry.smith@SNOW.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Thursday, February 07, 2002 12:14 PM
Subject: definition of weight (again)
This debate seems to resurface every year (or is it every semester?),
body,it would be nice if we could all agree on the definition of weight.
Here's what Hewitt says on page 159 of Conceptual Physics (9e): "In
Chapters 2 and 4 we defined weight as the force due to gravity on a
bemg. Your weight does have the value mg if you're not accelerating. To
weighingmore general we now refine this definition and say that the weight of
something is the force it exerts against a supporting floor or a
scalar,scale. According to this definition you are as heavy as you feel."
Kirkpatric and Wheeler (4e) generally agree with Hewitt's _revised_
definition, with the caveat that they say weight is the support force
itself (the "reaction" force of Hewitt's weight, equal, of course, in
magnitude, but opposite in direction).
But Serway and Beicher (5e) say on page 119: "the weight of an object,
being defined as the magnitude of F_g, is mg." Other than being a
definition.Serway's definition agrees with Hewitt's original "unrefined"
"TheGriffith
Hobson, in Physics: Concepts and Connections, p. 99, seems to agree with
Serway but keeps the vector: "The weight of an object refers to the net
gravitational force exerted on the object by all other objects."
(The Physics of Everyday Phenomena p. 64) agrees that it is a vector:
toforce of gravity acting on an object is what physicists commonly refer
talkas the weight of the object."
This issue is important enough by itself, but it also affects how we
toabout weightlessness and the definition of g.
I guess I'm looking for a little closure on this, folks. Is there any
be had? Consensus, please.
Thanks,
Larry