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Re: [Phys-l] g...



On Nov 18, 2006, at 3:29 PM, Anthony Lapinski wrote:

There's confusion with this in many places (esp. "g" forces). I tell my
students that "g" is an acceleration, not a force. When I ask for free
body diagrams, they usually say one of the forces is "gravity" downward.
Well, it's technically "weight" downward. Forces always involve two
objects, so the Earth's gravity is not a force.

The more difficult concept is teaching students that the shuttle
astronauts are not truly weightless (i.e., g ? 0 up there in orbit)...

Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu> writes:
I have a problem with g. Despite my best efforts many of my students
still want to say that g is a force. I can see where the whole thing
would be confusing for them. Our first experience with g is to call it
the acceleration of gravity on Earth. I stress over and over that it is
not "gravity", it is not "the force of gravity", it is the acceleration
that objects undergo as they fall to the Earth. However, bout the time I
think that I have that idea pumped into their heads we start using g to
find weights w=mg. Now the object is not accelerating at all so what's
g? I usually only address the idea that g is now being used as a
constant that represents the strength of the gravitational field if one
of my students brings it up, and that's not very often. Maybe that's a
mistake. Now g is appearing once again in Newton's Universal Law of
Gravity this time twice... Fg = G m1 m2 / d^2 I asked my students to
describe what g was on a quiz yesterday and got back answers that
were all over the map. Any advice?

Yes, allow them to say that g is weight par unit mass. It is not the entire weight of an object on earth. The total weigh is often different for different objects, but the weight per unit mass happens to be the same for all of them. Guide them to discover this fact in a lab. To them it is much less obvious than to us.

And then emphasize that g is also an acceleration. What unit do we obtain when we divide N by kg? How come, you might then ask, that a number we get, when we measure a free-fall acceleration of any object is always so close to the weight per unit mass?

Ludwik Kowalski
Let the perfect not be the enemy of the good.