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Re: Newton's third and centrifugal forces.

Main point here is that the contact force between the ground and me IS NOT
the same as the gravitational force of the earth (a hard point for students
as well). I still contend that the gravitational forces between myself and
the earth are equal and opposite (I'll stand still so we don't have to
consider propagation times.)

The force of the ground on me is equal to the force of me on the ground, but
that is an E&M contact force and at the equator is LESS THAN the
gravitational force. When testing for 3rd law understanding I
expect both the 'weight' pair and the 'contact' pair to be identified.

Let's call the gravitational force on me W (weight ;-) and the ground on me
force N. Let towards the center of the earth be the + direction. Then W-N
= MV^2/R. Just two forces. I stand by my reasons for the scale reading
less and the _sensation_ of being lighter (versus no rotation).

To be explicit only the difference W-N provides the needed downward force,
but that difference comes from the gravitational force being larger than the
contact force--this is what I meant by the fact that the earth's pull
provides the needed net force.

The tendency to want to fly off is just a result of N1 and not a force at
all. Rather, it requires a force to prevent this from happening--again all
from the external viewpoint.

Rick



----- Original Message -----
From: John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Thursday, October 21, 1999 4:12 PM
Subject: Re: Newton's third and centrifugal forces.


On Thu, 21 Oct 1999, Rick Tarara wrote:

Since I don't fully follow Leigh's reasons for his approach, let me
detail
how I would deal with me standing at the equator--maybe it will help to
be
explicit about this.

There are only two forces acting on me. The Earth's gravitational force
and
the ground pushing up. The earth's force is GMm/R^2

Subject to appropriate caveats about corrections due to asphericity I will
accept this as being the usual Newtonian meaning of gravitational force.

and I do exert a force of GmM/R^2 on the earth.

Not true. The force you exert on the ground is less than that.

I am accelerating and therefore need a net force towards the center of
that rotation of magnitude v^2/R. That force is provided by the earth's
pull on me.

Well... Far *more* than that net force is provided by the "earth's pull"
on you. But I think you do understand that.

In order that there be this net force towards the center, the force of
the ground on me must be less than GMm/R^2. It is.

Right. And so, by Newton's third law, is the force of you on the ground
in contradiction to your earlier statement.

Thus there is a net force towards the center. Why doesn't the ground
push up with as much force as the earth pulls down? Because of my
tendency to want to fly off in a straight line (N1) which could be made
more obvious if we sped up the earth's rotation considerably.

So now you have switched frames. You are no longer looking from the
outside but are now are working *in* your own natural and "noninertial
frame" and you are (properly) taking the centrifugal force into account.
The two upward forces balance the single downward force since your
acceleration in your own frame is zero. In any event we find that the
force of contact between you and the earth is less (in both directions!!)
than the "earth's pull."

There are no centrifugal forces here. The net downward force (from
gravity) provides the needed centripetal acceleration.

?? O.K. Now I *am* confused. Unless you are simply avoiding the word
"centrifugal" in favor of the more generic "inertial force"; a position I
endorse since there is no local way to distinguish inertial forces of any
kind from each other.

Now I feel somewhat lighter and a scale would read my weight to be less
than
GMm/R^2. The latter because if I am standing on the scale, IT is
providing
the upwards push which is less due to my tendency to fly off. I feel
lighter because I am accustomed to EXPERIENCING my weight as the upward
resistance to the downwards motion I would have due to the earth's pull
if
the ground/floor/scale were not there.

... the downwards *acceleration* you would have ... Otherwise, right.

These same arguments can be extended to other rotating frames where ONLY
a
net force towards the center is actually acting. I realize that the
problem
is that if you are IN such a frame you would 'experience' an outward
force,
but it is customary in intro courses (and I think most Newtonian
approaches)
to always look at these rotating frames from outside.

That is true. Which is why I was surprised to see you doing the opposite
in talking about your "tendency to want to fly off" (known to other
folks as "centrifugal force") above.

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