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Re: [Phys-L] treating force as a vector ... consistently

You use the phrase "pulling the student along". But the rope is stationary - it does not pull the student along. The rope exerts a force on the hands and the hands exert a force on the rope. But there is no more motion from that than if the student were hanging from a rope. It is the muscles in the arm that are "pulling the student along". This muscular motion is very complicated. The muscles will do real work - the rope does "virtual" work. I applaud you for being able to use that example as an introduction to force pairs and related motions.

Bob at PC
________________________________________
From: Phys-l <phys-l-bounces@www.phys-l.org> on behalf of Todd Pedlar <pedlto01@luther.edu>
Sent: Sunday, August 28, 2016 3:43 PM
To: Phys-L@phys-l.org
Subject: Re: [Phys-L] treating force as a vector ... consistently

My students don't seem to get confused by this. What you definitely in my
opinion do NOT want is another student on the end - but what you want
instead is the agent of the force that is pulling the student along to be
inanimate (just as the agent of the force that keeps the box from falling
through the table is inanimate). The sequence of action-reaction pairs
isn't that difficult to describe - and all that is needed is to deal with
the one end, since the only force that can pull the student is one that
acts directly on the student. If the exercise devolves into talking about
friction between the hand of the student and the rope, then it's gotten too
complicated.

An alternative is to have the student jump, and have her describe what
force she has to exert, on what, and in what direction, in order to make
herself leave the floor (and better if one has a force platform for this
exercise).

Todd

On Sunday, August 28, 2016, LaMontagne, Bob <RLAMONT@providence.edu> wrote:

As a student, I would be a little confused by this. It is one thing if you
sit in a chair holding one end of a rope and another student pulls you. It
is another (and more confusing) thing for a stationary rope to somehow
magically pull you along. You now have to consider the motion of you
muscles and all kinds of complicated "action-reaction" pairs.

Bob at PC
________________________________________
From: Phys-l <phys-l-bounces@www.phys-l.org <javascript:;>> on behalf of
Jeff Bigler <jcb@alum.mit.edu <javascript:;>>
Sent: Sunday, August 28, 2016 1:58 PM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] treating force as a vector ... consistently

I agree with pretty much all of this, but it begs the question of how you
would describe the N3 force pair in the example of the book on the table.

One of the demos I do for N3 is to put a student in an office chair with
wheels, tie a rope to a lab table, and ask the student to apply a force to
the rope (and say which direction the applied force is in), and then ask
the student to explain, referencing the forces involved, why they moved in
the direction they did.

Jeff Bigler
Lynn English HS; Lynn, MA

Sent from my phone, probably while multitasking and possibly using
voice-to-text. Please pardon any typos, speakos, autocorrect errors and
other vagaries, and ask about anything that doesn't make sense.

On Aug 28, 2016, at 1:24 PM, Scott Orshan <sdorshan@aol.com
<javascript:;>> wrote:

I thought I'd drop my two cents in, and describe how I teach Newton's
3rd.

As a preface, I'm not big on using momentum as a beginner concept. I
can't see or feel momentum, and it changes with the frame of reference.

I *can* feel forces and their effects, so I prefer presenting physics
from the force model with beginners.

The first part of the approach is to get rid of the words "action" and
"reaction". They are filled with varied and contradictory meanings from all
walks of life. "I hit Jimmy, so he hit me back." Not a N3 force pair. "I
pull on one end of a rope, and you pull on the other end." Not a N3 force
pair. "The stuff shoots out of the rocket so the rocket moves in the
opposite direction." Not a N3 force pair. The N3 force pair(s) in a rocket
are too complex to contemplate, and result from the pressure gradient in
the combustion chamber and hollow areas of the rocket. The rocket formula,
on the other hand, is a simple conservation of momentum formula.

The question/answer pair "Q: How does a rocket work? A: Newton's 3rd
Law." says nothing about either rockets or Newton's 3rd Law.

The second part is to define the nature of N3 force pairs.

They happen between the same objects.

They are equal in magnitude.

They are opposite in direction.

They are the same type of force (gravity and contact [EM] are of most
concern in the beginner classroom).

This next one is not one I've seen mentioned, but I think it's an axiom
as well:

They start and end together. Same start time, same duration, or the
slightly more complicated parallel for variable forces. (In the real world,
reactions tend to follow actions.)

That last axiom means that the weight of a book can not be a force pair
to the table's normal force. The book was interacting with the Earth long
before it was placed on the table.

Two vectors pointing in the same direction with the same magnitude may
be mathematically equivalent, but in Physics, the vector has units and
other meanings.

In math, 3=3, but in Physics, 3 only equals 3 if the units are the same.

Real world forces exist as pressures (not applied at a point), and real
world objects are deformable and/or elastic. That's what differentiates an
engineer from a scientist. Two elephants pulling on you from opposite sides
is very different from two elephants pushing on you from opposite sides.
However, the net force is zero in both cases.

Scott Orshan


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_______________________________________________
Forum for Physics Educators
Phys-l@www.phys-l.org <javascript:;>
http://www.phys-l.org/mailman/listinfo/phys-l
_______________________________________________
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Phys-l@www.phys-l.org <javascript:;>
http://www.phys-l.org/mailman/listinfo/phys-l



--
Todd K. Pedlar
Associate Professor of Physics
Luther College, Decorah, IA
pedlto01@luther.edu
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