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Re: [Phys-l] bound vectors ... or not

And consider two ropes attached to a rigid object--one is being pushed eastward and another is being pushed northward. To find the resultant force we slide vectors along the lines of their action, till their tales are superimposed. Yes, ropes represent reality while vectors are mathematical abstractions, used to solve problems.

Ludwik



On Sep 7, 2010, at 10:10 AM, Philip Keller wrote:

Not to get all philosophical, but the rope is not the force. It is the rope that is attached at a given point. The force is a vector that has magnitude and direction -- but we are permitted to translate the force vector, as we do when we make tip-to-tail diagrams.

I also disagree with those who say "you can't move the force or the torque will change". If you move the ROPE, the force torque will change. But in a given situation, you can move or not move the force vector as you feel like. Here's what I mean"

You want to calculate a torque associated with a given force. You need to know two vectors:

1. The force vector, which has magnitude and direction only
2. The lever arm vector, which has magnitude and direction only. It is the displacement vector from the chosen pivot to the point where the force is applied. That point is fixed by the physical situation (but the force itself is not a "bound" vector, whatever that may be).

The magnitude of the torque is always |r| |f| sin (angle between them).

None of these quantities change if you choose to translate your r or F vector. So they are still "free" and their freedom does not affect the value of the torque.

Now it is true that it is often more convenient to calculate the torque by using the perpendicular distance to the line of action of the force as the lever arm. If you are using that particular mathematical shortcut, then you can't move the force vector around willy-nilly. But that's just your choice of mathematical method. I don't think it should create a new category of vector type.


-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of Folkerts, Timothy J
Sent: Tuesday, September 07, 2010 9:02 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] bound vectors ... or not

I'm saying if I tie Rope B to a Object A, the force F(on A due to B)
*is* applied to a specific point (or can at least be integrated to one
effective point).

The vector we use to describe the magnitude and direction of the force
is not "localized' or physically bound anywhere. That same vector could
be used to describe the magnitude and direction of any number of other
forces (perhaps the force of you pushing on the other side).

These two forces are different forces with the SAME magnitude and
direction. As such they produce the same result -- the same
acceleration (when applied independently)-- but they are not the "same
force". To do vector addition, I can move the vectors around wherever I
want (but that doesn't move where the rope is tied).

Is kind of like the $1 bill in my pocket and the $1 bill in your pocket
have the same value, but they are not the "same $1 bill".

Tim Folkerts


-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Herbert
Schulz
Sent: Tuesday, September 07, 2010 7:50 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] bound vectors ... or not


On Sep 7, 2010, at 7:41 AM, Folkerts, Timothy J wrote:

It seems that "bound vector" relates to a "specific force"
and "free vector" relates to a "general force".

If I talk about F(AB) = the force on "A" due to "B", then I am talking
about a push with a specific magnitude and a specific direction AND
located at a specific point on "A" (assuming a "point force" for
simplicity).

As far as solving the equations of motion, this is an over-specified
condition. Any OTHER force of the same magnitude and direction would
cause the same acceleration of "A". It doesn't need to be this
specific
F(AB) (ie "bound vector"). It could any other similar force (ie free
vector).


So F(net) = ma works for ANY (free) vector, and it works in particular
for this specific (bound) vector F(AB).


Just my $0.02 ....

Tim Folkerts

Howdy,

Are you meaning to say that those vectors are physically attached to the
point? I can't ``move'' the vectors so they are tail to head to get the
sum of the vectors?

I'm not getting the concept of a ``bound vector'' at all.

Good Luck,

Herb Schulz
(herbs at wideopenwest dot com)



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Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
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_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
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Ludwik

http://csam.montclair.edu/~kowalski/life/intro.html