Contradiction in Couple Force: Can Its Point of Application Matter?

[klench]

We read that the a couple is a free vector, i.e it does not depend on the point of application.

However, please consider the following example: If you apply two equal magnitude forces anti-parallel to each other, one on each side of the wheel, a couple is formed, and so the wheel will rotate about the axis as expected. Now if take this couple and move it elsewhere (e.g., not at the wheel axis), the wheel will not turn in the same way, even though we did not change the magnitude or the direction of the couple moment. This implies that the external effect of a couple does depend on its point of application.

What am I missing? How do we resolve this apparent contradiction?

Thanks very much for all your help!

Kevin

P.S. This is my first post on this forum.

 

[jbriggs444]

You assert that a couple applied to the wheel at some pair of points far removed from the axis will not have the same effect. Have you tested that assertion?

 

[klench]

I'm phrasing the example differently to try to avoid misunderstanding:

We have a steering wheel. A force, F, is applied downward on the wheel handle at the 9 o'clock position; and there is another force, F, applied upward on the wheel handle at the 3 o'clock position. And say that the wheel diameter is d.

Now, find the resultant moment about the center of the wheel: F*d. It's clear to me that the wheel will rotate about its center.

Next, find the resultant moment of the forces about a point on the wheel handle at the 3 o'clock position (where one of the F is applied). The resultant moment is again F*d.

Here is where I'm confused/my intuition is wrong: it seems to me that if we apply the resultant moment at a point on the wheel handle at the 3 o'clock position, the wheel will not rotate about its axis in the same way it would if we apply the moment about the wheel's axis.

To answer your question, jbriggs, I haven't performed any experiment to test my assumption.

Thanks

Kevin

 

[jbriggs444]

Let me be sure that I understand the rephrased example. We have the same steering wheel in both cases. We have the same forces applied to the steering wheel in both cases. The computed moment is the same in both cases. The difference in the two cases is the choice of axis about which we choose to compute the moment.

The net force on the steering wheel is zero, so the center of the wheel must remain stationary in either case. What difference in the resulting motion do you anticipate?

 

[klench]

I agree with your first paragraph.

For the case of the couple being placed at the 3 o'clock position on the steering wheel, I would expect the couple to affect rotation of the wheel about that point (about the wheel handle at the 3 o'clock position). However, the reaction force in the axis of the wheel would oppose the couple, and so there would be not net movement of the wheel--no translation, and no rotation.

Or you could imagine this: grasp a round plate at one point near the edge with the one hand. Now, apply a couple--rotate the plate about the location of your fingers. The result is that the plate rotates about the location of your fingers (being sure not to translate the fingers up, down, left or right). But with the case of the steering wheel, the axis prevents the wheel from rotating about the point of the couple. So how can a couple be applied at different points of the object and still produce the same motion of the wheel?

What DOES happen to the wheel when we apply a couple at a point on the steering wheel handle? Would the wheel turn just the same as it would if the couple were placed at the center of the wheel (at its axis?).

Thanks for being patient with me.

Kevin

 

[Philip Wood]

What I'm not following is your talk (no pun intended) of placing a couple at a point. I want to think of a couple as a pair of equal but opposite (anti)parallel forces, and I don't see how it can make any sense to think of them "placed at a point".

 

[klench]

"Placed at a point" is just an indication of the location about which something rotates. Imagine a two dimensional object with a force acting on it (in the same plane as the object). Find the moment of the force about "a point" O on the object. The result is a vector pointing out of (or into) the paper (or computer screen, etc.). From this perspective, the object appears to rotate about "the point" O.

Practically speaking, I don't literally mean a zero-dimensional idealized point.

Kevin

 

[klench]

What DOES happen to the wheel when we apply a couple at a point on the steering wheel handle?

Maybe I should write: "What DOES happen to the steering wheel when we apply a moment about an axis passing through the steering wheel handle at the 3 o'clock position and perpendicular to the paper?"

 

[Philip Wood]

I'm probably being dense, but you seem to have segued from using couple to mean pair of equal and opposite forces not in the same straight line, to meaning moment due to a force. They are not interchangeable.

 

[klench]

I'm probably being dense, but you seem to have segued from using couple to mean pair of equal and opposite forces not in the same straight line, to meaning moment due to a force. They are not interchangeable.

I don't think you're being dense; I should have said "couple" in the quoted text in post #8.

In post #7, I was just using "moment" to illustrate what I meant by "placed at a point".

Maybe an assumption of mine is wrong: Imagine drawn on paper an object (unit thickness into the paper). If we have two equal magnitude, anti-parallel forces a distance d from each other acting on the object, and we take the moments of the forces (the result is a couple) about some location A on the object, I would expect all of the points of the object to rotate about A. And if we take the moments of the forces (again, the result is a couple) about some other location B on the object, I would expect all of the points of the object to rotate about B. I understand that in both cases, the magnitude of the couple is F*d, and the couples act out of (or into) the paper.

My understanding in the above paragraph seems to conflict with the statement, " In rigid body mechanics, force couples are free vectors, meaning their effects on a body are independent of the point of application." [from Wikipedia, "Couple (mechanics)]

I'm operating on some incorrect assumption(s); I'm just trying to figure out what it is (they are).

Thanks

Kevin

Kevin

 

[Dale]

Force couples are indeed "free vectors" in the sense above (I don't particularly like the term). Recall, that a couple is a pair of forces, equal in magnitude and opposite in direction. So, immediately from that it should be clear that a couple does not have a single point of application. Now, consider a clock face (neglect gravity) and the following couples:

1) a force F up at 9 o'clock and a force F down at 3 o'clock
2) a force 2F up at the center and a force 2F down at 3 o'clock
3) a force F right at 12 o'clock and a force F left at 6 o'clock

Can you see that all of these produce the same motion? That is what is meant by couples being independent of location.

Now, notice in example 2) how the force increased as the distance decreased. Would it surprise you to find out that the following couples also have the same effect as the above three forces:

4) a force 4F up at the center and a force 4F down halfway between the center and 3 o'clock
5) a force 4F up at halfway between the center and 3 o'clock and a force 4F down at 3 o'clock

The details of how large and where the equal and opposite forces are each applied do not matter, only the couple matters.

 

[klench]

DaleSpam,

Thank you for taking the time to write those examples; they were enlightening, and I agree with your statements.

Originally, regarding couples, I didn't want to trust the math because my naïve gut feeling of how a system would behave was wrong. I accept now that my gut feeling is wrong, and now I have to ingrain in myself a new (the correct) intuition, through study of examples, thought experiments and real experiments when I get the time and materials to do so.

Kevin