How are force couples free vectors?

Question

According to wiki,

In rigid body mechanics, force couples are free vectors, meaning their effects on a body are independent of the point of application.

I'm having some trouble understanding this statement. Consider the following scenarios:

Here, $O$ is the center of mass of the solid and uniform bar. Here, $|\underline{u}|=|\underline{v}|=5N$. Here, $\underline{u}$ and $\underline{v}$ constitute a force couple.

Scenario 1:

The two forces will have no resultant force, but they will have a resultant torque.

Scenario 2:

Here, since $\underline{u}$ and $\underline{v}$ are free vectors according to wiki, I've moved them such that they are facing each other directly. Now, there will be no resultant force nor will there be a resultant torque.

My question:

According to wiki, scenario 1 and scenario 2 should have the same effect on the bar. However, the effects are different. So, isn't wiki wrong?

Please write $u$ symbols such that they don't look like $\mu$ or $M$. — leftaroundabout, Mar 09 '22 at 15:10
Great how you're researching this. I have seen your post on MSE as well — tryst with freedom, Mar 09 '22 at 18:57
I myself had a similarish doubt. It seems like a full description of the issue is beyond what we learn in School. If you still wish to learn, look up the term Screw theory. It basically comes under something called Projective geometry. I found this video a good introduction to the ideas ofi t — tryst with freedom, Mar 10 '22 at 14:38
There is a user named John Alexiou on PSE who has written some answers to this. You can check that out as well for some information. — tryst with freedom, Mar 10 '22 at 14:40

Farcher · Accepted Answer · 2022-03-09T11:11:57.993

6

In rigid body mechanics, force couples are free vectors, meaning their effects on a body are independent of the point of application.

You have misinterpreted the Wiki statement.

Given a couple, two equal magnitude, parallel, non-linear forces, it does not matter where those two forces act on the body as long as the couple, $Fd$ anticlockwise in this case, stays the same.
You to deal with the couple as a whole.

One last point.
The torque on the body as a vector in all cases is $\tau_{\rm couple} = F\,d\,\hat z$ where $\hat z$ is a unit vector pointing out of the screen and note there is no mention of the forces, separation, position etc.

edited Mar 09 '22 at 11:11

answered Mar 09 '22 at 09:41

Farcher

95,680

"note there is no mention of the forces, separation, position etc."- sir I didn't understand your last point. Could you please clarify it? – tryingtobeastoic Mar 09 '22 at 12:29
1

It just means that without any extra information you can say that the couple acting was $3,\rm Nm$ (Plus a direction) without having to say anything more and that specifies the couple completely. – Farcher Mar 09 '22 at 14:58

Ethan Dandelion · Answer 2 · 2022-03-09T06:47:27.997

3

I'm currently studying statics of rigid bodies as well. I think this is a matter of addressing what kind of motion we're dealing with.

The act of adding the forces acting on a body is in an effort to determine whether or not the body will undergo any translational motion. If there is a net force on the object, then it should move in the direction of the line of action of the net force. As you can see in this example, as there is no net force, the rigid body does not undergo any translation – we could say that the center of mass of the body does not undergo displacement.

What we're dealing with here is the moment of the force. This describes the rotational, not translational motion of the object. In Scenario 1 there is a "net" moment because summing the moment vectors will show us that they do not cancel out. However, in Scenario 2 computing the moments, then adding them vectorially will show us that they sum to zero, and thus there is no rotational motion in the object as well.

As you say, the Wikipedia article is not correct because it lacks precision. The statement, "their effects on a body are independent of the point of application" only addresses translational motion and not rotational motion, which is a part of how rigid bodies are affected by forces.

edited Mar 09 '22 at 06:47

answered Mar 09 '22 at 06:25

Ethan Dandelion

725

According to wiki, "In rigid body mechanics, force couples are free vectors, meaning their effects on a body are independent of the point of application."- however, the effect on the bar in scenario 1 and 2 aren't the same, so isn't wiki wrong? – tryingtobeastoic Mar 09 '22 at 06:31
2

@tryingtobeastoic I see what you're saying. The Wikipedia page, as far as I understand, is not precise enough: It fails to specify that "their effects on a body are independent of the point of application" only insofar as translational motion goes. I noticed that the article is not locked, so I would make an edit if I were you! :) – Ethan Dandelion Mar 09 '22 at 06:35
If you include this comment in your answer, I'll be glad to upvote your answer! – tryingtobeastoic Mar 09 '22 at 06:42
2

The wiki quote says that "force couples" (my emphasis) are free vectors. It does not say that the force vectors separately are free vectors. You can't move the two points of application freely you must maintain the same couple.. – mike stone Mar 09 '22 at 12:56

score 3 · Answer 3 · answered Mar 09 '22 at 06:32

I agree with you. Their locations on the geometry is only irrelevant for their effect on translational motion. For rotational motion their points of application obviously matter. The Wikipedia article may be sloppy writing in this case.

In neither of your two cases there will be translational motion (of the centre-of-mass) since the two forces cancel out when considered linearly. But rotationally, as you describe yourself, the former example will show rotation.

How are force couples free vectors?

3 Answers3