I How Does the Position of a Free Vector Couple Affect Its Behavior?

[Trying2Learn]

TL;DR Summary

If couple is a free vector, then I can move it. However, I think the response is different

Hello

(Everyone here has been so helpful -- thank you. Things I thought I knew, I now doubt; and this is so helpful to have this group.)

There is an current discussion on Yaw. I am enjoying that. And that raised an issue for me.

However, I do NOT want to hijack that thread, so I am posting my question here.

I know that a couple is a free vector -- that means it can be moved. However, as I move the couple around, I see (in my mind) the object behaving differently, depending on where I "situate" the couple.

Could someone explain this to me?

See the attached PDF

BTW: I do understand how it is only the distance between the application points that matters. However, I cannot reconcile what I know to be true, with what I "imagine" happens.
https://www.quora.com/Why-is-couple-moment-a-free-vector

 

[wrobel]

Couple/sliding vector/free vector etc are just outdated and archaic terms. They are not used now. Find a good textbook

 

[Trying2Learn]

Couple/sliding vector/free vector etc are just outdated and archaic terms. Find a good textbook

Well, this is the most fascinating comment (and I am NOT being facetious). PLEASE take a few minutes to explain this. You have hit a nerve and i want to understand.

 

[BvU]

I know that a couple is a free vector

Leads to inconsistency $\Rightarrow$ must be incorrect.

How do you 'know' that ?

Even a regular force is not a free vector. It has a line along which it acts. Moving the force along the line of action has no consequences.
Moving a force perpendicular to the line of action means adding a torque :

 

[etotheipi]

You have some misconceptions. The torque of a couple is indeed coordinate independent, i.e. $$\bar{\boldsymbol{\tau}} = \sum_{i=1}^2 \bar{\mathbf{x}}_i \times (-1)^i \mathbf{F} = \sum_{i=1}^2 (\mathbf{x}_i - \mathbf{X}) \times (-1)^i \mathbf{F} = \sum_{i=1}^2 \mathbf{x}_i \times (-1)^i \mathbf{F} = \boldsymbol{\tau}$$That means, the torque of the two forces $\mathbf{F}$ and $-\mathbf{F}$ on a body $\mathcal{B}$ is the same with respect to any frame $\mathcal{O}xyz$, i.e. whether you choose $\mathcal{O}$ to be at the centre of the rod, the end of the rod, whatever. But if you want to analyse the dynamics, note that the centre of mass ($\mathcal{S}$) acceleration is zero,$$\ddot{\mathbf{x}}_{\mathcal{S}} = m^{-1} \sum_{i=1}^2 (-1)^i \mathbf{F} = \mathbf{0}$$i.e. the centre of mass moves at constant velocity with respect to an inertial frame. Meanwhile, angular velocity at time $t$ is determined fully by $\boldsymbol{\tau}$, i.e. it's easiest to just pick $\mathcal{O} = \mathcal{S}$ and write $\boldsymbol{\tau} = \dot{\mathbf{L}}$.

What the motion looks like depends on which frame you choose. In lab frame, for setup in your picture it just rotates about its mass centre $\mathcal{S}$, whilst for [non-inertial] frame with $\mathcal{O}$ at end of rod it rotates about the end of rod.

 

[wrobel]

Well, this is the most fascinating comment (and I am NOT being facetious). PLEASE take a few minutes to explain this. You have hit a nerve and i want to understand.

To write equations of mechanics and to solve any problem it is sufficient to know only standard geometric vectors. See for example Classical Dynamics by Donald Greenwood.

 

[kuruman]

Summary:: If couple is a free vector, then I can move it. However, I think the response is different

However, as I move the couple around, I see (in my mind) the object behaving differently, depending on where I "situate" the couple.

We are not mind readers, just text readers. To answer your question directly, one needs to understand what you see in your mind. So please explain in your own words what that is and, specifically, what kind of "different" behavior you think the object exhibits "depending on where you 'situate' the couple."

 

[BvU]

How do you 'know' that ?

Found it (via: quora $\rightarrow$ PF $\rightarrow$ Wiki (add a right bracket) ) .

 

[Trying2Learn]

We are not mind readers, just text readers. To answer your question directly, one needs to understand what you see in your mind. So please explain in your own words what that is and, specifically, what kind of "different" behavior you think the object exhibits "depending on where you 'situate' the couple."

Ah, no need...

etotheipi nailed it when he mentioned which frame one views it in. I see now.

Sorry to bother everyone.

But thank you!

 

[etotheipi]

@Trying2Learn just to make sure, do you understand that the moment vector itself is an object that can be "moved around" however you like, but in no way does "where you choose to visualise it" affect the evolution of the system? Or in other words, that it has no relation to the axis about which the system is rotating, or anything like that?

 

[Trying2Learn]

@Trying2Learn just to make sure, do you understand that the moment vector itself is an object that can be "moved around" however you like, but in no way does "where you choose to visualise it" affect the evolution of the system? Or in other words, that it has no relation to the axis about which the system is rotating, or anything like that?

YES! Thank you so much for that. It is obvious, right? And I am aware of how to visualize it.

I just never... visualized it. :-)
Thank you!