Why are couple moments free vectors?

Question

I don't understand how couple moments are free vectors. If I applied a couple moment to different points on a stick, it would rotate very different.

I drew a picture

Answer 1

They are the same situtation! Your intuition is getting in the way though. You are probably envisioning holding a stick at those exact locations and then applying a force to make it rotate. When you do that, the forces that are produced by your hands are not the forces that you draw. What you have to envision is two rockets strapped to a stick at the locations that you specify, and the stick is also free to rotate in space. We don't have intuition for this situtation because we rarely see two rockets strapped to a stick in our daily lives, so we have to resort to the math.

The math says that the center of mass will accelerate according to the net force. The net force is zero, so the center of mass will stay in place.

The math also says that the object will start to rotate according to its net torque. Here the situation becomes more subtle, because we have to pick the rotation center around which to calculate the torque ourselves. A straight line acceleration can also be seen as rotation, if we pick the center point outside the path of acceleration. But if we pick our point smartly, the motion becomes a rotation around that point. We know the center of mass stays in place, so this is a perfect point to pick. If we now calculate the torque, it becomes $\tau=Fd$ in both cases, where $F$ is the force produced by a single rocket and $d$ is the distance between the two rockets. For the force couple, it turns out that our answer does not depend on the point that we pick.

Answer 2

This is about the abstraction that is involved whenever we apply maths to physics. When we draw an arrow on a diagram, to represent a force, it is good practice to place the arrow at the location where the force acts, and that is what you have done, correctly. But some aspects of the response of the object to the force do not depend on where the force acts. For example, the acceleration of the centre of mass of a rigid object does not depend on the location of the applied force.

There is a similar result for rotations, but it is less intuitive.

If you have a rigid body which is otherwise free, and apply to it a pair of forces so as to produce a couple but no net acceleration, the body will start to rotate about its centre of mass, not about a point in between the locations of the place where the forces are applied. That is what may strike us as surprising. This is because our physical intuition is influenced greatly by contact forces---the kind we apply with our fingers or hands. If we hold a stick in two hands and then make the stick rotate, for example, then me might unconsciously adjust the forces we apply to it so as to make the stick rotate about a point chosen by ourselves. In this case it may be that we make the centre of mass of the stick accelerate. In that case we would not by applying a pure moment, but an overall force as well.

To free your intuition a bit, imagine a stick with two opposite electric charges attached to it. Place the charges at some fixed separation, but anywhere along the stick. Now place the stick in an electric field, at right angles to the field. Finally, ask yourself what will now happen. The stick has no net force on it, so its centre of mass will not accelerate. But it does have a moment acting on it, so it must begin to rotate. Therefore it must rotate about its centre of mass! This result holds, no matter where you placed the electric charges. That is why moments can be regarded as vectors.

Note that physical location is not entirely irrelevent here, because the size of the moment will depend on the distance between the charges. But once you have a moment $\bf C$ of given size and direction, the rotational response is related to it by $$ \frac{d{\bf L}}{dt} = {\bf C} $$ where $\bf L$ is angular momentum. (In applying this equation you have to pick an axis in order to define both $\bf C$ and $\bf L$, but the equation applies no matter which axis you pick.)

(Postscript. By the way, I have never encountered the terminology "free vector" and I am an experienced physicist. This means that this terminology is not as widely established as you might think. The reason is that all vectors are "free" vectors, from a mathematical standpoint. A (mathematical) vector only has magnitude and direction, and no other property (such as location in space). This does not change the fact that I am aware the response to a force can depend on the location of the force. The idea is that the force vector represents the magnitude and direction of the force, but not the location of the place where the force acts. The latter information is represented by some other mathematical quantity, such as a position vector.)

Answer 3

I don't understand how couple moments are free vectors. If I applied a couple moment to different points on a stick, it would rotate very different.

There are two reasons why it might rotate differently if you attempted to simulate the effect of a couple.

One is that you may not be applying a pure couple. You need to be sure you are applying two precisely equal and opposite parallel forces. That's not so easy to insure (especially without precise measurements).

The other is even if you could apply a true couple, it is likely your forces are not the only ones acting on the stick. That requires the stick to be perfectly isolated from all other possible external influences, notably friction and gravity.

The bottom line is physics tells us that unless there is a net force acting on the stick, the center of mass (COM) should remain at rest (not accelerate). Any rotation about a point other than the COM will initiate movement of the COM.

Finally, a word of caution about a couple being a free vector (the proper term is a free moment vector). You are free to move it for the purposes of studying the dynamics of rigid bodies and for establishing equilibrium requirements in statics problems. But for non rigid bodies (which all macroscopic objects are) the location of the couple is critical in the analysis of deformation.

In summary, a couple is a free moment vector because it can be moved to any location without changing its external effect on a rigid body. I've emphasized "external" because, as indicated above, the internal effect on deformable bodies depends on the actual location of the couple.

i can't see how the second couple would rotate the beam around its COM, unless the COM happens to be the same as point d/2 , d being the distance between the two opposite forces.

While intuitively one would think the center of rotation (COR) of the stick (beam) would be the mid point between the forces that act as a couple, that would result in the COM having angular acceleration about the COR. The laws of physics precludes this because the net force on the stick is zero.

What actually happens if the stick is isolated from all external influences except for the couple is probably best illustrated by the example given in the answer by @AccidentalTaylorExpansion of two rockets attached to the stick in outer space. I've attempted to illustrate this in FIG 1 below.

In FIG 1 the two rockets are affixed to the stick such that the propulsion is always perpendicular to the major axis of the stick. Under these conditions the stick should rotate about the COM as shown in FIG 1B. To illustrate that the couple at the end of the stick is a free moment vector, it is moved to the COM in FIG 1C (shown in colored pairs as the stick rotates) where its effect on the stick is identical to FIG 1B.

Now let's consider how the stick would behave in space if we only had one such rocket. The situation is depicted in FIG 2A. A force system equivalent to FIG 2A is shown in FIG 2B, where the force $F$ is moved to the COM and a force couple included (individual forces not shown for clarity) to account for the torque the force $F$ produced about the COM at its original location.

There are two effects. 1. Since the moment arm of $F$ about the COM is constant, it produces a constant torque (and therefore constant angular acceleration) of the stick about its COM. 2. Since the applied force is always perpendicular to the COM it, it will cause the COM to move in a curved path with angular acceleration (as opposed to linear path with translational acceleration which would occur if the orientation of the applied force were somehow maintained horizontal throughout).

It turns out that the details of rotation of the COM in FIG 2 are rather complicated, as evidenced in the answers in the following link: Does rod in free space rotate around CoM if continuous force act at one end?

Hope this helps.

Answer 4

Choose the end of the stick $A$ and $B$ with the forces applied at differing points along the stick.

Now work out the torque appled by each of the couples about the end of the stick.

You should find that the torque in both cases to be $F\,c$ clockwise.

The torque applied by a couple is independent of the position you choose to evaluate the torque about.

Answer 5

Your vector is free because you are free to move it along the stick. This should be compared to a tangent vector which is bound to the point on the stick where it is tangent at - being bound, it isn't free.