I've had some time today to get an answer together.
Right’ let’s go back a bit first.
We are talking about rotation. What hasn't been made clear in this discussion is that to produce rotation in a rigid body you need to identify two forces
, called a “couple”. A couple is two equal parallel forces acting in opposite directions but with different lines of action
. The torque produced by that couple (its turning effect or moment) is equal to one of those forces times the perpendicular distance between the lines of action.
A single force produces translation, not rotation. The simplest case is a single force acting on a point mass. There is no rotation. The mass accelerates. F=ma
A rigid body acts as if all its mass is concentrated at a single point, the centre of mass. You see this with gravitation where the force acts as if the mass of the Earth was all at the centre. Applying a single force to a rigid body such that the line of action acts through the c of m has exactly the same effect as if all the mass of the body was at the centre and you were just applying it to a point mass. There is no rotation and the body just has linear motion like a point mass. There is no couple, no torque, no rotation.
In the typical questions where you have moments of forces, you only show one force producing rotation. This is strange as I just said you need two forces
to produce rotation. What is happening in these questions using moments is that you are simplifying the diagram by removing half of the couple.
So on a beam where you have two supports (at A and B) applying upwards forces and the weight of the beam W downwards at the c of m as shown here in fig 1
only one of the forces in each couple is shown
. The other force acts at the point you take the moment of the force about. To understand this you need to think of this beam floating in space and the only force on it is the one originally labelled W downwards. fig 2. (The beam wouldn't have weight in space so just imagine someone pushing it there.) To make the beam rotate about the point B you need to apply an equal and opposite force to it at that point and hold B steady. Then you have a couple and rotation is possible. The torque (turning effect) is W x L (the same as if you took the moment of the force W about B). If there is no force acting at B, there is no couple and no turning effect. The object doesn't rotate.
rigid object can only rotate about its centre of mass. This is what the c of m is by definition. It cannot rotate about some other arbitrary point unless you constrain it to do so by applying a force at that point and fixing that point somehow, and applying a couple with a moment about that point.
So in your original question, when you say that the force through the c of m has a moment about some other point in the mass, and therefore the mass can’t be in rotational equilibrium, the fallacy is that in order to have a moment about that point, and a turning effect, there must be a couple. But where is the other force? If there is only one force, there is no couple, no moment and no turning effect.
So the problem arose because you applied the techniques of moments questions (where they work just fine) to the situation where there is only one force. The moment technique is only half the story (literally!) so when you apply it to the case of the force through the centre of mass, you are missing half the story. A single force acting through the c of m doesn't have a moment about some other point in the mass if it’s the only force acting
. You need a second force (at that point) and a couple, to provide a moment (and rotation) about that other point.
I ask that you take some time to read through this carefully a couple of times.
It would also be worth while looking up some of the theory on couple, torque and rotation of a rigid body.
I really don't have the time to go through all of that here.