What is the correct statement of Varignon's theorem?

[Hak]

What is the correct statement of Varignon's theorem?
On the net I find some discrepancies between the various statements: in some cases the vectors of the system referred to by the theorem must be applied at the same point or such that their lines of action pass through the same point, in other cases the vectors are generic; in some cases the theorem concerns the equality of momentum vectors, in other cases it concerns the equality of the magnitudes of the momentum vectors only...
I'm a bit confused. Thank you very much for any reply.

 

[PhanthomJay]

What is the correct statement of Varignon's theorem?
On the net I find some discrepancies between the various statements: in some cases the vectors of the system referred to by the theorem must be applied at the same point or such that their lines of action pass through the same point, in other cases the vectors are generic; in some cases the theorem concerns the equality of momentum vectors, in other cases it concerns the equality of the magnitudes of the momentum vectors only...
I'm a bit confused. Thank you very much for any reply.

I understand your confusion, because all are 'sort of' correct.

The theorem states essentially that the moment sum of 2 or more concurrent and coplanar forces (that is, acting in the same plane and meeting or tending (line of action) to meet at a point) is equal to the moment of the resultant of those forces about that point.

If the force vectors are not concurrent, but rather, parallel, then the theorem still applies, however, the resultant (non zero) of the parallel forces changes the location of that resultant, which can be calculated using the theorem, but that is a circular argument.

You then mention 'momentum vector' but you meant to say 'moment vector', the direction of which is out of plane using the right hand rule. The sum of each of the force moments about a point is equal to the resultant moment vector about that point. Now since moment vectors are often considered as plus or minus depending on if they are clockwise or counterclockwise, you might say that the magnitudes are equal, but that is a bit weak since moments are vectors.

The theorem can be extended to three dimensions, but then you are talking moments about an axis instead of a point.

 

[Hak]

I understand your confusion, because all are 'sort of' correct.

The theorem states essentially that the moment sum of 2 or more concurrent and coplanar forces (that is, acting in the same plane and meeting or tending (line of action) to meet at a point) is equal to the moment of the resultant of those forces about that point.

If the force vectors are not concurrent, but rather, parallel, then the theorem still applies, however, the resultant (non zero) of the parallel forces changes the location of that resultant, which can be calculated using the theorem, but that is a circular argument.

You then mention 'momentum vector' but you meant to say 'moment vector', the direction of which is out of plane using the right hand rule. The sum of each of the force moments about a point is equal to the resultant moment vector about that point. Now since moment vectors are often considered as plus or minus depending on if they are clockwise or counterclockwise, you might say that the magnitudes are equal, but that is a bit weak since moments are vectors.

The theorem can be extended to three dimensions, but then you are talking moments about an axis instead of a point.

Thank you very much.

 

[PhanthomJay]

You’re welcome

 

[Lnewqban]

Copied from:
https://www.uobabylon.edu.iq/eprints/publication_12_18868_684.pdf

 

[Hak]

Copied from:
https://www.uobabylon.edu.iq/eprints/publication_12_18868_684.pdf
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Thanks.