Why if net force=zero, moment at any point is same?

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gen x
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Why as long net force= zero, moment at any point is the same but if net force= non zero, that is not case anymore?

Here is simple example:
Untitled.webp


Net force=zero case: F1=10, F2=40, N=50

Moment at A= +20Nm

Moment at B=+20Nm

Any point you choose you will get same moment..

Net force = non zero: F1=10, F2=40, N=30

Moment at A=+20Nm

Moment at B=+100Nm

Now, moment is not the same in every point.
 
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gen x said:
Why as long net force= zero, moment at any point is the same but if net force= non zero, that is not case anymore?
If ##\vec{F_{net}}## is the net force, and ##\vec{M_{net,a}}## is the net moment around the point ##\vec{a}##, then the net moment around any point ##\vec{b}## is given by:
$$\vec{M_{net,b}} = \vec{M_{net,a}} + (\vec{a} - \vec{b}) \times \vec{F_{net}}$$
Which for ##\vec{F_{net}}=0## simplifies to:
$$\vec{M_{net,b}} = \vec{M_{net,a}}$$
 
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Baluncore said:
Please show the equations that you use to compute the moments, for the cases you present.
Net force=zero case: F1=10, F2=40, N=50

Moment at A=-10x2+40x1=+20Nm

Moment at B= 10x2-50x4+40x5=+20Nm



Net force = non zero: F1=10, F2=40, N=30

Moment at A=-10x2+40x1 =+20Nm

Moment at B= 10x2-30x4+40x5 =+100Nm


A.T. said:
$$\vec{M_{net,b}} = \vec{M_{net,a}} + (\vec{a} - \vec{b}) \times \vec{F_{net}}$$
How do you get this equation?
 
##\vec{F}\times(\vec{a}+\vec{b})=\vec{F}\times\vec{a}+\vec{F}\times\vec{b}##
 
A.T. said:
This is basic transfer of moments to another reference point:


Why is vector r a/p (lever arm) not perpendicular to force F?
moment of point P must be: Mp= lever arm perpendicular to force F x force F