Eliminating forces by taking a moment

[yoleven]

Homework Statement

If I take a moment around a point with a rigid connection what forces do I eliminate at that connection?
If I have a reaction force in the Y direction, a reaction force in the X direction and a moment at that point can I eliminate Ry, Rx and Mx or does Mx remain.
Can you explain this a bit to me?
This lack of understanding is a stumbling block for me and it keeps tripping me up.

Homework Equations

The Attempt at a Solution

 

[quantumdude]

Homework Statement

If I take a moment around a point with a rigid connection what forces do I eliminate at that connection?

All of them. The moment \vec{M} about a point O due to a force \vec{F} acting at a displacement \vec{r} from O is given by \vec{M}=\vec{r}\times\vec{F}. If \vec{F} acts at O then \vec{r} is the zero vector, hence the moment is zero.

 

[tiny-tim]

hi yoleven!

a (linear) force which goes through the point has zero moment about that point, so (to use your words) it can be ignored

a couple, ie a pair of equal and opposite forces which are not in-line and so result in a pure moment or torque (is that what you meant?), has the same moment about any point …

you can check this by just drawing it as two offset equal forces, then calculating the moment about a random point in the same plane

 

[yoleven]

Okay. Bear with me...
If the force I'm finding doesn't act through the point I'm taking the moment about.

Say A is rigid connection... I'll have Ax, Ay and Mx. I have a known point load acting somewhere along the beam and I want to find 'By' located a certain perpendicular distance from A.

When I take the moment about point A, I know I eliminate the Ax and Ay but does Mx remain.

When I take moments about point A will I have 2 unknowns in the equation or just one?

 

[tiny-tim]

is Mx a couple (in which case, shouldn't it be Mz)? … if so, it definitely stays in the equation

(i don't follow what you're saying about 2 unknowns )