I think you meant FMx=5.5, not FMy in your last post.

But I think there is a miscommunication here.

3 equations:

1. Sum or torques

2. Sum of vertical forces

3. Sum of horizontal forces

3 unknowns:

1. FL

2. FM (though we do know the vertical component FMv is 1000N)

3. FR

We agree I did the torque equation right so:

FL = 1.91FR

In post #16, I use sum of vertical forces to find FR = 434.78N. Because only vertical forces are used in sum of forces equation, I don't care about the horizontal force from FM at that point. I KNOW FMv (not FM) is 1000N. It's a given.

So: FR = 434.78 and so FL = 830.43. If you disagree, can you please tell me why my equation in post #16 is wrong?

FR was my goal so I actually don't need to know FMx in this case but I make the attempt to be sure I am not missing something in your posts.

Now that I know FL and FR, I can use sum of horizontal forces to find FMx and I do that in post #14. I find there that FMx = -5.21. Again, please tell me if my equation there is wrong.

Finally, if I cared, I could calculate FM as:

FM = sq root ( 5.21sq + 1000sq ) = 1000.1

And even the angle of FM as:

acos(1000/1000.01) = 26°

So, using sum of torques about M and sum of vertical forces, I am able to find FR, without doing any sum of horizontal forces or finding FMx. This is because I already know FMv (the vertical component of unknown FM) as a given and I know the angles of FL and FR.

I only needed 2 equations to find 2 unknowns in getting FR.

Now I can do this for multiple FM forces into differing rotations of the triangle and linkage at different angles and for each case I would need to consider the angle of FM to point M in order to find FMv to be applied to the same steps of then finding FL and FR.