Let’s first review the conditions for rigid body equilibrium. 
For a rigid body to be in equilibrium, there are two vector equations that need to be satisfied. The resultant force acting on the body needs 
to be zero, and the resultant moment summarized about any point needs to be zero as well. 
Or, we can write scalar equations instead. For 2D problems, there are a maximum of 3 scalar equations, enabling us to solve for 3 unknowns;
and for 3D problems, there are a maximum of 6 scalar equations, enabling us to solve for 6 unknowns. 
So again, in order to complete the free body diagram, we need to identify all the external forces and couple moments acting on the body, 
and therefore we need to identify all the support reactions first. 
Once again, the rule of thumb is:
if the support prevents the translational motion in a certain direction, then the support exerts a force along that direction against the motion. 
And if the support prevents the rotational motion about an axis, then it exerts a couple moment about that axis against the tentative rotation. 
Let’s look at this example: 
we are asked to determine the support reactions at the fixed support, point A. When we analyze the situation, we realize that since
this is a fixed support at point A, it prevents all kinds of motion. Therefore, there must be force support reactions 
along all three axes and moment support reactions about all three axes as well. 
Let’s first try to solve this problem using the vector formulation and represent the support reactions at point A as vectors, 
Force vector F_A and couple moment vector M_A, currently in arbitrary directions. 
And the two applied forces can be expressed as vectors as well. 
Therefore, the resultant force equals to 
F_A plus 320 j minus 240 k and plus 200 i, all in the unit of pound, and it equals to zero. Therefore, 
the force vector F_A is solved to be negative 200 i, minus 320 j plus 240 k pounds.
To summarize the resultant moment, once again, you can do it about any arbitrary point, 
but you want to do it in a way to eliminate the most unknowns in your equations.
Therefore, here we summarize the resultant moment about point A, since force F_A has line of action passing through point A 
and therefore creates no moment about point A and will not show up in your equations. 
So to summarize the moment about point A in vector formulation, we need to first find the position vectors from point A to, again, 
any point on the lines of action of the two applied forces. You can do it a million ways, 
but these are probably the best, since they are the easiest.
And now the resultant M_A equals to 
the support couple moment reaction at point A 
plus the moment caused by the first applied force
plus the moment caused by the second applied force.
And that’s all the external moments. From the calculation, the couple moment vector M_A equals to 
1400 i, plus 480 j plus 1840 k,
in the unit of pound foot.
And of course, alternatively, we can solve this problem using scalar formulation. Now at point A we have overall six scalar unknowns, 
three forces along x, y and z axes and three couple moments about these three axes as well. 
Now we summarize the forces in each direction,
and solve for the three force components.
And we summarize the resultant moment about each axis,
and solve for the three moment components.
And as you can see, the vector formulation and the scalar formulation yield the same results, 
and it is your own preference which method to choose. 
Now let’s look at several types of bearings. First one is a single journal bearing. Let’s analyze the reactions. Because the bearing prevents 
translational motions in these two radial directions,
and only allows movement along the axial direction.
And also, it prevents any rotation except for the rotation about the central axis of the shaft.
Therefore, this single journal bearing exerts 
force support reactions along the x and z direction, and moment supports about the x and z axes as well. 
The next one
is a single thrust bearing. As you can see, this looks just like the previous one, except for the thrusts, 
which now prevent the translational motion along the axial direction.
Therefore, now this bearing also exerts a force reaction along the y direction.
And lastly let’s look at the single journal bearing with the square shaft.
And again, it looks similar to the first bearing, but now it has a square shaft which won’t rotate.
Therefore, when compared to the single bearing with a circular shaft in the first image,
This one exerts an additional couple moment about the y axis. 
Please note, if these bearings are used in series or in combination with other supports, 
then the couple moment support reactions might not need to be considered. Just consider the force reactions.
Lastly let’s look at another example, with a ball and socket support.
Ball and socket support allows rotation about all axes, but prevents translational motions along all three directions, 
and therefore it exerts force reactions in all three directions but no moment reaction. 
Actually, the joints at our shoulders can be considered as ball and socket supports.
And we know that
the roller support exerts a force reaction only perpendicular to the contacting surface.
And since this journal bearing is used in combination with the other supports, 
it only exerts force reactions along the y and z axes.
Therefore, we note all forces and moments acting on the member, 
and we have six unknowns,
And we write the six scalar equations for equilibrium, 
summarizing forces in three directions, and moments about the three axes. Everything equals to zero.
Six equations, six unknowns, we can now solve for all of them.
