forcesfree-body-diagramnewtonian-mechanics
A rod is attached to a wall in such a way it can swivel.
In this case: In which direction does the force (of the wall on the rod) point to? I drew the blue force as I would make a force diagram. Am I wrong?
Here is an example in which the rod can swivel, but now the normal force is perpendicular to the wall. The direction of the force here is different. Why? Is maybe one of the pictures wrong?
Also: What is the recipe here? How do we determine the direction?
The normal force acting on the incline by the block does do work, but the normal force acting on the block by the incline does negative work, and the total work done by all normal forces in the system is zero (edit: see below for proof).
Therefore, the normal force can be considered a "constraint force", i.e. a force that does no work and is neither conservative nor non-conservative.
The work vanishes only when looking at all the normal forces in the system, since the normal force acts here as a mediating force, transferring the gravitational force from the block to the incline.
This example may be confusing since there are additional forces in different directions, consider the simpler setting of a force pushing two blocks on a horizontal plane:
Here the left block applies a normal force to the right block and vice versa, and again the total work done by the two normal forces cancels, since the normal force mediates the pushing force between the left block and the right block.
Another interesting example is the tension force of a string holding two weights over a pulley:
In this system the string pulls the lighter mass and does work on it, but it does negative work on the heavier mass and so the total work the tension forces do is zero. The string acts as a mediator that transfers the gravitational force between the two blocks.
Edit - corrected proof (credit to @DSinghvi for pointing out the error in the previous version of the proof in the comment below):
Here's how we can see the work done by the two normal forces cancel (and this proof can be easily generalized to any other problem with normal forces):
According to Newton's second law, the force acting on the incline by the block, $\mathbf{N}_{bi}$, is equal in size and opposite in direction to the force acting on the block by the incline, $\mathbf{N}_{ib}$, i.e.:
$$
\mathbf{N}_{bi} = - \mathbf{N}_{ib}.
$$
In the axis parallel to the normal force, the incline and the block move together, so if the incline travels an infinitesimal distance of $dx$, then the block at the same time travels the same distance $dx$. The total work done by both forces while this distance is traveled cancels:
$$
N_{bi} dx = - N_{ib} dx ~~~\Rightarrow~~~ N_{bi} dx + N_{ib} dx = 0
$$
In the elevator example, you are correct. The elevator floor does work on you as you stand in a rising elevator.
In the jumping example, the floor is not doing work. After all, it is just a floor, it has nowhere to get the energy from to do work on you. (The floor of the elevator gets it from the motor running the elevator.) When you jump, the force from the floor on your shoes increases, but there is no displacement between the floor and your shoes until your shoes leave the floor, and at that point there's no force any more. Your torso is moving upwards while you jump, but there's no force from the floor on your torso, so the floor isn't doing work. All the energy of a jump comes from internal energy, not external work.
In fact, the floor does a bit of negative work when you jump, taking energy away from you, because it deflects a bit, moving down as you jump up.
When you lift an object off a table, the table does a tiny bit of work on the object for a similar reason. When the object sits on the table, the table deforms a bit. As you lift, the table goes back to its original shape, so there's a small bit of force*distance work being done. The deflection is usually so small as to be unnoticeable. On a trampoline it is much larger; enough to make jumping a very different experience.
Best Answer
The pin forces can point in any direction, since all directions are constrained for motion.
You just can't have a normal force in the same direction as sliding is allowed because that would mean the joint can do/consume work.
Now for any example the actual direction is such that all forces converge to a single point. Slide the force vectors along their line of action such that they meet at a single point.
At this location the forces must balance, and that is how the magnitude and direction of the pin force (pink) is found, as well as the magnitude of the tension (black).