Static equilibrium question: Horizontal rod attached to a wall I'm doing a problem on static equilibrium and I'm unclear whether a force exists or not. This is my force diagram:

The setup is: a homogenous rod with a certain mass is attached to a vertical wall on one side, an object hangs on the other side, and a cable connects the rod to the wall with a tension T. I understand which forces are needed for static equilibrium. What I'm confused about is what's happening at point A. Since the wall exerts a force on the rod which points upwards (Fy, rod), per Newton's 3rd Law does the rod then exert a force on the wall which points downwards (Fy, wall)?
 A: You can simplify your force diagram by highlighting forces that is more important. On the horizontal component, you have $T_x$ from the rod acting to the wall, thus the wall also exert an equal force to the rod $F_{x,wall}=-T_x$. The force acting upwards that you described is better perceive as the (static) friction force. Now the vertical force component is balanced by force acting upwards against force against downwards. So, you can write it as $F_y+T_y = -(F_1+F_2)$. Here you already established the force equilibrium. You can further solve this by incorporating equilibrium in the moment to the equation too.
A: At point A, the rod pushes on the wall horizontally, to the left, due to Tx and downward, due to F1. 
These forces, Fx,wall and Fy,wall are countered by the reactive forces of the wall acting on the rod, Fx,rod and Fy,rod, which are equal in magnitude and opposite in direction. 
Also, Fwall=Frod, since their horizontal and vertical components are equal. Of course, all of the above observations follow from the Newton's third law.
So, you've pretty much have answered your question.
A: By your force diagram, 
The point where the load or $F_2$ is acting is C, where $F_1$ is acting is B and the point on wall is C
At A,
$T_y=F_2$
So A is fixed and does not move
Taking moments about A,
$F_1×(AB)+F_{y,rod}×(AC)=0$(For rotational equilibrium of the rod)
$F_{y,rod}$ is the force acting on the rod by the wall at C 
Note that we have taken downward direction to be positive .
$F_{y,rod}=-F_1×\frac{AC}{AB}$
Being negative the force on the rod at C is upwards . This is the reaction force produced at C due to some downward force or rather $F_{y,wall}$ applied at C on the wall by the rod.
So $F_{y,wall}$ is action and $F_{y,rod}$ is it's reaction.Basically these are unwanted forces which  as per my opinion will not aid you to your results.
A: Your diagram is pretty much clear, wall normal force at contact point A is :
$$ \vec{N}_A = - (\vec{F}_1+\vec{F}_2+\vec{T})_{\perp\,\text{wall}} $$
