Question: I am trying to understand this problem:

[![enter image description here](https://i.stack.imgur.com/AaHM5.jpg)](https://i.stack.imgur.com/AaHM5.jpg)

If we break down the 10T, they behave like this:

[![enter image description here](https://i.stack.imgur.com/R4TRl.jpg)](https://i.stack.imgur.com/R4TRl.jpg)

The right end of the beam can move as much as it wants, but the left end is pinned.

Why is there no horizontal reaction at A, if there is a horizontal part of the forces being applied on the beam axis?

I imagine, the force will push one section, and each section will push the next one, until the end, where it will create a reaction, just like the vertical forces are doing?

Thank you a lot.

Answer: 
Think about the three members that form the triangle EBC.

There is no horizontal load applied at E.

You know there is no horizontal load applied at C by beam CD, because there is a roller bearing at D.

Therefore by equilibrium, there is no horizontal load applied at B.

Therefore there is no horizontal tension in beam AB and no horizontal reaction at A.

Of course there will be horizontal tension in BC, but that will be balanced at each end by the horizontal components of tension in EB and EC.
You are more than overthinking this. Keep it simple. For an object at rest:

* ΣF(x) = 0 [sum of vertical forces = zero]
* ΣF(y) = 0 [sum of horizontal forces = zero]
* ΣMoments() = 0 [sum of moments about any point = zero]

Free body diagram of beam.

[![Free Body Diagram](https://i.stack.imgur.com/2CmYj.jpg)](https://i.stack.imgur.com/2CmYj.jpg)

Again, ΣF(x) = 0. With the diagram drawn this way, what do you thing Force A (in the x direction) is?