The figure from below shows a system at equilibrium. The bar is homogeneous and uniform and has a weight of $14\,N$ and the block which is labeled $Q$ has a weight equal to $28\,N$. Find the reaction force in Newtons experienced by the joint attached to the wall.

The alternatives are given as follows:

$\begin{array}{ll}

1.&14\sqrt{2}\,N\\

2.&7\sqrt{2}\,N\\

3.&14\,N\\

4.&21\,N\\

5.&28\,N\\

\end{array}$

I'm not sure exactly how to assign the equilibrium condition in this problem and where the forces are acting in the bar. Can someone help me with the right approach and and free body diagram for this thing?.

What I've attempted to do was to use the equilibrium condition as:

$\sum ^n_{i=1}\tau_{i}=0$

By following this

$T\cos53^{\circ}\cdot 6a + (T-mg)\cdot 2a - (14 + 28) = 0$

Then the reaction in the joint would be given by:

$R=T\sin 53^{\circ}$

or will be given by a vertical reaction?

$R= mg-T-T\cos53^{\circ}$

Which of these reactions are referring in this problem?

Am I understanding this correctly?. How exactly should be understood the reaction. Can someone help me here?. I would like someone can explain me the meaning of the reaction experienced by the joint

**this part is very important**hence a

**free body diagram**will help me a lot here.