First let's find the tension generated by the mass of 15 kg.
We can use the following formula to find this tension component:
[tex]\begin{gathered} T_{y1}=m\cdot g\cdot\sin (\theta) \\ T_{y1}=15\cdot9.81\cdot\sin (37\degree) \\ T_{y1}=15\cdot9.81\cdot0.602 \\ T_{y1}=88.58\text{ N} \end{gathered}[/tex]
Now, to find the tension generated by the beam, first let's find the angle from the connection point of the rope with the wall and the center of gravity of the beam:
[tex]\begin{gathered} \tan (37\degree)=\frac{x}{5} \\ 0.754=\frac{x}{5} \\ x=3.77 \\ \\ \tan (\theta)=\frac{x}{2.5} \\ \tan (\theta)=\frac{3.77}{2.5} \\ \text{tan(}\theta)=1.508 \\ \theta=56.45\degree \end{gathered}[/tex]
Now, let's use the same formula we used in the beggining:
[tex]\begin{gathered} T_{y2}=m\cdot g\cdot\sin (\theta) \\ T_{y2}=50\cdot9.81\cdot\sin (56.45\degree) \\ T_{y2}=50\cdot9.81\cdot0.833 \\ T_{y2}=408.59 \end{gathered}[/tex]
Adding both tensions, we have:
[tex]\begin{gathered} T=T_{y1}+T_{y2} \\ T=88.58+408.59 \\ T=497.17 \end{gathered}[/tex]
