The free-body diagram of the configuration shown is the following:
Now, we add the forces in the vertical direction:
[tex]\Sigma F_v=T_y+T_y-mg[/tex]
Since the system is in equilibrium the sum of forces in the vertical direction is zero:
[tex]T_y+T_y-mg=0[/tex]
Now, we add like terms:
[tex]2T_y-mg=0[/tex]
Now, the y-component of the tension is determined using the following right triangle:
Now, we use the trigonometric function cosine:
[tex]cos(26.3)=\frac{T_y}{T}[/tex]
Now, we multiply both sides by "T":
[tex]Tcos(26.3)=T_y[/tex]
Now, we substitute in the sum of vertical forces:
[tex]2Tcos(26.3)-mg=0[/tex]
Now, we solve for "T". First, we add "mg" to both sides:
[tex]2Tcos(26.3)=mg[/tex]
Now, we divide both sides by "2cos(26.3)":
[tex]T=\frac{mg}{2cos(26.3)}[/tex]
Now, we plug in the values of mass and the acceleration of gravity:
[tex]T=\frac{(13.7kg)(9.8\frac{m}{s^2})}{2cos(26.3)}[/tex]
Now, we solve the operations:
[tex]T=74.88N[/tex]
Therefore, the tension is 74.88 Newton
