Respuesta :

Here in such type of problems we will say that for the equilibrium of rod torque as well as net force must be balanced

So first we will use the torque balance on the rod about the hinge point

[tex]F_T_y \times L = 545\times 1.5 + 315 \times \frac{L}{2}[/tex]

we know that length of the rod is L = 5 m

now by solving above

[tex]F_T_y \times 5 = 817.5 + 787.5[/tex]

[tex]F_T_y = 321 N[/tex]

now we also know that

[tex]F_T_y = F_T sin53[/tex]

[tex]321 = F_T 0.8[/tex]

[tex]F_T = 401.25 N[/tex]

now by force balance in Y direction

[tex]F_T_y + R_y = 545 + 315[/tex]

[tex]321 + R_y = 860[/tex]

[tex]R_y = 539 N[/tex]

now force balance in X direction

[tex]R_x = F_T_x[/tex]

[tex]R_x = F_T cos53[/tex]

[tex]R_x = 401.25 cos53 = 240.75 N[/tex]

now the net reaction force is given as

[tex]R = \sqrt{R_x^2 + R_y^2}[/tex]

[tex]R = \sqrt{240.75^2 + 539^2}[/tex]

[tex]R = 590 N[/tex]