Problem 350 | Equilibrium of Non-Concurrent Force System

$7R_A + 4(30) + 4(50) = 10(60) + 4(120)$

$R_A = 108.57 \, \text{ kN}$           answer
 

$\Sigma M_A = 0$

$7B_V + 3(60) = 3(120) + 4(30) + 11(50)$

$B_V = 121.43 \, \text{ kN}$
 

$\Sigma F_H = 0$

$B_H = 30 \, \text{ kN}$
 

$R_B = \sqrt{{B_H}^2 + {B_V}^2} = \sqrt{30^2 + 121.43^2}$

$R_B = 125.08 \, \text{ kN}$
 

$\tan \theta_{Bx} = \dfrac{B_V}{B_H} = \dfrac{121.43}{30}$

$\theta_{Bx} = 76.12^\circ$
 

Thus,   $R_B = 12.08 \, \text{ kN}$   up to the left at   $76.12^\circ$   from horizontal.           answer