Problem 337 | Equilibrium of Parallel Force System

$10R_D + 4(60) = 6(190)$

$R_D = 90 \, \text{ kN}$           answer
 

$\Sigma M_D = 0$

$10R_C = 14(60) + 4(190)$

$R_C = 160 \, \text{ kN}$           answer
 

$\Sigma M_A = 0$

$10R_B = 4(400) + 14(160)$

$R_B = 384 \, \text{ kN}$           answer
 

$\Sigma M_B = 0$

$10R_A + 4(160) = 6(400)$

$R_A = 176 \, \text{ kN}$           answer