Problem 354 | Equilibrium of Non-Concurrent Force System

$\Sigma M_A = 0$

$24R_B + 3(20) + 3(\frac{1}{\sqrt{5}})(22.4) = 18(\frac{2}{\sqrt{5}})(22.4) + 18(30) + 12(20) + 6(10)$

$R_B = 46.27 \, \text{ kN}$           answer
 

 

$\Sigma M_B = 0$

$24A_V = 3(20) + 3(\frac{1}{\sqrt{5}})(22.4) + 6(\frac{2}{\sqrt{5}})(22.4) + 6(30) + 12(20) + 18(10)$

$A_V = 33.76 \, \text{ kN}$
 

$\Sigma F_H = 0$

$A_H = 20 + \frac{1}{\sqrt{5}}(22.4)$

$A_H = 30.02 \, \text{ kN}$
 

$R_A = \sqrt{{A_H}^2 + {A_V}^2}$

$R_A = \sqrt{30.02^2 + 33.76^2}$

$R_A = 45.18 \, \text{ kN}$
 

$\tan \theta_{Ax} = \dfrac{A_V}{A_H}$

$\tan \theta_{Ax} = \dfrac{33.76}{30.02}$

$\tan \theta_{Ax} = 48.36^\circ$
 

Thus,   $R_A = 45.18 \, \text{ kN}$   up to the right at   $48.36^\circ$   from horizontal.           answer