Problem 355 | Equilibrium of Non-Concurrent Force System

$\tan \theta = 4.5/9$

$\theta = 26.56^\circ$
 

$\cos \theta = 4.5/AC$

$AC = 4.5/\cos \theta = 4.5/\cos 26.56^\circ$

$AC = 5.03 \, \text{ m}$
 

 

$\cos \theta = AC/AD$

$AD = AC/\cos \theta = 5.03/\cos 26.56^\circ$

$AD = 5.626 \, \text{ m}$
 

$FB = AD$

$FB = 5.626 \, \text{ m}$
 

$\Sigma M_B = 0$

$18(R_A \cos 30^\circ) = 12(18 - 4.5) + (40 \cos \theta)(FB) + 40(FB) + 20(18 - AD)$

$18(R_A \cos 30^\circ) = 12(13.5) + (40 \cos 26.56^\circ)(5.626) + 40(5.626) + 20(18 - 5.626)$

$15.59R_A = 835.81$

$R_A = 53.61 \, \text{ kN}$           answer
 

$\Sigma F_H = 0$

$B_H + 40 \sin \theta = R_A \sin 30^\circ$

$B_H + 40 \sin 26.56^\circ = 53.61 \sin 30^\circ$

$B_H = 8.92 \, \text{ kN}$           answer
 

$\Sigma M_A = 0$

$18B_V = 12(4.5) + (40 \cos \theta)(18 - FB) + 40(18 - FB) + 20(AD)$

$18B_V = 12(4.5) + (40 \cos 26.56^\circ)(18 - 5.626) + 40(18 - 5.626) + 20(5.626)$

$18B_V = 1104.20$

$B_V = 61.34 \, \text{ kN}$           answer