Problem 351 | Equilibrium of Non-Concurrent Force System

$\Sigma M_A = 0$

$4(\frac{2}{\sqrt{5}}R_B) = 3(40)$

$R_B = 33.54 \, \text{ kN}$
 

$\Sigma M_B = 0$

$4A_V = 1(40)$

$A_V = 10 \, \text{ kN}$
 

$\Sigma F_H = 0$

$A_H = \frac{1}{\sqrt{5}}R_B = \frac{1}{\sqrt{5}}(33.54)$

$A_H = 15 \, \text{ kN}$
 

$R_A = \sqrt{{A_H}^2 + {A_V}^2} = \sqrt{15^2 + 10^2}$

$R_A = 18.03 \, \text{ kN}$
 

$\tan \theta_{Ax} = \dfrac{A_V}{A_H} = \dfrac{10}{15}$

$\theta_{Ax} = 33.69^\circ$
 

Thus,   $R_A = 18.03 \, \text{ kN}$   up to the right at    $33.69^\circ$   from horizontal.           answer

From Equilibrium of Concurrent Force System, three coplanar forces in equilibrium are concurrent.
 

$\dfrac{y}{1} = \dfrac{2}{1}$

$y = 2 \, \text{ m}$
 

$\tan \theta_{Ax} = \dfrac{y}{3}$

$\tan \theta_{Ax} = \dfrac{2}{3}$

$\theta_{Ax} = 33.69^\circ$       okay
 

$\tan \theta_{Bx} = \dfrac{2}{1}$

$\theta_{Bx} = 63.43^\circ$
 

$\alpha = 90^\circ - \theta_{Ax} = 56.31^\circ$

$\beta = 90^\circ - \theta_{Bx} = 26.57^\circ$

$\phi = \theta_{Ax} + \theta_{Bx} = 97.12^\circ$
 

$\dfrac{R_A}{\sin \beta} = \dfrac{R_B}{\sin \alpha} = \dfrac{40}{\sin \phi}$

$\dfrac{R_A}{\sin 26.57^\circ} = \dfrac{R_B}{\sin 56.31^\circ} = \dfrac{40}{\sin 97.12^\circ}$

$R_A = 18.03 \, \text{ kN}$       okay

$R_B = 33.54 \, \text{ kN}$       okay