Problem 323 | Equilibrium of Force System

$9R_B = (3 + 1.5)(20 \cos 30^\circ) + (9 + 3)(20 \sin 30^\circ)$

$9R_B = 197.94$

$R_B = 21.99 \, \text{ kN}$           answer
 

$\Sigma F_H = 0$

$A_H = 20 \cos 30^\circ$

$A_H = 17.32 \, \text{ kN}$
 

$\Sigma M_B = 0$

$9A_V = 1.5A_H + 3(20 \cos 30^\circ) + 3(20 \sin 30^\circ)$

$9A_V = 1.5(17.32) + 3(20 \cos 30^\circ) + 3(20 \sin 30^\circ)$

$9A_V = 107.94$

$A_V = 11.99 \, \text{ kN}$
 

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

$R_A = \sqrt{17.32^2 + 11.99^2}$

$R_A = 21.06 \, \text{ kN}$
 

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

$\tan \theta_{Ax} = \dfrac{11.99}{17.32}$

$\theta_{Ax} = 34.7^\circ$
 

Thus,
R_A = 21.06 kN down to the left at 34.7° with the horizontal.           answer

$y = 1.732 \, \text{ m}$
 

$\tan \theta_{Ax} = \dfrac{y + 3 + 1.5}{9}$

$\tan \theta_{Ax} = \dfrac{1.732 + 3 + 1.5}{9}$

$\theta_{Ax} = 34.7^\circ$           (okay!)
 

$\alpha = 90^\circ - \theta_{Ax} = 90^\circ - 34.7^\circ$

$\alpha = 55.3^\circ$
 

$\beta = 180^\circ - \alpha - 60^\circ = 180^\circ - 55.3^\circ - 60^\circ$

$\beta = 64.7^\circ$
 

$\dfrac{R_A}{\sin 60^\circ} = \dfrac{R_A}{\sin 60^\circ} = \dfrac{20}{\sin \alpha}$

$\dfrac{R_A}{\sin 60^\circ} = \dfrac{R_B}{\sin 64.7^\circ} = \dfrac{R_A}{\sin 55.3^\circ}$

$R_A = 21.07 \, \text{ kN}$           (okay!)

$R_B = 21.99 \, \text{ kN}$           (okay!)