Problem 322 | Equilibrium of Force System

$R = 2(1000) + 3(2000)$

$R = 8000 \, \text{ lb}$
 

 

$R_x = R \sin 30^\circ$

$R_x = 8000 \sin 30^\circ$

$R_x = 4000 \, \text{ lb}$
 

$R_y = R \cos 30^\circ$

$R_y = 8000 \cos 30^\circ$

$R_y = 6928.20 \, \text{ lb}$
 

$\Sigma M_B = 0$

$60R_A = 40R_y$

$60R_A = 40(6928.20)$

$R_A = 4618.80 \, \text{ lb}$           answer
 

$\Sigma M_A = 0$

$60B_V = 20R_y$

$60B_V = 20(6958.20)$

$B_V = 2309.40 \, \text{ lb}$
 

$\Sigma F_H = 0$

$B_H = R_x$

$B_H = 4000 \, \text{ lb}$
 

$R_B = \sqrt{{B_H}^2 + {B_V}^2}$

$R_B = \sqrt{4000^2 + 2309.40^2}$

$R_B = 4618.80 \, \text{ lb}$
 

$\tan \theta_{Bx} = \dfrac{B_V}{B_H}$

$\tan \theta_{Bx} = \dfrac{2309.40}{4000}$

$\theta_{Bx} = 30^\circ$
 

Thus,
R_B = 4618.80 lb at 30° with horizontal           answer

$\tan 30^\circ = \dfrac{20}{y}$

$y = 34.64 \, \text{ ft}$
 

 

$\tan \theta_{Bx} = \dfrac{y}{60}$

$\tan \theta_{Bx} = \dfrac{34.64}{60}$

$\theta_{Bx} = 30^\circ$           (okay!)
 

From the Force Polygon
$\dfrac{R_A}{\sin 30^\circ} = \dfrac{R_B}{\sin 30^\circ} = \dfrac{8000}{\sin 120^\circ}$

$R_A = 4618.80 \, \text{ lb}$           (okay!)

$R_B = 4618.80 \, \text{ lb}$           (okay!)