Problem 352 | Equilibrium of Non-Concurrent Force System

From FBD of pulley
$T = 200 \, \text{ lb}$
 

 

$\Sigma F_V = 0$

$B_V + T \sin 30^\circ = 200$

$B_V + 200 \sin 30^\circ = 200$

$B_V = 100 \, \text{ lb}$
 

$\Sigma F_H = 0$

$B_H = T \cos 30^\circ$

$B_H = 200 \cos 30^\circ$

$B_H = 173.20 \, \text{ lb}$
 

From FBD of beam
$\Sigma M_A = 0$

$8R_C = 4B_V$

$8R_C = 4(100)$

$R_C = 50 \, \text{ lb}$           answer
 

 

$\Sigma M_C = 0$

$8A_V = 4B_V$

$8A_V = 4(100)$

$A_V = 50 \, \text{ lb}$
 

$\Sigma F_H = 0$

$A_H = B_H$

$A_H = 173.20 \, \text{ lb}$
 

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

$R_A = \sqrt{173.20^2 + 50^2}$

$R_A = 180.27 \, \text{ lb}$
 

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

$\tan \theta_{Ax} = \dfrac{50}{173.20}$

$\theta_{Ax} = 16.1^\circ$
 

Thus,   $R_A = 180.27 \, \text{ lb}$   up to the right at   $16.1^\circ$   from horizontal.           answer