Problem 536 | Friction on Wedges

From the FBD of block A
$\Sigma F_V = 0$

$R_1 \cos 45^\circ = 100$

$R_1 = 141.42 \, \text{ lb}$
 

From the FBD of block B
$\Sigma F_H = 0$

$R_2 \sin 15^\circ = R_1 \sin 45^\circ$

$R_2 \sin 15^\circ = 141.42 \sin 45^\circ$

$R_2 = 386.37 \, \text{ lb}$
 

$\Sigma F_V = 0$

$W_B + R_1 \cos 45^\circ = R_2 \cos 15^\circ$

$W_B + 141.42 \cos 45^\circ = 386.37 \cos 15^\circ$

$W_B = 273.20 \, \text{ lb}$           answer