Sum up vertical forces in block A
$N_A = 400 + C \sin 30^\circ$
Friction force at block A
$f_A = \mu N_A = 0.30(400 + C \sin 30^\circ)$
$f_A = 120 + 0.15C$
Sum up horizontal forces in block A
$f_A + C \cos 30^\circ = 500$
$(120 + 0.15C) + C \cos 30^\circ = 500$
$1.016C = 380$
$C = 374 \, \text{ lb}$
Sum up forces normal to the incline in block B
$N_B = C \sin 30^\circ + W_B \cos 60^\circ$
$N_B = 374 \sin 30^\circ + W_B \cos 60^\circ$
$N_B = 187 + 0.5W_B$
Amount of friction force under block B
$f_B = \mu N_B = 0.30(187 + 0.5W_B)$
$f_B = 56.1 + 0.15W_B$
Sum up forces parallel to the inclined plane in block B
$f_B + W_B \sin 60^\circ = C \cos 30^\circ$
$(56.1 + 0.15W_B) + W_B \sin 60^\circ = 374 \cos 30^\circ$
$1.016W_B = 267.79$
$W_B = 263.57 \, \text{ lb}$ answer
