$\Sigma F_H = 0$
$T_{AB} \cos 30^{\circ} = R_D \sin 50^{\circ}$
$R_D = 1.1305T_{AB}$
$\Sigma F_V = 0$
$T_{AB} \sin 30^{\circ} + T_{AB} + T_C + R_D \cos 50^{\circ} = W$
$T_{AB} \sin 30^\circ + T_{AB} + T_C + (1.1305T_{AB}) \cos 50^\circ = W$
$2.2267T_{AB} + T_C = W$
$T_C = W - 2.2267T_{AB}$
$\Sigma M_D= 0$
$6(T_{AB} \sin 30^\circ) + 4T_{AB} + 2T_C = 3W$
$7T_{AB} + 2(W - 2.2267T_{AB}) = 3W$
$2.5466T_{AB} = W$
$T_{AB} = 0.3927W$
$T_C = W - 2.2267T_{AB}$
$T_C = W - 2.2267(0.3927W)$
$T_C = 0.1256W$
Based on cable AB:
$T_{AB} = \sigma_{AB} A_{AB}$
$0.3927W = 100(250)$
$W = 63\,661.83 \, \text{N}$
Based on cable at C:
$T_2 = \sigma_C A_C$
$0.1256W = 100(300)$
$W = 238\,853.50 \, \text{N}$
Sfave value of W
$W = 63\,661.83 \, \text{N}$
$W = mg$
$63\,661.83 = m (9.81)$
$m = 6\,489.5 \, \text{kg}$
$m = 6.49 \, \text{Mg}$ answer
