Answer
The tension in cable C is 1200 N.
The tension in cable B is 4970 N.
Work Step by Step
The horizontal component of the tension $T_c$ in cable C is equal in magnitude to the tension in cable A.
$T_c~sin(37.1^{\circ}) = 722~N$
$T_c = \frac{722~N}{sin(37.1^{\circ})}$
$T_c = 1200~N$
The tension in cable C is 1200 N.
The tension $T_b$ in cable B is equal in magnitude to the sum of the vertical component of cable C and the tension in cable D (which is equal to the weight of the engine).
$T_b = 1200~cos(37.1^{\circ}) + (409~kg)(9.80~m/s^2)$
$T_b = 4970~N$
The tension in cable B is 4970 N.
