Answer
(a) Please refer to the free-body diagram. The tension of $T_1$ (with the angle of $60^{\circ})$ must be greater than $T_2$ (with the angle of $40^{\circ})$.
(b) The maximum value of the hanging weight is 6430 N.
Work Step by Step
(a) Please refer to the free-body diagram. The horizontal components of $T_1$ and $T_2$ must be equal. Therefore, the tension of $T_1$ must be greater than $T_2$ because $cos(60^{\circ})$ is less than $cos(40^{\circ})$.
(b) Let's suppose the the tension of $T_1$ is 5000 N.
horizontal forces:
$T_2~cos(40^{\circ}) = T_1 ~cos(60^{\circ})$
$T_2 = \frac{T_1 ~cos(60^{\circ})}{cos(40^{\circ})}$
vertical forces:
$mg = T_1 ~sin(60^{\circ}) + T_2~sin(40^{\circ})$
$mg = T_1 ~sin(60^{\circ}) + (\frac{T_1 ~cos(60^{\circ})}{cos(40^{\circ})})~sin(40^{\circ})$
$mg = (5000~N) ~sin(60^{\circ}) + (5000~N) ~cos(60^{\circ})~tan(40^{\circ})$
$mg = 6430~N$
The maximum value of the hanging weight is 6430 N.
