Answer:
Maximum weight = 76.64 lb
Explanation:
In the second figure attached it can be seen the free body diagram of the problem.
The equation of equilibrium respect x-axis is
[tex] \sum F_x = 0 [/tex]
[tex] F_{AC} \times sin 30 - F_{AB} \times \frac{3}{5} = 0 [/tex]
[tex] F_{AC} = 1.2 \times F_{AB} [/tex]
So, cable segmet AC supports bigger tension than segment AB. If [tex] F_{AC} = 50 lb [/tex] then
[tex] F_{AB} = \frac{F_{AC}}{1.2} [/tex]
[tex] F_{AB} = 41.67 lb [/tex]
The equation of equilibrium respect y-axis is
[tex] \sum F_y = 0 [/tex]
[tex] F_{AB} \times \frac{4}{5} + F_{AC} \times cos 30 - W = 0 [/tex]
[tex] 41.67 lb \times \frac{4}{5} + 50 lb \times cos 30 - W = 0 [/tex]
[tex] W = 76.64 lb [/tex]
