Answer :
Explanation:
Let us assume that forces acting at point B are as follows.
[tex]\sum F_{x} = 0[/tex]
[tex]T + F_{AB} Sin 60[/tex] = 0 ...... (1)
[tex]\sum F_{y}[/tex] = 0
[tex]F_{AB} Cos 60 + W[/tex] = 0 .......... (2)
Hence, formula for allowable normal stress of cable is as follows.
[tex]\sigma_{allow} = \frac{T}{A}[/tex]
T = [tex](20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}[/tex]
= 3925 kip
From equation (1), [tex]F_{AB}Sin (60^{o})[/tex] = -3925
[tex]F_{AB} \times -0.304 8[/tex] = -3925
[tex]F_{AB}[/tex] = 12877.29 kip
From equation (2), -12877.29 (Cos 60) + W = 0
[tex]-12877.29 kip \times \frac{1}{2} + W[/tex] = 0
W = 6438.64 kip
Thus, we can conclude that greatest weight of the crate is 6438.64 kip.