If the tension developed in cables ab and ac cannot exceed 460 lb , determine the largest weight of the crate that can be supported.

The point of interest is A. Begin with setting up the free-body diagram of A, with the four forces acting at A. There are no moments to calculate, as all the forces pass through A. 

The four forces are as follows: 
Weight of the Box, down. (Let's call this W) 
Tension of AB, toward B. (We'll call it Tb) 
Tension of AC, toward C. (We'll call this Tc) 
"Tension" from AD, from D. (And we'll call this D) 

To begin with, calculate the unit vectors of AB, AC, and AD. The unit vector for W is <0,0,-1>. The unit vectors for the others are, as previously ordered, <-4/13,-12/13,3/13>, <2/7,-6/7,3/7>, and <0,12/13,5/13>. 

Breaking the force vectors into their components, we are left with the following equations: 
(1) Fx = 0 = -4/13*Tb + 2/7*Tc 
(2) Fy = 0 = -12/13*Tb + -6/7*Tc + 12/13*D 
(3) Fz = 0 = 3/13*Tb + 3/7*Tc + 5/13*D - W 

From (1), we can solve for Tb in terms of Tc, such that Tb = 13/14*Tc 
From (2), we can substitute our solution from (1) into Tb and then solve D in terms of Tc, D = 13/7*Tc 
Then from (3), we can substitute (1) and (2) for Tb and D and put W in terms of Tc, W = 19/14*Tc. 

From (1), we can see that Tb will be less than Tc, so Tc shall be equal to 340 lbs. 
W thus shall equal 460 lbs 
D will equal 630 lbs

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