What are the tensions in BC and BD?

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The discussion focuses on determining the tensions in cables BC and BD in a static equilibrium scenario involving a bar and a weight. The moment about point A from the forces acting on the bar and the weight is set to zero, leading to an equation that includes the tensions Tbc and Tbd. The user is attempting to solve for the magnitudes of these tensions using unit vectors in their respective directions. They are advised to apply vector algebra to calculate the vectors for AB, DB, and CD, while ensuring that the structure remains in static equilibrium. The conversation emphasizes the importance of using the provided position vectors and equilibrium equations to find the solution.
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mass of e is 100 kg
mass of bar is 20 kg and acts at midpoint
sume of moments about point a due to weight of bar and forces on b by bc, bd, and be, are 0, determine tensions in bc and bd

so i continued like this.
The moment about point A from all those forces is 0.
0 = (Raf x Wab) + (Rab x Wbe) -(Rab x Tbc) - (Rab x Tbd)

I can figure out the first two terms of this eq, but i have to find magnitude of the vectors of Tbc and Tbd. But i only have this equation and these two equations:

Tbc= |Tbc|*Ebc where Ebc is a unit vector in BC direction
Tbd= |Tbd|*Ebd where Ebd is a unit vector in the BD direction

Did i set this up incorrectly or how do i proceed?

a picture has been included.

Thanks for your help

Pt A is at the origin
Pt D is (0,5,5)
Pt C is (0,4,-3)
Pt B is (4,3,1)
Pt E is where the 100 kg weight is
AB is the bar where 20 kg acts down the midpoint
any help is APPRECIATED!
 

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You know the position vectors of all the points.
Use vector algebra to work out the vectors AB, DB, CD.
Thes structure is in static equilibrium, therefore
\sum F_i = 0 \sum F_j = 0 \sum F_k = 0

\mbox{If}\ \bf{AB} = (x_1,y_1,z_1)\ \mbox{then}\ F_{AB} = \lambda(x_1,y_1,z_1)

You should be able to use the above two lines to solve the problem.
 
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