The weight on the drawing is 600N. In the text it is 900N. Which one?
Let us say it is 900N.
First get the vertical, Fy, and horizontal , Fx, components of the two tensions.
Say T is the tension at BE, and U is the tension at CF.
For the T:
T is acting along BE, so get the length of BE.
BE = sqrt[(200^2 +80^2) = 215.4066 mm
Tx = T(200/215.4066) = (0.92848)T
Ty = T(80/215.4066) = (0.37139)T
For the U:
CF = sqrt(100^2 +80^2) = 128.0625 mm
Ux = U(100/128.0625) = (0.78087)U
Uy = U(80/128.0625) = (0.62469)U
Now, to solve for T, cut the system vertically between points C and D to create a FBD. Take moments at point C. Break the T into its Tx and Ty components.
The Tx has no moment at point C because its line of direction passes through point C.
Ty(100) = 900(200)
TY = 1800 N
So, T = 1800 / 0.37139 = 4847 N tension --------------answer.
For the U,
in the same FBD above, break the U into its Ux and Uy components.
Take moments at point B. Ux has no moment there
Uy(100) = 900(100)
Uy = 900 N
U = 900 / 0.62469 = 1441 N tension ----------answer.
For the reaction at point D:
Since the beam is supported by a roller there, there will be no vertical restraint there. It will all be horizontal.
It will be Ux.
It will be 1441(0.78087) = 1125 N compression -----answer. (Oppss, my mistake here. Let me correct it.)
I forgot the the Tx.
The reaction at D should be R = Tx +Ux
R = 4847(0.92848) +1125 = 5625 N compression ---------corrected answer.
Which cable will break first?
Cable BE, because it carries more stress than cable CF. ------answer.
Or, because the T is greater than the U.