A load of lumber weighing W=25kN is being raised as shown: http://i29.tinypic.com/2564idt.jpg by a mobile crane. Knowing that the tension is 25 kN in all portions of cable AEF and that the weight of boom ABC is 3 kN, determine:
a. the tension in rod CD
b. the reaction at pin B

answers: part a: tension in CD=81.1kN part b: 134.1kN up.

I don't understand this problem much, but here's my attempt to solve it. I've spent about 3 hours looking at it trying to figure out ways to do it...
Sum of Forces in x-direction= 0. at Pin B, there is no force in the x-direction
Sum of forces in y-direction=0.
therefore, -W-3-25-(CD)=0, but that's not right.
Sum of moments at any point (lets say B)=0.
Therefore, 25(2.6)+.6(3)+.4(25)+.7(CD)=0 and so CD=109.7, which is also incorrect.

Can someone PLEASE help? Thanks.

Part A
the summation Mb=0
(25kN*2.6) + (3kN*0.6) - (25kN*0.4)- Tcd*(0.7) = 0
0.7Tcd= 56.8kN
Tcd = 81.1kN

Part B

for Summation F= 0
SumFx=0
Bx=0
SumFy=0
By= 25kN+3kN+25kN+81.1kN=134.1 kN up

hope that help