1. The problem statement, all variables and given/known data I´ll show the problem using a picture: 2. Relevant equations Sum Forces = 0 Sum Moments = 0 3. The attempt at a solution Well, I´m a bit confused because I need to take care of the weight of the own structure. 1.- First attempt to solve to problem, since the structure is symmetric I think that the reactions at A and C are equal. And at C there is not any reaction on the horizontal axis X. So the total weight of the structure is = 7 members * 400 lb = 2800 lb ( I don´t multiply by the gravity because the answers in the book are not given in Newtons, they are given in lb too). Reaction at A Ra= 2800/2= 1400 lb Now I try to use the method of joints at joint A, and I see three forces acting on it: Ra, the compression force AE, and the tension force AB. If I´m right you don´t need to consider the weights mg of the bars, because that external force is acting at the bars, not at the joints, I´m not sure about this. IF I only consider these three forces I get these equations: Y axis----> Ra - AE sin(60) = 0; AE= 1400/ sin(60) = 1616,58 lb But the book gives me AE= 2000√3 lb C = 3464 lb C Well using that result I went backwards to guess the value of Ra and it looks like it´s 3000 lb. What I´m doing wrong? Even if I try to consider the weights of the two bars AE and AB I don´t get the correct result. I used the moment around A to check if I was wrong: -the moment arm of the bar AE is equal to 10 feet(since the cg is at the middle point) * sin (60)= 5 feet. -the horizontal bar AB has a moment arm of 10, since it is in the middle too -bar EB has a moment arm of 15 feet ... -the bar DC has a moment arm of 35 feet -Finally reaction force Rc has a moment arm of 40 feet so: Ma= -mg (5+10+15+20+25+30+35) + Rc*40= 0 Rc = 140 * 400 /40= 1400 lb which confirms the previous result. So I believe that I am not considering correctly other forces acting at joint A.