1. The problem statement, all variables and given/known data After reviewing a quiz I had in Statics, I realized I did not understand why a certain force was considered in the moments equation. If a force goes directly through the point of the moment, it is not considered when solving for a force using moments. The question is asking us to use the method of sections by splitting the truss from section 1-1 and find the force in member BC. 2. Relevant equations ΣMH=0 By the way, I substituted kips for k. 3. The attempt at a solution ΣME=0 → -Ay(60 ft) +(40k)(45 ft)+ (40k)(30 ft)+ (40k)(15 ft)-(30k)(20 ft)=0. Ay=50k. The equation I would have used to find FBC: ΣMH=0 → -50k(15 ft) + FBC(20 ft)=0. The equation the professor used to get the correct answer for the force along member BC: ΣMH=0 → (-30k)(20 ft) -50k(15 ft) + FBC(20 ft)=0 Why is the 30k force considered when it points directly into H? I understand it being used to find the moment about E, but not for the moment about H. Also, why is the lever arm distance 20 ft when the force is right at H?