What about that 2.5kN force? Why do we disregard that?Just cut the beams AB, BC and CD diagonally. Then if you take the moments about A, F(AB) and F(CD) go through A so they don't count. The only force that counts is F(BC), but we know the moment about A must be 0, meaning F(BC) = 0.
This.When you do the method of sections, I believe you're only analysing the forces at where the cuts are made. You're cutting through the members, and hence you're only looking at the axial forces along those members. You're not cutting through point B, so the force at B isn't taken into consideration.
I could be wrong (it's been a decade since I've taken an engineering statics course). I always prefer to use the method of joints since its more intuitive.
When you do the method of sections, I believe you're only analysing the forces at where the cuts are made. You're cutting through the members, and hence you're only looking at the axial forces along those members. You're not cutting through point B, so the force at B isn't taken into consideration.
I could be wrong (it's been a decade since I've taken an engineering statics course). I always prefer to use the method of joints since its more intuitive.
Ok wow that helps a lot thanks!This.
Since we are analysing the left hand side of the diagonal cut. I'll attach an image to make it easier.