To be clear,here is the answer they have given but i don't understand
AB ~ Y(2a cos 60)-x(2a Sin 60)-w(a cos 60) = 0
~ Y-(√3)x = w/2 ------- (1)
CD ~ y(2a cos 60)+x(2a Sin 60) + w(a cos 60)=0
~ Y+(√3)x=-w/x -----------(2)
and they solve it and I am like :O
consider this force too in the FBD. It will have two components. same story will be there with point C on rod BC. again two more unknown forces are introduced.
we have a total of 6 unknowns now. x,y, two components of force on point A, and two components of force on point B.
you must be aware of the fact that we can have three equations of statics for a rigid body in two dimensions.Consider rod AB, these three equations can be:
1) summation of torques about point A is zero.
2)summation of torques about point B is zero.
3)vector sum of the forces acting on AB is zero.
similarly we will have 3 more equations when we consider rod BC's statics.
hence 6 equations and six unknowns.
Did this help?