Question

A 2.0 – m uniform rod is 2.0 N. on one end is weight of 4 N (point A) while on

the other is a loads of 3 N (point B). A load of 1.0 N is suspended 0.5 m form

point B and another 1.0 N load is suspended 0.75 m from the same point. If a

1.5 N load is suspended 0.25 m from point A, what is the amount of force

needed to balance the rod? How far from point B shall the rod be balanced?

Answer 1

$$4\cdot1+1(0.25)+1.5(0.25)\\=3\cdot2\cdot 0.5+1(1-0.5)+M\\M=1.125\ Nm\\x=\frac{1.125}{2}=0.5625\ m$$