Question

the mid point of a non- uniform rod 1.0m is placed over a knife edge. it balances when a downward force of 2.0N is applied to one end . if the weight of the rod is 10N calculate the position of its center of gravity

Answer 1

Taking moments about pivot:

$$F(0.5l)=Gx$$

$$(2)(0.5)=(10)x$$

$$x=0.1\ m$$