Question

A non-uniform bar 2 meters long is supported by means of a string tied at its midpoint. To keep the bar in a horizontal position, a downward force of 60 Newton must be applied at the thinner end of the bar. If the weight of the bar is 120 Newton, Locate its center of gravity. 

Answer 1

$$0.5FL=Wx\\60(1)=120x\\x=0.5\ m$$

The center of gravity located at 0.5 m from another end of the bar.