Question

A uniform bar of weight 40 N is 4 m long. Weights of 60N and 100N are placed on the bar at points which are 1.5m and 3m, respectively, from the left end. At what point on the bar must a single force be applied to support it in a horizontal position? 

Answer 1

$$F=40+60+100=200\ N\\ 200(x-2)=100(3-2)-60(2-1.5)\\x=2.35\ m$$