Question

A uniform meterstick is supported at the 50 cm mark. A 25N weight is placed at the 10 cm mark. Where should a 200N weight be placed for the meterstick to be in equilibrium?

Answer 1

The equilibrium condition

$$M_1=M_2$$

$$F_1d_1=F_2d_2$$

$$d_2=\frac{F_1}{F_2}d_1=\rm\frac{25\: N}{200\: N}*40\: cm=5\: cm$$

Answer: the second weight should be placed at 55 cm mark.