Ameter stick is supported by a knife-edge at the 50-cm mark. doug hangs masses of 0.40 and 0.60 kg from the 20-cm and 80-cm marks, respectively. where should doug hang a third mass of 0.30 kg to keep the stick balanced?

30 cm

Explanation:

From the principle of moment,

when a body is in equilibrium,

Sum of clock wise moment = sum of anti clock wise moment.

Note: That before Doug hangs the last mass, The clock wise moment is greater than the anti close wise moment.

As such the third mass will be placed before the knife edge. As  shown in figure 1 on the attached file.

From the diagram,

W₁(50-30) + W₃(x-50) = W₂(80-50)

were, W₁ = m₁g = 0.4×9.8 = 3.92 N ( Where m₁ = 0.4 kg and g = 9.8 m/s²)

          W₂ = m₂g =0.6×9.8 = 5.88 N ( where m₂ = 0.6 kg)

           W₃ = m₃g = 0.3×9.8 = 2.94 N ( where m₃ = 0.3 kg)

Therefore,

3.92(30) + 2.94(50-x) = 5.88(30)

117.6 + 147-2.94x = 176.4

2.94x+264.6 = 176.4

-2.94x = 176.4-264.6

2.94x = 88.2

x = -88.2/-2.94

x = 30 cm

Thus the third mass must be hung 30 cm

i think it would be 5