a uniform meter stick of mass 100g is pivoted about the 30 cm mark. a mass of 200 g is placed at the 10 cm mark. where should a mass of 50 g be placed so that the meter stick balance horizontally?

Answer 1

The mass of the stick where M resides is 30g, centered at distance 15.

The mass of the stick where m resides is 70, centered at distance 35

Adding up all the mass*distance values, they must balance on both sides of the fulcrum.

30(15) + 200(20) = 70(35) + 50(d)
450 + 4000 = 2450 + 50d
50d = 2000
d = 40

so, m sits at 40cm from the 30cm mark, or at the 70cm mark.