posted by visoth .
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?
200•20 +30•15 = 70•35 +50•x
x = 4000+450 -2450/50 = 40 cm ( from the pivot point)