You attempt to put a meter stick of mass 1 kg into rotationa

You attempt to put a meter stick of mass 1 kg into rotational equilibrium. If you place a fulcrum at the 40 cm mark and a mass of 10 kg at the 20 cm mark, where should you place a mass of 5 kg? Give the answer in terms of the actual cm mark on the meter stick.

Solution

Sum of Moment of forces with respect to fulcrum = 0

[10 kg x(40-20)] = [1kg x(50-40)]+[5kgx(X-40)]

            200 = 10 + 5(X-40)

             190 = 5(X-40)

        X-40 = 190/5

               = 38

           X = 40 + 38

             = 78 cm