a uniform meter stick is at rotational equilibrium when 220 g is suspended at 5 com, 120 g is suspended at 90, and the support stand is placed at the 40 cm mark what is the mass of the meter stick?

Answer 1

220*(40 - 5) - 120*(90 - 40) - m*(50 - 40) = 0 (the center of mass of the stick is at the 50 cm point).

m = [220*(40 - 5) - 120*(90 - 40)]/(50 - 40) = 170 g