The center of mass of a 0.3 kg (non-uniform) meter stick is located at its 46-cm mark. what is the magnitude of the torque (in nm) due to gravity if it is supported at the 25-cm mark? (use g = 9.79 m/s2)

suppose that a meter stick is balanced at its center(50 cm mark). a 0.27 kg mass is then positioned at the 20 cm mark. at what cm mark must a 0.30 kg mass be placed to balance the 0.27 kg mass?