Question

Suppose that a meter stick is balanced at its center. A 0.18 kg mass is then positioned at the 8-cm mark. At what cm mark must a 0.22 kg mass be placed to balance the 0.18 kg mass?

Please help

Answer 1

Letx be the cm mark at which placed 0.22 kg mass on right side to balances the system. At equilibrium state of rotation, the net torque acting on the meter scale is zero. n on the meter scale is zero. distance of the first mass from the center mark =50 cm-8 cm and distance of the second mass from the center mark -x-50 cm mg (50 cm -8 cm)-m,g(x -50 cm) (0.18 kg) (50 cm -8 cm)-(0.22 kg)(x-50 cm) (0.18 kg) (50 cm -8 cm) 0.22 kg x-50 cm x=34.4 cm