Question: A uniform meter stick off mass Ms = 100g is supported by a pivot at its 35 cm mark. When an unknown mass is suspended from the 10 cm mark, the system is in equilibrium. What is the value of the unknown mass? Given: Pivot point at 35 cm, and mass of meterstick = 100g. Attempt: Do I find the Torque in CW & CCW motion from 35cm and 65 cm, and then subtract my CW torque from CCW to get net torque. And then use the Torque = weight * r (distance from pivot) -> weight=T/r? (100*9.8 / 1000) = 0.980 T = w*r = (0.98 * 0.35) = 0.343N Left T = (0.98 * 0.65) = 0.637N Right Left Torque = (0.343/2)* 35 = 6.003 N/cm Right Torque = (0.637/2) * 65 = 20.703 N/cm Net Torque = 20.703 - 6.003 = 14.7 N/cm T = W*r 14.7 / .25 = 58.8 g?