Final answer:
A mass of approximately 61.7 grams is needed to be hung at the 18.0 cm mark of the meter stick to keep the meter stick in equilibrium when it is suspended at its 39.0 cm mark.
Step-by-step explanation:
To determine the mass required to be hung at the 18.0 cm mark to keep the system in equilibrium, we use the principle of torque balance. Torque is the product of force and the perpendicular distance from the pivot to the line of action of the force. For the meter stick to be in equilibrium, the sum of the torques around the pivot (39.0 cm mark) must be zero.
Let's denote the mass to be hung at the 18.0 cm mark as 'm'. The torque due to this mass will be m × (39.0 cm - 18.0 cm) = m × 21.0 cm. The torque due to the weight of the meter stick on the other side can be considered as if all its mass were concentrated at its center of mass, which is at 50.0 cm from the pivot when the stick is balanced. Thus, the torque due to the meter stick's mass (0.12 kg) is 0.12 kg × 9.8 m/s² (gravitational acceleration) × (50.0 cm - 39.0 cm) = 0.12 kg × 9.8 m/s² × 11.0 cm.
In equilibrium, these torques must be equal: m × 21.0 cm = 0.12 kg × 9.8 m/s² × 11.0 cm. Solving for 'm' gives:
m = (0.12 kg × 9.8 m/s² × 11.0 cm) / (21.0 cm) = approximately 0.0617 kg.
Therefore, a mass of approximately 61.7 grams is required to be hung at the 18.0 cm mark to balance the system.