Answer:
1.03 kg.
Step-by-step explanation:
To prevent the meter stick from tipping when the weight is placed at 39 cm, we need to ensure that the torque due to the weight is balanced by the torque due to the mass of the meter stick itself.
The torque due to the weight is:
τ_w = F_w * r_w
where F_w is the force due to the weight (which is equal to the weight of the mass, W = m*g), and r_w is the distance between the weight and the fulcrum (which is 39 cm).
Plugging in the given values, we get:
τ_w = (0.150 kg * 9.81 m/s^2) * 0.39 m
τ_w = 0.571 J
The torque due to the mass of the meter stick is:
τ_m = m_m * g * r_m
where m_m is the mass of the meter stick (which is 0.080 kg), g is the acceleration due to gravity (which is 9.81 m/s^2), and r_m is the distance between the center of mass of the meter stick and the fulcrum (which is 50.3 cm).
Plugging in the given values, we get:
τ_m = (0.080 kg * 9.81 m/s^2) * 0.503 m
τ_m = 0.397 J
To prevent the meter stick from tipping, the torques due to the weight and the meter stick must be equal, so we have:
τ_w = τ_m
Solving for the required mass, we get:
m = (τ_m / r_w) / g
Plugging in the values for τ_m, r_w, and g, we get:
m = (0.397 J / 0.39 m) / 9.81 m/s^2
m ≈ 1.03 kg
Therefore, the minimum mass required to prevent the meter stick from tipping when the weight is placed at 39 cm is approximately 1.03 kg.