To solve this problem, we can use the principle of moments, which states that the sum of the moments of the forces acting on an object is zero when the object is in rotational equilibrium.
In this case, the forces acting on the system are the weight of the 50-gram mass (0.5 kg) acting downward and the weight of the meter stick acting downward at its center of mass. When the system is at the tipping point, these forces must balance each other out to keep the system in equilibrium.
Let's denote the mass of the meter stick by m. The moment of the weight of the 50-gram mass about the tipping point is:
(0.05 kg) x (0.01 m) = 0.0005 N·m
The moment of the weight of the meter stick about the tipping point is:
m x (0.36 m/2) = 0.18m N·m
(we use 0.36 m/2 because the center of mass of the meter stick is at its midpoint)
Since the system is in equilibrium, these moments must be equal:
0.0005 N·m = 0.18m N·m
Solving for m, we get:
m = 0.0028 kg or 2.8 grams
Therefore, the mass of the meter stick is approximately 2.8 grams.