The mass of the meterstick is found by applying the principle of torque balance. With a rock weighing 1 kg placed at the 0-cm mark and the meterstick supported at the 25-cm mark, the meterstick's mass is determined to be 1 kg, based on the equality of torques.To solve for the mass of the meterstick, we can use the concept of torque balance.
Torque (moment of force) is the product of force and the perpendicular distance from the pivot point to the line of action of the force. In this case, the meterstick balances when a 1-kg mass is placed at the 0-cm mark, and the meterstick is supported at the 25-cm mark. Applying the principle of moments (or torque balance), where the sum of clockwise torques equals the sum of counter-clockwise torques, we can calculate the mass of the meterstick.
The torque due to the 1-kg mass on one side is the product of its weight (force due to gravity) and its distance from the fulcrum, which is 25 cm (0.25 m). Assuming the mass of the meterstick acts at its center of gravity, which would be at the 50-cm mark (0.5 m from the 0-cm end), we need to balance the torque produced by the 1-kg mass with the torque produced by the meterstick itself. Using the equation (1 kg * 9.8 m/s² * 0.25 m) = (mass of meterstick * 9.8 m/s² * 0.25 m), the force of gravity cancels out on both sides. The mass of the meterstick can thus be calculated as being equal to the 1-kg mass.
So, the mass of the meterstick is the same as the mass of the rock, which is 1 kg. This balances out because equal torques act on both sides of the pivot point.