A meterstick is balanced on a fulcrum placed at the 75 cm mark by 0.23 kg mass hanging at the 85 cm mark. Find the mass of the meterstick.

To solve this problem, we need to use the principle of moments or torques, which states that for an object to be in equilibrium (balanced), the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.

Step 1: Identify the forces and their distances from the fulcrum.

• The weight of the meterstick acts at its center of mass, which is at the 50 cm mark since it is a uniform meterstick. The distance from the fulcrum (75 cm mark) to the center of mass is 25 cm.
• The weight of the 0.23 kg mass acts at the 85 cm mark. The distance from the fulcrum to this mass is 10 cm.

Step 2: Set up the equation for the principle of moments.

• The weight of the meterstick provides an anticlockwise moment, while the weight of the 0.23 kg mass provides a clockwise moment. Therefore, we can set up the equation as follows:

  (Weight of meterstick) * (Distance of meterstick's weight from fulcrum) = (Weight of 0.23 kg mass) * (Distance of 0.23 kg mass from fulcrum)

Step 3: Substitute the known values into the equation.

• The weight of an object is given by the product of its mass and the acceleration due to gravity. We can take the acceleration due to gravity as approximately 9.8 m/s². Therefore, we can let M be the mass of the meterstick and set up the equation as follows:

  M * 9.8 * 25 = 0.23 * 9.8 * 10

Step 4: Solve the equation for M.

• Simplify the equation to find the value of M:

  M = (0.23 * 9.8 * 10) / (9.8 * 25)

  M = 0.092 kg

Therefore, the mass of the meterstick is approximately 0.092 kg.