Finding the Mass of a Meter Stick Without Using a Scale

     In the first step, my group and I simply drew three different scenarios in which the meter stick was balancing. We then labeled the force and lever arm that caused the torque on the first drawing, and the center of gravity, force, and the absence of a lever arm on the second diagram. On the last drawing, we drew a picture with the forces and lever arms that were causing the clockwise and counterclockwise torques. For step 2, we simply wrote out the equation FxLA=FxLA to start planning before we actually plugged numbers into it. This picture to the right accurately shows how we setup the meter stick with the 100g weight on the end.

      To start our plan, we figured out that the center of gravity with no weight was dead center of the meter stick, at 50 cm. We then figured that the center of gravity with the weight was at 75.5 cm. Knowing that the torques were equal on both sides, we started plugging in numbers to our earlier equation, and we got .98x24.5=fx25.5. The .98 is the force of the weight on the weighted end, and the 24.5 is the lever arm we found from the end of the weighted side to the center of mass. The 25.5 was the other lever arm we found, and we were solving for the weight, which we calculated was .94N. We then had to convert this to kg, which would be 95.9kgm/s^2.
      Using this method worked extremely well, because the actual weight of our meter stick was 95.8kgm/s^2, so we were only .1 off of the real weight! The meter stick balanced because the torques were equal, and since the weighted end had a greater force and a smaller lever arm, the unweighted end had a longer lever arm and a lesser force, making them balanced. On the right is a fully labeled drawing of the diagram we used in order to figure out some of the parts.