Question

# A student places a mass of 25 grams near the end of the meter stick (at...

A student places a mass of 25 grams near the end of the meter stick (at the 5 cm mark) and moves the balance point (the fulcrum) from the former center of gravity of the meter stick at 50 cm to the 42 cm meter mark. There is no additional weight on the opposite side of the meter stick, but it is now in balance.

What conclusions and generalizations can you make from the following data collected by a student to answer the question of “What is the weight of the meterstick?”

Explain your reasoning and provide specific evidence from the data given, with sketches of the free-body diagram if necessary, to support your reasoning.

Initially the fulcrum was at the center of gravity, but when a mass was added near one end, the fulcrum got shifted to maintain equilibrium or to keep the net torque zero.

Let the weight of meter scale be W,

Torque at equilibrium = 0

0.025*9.8*(0.42 - 0.05) - W*(0.50 - 0.42) = 0

W = 0.025*9.8*(0.42 - 0.05)/(0.50 - 0.42)

Here when we shifted the fulcrum, the weight of meter stick will also exert some torque as now the weight is not passing through fulcrum which is because the fulcrum is shifted. This torque is opposite in direction to the the torque produced by the mass placed at the end, thus the net torque becomes zero.