Question

A 0.12 kg, 50 cm long uniform bar has a small 0.055 kg mass attached to it. The system is balanced horizontally on a fulcrum. Where should the fulcrum be placed for the system to be in equilibrium?

A) 10 cm from the small mass
B) 20 cm from the small mass
C) 30 cm from the small mass
D) 40 cm from the small mass

Answer 1

The position of the fulcrum for equilibrium involves calculating the torques due to both the uniform bar's weight and the attached mass, taking into account the concept of lever arm and static equilibrium.

Step-by-step explanation:

The question asks where the fulcrum should be placed for a system to be in equilibrium when a 0.12 kg, 50 cm long bar with a small mass of 0.055 kg attached to it is balanced horizontally. This is a problem that deals with torques and balancing forces in static equilibrium.

For the system to be in equilibrium, the clockwise torque must equal the counterclockwise torque about the fulcrum. Since the bar is uniform, its weight acts at its center, which is 25 cm from either end. The torque by the bar's weight is thus: Torquebar = 0.12 kg × 9.8 m/s² × 0.25 m. The torque by the small mass is Torquemass = 0.055 kg × 9.8 m/s² × distance from the fulcrum to the small mass.

Setting Torquebar equal to Torquemass and solving for the distance will give the correct positioning of the fulcrum. This problem demonstrates the concept of lever arm and torque in physics, specifically within the study of statics, which is part of classical mechanics.