A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30 kg mass hanging from it at the 60-cm mark. Determine the position on the meter stick at which one would hang a third mass of 0.60 kg to keep the meter stick balanced.
A) 74 cm
B) 70 cm
C) 65 cm
D) 86 cm
E) 62 cm

Question

Quizplus · Accepted Answer

To balance the meter stick, the torques around the pivot must sum to zero. The torque from the 0.50 kg mass is (0.50 kg * 9.8 m/s² * 30 cm), and the torque from the 0.30 kg mass is (0.30 kg * 9.8 m/s² * 10 cm). The third mass of 0.60 kg should be placed such that its torque balances the other two, which occurs at 70 cm.

A Horizontal Uniform Meter Stick Supported at the 50-Cm Mark