A meter stick is supported by a knife-edge at the 50-cm mark.Doug hangs masses of 0.40 and 0.60 kg from the 20-cm and 80-cm marks,respectively.Where should Doug hang a third mass of 0.30 kg to keep the stick balanced?

Question

Quizplus · Accepted Answer

To balance the meter stick, the torques around the knife-edge must sum to zero. The torque due to the 0.40 kg mass is (0.40 kg)(9.8 m/s²)(30 cm) = 11.76 kg·cm/s² counterclockwise, and the torque due to the 0.60 kg mass is (0.60 kg)(9.8 m/s²)(30 cm) = 17.64 kg·cm/s² clockwise. To balance, the third mass must create a counterclockwise torque of 5.88 kg·cm/s². Placing the 0.30 kg mass at 30 cm gives (0.30 kg)(9.8 m/s²)(20 cm) = 5.88 kg·cm/s², balancing the stick.

A Meter Stick Is Supported by a Knife-Edge at the 50-Cm