Question

A meter stick is supported by a knife-edge at the 50-cm mark. Arif hangs masses of 0.4 kg and 0.6 kg from the 20-cm and 80-cm marks, respectively. Where should arif hang a third mass of 0.3 kg to keep the stick balanced?
Please Answer.​

Answer 1

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The torque on both sides of the 50 cm mark must be balanced for the meter stick to balance horizontally.

The 0.4 kg mass is hanging 30 cms from the center of the meter stick, on the left side of the 50 cm mark. It exerts a torque equal to 0.4 kg x 30 cm = 12 kg-cm.

The 0.6 kg mass is hanging 30 cms from the center of the meter stick, on the right side of the 50 cm mark. It exerts a torque equal to 0.6 kg x 30 cm = 18 kg-cm

The 0.3 kg mass is hanging ?? cm from the center of the meter stick, on the left side of the 50 cm mark. It exerts a torque equal to 0.3 kg x ?? cm

18 kg-cm = 12 kg-cm + 0.3 kg x ?? cm

18 kg-cm – 12 kg-cm = 0.3 kg x ?? cm

6 kg-cm = 0.3 kg x ?? cm

?? cm = 6.0 kg-cm/0.3 kg

?? cm = 20 cm

?? = 20

The 0.3 kg mass is hanging 20 cm from the center of the meter stick, on the left side of the 50 cm mark. It exerts a torque equal to 0.3 kg x 20 cm = 6 kg-cm

Left side torque = 12 kg-cm + 6 kg-cm = 18 kg-cm

Right side torque = 18 kg-cm

Left side torque = right side torque

METER STICK IS BALANCED HORIZONTALLY.

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$\textit{Hope this helps you}$