Ameter stick is balanced at the 50 cm mark. you tie a 20 n weight at the 20 cm mark. where should a 30 n weight be placed so the meter stick will again be balanced?

20 cm to the right of the center or 20+50 = 70 cm from the left side.

Explanation:

The length of meter stick is 1 m = 100 cm

Balance point on 50 cm

From the center the 20 N weight is 50-20 = 30 cm

Torque is obtained when force is multiplied with the distance

As the force is conserved we have

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.

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