Question

A meter 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?

Answer 1

Answer:

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

$\dfrac{20}{30}\times 30=20\ cm$

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