Ameter stick is pivoted at the 0.50-m line. a 3.0-kg object is hung from the 0.15-m line. where should a 5.0-kg object be hung to achieve equilibrium (the meter stick oriented horizontal and motionless)?

0.71 m

Explanation:

Let the 5 kg is hang at a distance d from the midpoint.

use the concept of moments.

take moments about the mid point.

Anticlockwise moments = clockwise moments

3 x (0.5 - 0.15) = 5 x d

1.05 = 5 d

d = 0.21 m

So, the distance of 5 kg from 0 mark on the stick is 0.5 + 0.21 = 0.71 m

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