Question

A uniform meter stick of mass 0.29 kg is pivoted at the 40 cm mark. Where should one hang a mass of 0.50 kg to balance the stick?

Answer 1

The stick is in equilibrium when:
net force = 0
net moment = 0
total clockwise moment = total anticlockwise moment
\[W _{hanging mass}\times d = W _{stick}\times 10 cm\]
\[d=\frac{ W _{stick}\times 10 cm }{ W _{hanging mass} }\]
\[d=\frac{ 0.29kg \times 10 cm }{ 0.50 kg }=5.8 cm\]
So the distance of hanging mass from the pivot is 5.8 cm.