Question

 A uniform meter stick of mass 0.20 kg is pivoted at the 40 cm mark. Where should one hang a mass of 0.50 kg to balance the stick?​

Answer 1

Answer:

36 cm

Explanation:

Mass of stick; m1 = 0.20kg at midpoint.

Total length; L=1.0 m

Pivot at 0.40m

Atached mass m2 = 0.50kg

Applying rotational equilibrium we have;

Ʈnet = 0

(m1g) • r1 = (m2g) • r2

(0.2) (0.1m) = (0.5)(x)

x = 0.04m =4cm

measured away from 40cm mark gives a position on the stick of; 40cm - 4cm = 36 cm