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

From the figure attached;

0.5 kg = 0.5 x 10 N = 5N
0.2 kg = 0.2 x 10 N = 2N

The weight of the stick acts at its centre.

Taking moment about the pivot

Anticlockwise moments = Clockwise moments

5x = 10 x 20
5x = 200
x = 40

the 0.5 kg mass should be hanged at = (40 - 40) cm mark

It should be hanged at the 0 cm mark

Answer 2

Answer:

36 cm

Step-by-step explanation:

Mass of stick m1 = 0.20 kg at mid point, Total length L= 1.0 m, pivot at 0.40 m, attached mass m2 = 0.50 kg

Applying rotational equilibrium τ(net) =0

(m1 g)* r1 = (m2 g) *r2

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

x = 0.40 m (measured away from 40 cm mark)

---> gives a position on the stick of 36 cm