Question

A uniform metre scale of weight 50 gf, is balanced at 60 cm mark, when a weight of 15 gf is suspended at the 10 cm mark. Where must a weight 100 gf be suspended to balance the metre scale?

Answer 1

Let x be the point where 100 gf weight is suspended.

Anticlockwise moment = 15 gf × 50 cm + 50 gf × 10 cm = 1250 gfcm

Clockwise moment = 100 gf (x − 60 cm) = 100 x gf − 6000 gfcm

∴ By the law of moments,

100 x gf − 6000 gfcm = 1250 gfcm

∴ 100 x gf = 7250 gfcm

\[ \therefore \text{x} = \frac{7250}{100}\]

= 72.50 cm