Question

A uniform metre scale of weight 50 gf is balanced at the 40 cm mark, when a weight of 100 gf is suspended at the 5 cm mark. Where must a weight of 80 gf be suspended to balance the metre scale?

Answer 1

Let x be the point where 80 gf weight is suspended.

Anticlockwise moment = 100 gf × 35 cm = 3500 gfcm

Clockwise moment = 50 gf × 10 cm + 80 gf (x − 40 cm)

= 500 gfcm + 80 x gf − 3200 gfcm

= 80 x gf − 2700 gfcm

∴ By the law of moments,

80 x gf − 2700 gfcm = 3500 gfcm

∴ 80 x gf = 6200 gfcm

\[ \therefore x = \frac{6200}{80} \]

= 77.5 cm