A metre rule is pivoted at the 40 cm mark. A 15 g mass is suspended at the 15 cm mark. The rule is balanced. Calculate the mass of the rule.

The metre rule is supported at the $40 \text{ cm}$ mark, which is the pivot point.
There are two forces creating moments:

• The $15 \text{ g}$ mass suspended at the $15 \text{ cm}$ mark.
• The mass of the metre rule, acting at its centre of mass at the $52.5 \text{ cm}$ mark.

For the $15 \text{ g}$ mass:
Distance $d_1 = 40 \text{ cm} - 15 \text{ cm} = 25 \text{ cm}$.
This mass creates an anticlockwise moment.

For the mass of the metre rule ($M$):
Distance $d_M = 52.5 \text{ cm} - 40 \text{ cm} = 12.5 \text{ cm}$.
This mass creates a clockwise moment.

For the metre rule to be in equilibrium, the sum of the clockwise moments must equal the sum of the anticlockwise moments.
Anticlockwise moment = Clockwise moment

$$(15 \text{ g}) \times g \times d_1 = M \times g \times d_M$$ $$(15 \text{ g}) \times g \times (25 \text{ cm}) = M \times g \times (12.5 \text{ cm})$$

The acceleration due to gravity ($g$) cancels out from both sides.

$$15 \text{ g} \times 25 \text{ cm} = M \times 12.5 \text{ cm}$$ $$M = \frac{15 \text{ g} \times 25 \text{ cm}}{12.5 \text{ cm}}$$ $$M = \frac{375}{12.5} \text{ g}$$ $$M = 30 \text{ g}$$

The mass of the metre rule is $\text{30 g}$.