Question

A uniform meter rule which is pivoted at the 20cm mark and balanced horizontally by an object of weight 3.15N placed at the 10cm mark calculate the weight of the ruler

Answer 1

To find the weight of the ruler, we will apply the principle of moments (torque balance). The rule is balanced at the pivot point (20 cm mark) and the weight of the ruler acts at its center of mass, which for a uniform meter rule is at the 50 cm mark.

Let's denote:
- Weight of the ruler = \( W \) N
- Distance from pivot (20 cm mark) to the center of the ruler (50 cm mark) = \( 30 \) cm (which we will convert to meters for calculations, \( 0.3 \) m).
- Weight of the object = \( 3.15 \) N
- Distance from pivot (20 cm mark) to the 10 cm mark = \( 10 \) cm (which we will convert to meters, \( 0.1 \) m).

Now we set up the balancing equation based on the moments (taking moments about the pivot point at the 20 cm mark):

The clockwise moment must equal the counter-clockwise moment for the system to be in equilibrium.

**Clockwise moment due to the weight of the ruler:**
\[
\text{Moment} = W \times 0.3 \quad (\text{the distance to the pivot is } 30 \text{ cm} = 0.3 \text{ m})
\]

**Counter-clockwise moment due to the weight of the 3.15 N object:**
\[
\text{Moment} = 3.15 \times 0.1 \quad (\text{the distance to the pivot is } 10 \text{ cm} = 0.1 \text{ m})
\]

Setting the clockwise moment equal to the counter-clockwise moment gives us:
\[
W \times 0.3 = 3.15 \times 0.1
\]

Calculating \( 3.15 \times 0.1 \):
\[
3.15 \times 0.1 = 0.315 \text{ N m}
\]

Thus, the equation is:
\[
W \times 0.3 = 0.315
\]

Now, solve for \( W \):
\[
W = \frac{0.315}{0.3} = 1.05 \text{ N}
\]

Therefore, the weight of the ruler is \( 1.05 \) N.