You are provided with a metre rule, a knife edge, two pieces of thread and two masses m\(_{1}\) and m\(_{2}\)
(b}i. With the aid of a diagram, indicate the forces acting on the metre rule in the experimental set-up above.
ii. Define moment of a force about a point and state its S.1. unit.
(a) Metre-rule balance experiment
With the knife edge (pivot) at the 60 cm mark P, mass \(m_1\) hangs at Y and mass \(m_2\) at Q, and the rule balances. For each position of \(m_1\) the distances \(l = P - Y\) and \(d = Q - P\) are found. Taking moments about the pivot, \(m_1 g\, l = m_2 g\, d\), so \(l\) and \(d\) are directly proportional and the graph is a straight line through the origin. A specimen table (readings depend on your masses) is:
| Position of m1 (cm) | l = P - Y (cm) | d = Q - P (cm) |
|---|
| 20 | — | — |
| 18 | — | — |
| 16 | — | — |
| 14 | — | — |
| 12 | — | — |
Graph: plot l (vertical) against d (horizontal); a straight line through the origin is obtained.
Slope: \[ s = \frac{l_2 - l_1}{d_2 - d_1} = \frac{m_2}{m_1} \] (dimensionless).
Two precautions:
- Ensure the rule is exactly horizontal (balanced) before taking each reading.
- Use light loops of thread and read positions with the eye vertically above the mark to avoid parallax.
(b)(i) Forces on the rule
Acting on the rule are: the upward normal reaction R at the knife edge (60 cm mark); the downward weight \(m_1 g\) at Y; the downward weight \(m_2 g\) at Q; and the downward weight W of the rule acting at its centre of gravity G. For equilibrium the anticlockwise moment equals the clockwise moment about the pivot, and R equals the total downward weight.
(b)(ii) The moment of a force about a point is the product of the force and the perpendicular distance from that point to the line of action of the force: \[ \text{moment} = F \times d \] Its S.I. unit is the newton metre (N m).