(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces (b)(i) Describe an experiment, using the principle of moments, to...

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(a) Conditions for equilibrium of a rigid body under parallel forces:

1. The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force (no net force).
2. The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment (principle of moments).

(b)(i) Experiment to find the mass of a metre rule (principle of moments):

• Balance the metre rule alone on a knife-edge and note the balance point; this is its centre of gravity \(G\) (near the 50 cm mark).
• Now shift the knife-edge to one side of \(G\) and suspend a known mass \(m\) from the short arm; adjust its position until the rule balances horizontally.
• Measure the distance \(a\) of the known mass from the knife-edge and the distance \(b\) of \(G\) from the knife-edge.
• By the principle of moments, \( m \times a = M \times b \), so the mass of the rule \( M = \dfrac{m\,a}{b} \).

(ii) Two precautions:

1. Read the balance positions with the eye directly above the scale to avoid parallax error.
2. Ensure the knife-edge is sharp and the rule is horizontal (balanced) before taking readings.

(c) Bullet: \( m = 120\,\text{g} = 0.12\,\text{kg}, \; u = 20\,\text{m s}^{-1}, \; v = 0, \; t = 0.1\,\text{s}. \)

(i) Resistance (magnitude of the constant force):

\[ F = ma = m\frac{(u - v)}{t} = 0.12 \times \frac{20}{0.1} = 0.12 \times 200 = 24\,\text{N} \]

(ii) Distance moved in the wood:

\[ s = \left(\frac{u + v}{2}\right)t = \left(\frac{20 + 0}{2}\right)(0.1) = 10 \times 0.1 = 1.0\,\text{m} \]

Answers: resistance \(= 24\,\text{N}\); distance \(= 1.0\,\text{m}\).