(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces. (b)(i) Describe an experiment to determine the mass of a metre r...

Answer Details

(a) Conditions for equilibrium under parallel forces:

• The sum of the forces in one direction equals the sum of the forces in the opposite direction (net force = 0).
• The sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point (net moment = 0).

(b)(i) Experiment to find the mass of a metre rule: Suspend the metre rule from a knife-edge and find its centre of gravity \(G\) (the balance point with no load). Hang a known mass \(m\) at a distance \(d_1\) on one side of a chosen pivot, and adjust the position of the pivot until the rule balances horizontally. Measure the distance \(d_2\) from the pivot to \(G\) (through which the whole weight \(W = Mg\) of the rule acts). Taking moments about the pivot:

\[ m g \times d_1 = M g \times d_2 \Rightarrow M = \frac{m \, d_1}{d_2} \]

which gives the mass \(M\) of the rule.

(ii) Two precautions:

• Ensure the rule is horizontal (balanced) before taking readings.
• Avoid parallax error when reading the ruler; ensure the knife-edge is sharp and the mass is firmly fixed.

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

(i) Resistance (retarding force):

\[ F = \frac{m(u - v)}{t} = \frac{0.12 \times (20 - 0)}{0.1} = \frac{2.4}{0.1} = 24\ \text{N} \]

(ii) Distance moved in the wood (average velocity method):

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

The resistance is \(24\ \text{N}\) and the bullet travels \(1.0\ \text{m}\) into the block.