(a)(i) Explain what is meant by a machine (ii) Define the terms: mechanical advantage, velocity ratio and efficiency as applied to a machine. Derive the equ...

(a)(i) A machine is a device by means of which a force (the effort) applied at one point is used to overcome another force (the load) at some other point, usually giving a mechanical advantage or a change in the direction of the force.

(a)(ii)

• Mechanical advantage (M.A.) is the ratio of the load to the effort: \(M.A.=\dfrac{L}{E}\).
• Velocity ratio (V.R.) is the ratio of the distance moved by the effort to the distance moved by the load in the same time: \(V.R.=\dfrac{\text{effort distance}}{\text{load distance}}\).
• Efficiency is the ratio of the useful work output to the work input, expressed as a percentage.

Derivation. \(\text{Efficiency}=\dfrac{\text{work output}}{\text{work input}}=\dfrac{L\times d_L}{E\times d_E}=\dfrac{L/E}{d_E/d_L}=\dfrac{M.A.}{V.R.}\). Hence \(\text{Efficiency}=\dfrac{M.A.}{V.R.}\times100\%\).

(b) Efficiency is less than 100% because part of the work input is used to overcome friction between moving parts and to lift the useless (movable) parts of the machine; this energy is lost mainly as heat and is not delivered to the load.

(c) Pitch = 4.4 mm = \(4.4\times10^{-3}\,\text{m}\); tommy bar L = 70 cm = 0.70 m; height h = 20 cm = 0.20 m; mass = 8000 kg.

(i) Velocity ratio: \(V.R.=\dfrac{2\pi L}{\text{pitch}}=\dfrac{2\times\frac{22}{7}\times0.70}{4.4\times10^{-3}}=\dfrac{4.4}{4.4\times10^{-3}}=1000\).

(ii) Mechanical advantage: \(M.A.=\text{efficiency}\times V.R.=0.80\times1000=800\).

(iii) Effort: Load \(=mg=8000\times10=80000\,\text{N}\). \(E=\dfrac{L}{M.A.}=\dfrac{80000}{800}=100\,\text{N}\).

(iv) Work done by the effort: \(=\dfrac{\text{work output}}{\text{efficiency}}=\dfrac{mgh}{0.80}=\dfrac{80000\times0.20}{0.80}=\dfrac{16000}{0.80}=20000\,\text{J}=20\,\text{kJ}\).