(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 equation connecting the three terms.
(b) Explain why the efficiency of a machine is usually less than 100%
(c) A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000 kg through a height of 20cm. The length of the tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calcuate the: (i) velocity ratio of the jack; (ii) mechanical advantage of the jack (iii) effort required in raising the body, (iv) work done by the effort in raising the body \((g = 10ms^{-2}; \pi = \frac{22}{7}\))
(a)(i) A machine is a device used to overcome a load at one end by the application of an effort at another end. Examples of simple machine include the screw jack, can opener, pulley system. (ii) M.A. is the ratio of the load (L) to the Effort (E).
\(M.A = \frac{Load}{Effort}\)
V.R. is the ratio of the distance moved by the effort tc the distarice moved by the load (at the same time).
\(V.R = \frac{\text{distance moved by effort}}{\text{distance moved by load}}\)
Efficiency (E) is the ratio of the useful work done by the machine (work output) to the total work put into the machine (work input).
\(Efficiency = \frac{\text{work done on load}}{\text{work done by effort}} \times 100%\)
\(Efficiency = \frac{M.A}{V.R} \times 100%\);
\(Work = Force \times Distance\)
\(Efficiency = \frac{\text{work output}}{\text{work input}} \)
= \(\frac{Load \times \text{distance moved by load}}{Effort \times \text{distance moved by effort}}\)
= \(M.A \times \frac{1}{V.R} \times 100%\)
(b) The efficiency of a simple machine is not 100% because of the work done in overcoming friction.
(c) length of tommy bar. L = 70cm = 0.7m; Pitch = 4.4mm = 0.0044m.
(i) \(V.R = \frac{2\pi L}{pitch} = \frac{2 \pi \times 0.7}{0.0044}\)
= \(2 \times \frac{22}{7} \times \frac{0.7}{44 \times 10^{-4}}\)
\(V.R = 1000\)
(ii) \(Efficiency = \frac{M.A}{V.R} \times 100%\)
\(\frac{80}{100} = \frac{M.A}{1000} \implies M.A = 800\)
(iii) \(M.A = \frac{Load}{Effort}\)
\(Effort = \frac{8000 \times 10}{800} = 100N\)
(iv) Work done = \(\frac{load \times \text{distance moved by load}}{effort \times \text{distance moved by effort}}\)
= \(\frac{8000 \times 10 \times 0.2}{0.8} = 20,000J\)
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