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Physics

Derive a relationship between mechanical advantage, velocity ratio and efficiency of a machine.

Machines

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Answer

Suppose a machine overcomes a load L by the application of effort E.

Let,

the displacement of effort = dE and

the displacement of the load = dL in time t.

Work Input=effort×displacement of effort=E×dEWork Output=load×displacement of load=L×dLEfficiency η=work outputwork inputη=L×dLE×dEη=LE×dLdE=LE×1dEdL\text{Work Input} = effort \times \text{displacement of effort} \\[0.5em] = E \times dE \\[0.5em] \text{Work Output} = load \times \text{displacement of load} \\[0.5em] = L \times dL \\[0.5em] \text{Efficiency η} = \dfrac{\text{work output}}{\text{work input}} \\[0.5em] η = \dfrac{L \times dL}{E \times dE} \\[0.5em] η = \dfrac{L}{E} \times \dfrac{dL}{dE} \\[0.5em] = \dfrac{L}{E} \times \dfrac{1}{\dfrac{dE}{dL}} \\[0.5em]

But we know that,

LE=M.A.\dfrac{L}{E} = M.A. \\[0.5em]

and

dEdL=V.R.\dfrac{dE}{dL} = V.R. \\[0.5em]

Substituting the above two values in the formula for efficiency we get

η=M.A.V.R.M.A.=V.R.×ηη = \dfrac{M.A.}{V.R.} \\[0.5em] \Rightarrow M.A. = V.R. \times η \\[0.5em]

Hence, the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

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