Machines — Question 1

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

Let,

the displacement of effort = d_E and

the displacement of the load = d_L in time t.

$\text{Work Input} = effort \times \text{displacement of effort} \\[0.5em] = E \times d_E \\[0.5em] \text{Work Output} = load \times \text{displacement of load} \\[0.5em] = L \times d_L \\[0.5em] \text{Efficiency η} = \dfrac{\text{work output}}{\text{work input}} \\[0.5em] η = \dfrac{L \times d_L}{E \times d_E} \\[0.5em] η = \dfrac{L}{E} \times \dfrac{d_L}{d_E} \\[0.5em] = \dfrac{L}{E} \times \dfrac{1}{\dfrac{d_E}{d_L}} \\[0.5em]$

But we know that,

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

and

$\dfrac{d_E}{d_L} = V.R. \\[0.5em]$

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

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

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