(a) Derive a relation between mechanical advantage (MA), velocity ratio (VR) and efficiency (η).

(a) Let a machine overcome a load L by the application of an effort E. In the process, let the displacement of the effort be d_E and the displacement of the load be d_L.

Then,

Work input = effort x displacement of effort = E x d_E ...(i)

Work output = load x displacement of load = L x d_L ...(ii)

By definition,

$\text{Efficiency }\text{η} = \frac{\text{work output}}{\text{work input}}$

From eqns. (i) and (ii),

$$\begin{aligned}\text{η} = \frac{\text L \times \text d_\text L}{\text E \times \text d_\text E} = \frac{\text L}{\text E} \times \frac{\text d_\text L}{\text d_\text E} \\[1em] = \frac{\text L}{\text E} \times \frac{1}{\frac{\text d_\text E}{\text d_\text L}}\end{aligned}$$

But,

$\frac{\text L}{\text E}$ = M.A. and $\frac{\text d_\text E}{\text d_\text L}$ = V.R.

$$\begin{aligned}\therefore \text{η} = \frac{\text{M.A.}}{\text{V.R.}} \\[1em] \Rightarrow \text{M.A.} = \text{V.R.} \times \text{η}\end{aligned}$$

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

(b)

1. It is a speed multiplier.

2. Reason — In a class III lever the effort is applied in between the fulcrum and the load, so the effort arm is always shorter than the load arm. Hence its mechanical advantage and velocity ratio are always less than 1. Such a lever does not give a gain in force, but it gives a gain in speed, i.e., a larger displacement of the load is obtained by a smaller displacement of the effort.