Question

The handle of a water pump is 90 cm long from its piston rod. If the pivot of handle is at a distance of 15 cm from the piston rod, calculate : (a) mechanical advantage of handle (b) least effort required at its other end to overcome a resistance of 60 kgf.

Answer 1

The handle is a lever of the 1st order.

Given: Total length of handle = 90 cm

Load arm (pivot to piston rod) = 15 cm

∴ Effort arm = (90 − 15) cm = 75 cm

\[ \text{(a) Mech. advantage} = \text{velocity ratio} = \frac{\text{effort arm}}{\text{load arm}}\]

\[= \frac{75\ \mathrm{cm}}{15\ \mathrm{cm}}\]

= 5

(b) Taking moments about the pivot:

Effort × effort arm = Load × load arm

E × 75 cm = 60 kgf × 15 cm

\[ \therefore\ \mathrm{E} = \frac{60\ \mathrm{kgf} \times 15\ \mathrm{cm}}{75\ \mathrm{cm}}\]

= 12 kgf