Question

A crowbar of length 3 m is pivoted at a point 15 cm from its tip. Calculate (a) mechanical advantage of the crowbar (b) least effort required at its other end to displace a load of 150 kgf.

Answer 1

Crowbar is a lever of 1st order.

Given: Total length of crowbar = 3m = 300 cm

Load arm of crowbar = 15 cm

∴ Effort arm of crowbar = (300 − 15) cm = 285 cm

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

\[= \frac{285\ \mathrm{cm}}{15\ \mathrm{cm}}\]

= 19

(b) Taking moment about pivot,

Effort × effort arm = Load load arm

E × 285 cm = 150 kgf × 15 cm

\[ \therefore\ \mathrm{E} = \frac{150\ \mathrm{kgf} \times 15\ \mathrm{cm}}{285\ \mathrm{cm}}\]

= 7.89 kgf