Question

A force of 5 Kg is required to cut a metal sheet.A shears used for cutting the metal sheet has its blades 5 cm long, while its handle are 10 cm long.What effort is needed to cut the sheet?

Answer 1

Given that,

Load resistance $L=5$ kg

Load arm $(L.A)=5$ cm

Effort arm $(E.A)=10$ cm

If, the effort $E$ is

By principle of lever,

$L\times (L.A)=E\times (E.A)$

$E=\frac{L\times (L.A)}{(E.A)}$

$E=\frac{5\times 5\times {10}^{-2}}{10\times {10}^{-2}}$

$E=2.5kg$

Hence, the effort is needed to cut the sheet is $2.5$ kg.