Question

The handle of a nutcracker is 16 cm long and a nutis placed 2cm from its hinge.If a force of 4 kgf is applied at the end of handle to crack it, what weight if simply placed on the nut to crack it?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Distance}\mathrm{of}\mathrm{nut}\mathrm{from}\mathrm{fulcrum},\mathrm{d}=2\mathrm{cm} \\ \mathrm{distance}\mathrm{of}\mathrm{effort}\mathrm{from}\mathrm{fulcrum},\mathrm{D}=16\mathrm{cm} \\ \mathrm{Effort}=4\mathrm{kgf} \\ \mathrm{Load}\times \mathrm{d}=\mathrm{effort}\times \mathrm{D} \\ \mathrm{Load}\times 2=4\times 16 \\ \mathrm{Load}=\frac{4\times 16}{2}=32\mathrm{kgf} \end{gathered}$$