Question

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

Answer 1

Given:

Load arm = 2 cm

Effort arm = 16 cm

Effort (Applied Force) = 4 kgf

Formula:

Load × Load arm = Effort × Effort arm

Load × 2 = 4 × 16

Load = `(4 × 16)/2`

= 32 kgf

Therefore, a weight of 32 kgf placed on the nut will crack it.