Question

if the handle of a nutcracker is 18cm long and a nut is placed 2cm from its hinge. If a force of 36kgf is placed on the nut, it will crack. calculate the force which has to be applied at the end of the handle to crack the nut

Answer 1

We know that :

Sum of clockwise moments = sum of anticlockwise moments

Load × load arm = effort × effort arm

Given :

Load = 36 kgf

Load arm = 2 cm

Effort arm = 18 cm

Effort = ?

36 kgf × 2 cm = E × 18 cm

==> E = 72 kgf cm / 18 cm

==> E = 4 kgf

The force to be applied is 4 kgf

Hope it helps mate :)

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