The Nut Cracker Problem

Let X be the minimum force required to crack the nut. The nut cracker can be visualised as a lever with the hinge as the pivot. To break the nut, according to the principle of moments, total clockwise torque equals total anti clockwise torque. Therefore:

$30 \times 2$ = $X \times 25$ Theorfore,

$X$ = $60 \div 25$ = $2.4N$

This is a statics torque question. Imagine one arm of the nutcracker on a table with the other arm free to move. Just prior to the nut cracking, you are applying a downward force $F$ perpendicular to the upper arm at a distance of $0.25$ m from the hinge of the nutcracker, while the nut is applying an upward force of $30$ newtons perpendicular to the upper arm at a distance of $0.02$ m from the hinge.

So just prior to the nut cracking, these two torques are equal and opposite, so

$0.25 * F = 0.02 * 30 \Longrightarrow F = \boxed{2.4}$ newtons.