How does a nutcracker work

Question

For a nutcracker looks like the follows:

1. I read that "Torque of force we act must be larger magnitude than the torque of the reaction force of the nut." I don't know why this is helpful for cracking the nut. I think that statement is true as the reaction force from the nut always has the same magnitude as the force we exert on the cracker. This is because the forces between us and the cracker is a pair of Newton's 3rd law forces, while the forces between the cracker and the nut is another 3rd law pair. Then, the force the cracker exerts on the nut is the same as the force we exert on the cracker. However, the forces we exert are further from the pinpoint of the nutcracker, thus, it causes more torque than the reaction force of the nut. But why is this helpful for cracking the nut?

2. Why doesn't the nutcracker spin around an axis through its center of mass, but rather that pinpoint at the end of the cracker? I think this may be because of the fact that those 2 handles are joined at that pinpoint. However, how exactly does this ensure that the nutcracker only spins around the axis through that pinpoint?

3. Do the forces in this system balance out eventually? Specifically, the 2 forces one exerts on the upper and lower handles of the cracker and the 2 reaction forces from the upper and lower part of the nut exerts on the respective handles. It seems to me that they cancel out, if I join the forces on the cracker as in the leftmost diagram and the reaction forces in the rightmost diagram. The resultant orange vectors seem to me that they will cancel out: Is the resultant force from the leftmost diagram what eventually results in the nut being cracked? But why doesn't the nut end up accelerating in the direction of that force?

Answer 1

It is better to consider only the forces that are perpendicular to the symmetry plane.

Shown above at the forces acting on the top arm of the tool. To crush the nut (and to answer your first question) the torque about the pivot due nut must be less that the applied torque due to the hand

$$ x N < (x+y) F $$

But to estimate the crushing force, solve the above for $$N = \frac{x+y}{x} F$$

To find the pivot force, use the force balance $ N - P - F = 0 $ or $$P = \frac{y}{x} F$$

You do not expect the tool to rotate about the center of mass as it is constrained in place by the user. It is not free floating in the air. You can use the symmetry constraint to show that the pivot is going to remain fixed and the arms are going to close in.

At some point the forces do balance out, as either the nut is going to jam and provide $N$ as calculated above, or the nut is going to crush and the two arms of the tool are going to be in contact stopping the crushing.