Uncertainty principle for a sitting person

If a person is sitting on a chair his momentum is zero...

How close to zero?

The uncertainty principle says that if $\Delta x$ is the uncertainty in position and $\Delta p$ is the uncertainty in momentum, then $\Delta x\,\Delta p\sim \hbar$. So, consider an object with the mass of a person, say $M = 70\ \mathrm{kg}$. Suppose the uncertainty in this object's position is roughly the size of a proton, say $\Delta x = 10^{-15}\ \mathrm m$. The uncertainty principle says that the uncertainty in momentum must be $$ \Delta p\sim\frac{\hbar}{\Delta x}\approx\frac{1 \times 10^{-34}\ \mathrm m^2\ \mathrm{kg/s}}{10^{-15}\ \mathrm m}\approx 1\times 10^{-19}\ \mathrm{m\ kg/s}, $$ so the uncertainty in the object's velocity is $$ \Delta v=\frac{\Delta p}{M}\approx \frac{\approx 1\times 10^{-19}\ \mathrm{m\ kg/s}}{70\ \mathrm{kg}}\sim 1\times 10^{-21}\ \mathrm{m/s}. $$ In other words, the uncertainty in the person's velocity would be roughly one proton-radius per month.

This shows that the uncertainties in a person's position and momentum can both be zero as far as we can ever hope to tell, and this is not at all in conflict with the uncertainty principle.

If we pretend that person is a quantum mechanical particle of mass $m=75$ kg and we localize him in a box of length $L=1$ m, then the resulting uncertainty in his velocity would be about one Planck length per second. Are you sure you know his velocity to within one Planck length per second?

Applying quantum mechanical principles to classical systems is always a recipe for disaster, but this underlying point is a good one - in macroscopic systems, the uncertainty principle implies fundamental uncertainties which are so small as to be completely meaningless from an observational point of view. If you were moving at a planck length per second for a hundred quadrillion years, you'd be about halfway across a hydrogen atom.