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.