Why can you "suck in" cooked spaghetti? We all know that there is no "sucking", only pushing. So how are cooked spaghetti pushed into your mouth? The air pressure applies orthogonal on the spaghetti surface. Where does the component directed into your mouth come from?
 A: The component directed into your mouth comes form the different pressure between the outside and the inside of the mouth. If you create a difference in pressure of $\Delta P$ the force pushing the spaghetti in will be $\Delta P \cdot S$ where $S$ is the section of the spaghetto (or spaghetti.. depending how hungry you are;) )
If for example you create inside your mouth a void with a 10% efficiecy the difference in pressure will be:
$\Delta P\approx 1atm - 0.1atm= 10^4N/m^2$
If the cross section of your single spaghetto is $1mm^2$ the force pushing the food in your mouth will be 10 N: almost a kilogram!
You won't even need to be as efficient as 10% ;)
A: When you perform the sucking action, a pressure difference is clearly created and maintained by your lungs between the surrounding air and the air inside your mouth. An important point to notice here is that the mouth must not too far open (a bit lets it work still), else the pressure gradient between inside and outside of the mouth cannot be maintained by the expanding lungs over any significant time period.
Firstly, observe that there is a normal force on the surface of the spaghetti due to (some arbitrary) air pressure. As I have just explained, there is a pressure gradient between the inside and outside of the mouth, directed inwards, hence there is a driving force inwards. Combining the two effects gives a net force directed at some angle to the surface of the spaghetti. Hence, it is an angled force, with some component pointing inwards towards the mouth, that pushes it in.
A: Since I'm not entirely content with the answers to date, here's my take - everyone seems to agree on the basics of forces generated by pressure differentials. If you took a rigid uncooked spaghetto with a cross section of $A$, the case is quite clear - on the cylinder's base in your mouth, a force of $p_{in} A$ is trying to push the spaghetto out, and on the other end, a force of $p_{out} A$ pushes it inwards. The forces generated by the pressures on the sides even out.
It becomes more confusing when referring to cooked spaghetti since a) the cooked spaghetto won't stick straight out from your mouth and b) we intuitively don't want it to transmit forces because "you can't push on a string".
However, these complications do not change the underlying principle. Imagine cutting the spaghetto right outside your mouth while you keep sucking on the inner part. Clearly, the free spaghetto wouldn't feel any net force from air pressure after the cut and just float away/fall down. By cutting, you have changed the total picture by adding $p_{out} A$ orthogonal to the cut surface, but removed the push of $p_{in} A$, so before there must have been a net force on this surface before if there is none afterwards.
I know that it's not the qusestion, but in this case it is much simpler to look at the basic thermodynamics - the system is basically a microcanonical ensemble, with $E = p_{out} V_{out} + p_{in} V_{in}$ under the constraint that $V_{in} + V_{out} = \mathrm{const}$, and if $p_{in} < p_{out}$, the minimum energy (and thus equilibrium) state is clearly that with $V_{out}=0$.
A: I think he is envisaging the spaghetti as a long thin cylinder. Air pressure can only exert a force normal to the surface, so the "push" has to come from the end of the noodle, which 
would seem to be too far away to transmit stress through such a squishy medium. Mechanical engineers sometimes decompose stress into pressure (a component uniform if all directions), and deviatoric, whatever is left to fill out the stress tensor. The later deviatoric stress would indeed not transmit well
through the spaghetti. But as long as the medium is incompressible (like water) the pressure
will be easily transmitted. This would also apply to compressible spaghetti, but the volume must change with a change of pressure.
