How much force is required to break precession? 
Gravity, in this diagram, provides a downward force creating a torque  that causes the wheel to precess around the chain. This is cool but counter intuitive, and I am wondering what force is needed to push the wheel to the point that it is parallel with the ground if the force is applied to the free side of the wheel. Will increased force here only cause it to precess faster around the chain? And what equations would govern this process. 
 A: Given enough force the tendency to go into precessing motion can always be overpowered.
One way to see that is to consider the case of a relatively slowly spinning wheel. When the wheel is spinning relatively slowly, and you release it the wheel will mainly flop down, with just a hint of precessing motion.
If you try that a number of times, each time with a bit faster spinning wheel, you will notice that the faster the spin of the wheel the more vigorous the response on being released.
The faster the wheel spins, the slower the required precessing motion to keep the spin axis perpendicular to gravity.
Conversely, the slower the wheel spins the faster the required precession motion. The kinetic energy for that precessing motion has to come from somewhere. The source of this kinetic energy is a bit of sag of the wheel; gravitational potential energy is then converted to kinetic energy.
(This can also be seen from angular momentum considerations, the angular momentum of the precssing motion has to come from somewhere. The change of angular momentum of the spin axis as the wheel drops a little is the source of that angular momentum.)
If on release the wheel spins only slowly then the wheel has to drop a lot in order to arrive at the required precession speed. But the wheel can drop only so much.
As we know: if the wheel spins very fast then wheel's own weight is not enough to make it drop signicantly. So you increase the force. Increasing the force (in the direction of gravity) will make the wheel sag a bit more, but again, the wheel can drop only so much.
There is a widespread misconception that the precessing response to an applied torque happens instead of giving in to that torque. Again: simply considering the case of a slowly spinning wheel shows that this "instead" thing cannot be the case. A slowly spinning wheel simply flops over when you release it.
For further reading: my 2012 answer here on stackexchange to a question titled: What determines the direction of precession of a gyroscope?
