Physics of the Planche The planche is a gymnastics exercise where the athlete holds their body parallel to ground supported only by their hands.

People sometimes train for this exercise using a progression of ``planche leans''.  The athlete will move their hands progressively closer to their center of mass while maintaining a straight back and straight arms.  Over years, by steadily decreasing this distance, the athlete develops the requisite anterior deltoid strength to hold the final position.
I would like to have a mathematical model of this progression, but I think I lack the basic physics skills to develop such a model.
I am baffled by even basic questions like ``As the hands approach the center of mass, how does the force being exerted through the feet change?''
I would assume that the relevant variables would be the length of the arms, the location of the center of mass in relation to the shoulder, and the position of the hands relative to the center of mass.  I am not sure if those are the only variables to consider.
I would be interested in having a model which takes these variables, and returns the torque at the shoulder.  At least:  I think that is what I should be interested in!  Again, my physics knowledge is extremely poor.  If I am wrong, please let me know what quantity you think would best correlate with the "difficulty" of the exercise.
My goal is to be able to use this model to create a roughly linear difficulty progression.

Edit:  While the answer I have has taught me a lot, I do not think that the model produced is accurate.  In particular, having a small force F radically shifts the torque curve, as can be seen here:
https://www.desmos.com/calculator/npazao2nvc
The model when F=0 seems reasonable, but I am not sure whether summing the absolute values of these torques makes much sense as a "difficulty measure" or not.
 A: This is an interesting problem, and certainly a model for the forces and torques could be generated. More on this in a moment. However, I think you may run into some issues with the "linear difficulty" curve. The model would tell you the forces, but that doesn't directly correspond to "difficulty." The muscular system can have different efficiencies in converting energy to force dependent upon the orientation of the body. For example, I recall years ago reading about how soccer players throw the ball from the sidelines at something like 60 degrees, much higher than the 45 degrees kinematics would predict as optimal (unfortunately I no longer remember the source) just because the human body is able to throw harder at this higher angle and that extra oomf overcomes the drawback of throwing at a higher angle. Though, perhaps I should add the disclaimer that this is not my area of expertise, so grain of salt and all that.
As a result, while we can model this situation and we can find a progression of hand locations which will result in something like a linear torque progression on the shoulder, it's not clear to me that this will necessarily correspond to a linear progression in the difficulty of the exercise.
With all that said, we can create a simple model to get a quick set of estimates...models can always be made more complicated with additional bells and whistles. So for now, let's assume the person is essentially flattened and consider only 1 arm (so if we use 2 arms, each arm must supply half the forces and torques we find in this model, but simplifying to 1 arm will make the model much simpler).
Roughly speaking, the diagram should look something like this:

I have labeled the stick frame of the person in black, the measurements in yellow, and some forces in green. Clearly this is very simplified, but hopefully will be enough to get started with. Also note that I have labeled the diagram with more information than is strictly necessary to solve a statics problem of this type because it may be useful for your purposes to use somewhat different variables. For example, the angles can be expressed in terms of the height $h$ (which is the height only up to the shoulders by the way), the arm length $L$, and the distance $x_h$ between the hands and feet by using some trigonometric identities...law of sines, cosines and so on.
Now, note that I have not labeled any forces on the shoulder. These are internal to the body and I'll come back to what I think is the right way to get at the torque that the musculature has to apply (realistic models, at least what I found on a quick google search, of the human shoulder can get very complicated very quickly). The problem at hand here is to find the forces supplied by the hands and feet, labeled $F$ and $N$ with appropriate subscripts, such that the body doesn't move.
To do this, I found it convenient to write down Newton's equations for the entire body to impose the condition that the body does not sink through the floor or begin levitating away, and the torque equations to make sure the body doesn't rotate, we are supposed to be remaining still after all. For the torque equations, I chose the feet and the shoulders as my pivot points, though you could have chosen any pair of points you like. The resulting equations are
$$
0=N_h+N_f-mg,\ \ \ 0=F_h-F_f, \ \ \ 0=N_hx_h-mgx_{CM},\\
 0=mg(h\cos\phi-x_{CM})-N_fh\cos\phi-F_fh\sin\phi-N_hL\cos(\theta+\phi)+F_hL\sin(\theta+\phi).
$$
These are a little bit messy, but at the end of the day they are only linear equations for the forces we are asking about, and, assuming I've made no mistakes to this point, the solution is given by
$$
N_f=\frac{mg}{x_h}(x_h-x_{CM}),\ \ \ N_h=mg\frac{x_{CM}}{x_h},\\
F_f=F_h=mg\frac{x_{CM}}{x_h}\frac{x_h+L\cos(\theta+\phi)-h\cos\phi}{L\sin(\theta+\phi)-h\sin\phi}.
$$
Again, this is a little bit messy. Perhaps there's another set of variables which would make this look neater, such as eliminating the sines and cosines in favor of the other measures I included in the diagram. Not sure, haven't checked.
Now, we know all the forces acting on the body through the hands and feet and, as a result, the forces the hands and feet are applying to the floor. Assuming the shoulder is supplying all the torque to make this happen and maintain the angle $\theta$ in the diagram, I think it would be reasonable to define the "torque" that the shoulder has to supply as being the sum of the magnitudes of the torques supplied by the hands and by the feet. Admittedly, there could be something wrong with this thinking, and if anyone has another idea, I'd be interested to hear it. If we run with this idea though, we can now write the "torque" supplied by the shoulder (the quotes because it's not really a torque so much as a measure of torques supplied on either side of a joint) as
$$
\tau=mg\frac{x_{CM}}{x_h}\left(|x_h+h(F\sin\phi-\cos\phi)|+L|\cos(\theta+\phi)-F\sin(\theta+\phi)|\right),
$$
where I have used Mathematica to just plug everything in and simplify. I have also left $F\equiv F_f=F_h$ in this expression for reasons described in the edit below.
Finally, let me note again that this $\tau$ is some measure of the effort both shoulders need to exert. If you want a per-shoulder measure, then halving this seems reasonable.
Edit: Let me comment on something I entirely brushed over in the above comments, but is actually important to understanding how this model works. The attentive reader may have noticed that the expression obtained for $F_f$ and $F_h$ is actually ill-defined as both the numerator and denominator of the fraction appearing therein are both zero, though this is hidden by the fact that we are working with too many variables. Essentially, and this can be checked directly, once we have eliminated the normal forces and obtained an equation for, say, $F_f$, the equation takes the form $aF_f=b$ where $a$ and $b$ are zero. As a result, the above equations do not actually determine $F$.
How can we understand this? Aside from this point out answer seems to make sense: in the special case where $x_h=x_{CM}$ we have $N_f=0$ and $N_h=mg$, which is precisely what we would expect when the person is just holding themselves up with their hands alone. One way to imagine this is as follows. For the moment, let's remove from our minds the human body and suppose instead we are dealing with a pair of steel beams that we've welded together at what we have been calling the shoulder. Just placing such a thing on the floor, I think it might be intuitive to expect that there would be no need for horizontal forces acting on the ends of the beams.
At the same time, however, we could imagine ourselves getting on the floor to do, say a pushup or a plank. While doing so we always have the option to apply pressure to our hands and feet which is parallel to the floor. Doing so would lead to greater-than-necessary exertion, but we could do so without actually moving in any way. I think this is precisely what's happening here: the math is telling us that any amount of horizontal forces we would like to apply would work out fine, but we actually don't need any of them, at least in this simple model.
This is why I went back and edited the expression for $\tau$ above to just have $F$s in it. We can plug in any value we like for $F$: we are always free to exert more effort for no reason. There will, however, be a minima to this function which will correspond to the minimal amount of effort. And by the way, $F=0$ is not necessarily the point of minimal effort because we are dealing with a function whose basic form is $|aF+b|+|cF+d|$ where the constants $a,b,c$, and $d$ could be either positive or negative (we know it has a minima because $F\rightarrow\pm\infty$ causes the function to reach positive infinity on both ends, implying a minima somewhere in the middle by continuity). This is, I think, the mathematical realization of the observation that if we were to do a pushup or something, most people would agree that they do, in fact, apply some non-zero horizontal force...the minima of effort actually corresponds to applying some non-zero horizontal force in general.
A: At a glance the setup in Richard Myer's answer seems correct, but there are a couple of errors. That the net torque on the system is zero (the body as a whole is not rotating) amounts to a single equation --- if one examines them closely, the condition that the torque about the feet is zero should be identical to the condition that the torque about the shoulder is zero.
If I may borrow his free-body diagram, the statement that the net force is zero reads
$$ N_h + N_f = mg \\
F_h = F_f $$
The statement that the net torque is zero reads
$$ N_h x_h = mg x_c $$
Note that the frictional forces are not uniquely determined by the above equations --- this is to be expected, since we haven't specified the surface we're working out on. Let's assume we're performing the planche lean on a completely smooth surface, so that $F_h = F_f = 0$. Solving the above then gives $N_f = mg(x_h-x_c)/x_h$ and $N_h = mg x_c/x_h$.
The torque at the shoulder.
One has to be a bit careful with signs in calculating the torque at the shoulder. Taking the body as a whole, the net torque about the shoulder is zero (as indeed it is about any point). Note however that both the weight and the reaction force on the hands act to close the joint angle $\theta$, whilst the reaction force on the feet acts to open it. So the torque your shoulder joints need to supply is
$$ \tau = N_h L \cos(\theta + \phi) + mg  (L \cos(\theta + \phi) + x_h - x_c) - N_f (L \cos(\theta + \phi) + x_h)  $$
Using our previous equations to eliminate $N_f$ and $N_h$, we have that the torque is
$$ \tau/mg =  \frac{2x_c L}{x_h} \cos(\theta + \phi) $$
In order to finish the problem, we need to determine the relations between the lengths $x_h$, $x_c$, the angles $\theta$, $\phi$, and the dimensions of the person performing the planche. To do this exactly we need to work out how the location of the centre of mass depends on the orientation of the arms. If we assume the arms to be massless (we actually already made this assumption when taking the entire weight to act to close the joint), this will simplify our life greatly, as then the centre of mass will be fixed, somewhere around the upper pelvis. This is still pretty messy in my eyes, so to get a rough handle on the situation, we can imagine doing our planche leans with feet raised, so that our legs are parallel to the ground and $\phi = 0$. If we also put the centre of mass half way up the person's body (giving the relations $x_c = h/2$ and $x_h + L \cos(\theta) = h$), we get
$$ \tau(\theta) = \frac{mg h L \cos(\theta)}{h - L \cos(\theta)} $$
Check that this expression makes sense. When $\theta = \pi/2$, the above returns zero, which is good because we're just performing a bog-standard plank. When our hands are directly beneath our centre of mass (so $L \cos(\theta) = h/2$), the above returns $mg h$ --- this is a torque $mgh/2$ from our weight and a torque $mgh/2$ from the reaction force at the hands. So this looks good to me.
