What is the force on the arms in a pushup? What force do the arms have to generate to do a pushup? Let us look a this simplified model:

The body can be represented by the green plank of mass B. Its angle to the ground is $\theta$. 
This question was asked on fitness.stackexchange.com, although not phrased with physics in mind.
 A: A push up is a form of lever.  The athlete must exert roughly half her body weight (under some assumptions I'll clarify at the end of the post.)
We can solve this problem using the principle of virtual work.  Assume the athlete raises her body through a small angle $\textrm{d}\theta$.  Then her center of mass rises by $l \cos\theta \ \textrm{d}\theta$, with $l$ the distance from her feet to her center of mass.  The work done is
$$\textrm{d}W = mgl\cos\theta \ \textrm{d}\theta$$
This work is equal to the force multiplied by the distance over which the force is exerted.  If we call the distance from her feet to her shoulders $L$, then
$$mgl\cos\theta \ \textrm{d}\theta = F L \cos\theta \ \textrm{d}\theta$$
$$F = \frac{mgl}{L}$$
Your center of mass is roughly half way up your body, so a push up requires you to lift about half your weight.
This answer assumes the pushup is infinitely slow (i.e. no acceleration).  This is actually not so bad an approximation as it sounds, but people doing fast push ups will probably exert a higher force at the bottom of the push up, then let gravity do negative work to slow them down as they near the top.  This is the basis for "clapping push ups" (which I am too weak to perform).
We have only calculated the component of force in the direction of motion, so, we are assuming the athlete pushes directly in the direction of motion of her shoulders.  This means she doesn't push along the red line in your picture, but diagonally ahead of it.  In reality, she might push more straight down, increasing the force and decreasing the angle through with her elbows flex.  
The force may actually change throughout the push up due to this effect.  If you lean into the arm of a couch and do a push up off that, you'll find it's much easier at this high angle.  The reason is you aren't pushing straight down any more, so it's a little bit like going up a ramp (in that the distance over which the force is exerted is lengthened to accomplish the same amount of work).
I also assumed the athlete's center of mass doesn't move, which means I'm pretty much thinking of her arms as massless, and the rest of her is a rigid board.
Finally, I'm assuming her hands and feet don't slip, the floor is too big to move, etc.
An empirical way to determine the force would simply be to put scales underneath the hands, but the reading could be a little off depending on whether the scales are constructed to measure total force on them, or only the vertical component of the force.
Finally, an actual human is not simply one force pushing in one direction.  There are muscles, bones, ligaments, etc.  All sorts of different forces are going on in different places.  This calculation gives the force that the arms exert on the main body.  The force that the triceps exert on the elbow, for instance could be much higher.
A: As an independent check on Mark's post, I would solve this as a summation of moments problem, with the exerciser's toes as the center point.
I'm assuming you are more interested in how hard the exercise is, and not in how much weight the floor is supporting.  In performing a push-up, most of your arm weight does not count, since the lower arms do not go up and down, and the center of mass of the upper arms only moves half as far as your chest.  Ballpark this as a 10% credit against total body weight.  
That leaves 90% of your body weight, with a center of mass at approximately 4.5/9 body lengths from your toes.  The moment generated is 90% x ((4.5/9) body lengths) = 45 (% x BL). At the lowest point in the pushup, your chest muscles are generating enough force to counteract this moment, on a lever arm of about 8/9 body lengths.  the force is 45 (% x BL) / ((8/9) BL) = about 51% of your body weight.
A: Set $h$ the height of the training body of weight $W$ and $l_{cm}, l_{shoulder}$ the distance between feet and centre of mass and shoulder respectively. So there exist $α,β\in (0,1]$ such that $l_{cm}=αh$ and $l_{shoulder}=βh$. Let $F$ denote the perpendicular to the ground force acting on the shoulder, which we are searching for. We denote $F_{\bot}$ the perpendicular to the body component of $F$, which gives torque opposite to the torque of $W_{\bot}$. Fixing the training body at an arbitrary slope $φ$, there we get $Στ=0\iff (Wcosφ)l_{cm}-(F_{\bot}cosφ)l_{shoulder}=0\iff Wαhcosφ-Fβhcosφ=0\iff$ $F=\frac αβ W$, where $\fracαβ \in (0,1)$, because the centre of mass is lower than the shoulders.
