First i want to start saying that indeed. Biomechanics problems are really complicated as humans, and animals for that matter have too many moving parts and the relationship between forces and linkages is non-linear and does not have, most of the time. A closed form solution.

That being said I think you are over complicating the problem here Mr Moun.

As you point out drag is dependent on the object velocity squared, but also with the object shape, surface and interface. Normally an object and the medium, water in this case stay constant across a experiment but in this case Maddie is slowly getting out of water. So her contact surface is not constant. Hence this is a time-variant problem. Alas I doubt there is enough water, or speed for that matter, to have a significant effect on the mechanics of the situation. I think the friction between Maddie's lower body and the edge of the "pool" is more important, i think even the buoyancy has more effect on this than water drag which in any case would help Maddie not get pulled out of the pool as it works by opposing relative motion.

In this case we can simplify the issue, we can without loss of much precision, consider both Maddie and Karen as rigid bodies operating under the same relative plane. So we can reduce this to a two-dimensional problem, far easier than a general bio mechanics 3D problem.

The point of maximum torque is when Karen is perpendicular to the wall, as the distance is maximum to the fulcrum, her foot. If Maddie can generate enough force to pull her from that position she will be able to pull it from any other. The only forces on Karen are then Maddie's pull, her weight the normal from the wall and the friction. From there is is relatively issue to compute the strength on Maddie's arm by assuming equilibrium at that configuration by summing torque at Karen feet. This also allows us to compute the angle she is pulling from:

Code: Select all

```
F = m g x_l / (sin (a) (x_l + x_a)
a = arctan(x_l mu / x_a)
```

Even if you ignore my assumption of a horizontal Maddies since we have the picture is is trivial, although a bit time consuming. To compute the angles required. After all we know Karen height and there is negligible effect on perspective here so using pixel measurements you can obtain a metric representation of the whole scene.

This approach does ignore the inertia Karen had at the beginning but i think it is small enough to ignore. Besides if we include that then this turn into a dynamic problem which is much harder to model and solve.

Finally as Mr Bell points out even if Maddies is not strong enough to pull Karen all she had to do is to bend her knees. Lay flat on the wall and there is no torque anymore Maddie only needs to pull her up.