To solve this problem, let's break down the forces and solve for the force exerted by the floor on each foot:

1. Understanding the setup: The person is in a static position, so the sum of the forces and torques must be zero.

2. Free-Body Diagram:

   • The person's weight ($W = mg$) acts downward at the center of mass.
   • There are reaction forces at the hands ($F_h$) and the feet ($F_f$).
   • Since the person is in static equilibrium, the upward forces ($F_h + F_f$) must balance the downward weight ($mg$).

3. Setting up equations:

   • Let $d_h$ be the distance from the feet to the hands.
   • Let $d_{cm}$ be the distance from the feet to the center of mass.
   • Using torque equilibrium around the feet to eliminate $F_f$, we get: $F_h \cdot d_h = mg \cdot d_{cm}$
   • Then using vertical force balance to find $F_f$.

Let’s assume typical values for $d_h$ and $d_{cm}$, or if these are in the problem image, provide them for exact calculations.

After solving, let me know if you need a specific answer calculation or assumptions refinement.

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Would you like the full step-by-step calculations, or do you have specific values from the diagram to share? Here are some further explorative questions to expand your understanding:

1. How would changing the position of the hands affect the force distribution?
2. What happens to the force on each foot if the person shifts their weight forward?
3. How could this force analysis change for a heavier person?
4. If friction at the hands is negligible, how does that impact equilibrium?
5. Why does torque balance help us find individual forces in static equilibrium?

Tip: Always consider using torque equilibrium around one point to reduce unknowns when analyzing static systems.