Luccas Viana
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- Homework Statement
- In a push-up, what percentage of body weight is supported by each arm? How does this percentage decrease when performing a knee push-up, with support on the knees instead of the feet?
How does this force vary throughout the different stages of a push-up (that is, when the arms are more extended and when they are more flexed)?
How does this force vary when performing a push-up with the hands supported on a surface higher than the feet? (Consider the height of the support surface)
- Relevant Equations
- 2+2 = 4
I would really appreciate any help with this study. If necessary, I can try to create diagrams to better illustrate the situation described. Please do not hesitate to contact me if anything is unclear or if you need further clarification. English is not my native language, so I kindly ask you to excuse any grammatical errors.
I will present below what I have developed so far:
The human body is modeled as a rigid rod pivoting about a fixed axis located at the foot supports. The mass distribution is concentrated at a single point, the Center of Mass (C), while the reaction force exerted by the hands is applied along the vertical line passing through the Shoulder (S).
Description of the variables
Static Equilibrium
For the body to remain in static equilibrium, the sum of all forces and the sum of all moments about any point in the system must be zero:
$$\sum F = 0$$
$$\sum \tau = 0$$
The total gravitational force acting on the Center of Mass is calculated as follows:
$$P = W \cdot g$$
Percentage of Body Weight
The geometric relationship between the position of the Center of Mass (C) and the point where the force is applied at the shoulders (S) determines the exact percentage of body weight transferred to the arms:
$$K_{arms} = \left(\frac{C}{S}\right) \cdot 100$$
Upper Position (Arms Extended)
In the inclined plane configuration at the starting point, the condition for rotational equilibrium about the foot support point yields the following equation:
$$F_{hands} \cdot S - P \cdot C = 0 \implies F_{hands} = P \cdot \left(\frac{C}{S} \right)$$
From this point onward, I have not developed anything further yet.
How does one calculate the fraction of body weight supported by the arms in the lower position, with the elbows bent?
I will present below what I have developed so far:
The human body is modeled as a rigid rod pivoting about a fixed axis located at the foot supports. The mass distribution is concentrated at a single point, the Center of Mass (C), while the reaction force exerted by the hands is applied along the vertical line passing through the Shoulder (S).
Description of the variables
| Symbol | Description |
| H | Height of the person |
| S | Height of the shoulder |
| C | Position of the center of mass |
| W | Total mass of the person |
| g | Gravity |
Static Equilibrium
For the body to remain in static equilibrium, the sum of all forces and the sum of all moments about any point in the system must be zero:
$$\sum F = 0$$
$$\sum \tau = 0$$
The total gravitational force acting on the Center of Mass is calculated as follows:
$$P = W \cdot g$$
Percentage of Body Weight
The geometric relationship between the position of the Center of Mass (C) and the point where the force is applied at the shoulders (S) determines the exact percentage of body weight transferred to the arms:
$$K_{arms} = \left(\frac{C}{S}\right) \cdot 100$$
Upper Position (Arms Extended)
In the inclined plane configuration at the starting point, the condition for rotational equilibrium about the foot support point yields the following equation:
$$F_{hands} \cdot S - P \cdot C = 0 \implies F_{hands} = P \cdot \left(\frac{C}{S} \right)$$
From this point onward, I have not developed anything further yet.
How does one calculate the fraction of body weight supported by the arms in the lower position, with the elbows bent?
