To find how much the femurs in the lower performer’s upper legs stretch, we can treat this problem as a physics question involving Hooke’s Law and the Young’s modulus of the material (bone in this case). We'll break down the solution step by step.

Given:

• Mass of the lower performer, $m = 57.5 \, \text{kg}$
• Gravitational acceleration, $g = 9.81 \, \text{m/s}^2$
• Length of each femur (treated as a uniform rod), $L = 33.0 \, \text{cm} = 0.33 \, \text{m}$
• Radius of each femur, $r = 1.86 \, \text{cm} = 0.0186 \, \text{m}$
• The force acting on the lower performer is three times her weight, $F = 3 \times m \times g$
• Young's modulus for bone (approximation), $E = 1.5 \times 10^{10} \, \text{N/m}^2$ (standard value for human bone).

Step 1: Calculate the weight of the lower performer

The weight is given by: $W = m \times g = 57.5 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 563.175 \, \text{N}$

Step 2: Calculate the force applied on the femurs

Since the upward force is three times her weight, the force acting on her legs is: $F = 3 \times W = 3 \times 563.175 \, \text{N} = 1689.525 \, \text{N}$

This force is distributed over two femurs, so the force per femur is: $F_{\text{per femur}} = \frac{1689.525}{2} = 844.7625 \, \text{N}$

Step 3: Calculate the cross-sectional area of each femur

The cross-sectional area $A$ of a femur (which is cylindrical in shape) is given by: $A = \pi r^2 = \pi \times (0.0186 \, \text{m})^2 = 0.001086 \, \text{m}^2$

Step 4: Apply Hooke's Law (Young’s modulus equation)

Young’s modulus relates stress and strain as: $E = \frac{\text{stress}}{\text{strain}} = \frac{\frac{F}{A}}{\frac{\Delta L}{L}}$ Rearranging for $\Delta L$ (the amount the femurs stretch): $\Delta L = \frac{F \times L}{A \times E}$

Substitute the known values: $\Delta L = \frac{844.7625 \, \text{N} \times 0.33 \, \text{m}}{0.001086 \, \text{m}^2 \times 1.5 \times 10^{10} \, \text{N/m}^2}$

Step 5: Calculate the stretch

$\Delta L = \frac{278.771625 \, \text{N} \cdot \text{m}}{1.629 \times 10^7 \, \text{N}} = 1.711 \times 10^{-5} \, \text{m}$

Final Answer:

The bones in her upper legs (femurs) stretch by approximately $1.71 \times 10^{-5} \, \text{m}$, or 0.017 mm.

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Would you like any further details or clarifications? Here are a few follow-up questions:

1. What happens to the stretch if the performer's mass increases?
2. How would the result change if the radius of the femur were smaller?
3. How would the stretch change if the performer was hanging from a higher height, assuming no change in force?
4. How does the Young's modulus affect the stretch in materials other than bone?
5. What happens to the force on the femurs if the performer is swinging in motion?

Tip: In stress-strain calculations, the material's Young’s modulus is crucial to understand how different materials deform under force.