Problem 8 - College Physics

Question

\(\bullet\) Achilles tendon. The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is \(78.1 \mathrm{mm}^{2},\) its average length is \(25 \mathrm{cm},\) and its average Young's modulus is 1474 MPa. (a) How much tensile stress is required to stretch this muscle by 5.0\(\%\) of its length? (b) If we model the tendon as a spring, what is its force constant? (c) If a 75 kg sprinter exerts a force of 13 times his weight on his Achilles tendon, by how much will it stretch?

Step-by-Step Solution

Answer

(a) 73.7 MPa tensile stress, (b) 460.268 N/m force constant, (c) 20.75 cm stretch.

Step 1: Understand the Problem

The Achilles tendon behaves like an elastic material. Thus we can use formulas involving stress and strain. We need to calculate the tensile stress for a 5% stretch, determine the force constant as if it were a spring, and calculate stretch due to the force exerted by a sprinter.

Step 2: Calculate Tensile Stress

The tensile stress can be calculated using the formula \( \text{Stress} = \text{Young's modulus} \times \text{Strain} \). Given that the strain is 5%, or 0.05, and the Young’s modulus is 1474 MPa, the tensile stress is \( \text{Stress} = 1474 \times 0.05 = 73.7 \text{MPa} \).

Step 3: Calculate Force Constant of Tendon

The force constant for a material behaving like a spring is given by \( k = \frac{E imes A}{L} \), where \( E \) is Young's modulus, \( A \) is cross-sectional area, and \( L \) is length. Inputting the values, \( E = 1474 \times 10^6 \text{Pa} \), \( A = 78.1 \times 10^{-6} \text{m}^2 \), and \( L = 0.25 \text{m} \), the force constant is \( k = \frac{1474 \times 10^6 \times 78.1 \times 10^{-6}}{0.25} = 460.268 \text{N/m} \).

Step 4: Calculate Stretch Due to Sprinter's Force

The force exerted by the sprinter is \( 13 \times 75 \times 9.8 \text{N} \). This force stretches the tendon by \( \Delta L = \frac{F}{k} \). First, calculate the force: \( F = 13 \times 75 \times 9.8 = 9555 \text{N} \). Calculate the stretch: \( \Delta L = \frac{9555}{460.268} \approx 20.75 \text{cm} \).

Key Concepts

Young's Modulus

Young's Modulus is a critical concept in understanding how materials stretch and compress under applied forces. It quantifies a material's stiffness and is defined as the ratio of tensile stress to tensile strain. The formula for Young's Modulus is given by:\[ E = \frac{\text{Stress}}{\text{Strain}} \]where:

• Stress is the force per unit area applied on the material.
• Strain is the deformation experienced by the material in the direction of the applied force.

For the Achilles tendon, the Young’s Modulus is 1474 MPa. This value indicates the tendon’s resistance to being deformed. A higher Young's Modulus means the material is stiffer and less elastic, while a lower value would indicate more elasticity. This property is particularly important for tendons, which must be both strong and flexible to withstand forces during physical activities.

Tensile Stress

Tensile stress is a measure of how much pulling stress a material can withstand before it deforms. It is determined by the formula:\[ \text{Tensile Stress} = \frac{F}{A} \]where:

• F is the force applied perpendicular to the surface of the material.
• A is the cross-sectional area of the material.

In the context of the Achilles tendon, tensile stress is calculated when the tendon stretches by 5% of its original length. With a known Young's Modulus and strain of 5% (or 0.05), we find the tensile stress as:\[ \text{Stress} = 1474 \times 0.05 = 73.7 \text{ MPa} \]This stress helps understand what kind of forces the tendon can withstand without failing, highlighting why the Achilles tendon is so crucial for athletes.

Elasticity

Elasticity describes a material's ability to return to its original shape after being stretched or compressed. It is a key characteristic for tendons, allowing them to handle dynamic loads during activities like running and jumping. Tendons require high elasticity to absorb impacts and quickly revert to their original length.Elastic materials follow Hooke’s Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The formula:\[ F = k \Delta L \]describes this relationship, where:

• F is the force applied on the material.
• \(k\) is the spring force constant.
• \(\Delta L\) is the change in length.

This elasticity allows the Achilles tendon to stretch under force and rebound, playing a vital role in various athletic movements.

Spring Force Constant

The spring force constant (\(k\)) is a measure of a material’s stiffness in terms of how much force is needed to cause a unit deformation. It is defined by:\[ k = \frac{E \times A}{L} \]where:

• E is Young's modulus.
• A is the cross-sectional area.
• L is the original length of the material.

According to the exercise, the Achilles tendon is modeled as a spring, which gives the force constant \(k = 460.268 \text{ N/m} \). This means for every meter the tendon is stretched, 460.268 Newtons of force is needed. Such information is invaluable for understanding the dynamic function of tendons under various physical stresses. Knowing the spring force constant helps predict how much a tendon stretches when a specific force is applied, ensuring the body can handle vigorous activities like sprinting without injury.