Question

A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; the same muscle under maximum tension requires a force of 500 N for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm${}^2$.

Short answer

Young's modulus is 333.3 N/m^2 for relaxed tension and 6666.7 N/m^2 for maximum tension.

Step-by-step solution

Gather Known Values

We know the following values for both conditions:- Relaxed tension: Force, $F=25.0$ N.- Maximum tension: Force, $F=500$ N.- Elongation, $\mathrm{\Delta }L=0.03$ m (converted from 3.0 cm).- Original length of the muscle, $L=0.20$ m.- Cross-sectional area, $A=50.0\times {10}^{-4}$ mdue to conversion from $c{m}^2$ to $m^2$.

Understand Young's Modulus

Young's Modulus $E$ is calculated using the formula:$$E=\frac{F\cdot L}{A\cdot \mathrm{\Delta }L}$$where $F$ is the force applied, $L$ is the original length of the material, $A$ is the cross-sectional area, and $\mathrm{\Delta }L$ is the change in length (elongation).

Calculate Young's Modulus for Relaxed Tension

Substitute the known values into the Young's modulus formula for the relaxed state:$$E_{\text{relaxed}}=\frac{25.0\text{N}\times 0.20\text{m}}{50.0\times {10}^{-4}{\text{m}}^2\times 0.03\text{m}}$$Calculate to find:$$E_{\text{relaxed}}=\frac{5.0}{1.5\times {10}^{-2}}=\frac{5.0}{0.015}=333.3{\text{N/m}}^2$$

Calculate Young's Modulus for Maximum Tension

Substitute the known values into the Young's modulus formula for the maximum tension state:$$E_{\text{maximum}}=\frac{500\text{N}\times 0.20\text{m}}{50.0\times {10}^{-4}{\text{m}}^2\times 0.03\text{m}}$$Calculate to find:$$E_{\text{maximum}}=\frac{100.0}{0.015}=6666.7{\text{N/m}}^2$$

Key concepts

Elasticity

Elasticity is the property of materials that allows them to return to their original shape after being deformed by an external force. This characteristic is what makes materials like rubber bands snap back after being stretched. In the context of a biceps muscle, elasticity describes the muscle's ability to stretch and recoil. When a muscle is relaxed, it stretches with less force, but under maximum tension, it requires a significantly greater force to achieve the same amount of elongation. This behavior under different forces is a classic example of elasticity at work. A material with higher elasticity will return to its original shape more quickly and with less permanent deformation. Understanding elasticity is crucial for applications like designing sports equipment or medical prosthetics, where materials must endure repeated stress.

Stress-Strain Relationship

The stress-strain relationship is central to understanding material behavior under load. Stress refers to the force applied to a material, expressed per unit area. It is mathematically defined as $\text{Math input error}$, where $F$ is the force applied and $A$ is the cross-sectional area. Strain, on the other hand, measures the deformation of the material relative to its original length and is expressed as a ratio: $\text{Math input error}$.

In our example, both stress and strain were used to determine how the biceps muscle behaves under different levels of tension. By plotting stress against strain, we create a stress-strain curve, which provides insights into a material's elastic limit and ability to undergo plastic deformation. The slope of the initial, linear portion of this curve is what we refer to when calculating Young's modulus, providing a measure of a material's stiffness.

Material Properties

Material properties such as hardness, tensile strength, and elasticity define how substances respond to external forces. These properties are essential for engineers and scientists when they need to choose materials for specific applications. For example, the muscle tissue in our exercise exhibits different Young's modulus values under relaxed and maximum tension, indicating variable stiffness.

The intrinsic material properties information allows for understanding and predicting how a material will behave in a given situation, reducing the risk of failure in applications like medical implants or athletic equipment. Moreover, knowledge of these properties helps in creating models for analyzing how a muscle or any cylindrical structure might react under different stress conditions. This aids in biologically relevant studies, ensuring safety and efficacy in biological and mechanical systems. By studying these characteristics, professionals can push the limits of what materials can be used for various innovations.