Question

The Young's modulus (see Problem P2.40) of muscle fiber is approximately \(2.80 \times 10^{7}\) Pa. A muscle fiber \(3.25 \mathrm{cm}\) in length and \(0.125 \mathrm{cm}\) in diameter is suspended with a mass \(M\) hanging at its end. Calculate the mass required to extend the length of the fiber by \(10 \%\)

Short answer

The mass required to extend the length of the muscle fiber by 10% is approximately 1.401 kg.

Step-by-step solution

Define the variables and formulae involved

We are given: Young's modulus, E = \(2.80 \times 10^7\) Pa Initial length, L₀ = 3.25 cm = 0.0325 m Diameter, d = 0.125 cm = 0.00125 m The required percentage of extension = 10% We need to find the mass, M The formula that relates stress, strain, and Young's modulus: \(E = \frac{\text{stress}}{\text{strain}}\) Where stress = \(\frac{F}{A}\) (Force over Area) and strain = \(\frac{\Delta L}{L_0}\) (change in length over initial length)

Calculate the change in length of the fiber and required strain

We first need to find the change in length of the fiber, represented by ΔL, that must occur to achieve a 10% extension. ΔL = 10% × L₀ ΔL = 0.1 × 0.0325 ΔL = 0.00325 m Now we can find the strain in the fiber: strain = \(\frac{\Delta L}{L_0}\) strain = \(\frac{0.00325}{0.0325}\) strain = 0.1

Calculate the stress on the fiber using Young's modulus

Next, we will find the stress on the fiber using the Young's modulus and the strain we just calculated. E = \(\frac{\text{stress}}{\text{strain}}\) stress = E × strain stress = \(2.80 \times 10^7\) Pa × 0.1 stress = \(2.80 \times 10^6\) Pa

Convert stress to force

The stress on the fiber can be represented as the force exerted on it per unit area: stress = \(\frac{F}{A}\) Therefore, we need to find the area A of the cross-section of the fiber (which is a circle): A = π × \(\frac{d^2}{4}\) = π × (0.00125 / 2)² A = \(4.9087 \times 10^{-6}\) m² Now we can use the formula stress = \(\frac{F}{A}\) to find the force, F: F = stress × A = \(2.80 \times 10^6\) Pa × \(4.9087 \times 10^{-6}\) m² F = 13.744 N

Calculate the required mass

Now that we have the force required to extend the muscle fiber by 10%, we can find the mass required using the formula: \(F = m \times g\) Where F is the force, m is the mass, and g is the acceleration due to gravity (around 9.81 m/s²). m = \(\frac{F}{g}\) = \(\frac{13.744 N}{9.81 \, \text{m/s}^2}\) m = 1.401 kg The mass required to extend the length of the fiber by 10% is approximately 1.401 kg.

Key concepts

Stress and Strain Relationship

Understanding the connection between stress and strain is key when studying materials like muscle fibers and their elasticity. Stress is defined as the force applied per unit area of a material and is measured in Pascals (Pa). In mathematical terms, stress can be represented as \( \frac{F}{A} \), where \( F \) is the force applied and \( A \) is the cross-sectional area of the material.
This provides insight into how much force is being exerted over a specific area of the material.
Strain, on the other hand, measures the deformation of the material in response to applied stress. It's a ratio of the change in length \( \Delta L \) to the original length \( L_0 \) and is dimensionless. The formula for strain is \( \frac{\Delta L}{L_0} \).
Understanding strain helps us quantify how much a material stretches or compresses under stress.Young's modulus is a crucial concept that bridges stress and strain. It is a measure of the stiffness of a material and is calculated by the ratio \( E = \frac{\text{stress}}{\text{strain}} \).
A higher Young's modulus indicates a stiffer material which requires more stress to achieve the same strain.

Muscle Fiber Elasticity

Muscle fiber elasticity refers to the ability of muscle fibers to stretch and return to their original shape. This property is incredibly important for movement and recovery in living organisms.
The elasticity of muscle fibers can be described using Young's modulus. When a muscle fiber is exposed to stress, like the weight of a suspended mass, its elasticity determines how much it will extend under the given force.
In the provided exercise, Young's modulus is given as approximately \( 2.80 \times 10^7 \) Pa, implying that muscle fibers have a moderate stiffness, allowing for a certain amount of stretch but also resisting deformation to some extent.When calculating changes in a muscle fiber's length due to elasticity, it's essential to consider both stress applied and the inherent tensile properties of the fiber. This allows us to appreciate how muscle fibers endure forces while performing various functions, from basic stretching to complex contractions in athletic activities.

Calculation of Force and Mass

In physics, calculating the force and mass involved in material deformation is a critical skill. Given a Young's modulus and desired strain, you can compute the necessary force to achieve that strain.
First, calculate stress using \( \text{stress} = E \times \text{strain} \). With known Young's modulus and strain, this step helps determine stress, expressed in Pascals.
The next step involves determining the cross-sectional area, especially for cylindrical objects like muscle fibers. Use the formula \( A = \pi \times \left(\frac{d}{2}\right)^2 \) to find the area, then substitute to find force with \( F = \text{stress} \times A \).Finally, once force is known, the mass can be deduced using the relation \( F = m \times g \), where \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)). Mass \( m \) is then \( m = \frac{F}{g} \), which connects the physical forces acting on the fiber to the mass required to achieve the desired stretch.
This sequence of calculations provides a deeper comprehension of equations governing the mechanics of materials.