Problem 58 - Physics

Question

A piece of mohair taken from an Angora goat has a radius of \(31 \times 10^{-6} \mathrm{~m}\). What is the least number of identical pieces of mohair that should be used to suspend a \(75-\mathrm{kg}\) person, so the strain \(\Delta L / L_{0}\) experienced by each piece is less than \(0.010 ?\) Assume that the tension is the same in all the pieces.

Step-by-Step Solution

Answer

At least 276 pieces of mohair are needed.

Step 1: Understand the Physical Context

We have a fiber (mohair) used to support a mass. We need to ensure that the deformation (strain) on the fiber is less than 0.010 while supporting a 75 kg person. The strain is the ratio of the change in length to the original length.

Step 2: Calculate the Force on Each Fiber

The weight of the person (75 kg) is supported by multiple pieces of fibers, meaning the force on each fiber is the total weight divided by the number of fibers, \( n \). Use the formula \( F = mg \), where \( g = 9.8 \ \text{m/s}^2 \), giving \( F = 75 \times 9.8 = 735 \ \text{N} \). The force on each fiber is \( \frac{735}{n} \) N.

Step 3: Relate Strain to Stress and Force

Strain \( \frac{\Delta L}{L_0} \) is given by \( \frac{\sigma}{E} \), where \( \sigma \) is the stress, and \( E \) is Young’s Modulus. The stress \( \sigma \) is \( \frac{F}{A} \), where \( A \) is the cross-sectional area of one fiber.

Step 4: Calculate Cross-Sectional Area and Strain

Calculate the area \( A = \pi r^2 \), with \( r = 31 \times 10^{-6} \ \text{m} \), thus \( A = \pi (31 \times 10^{-6})^2 \ \text{m}^2 \). Substitute \( A \) and rearrange \( \frac{F}{n} = \sigma A \) into the strain formula: \( \frac{735}{nA} = 0.010 \times E \).

Step 5: Calculate Minimum Number of Fibers

Rearrange to find \( n \): \[ n = \frac{735}{0.010 \times E \times A} \]. Substitute \( E \) and \( A \) to solve for \( n \). Use the approximate value of \( E \) for mohair fibers, such as \( 5 \times 10^9 \ \text{Pa} \). Complete the calculation to find the smallest integer \( n \).

Step 6: Solve for n, the Number of Fibers

Solve \( n = \frac{735}{0.010 \times 5 \times 10^9 \times \pi (31 \times 10^{-6})^2} \). Compute to get \( n \approx 276 \). Since \( n \) must be an integer, we round up to the nearest whole number.

Key Concepts

Stress and Strain

Stress and strain are fundamental concepts in understanding how materials deform under force. Stress refers to the force applied over a specific area, determining how much an object is pushed or pulled. It is mathematically represented by \( \sigma = \frac{F}{A} \), where \( F \) is the force applied, and \( A \) is the cross-sectional area.Strain, on the other hand, represents the deformation and is the ratio of the change in length to the original length of the material. It is expressed as \( \frac{\Delta L}{L_0} \). This dimensionless measure shows how much the material stretches or compresses compared to its original state.Young's Modulus \( E \) is a vital parameter when dealing with stress and strain. It connects these two concepts through the formula \( \frac{\Delta L}{L_0} = \frac{\sigma}{E} \). A higher Young's Modulus indicates a stiffer material, meaning it doesn't deform easily.

Cross-Sectional Area

Cross-sectional area plays a crucial role in determining the stress experienced by a material. It refers to the area of the specific section of a material, perpendicular to the force applied. For a cylindrical object like a fiber, the cross-sectional area is calculated using the formula of a circle, \( A = \pi r^2 \), where \( r \) is the radius.For the mohair fiber in the problem, the radius is given as \(31 \times 10^{-6} \text{m}\). Hence, the cross-sectional area becomes \( A = \pi (31 \times 10^{-6})^2 \text{m}^2 \). A larger cross-sectional area would mean that more force is distributed over the area, resulting in lower stress.

Tension in Fibers

Tension in fibers is about how force is distributed across the fibers to maintain structural integrity. Each piece of mohair is subjected to a part of the total force required to hold up the weight it supports.Tension refers to the pulling force exerted by a material to hold an object in place. In our scenario, numerous fibers share the total weight of a person, which means each fiber experiences only a portion of the total force.For instance, if 275 pieces of fibers are individually subjected to the force of a 75 kg mass, the force experienced by each fiber is \( \frac{735}{275} \text{N} \). The goal is to ensure that this tension causes only minimal strain in each fiber.

Force Distribution

Force distribution helps in understanding how the load is spread across multiple structural components. When a weight is supported by numerous fibers, each fiber does not bear the entire load. Instead, the total force is equally distributed among them.This concept allows us to reduce the amount of stress each fiber experiences by increasing their number. Calculating the force experienced by one fiber involves dividing the total force by the number of fibers, \( n \), as shown in \( \frac{F}{n} \).Ensuring even force distribution is essential to avoid exceeding the stress limit of the material, thus preventing failure or excessive deformation.

Material Deformation

Material deformation occurs when a force exerts pressure on a material, causing it to change shape or size. This process is closely monitored to ensure the structural integrity of materials in engineering applications.The degree to which a material deforms is quantified by strain, as explained earlier, indicating the relative change in length.In our example, reducing material deformation is crucial, so each mohair fiber undergoes a maximum strain of \(0.010\), a conservative limit to maintain their functionality. Hence, enough fibers are used to ensure that individual deformation remains minimal, safeguarding against permanent damage.