Problem 7 - College Physics

Question

A steel wire \(2.00 \mathrm{~m}\) long with circular cross section must stretch no more than \(0.25 \mathrm{~cm}\) when a \(400.0 \mathrm{~N}\) weight is hung from one of its ends. What minimum diameter must this wire have?

Step-by-Step Solution

Answer

The minimum diameter of the wire must be approximately 0.45 cm.

Step 1: Understand the Problem

We are asked to find the minimum diameter of a steel wire needed to ensure it stretches no more than 0.25 cm when a 400 N weight is hung from it. We will utilize the formula for elongation in a material under a load, which is: \( \Delta L = \frac{FL}{EA} \) where \( \Delta L \) is the change in length, \( F \) is the force applied, \( L \) is the original length, \( E \) is the Young's modulus of the material, and \( A \) is the cross-sectional area of the wire.

Step 2: Rearrange the Formula and Identify Known Values

We want to find the diameter, and the cross-sectional area \( A \) of the wire is related to the diameter \( d \) by \( A = \frac{\pi d^2}{4} \). We need to solve for \( d \) in the equation: \( \Delta L = \frac{FL}{E \cdot \frac{\pi d^2}{4}} \). Known values: \( F = 400.0 \text{ N} \), \( \Delta L = 0.0025 \text{ m} \), \( L = 2.00 \text{ m} \), Young's modulus for steel \( E = 2.0 \times 10^{11} \text{ N/m}^2 \).

Step 3: Solve for Diameter d

Start by rearranging the equation to solve for \( d^2 \): \[ \Delta L = \frac{4FL}{\pi Ed^2} \] \[ d^2 = \frac{4FL}{\pi E \Delta L} \]. Substitute the known values: \[ d^2 = \frac{4 \times 400 \times 2}{\pi \times 2.0 \times 10^{11} \times 0.0025} \]. Simplify and calculate \( d \): \( d^2 = \frac{3200}{\pi \times 5 \times 10^8} \), which gives \( d^2 \approx 2.038 \times 10^{-5} \). Calculate \( d \): \( d \approx 0.0045 \text{ m} \).

Step 4: Convert Diameter to \( \text{cm} \)

Since we want the answer in centimeters, convert \( d = 0.0045 \text{ m} \) to centimeters: \( d = 0.45 \text{ cm} \).

Key Concepts

Young's Modulus

Young's Modulus is a fundamental property of materials that quantifies their ability to withstand changes in length when subjected to stress. It is denoted by the symbol \( E \). The higher the value of Young's Modulus, the stiffer the material. For example, steel, with a Young's Modulus of approximately \( 2.0 \times 10^{11} \text{ N/m}^2 \), is very resistant to deformation compared to materials like rubber.

Young's Modulus is defined as the ratio of tensile stress to tensile strain. The formula is given by:
\[ E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
where:

• \( F \) is the force applied,
• \( A \) is the cross-sectional area,
• \( \Delta L \) is the change in length,
• \( L \) is the original length.

For practical applications, Young's Modulus helps engineers and designers predict how much a material will stretch or compress when a specific load is applied. By using the known value of Young's Modulus for steel in our problem, we can determine the minimum diameter required for the wire not to stretch more than a specified length when a known load is applied.

Material Stretching

Material stretching refers to the deformation or elongation of a material when subjected to external forces. When a material like a wire is pulled by a force, it stretches by a certain amount governed by the material's elastic properties.

In physics, the elongation \( \Delta L \) of a wire under a load is given by the formula:
\[ \Delta L = \frac{FL}{EA} \]
This formula helps in determining how much a piece of material will elongate when a particular force is applied.

• \( F \) - force applied.
• \( L \) - original length of the material.
• \( E \) - Young's Modulus.
• \( A \) - cross-sectional area.

Using this equation, it's possible to rearrange and solve for any one of the parameters if the rest are known. In our exercise, this relationship allows us to compute the required diameter of the wire to ensure the force causes no more than a specific extension. Essentially, this involves ensuring the structural integrity by choosing the appropriate material size and properties.

Cross-Sectional Area Calculation

The cross-sectional area of a material, particularly in wires or rods, is crucial in determining its ability to handle stress without deforming excessively. The cross-sectional area \( A \) is related to the diameter \( d \) for circular cross-sections through the formula:
\[ A = \frac{\pi d^2}{4} \]
This formula expresses the area as a function of the diameter squared, showcasing the significant impact that the diameter has on the overall area - and subsequently - on its structural capacity.

In the process of solving for the diameter of a wire required to endure specific loads, the cross-sectional area is integral. The larger the area, the better the wire can distribute the load over its volume. However, the material's properties and the conditions required (such as maximum allowable stretch) dictate the minimal values for diameter to prevent excessive elongation. By rearranging the given equation and solving for \( d^2 \), you can subsequently determine the necessary diameter of the wire to meet the demands of the specific application.