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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?

Short Answer

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

Step by step solution

01

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.
02

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 \( E = 2.0 \times 10^{11} \text{ N/m}^2 \).
03

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 \( d \): \( d \approx 0.0045 \text{ m} \).
04

Convert Diameter to \( \text{cm} \)

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A steel wire has the following properties: $$ \begin{array}{l} \text { Length }=5.00 \mathrm{~m} \\ \text { Cross-sectional area }=0.040 \mathrm{~cm}^{2} \end{array} $$ Young's modulus \(=2.0 \times 10^{11} \mathrm{~Pa}\) Shear modulus \(=0.84 \times 10^{11} \mathrm{~Pa}\) Proportional limit \(=3.60 \times 10^{8} \mathrm{~Pa}\) Breaking stress \(=11.0 \times 10^{8} \mathrm{~Pa}\) The proportional limit is the maximum stress for which the wire still obeys Hooke's law (see point \(\mathrm{B}\) in Figure 11.12 ). The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

Shear forces are applied to a rectangular solid. The same forces are applied to another rectangular solid of the same material, but with three times each edge length. In each case, the forces are small enough that Hooke's law is obeyed. What is the ratio of the shear strain for the larger object to that of the smaller object?

A \(2.00 \mathrm{~kg}\) frictionless block is attached to an ideal spring with force constant \(315 \mathrm{~N} / \mathrm{m} .\) Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at \(12.0 \mathrm{~m} / \mathrm{s}\) Find (a) the amplitude of the motion, (b) the maximum acceleration of the block, and (c) the maximum force the spring exerts on the block.

Effect of diving on blood. It is reasonable to assume that the bulk modulus of blood is about the same as that of water ( \(2.2 \mathrm{GPa}\) ). As one goes deeper and deeper in the ocean, the pressure increases by \(1.0 \times 10^{4} \mathrm{~Pa}\) for every meter below the surface. (a) If a diver goes down \(33 \mathrm{~m}\) (a bit over \(100 \mathrm{ft}\) ) in the ocean, by how much does each cubic centimeter of her blood change in volume? (b) How deep must a diver go so that each drop of blood compresses to half its volume at the surface? Is the ocean deep enough to have this effect on the diver?

Find the period, frequency, and angular frequency of (a) the second hand and (b) the minute hand of a wall clock.

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