/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A metal wire \(75.0 \mathrm{~cm}... [FREE SOLUTION] | 91Ó°ÊÓ

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A metal wire \(75.0 \mathrm{~cm}\) long and \(0.130 \mathrm{~cm}\) in diameter stretches \(0.0350 \mathrm{~cm}\) when a load of \(8.00 \mathrm{~kg}\) is hung on its end. Find the stress, the strain, and the Young's modulus for the material of the wire. $$ \begin{array}{l} \sigma=\frac{F}{A}=\frac{(8.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{\pi\left(6.50 \times 10^{-4} \mathrm{~m}\right)^{2}}=5.91 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=5.91 \times 10^{7} \mathrm{~Pa} \\ \varepsilon=\frac{\Delta L}{L_{0}}=\frac{0.0350 \mathrm{~cm}}{75.0 \mathrm{~cm}}=4.67 \times 10^{-4} \\ Y=\frac{\sigma}{\varepsilon}=\frac{5.91 \times 10^{7} \mathrm{~Pa}}{4.67 \times 10^{-4}}=1.27 \times 10^{11} \mathrm{~Pa}=127 \mathrm{GPa} \end{array} $$

Short Answer

Expert verified
Stress: \(5.91 \times 10^{7} \text{ Pa}\), Strain: \(4.67 \times 10^{-4}\), Young's Modulus: \(127 \text{ GPa}\).

Step by step solution

01

Calculate the Cross-Sectional Area

The diameter of the wire is given as \(0.130\, \text{cm}\). Convert this to meters: \(0.130\, \text{cm} = 0.00130\, \text{m}\). The radius \(r\) is half the diameter: \(r = \frac{0.00130}{2} = 0.00065\, \text{m}\). The cross-sectional area \(A\) is \(\pi r^2\). Calculate it: \[ A = \pi (0.00065)^2 \approx 1.33 \times 10^{-6} \text{m}^2. \]
02

Calculate the Force

Given the mass \(m = 8.00 \text{ kg}\) and acceleration due to gravity \(g = 9.81 \text{ m/s}^2\), the force \(F\) is calculated using \(F = mg\). \[ F = 8.00 \times 9.81 = 78.48 \text{ N}. \]
03

Calculate the Stress

Stress \(\sigma\) is the force per unit area, given by \(\sigma = \frac{F}{A}\). Use the force from step 2 and area from step 1.\[ \sigma = \frac{78.48}{1.33 \times 10^{-6}} \approx 5.91 \times 10^7 \text{ Pa}. \]
04

Calculate the Strain

Strain \(\varepsilon\) is the relative change in length, given by \(\varepsilon = \frac{\Delta L}{L_0}\). The change in length \(\Delta L\) is \(0.0350\, \text{cm} = 0.000350\, \text{m}\) and the original length \(L_0\) is \(75.0\, \text{cm} = 0.750\, \text{m}\).\[ \varepsilon = \frac{0.000350}{0.750} = 4.67 \times 10^{-4}. \]
05

Calculate Young's Modulus

Young's modulus \(Y\) is the ratio of stress to strain, \(Y = \frac{\sigma}{\varepsilon}\). Use the stress from step 3 and strain from step 4.\[ Y = \frac{5.91 \times 10^7}{4.67 \times 10^{-4}} \approx 1.27 \times 10^{11} \text{ Pa}. \]

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Key Concepts

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

Understanding Stress
When we talk about stress in material science, we aren't discussing emotional pressure like we do with humans. Stress in this context refers to the internal forces acting within a material. Specifically, it's the force exerted per unit area within the material due to an external load. For instance, when a load is applied to a wire, stress occurs as the wire attempts to resist this load. The formula to calculate stress is given by:
  • \( \sigma = \frac{F}{A} \)
Here, \( F \) is the force applied, and \( A \) is the cross-sectional area of the wire. Stress is measured in Pascals (Pa). In practical terms, understanding stress helps engineers ensure that structures can withstand external pressures without deforming or breaking. It plays a crucial role in evaluating how materials perform under load.
Exploring Strain
Strain is closely related to stress but serves as a measure of deformation. While stress looks at force, strain measures how much a material changes in length when subjected to a force. It’s the ratio of the change in length to the original length. This dimensionless quantity captures the relative changes occurring in the material's dimensions:
  • \( \varepsilon = \frac{\Delta L}{L_0} \)
Where \( \Delta L \) is the change in length, and \( L_0 \) is the original length. Strain can be understood as a percentage or a fraction illustrating how much the material elongates or compresses from its original state. Understanding strain is fundamental for applications where flexibility and deformation are considerations, such as in the design of bridges, buildings, and even everyday objects like springs.
Elasticity and Young’s Modulus
Elasticity describes a material's ability to return to its original shape after being deformed by an external force. It's a key property that determines a material's behavior under stress and strain. The concept is often quantified by Young's Modulus, denoted as \( Y \). This modulus is a measure of a material's stiffness, capturing how much it will stretch or compress under stress:
  • \( Y = \frac{\sigma}{\varepsilon} \)
Young's Modulus is expressed in Pascals (Pa). A higher value indicates a stiffer material, which means it requires more stress to achieve the same strain. This concept is critical when selecting materials for construction and manufacturing, ensuring components are neither too stiff nor too flexible for their intended use. Rubber, for instance, has a low Young's Modulus, making it very elastic, whereas steel's high modulus makes it much stiffer.
Material Science Foundations
Material science is a field focused on understanding the properties and performance of different materials. It combines physics, chemistry, and engineering to explore how a material will behave under various conditions. Fundamental concepts such as stress, strain, and elasticity play a crucial role in this discipline, providing insight into material selection for specific uses.
A strong foundation in material science helps experts develop new materials and enhance existing ones. It's essential for innovations in industries such as construction, aerospace, electronics, and automotive engineering. By examining material characteristics like hardness, strength, ductility, and elasticity, scientists and engineers can predict how materials will perform, leading to safer, more efficient products and structures.

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Most popular questions from this chapter

A vertical wire \(5.0 \mathrm{~m}\) long and of \(0.0088 \mathrm{~cm}^{2}\) cross- sectional area has a modulus \(Y=200 \mathrm{GPa}\). A \(2.0-\mathrm{kg}\) object is fastened to its end and stretches the wire elastically. If the object is now pulled down a little and released, the object undergoes vertical SHM. Find the period of its vibration. The force constant of the wire acting as a vertical spring is given by \(k=F / \Delta L\), where \(\Delta L\) is the deformation produced by the force (weight) \(F\). But, from \(F / A=Y\left(\Delta L / L_{0}\right)\), $$ k=\frac{F}{\Delta L}=\frac{A Y}{L_{0}}=\frac{\left(8.8 \times 10^{-7} \mathrm{~m}^{2}\right)\left(2.00 \times 10^{11} \mathrm{~Pa}\right)}{5.0 \mathrm{~m}}=35 \mathrm{kN} / \mathrm{m} $$ Then for the period we have $$ T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{2.0 \mathrm{~kg}}{35 \times 10^{3} \mathrm{~N} / \mathrm{m}}}=0.047 \mathrm{~s} $$

Compute the volume change of a solid copper cube, \(40 \mathrm{~mm}\) on each edge, when subjected to a pressure of \(20 \mathrm{MPa}\). The bulk modulus for copper is \(125 \mathrm{GPa}\).

Battery acid has a specific gravity of \(1.285\) and is \(38.0\) percent sulfuric acid by weight. What mass of sulfuric acid is contained in a liter of battery acid?

Atmospheric pressure is about \(1.01 \times 10^{5} \mathrm{~Pa}\). How large a force does the atmosphere exert on a \(2.0-\mathrm{cm}^{2}\) area on the top of your head? Because \(P=F / A\), where \(F\) is perpendicular to \(A\), we have \(F=P A\). Assuming that \(2.0 \mathrm{~cm}^{2}\) of your head is flat (nearly correct) and that the force due to the atmosphere is perpendicular to the surface (as it is), $$ F=P A=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.0 \times 10^{-4} \mathrm{~m}^{2}\right)=20 \mathrm{~N} $$

A \(60-\mathrm{kg}\) woman stands on a light, cubical box that is \(5.0 \mathrm{~cm}\) on each edge. The box sits on the floor. What pressure does the box exert on the floor? $$ P=\frac{F}{A}=\frac{(60)(9.81) \mathrm{N}}{\left(5.0 \times 10^{-2} \mathrm{~m}\right)^{2}}=2.4 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} $$

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