/*! 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

91Ó°ÊÓ

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

Calculating the Area

First, calculate the cross-sectional area (\(A\)) of the wire. The diameter is given as 0.130 cm, which is equivalent to 0.00130 m. Hence, the radius \(r = \frac{0.00130}{2} = 0.00065\) m. The area \(A\) is \(\pi r^2 = \pi (0.00065)^2 = 1.327 \times 10^{-6} \text{ m}^2\).
02

Calculate the Force

Convert the load from kilograms to force in newtons using gravity. \( F = m \cdot g = 8.00 \text{ kg} \cdot 9.81 \text{ m/s}^2 = 78.48 \text{ N} \).
03

Calculating Stress (\(\sigma\))

Stress is calculated by dividing the force by the area. \( \sigma = \frac{F}{A} = \frac{78.48 \text{ N}}{1.327 \times 10^{-6} \text{ m}^2} = 5.91 \times 10^{7} \text{ Pa} \).
04

Calculate Strain (\(\varepsilon\))

Strain is the change in length (\(\Delta L\)) over the original length (\(L_0\)). \(\varepsilon = \frac{\Delta L}{L_0} = \frac{0.0350 \text{ cm}}{75.0 \text{ cm}} = 4.67 \times 10^{-4} \).
05

Calculate Young's Modulus (\(Y\))

Young's Modulus is calculated by dividing stress by strain. \( Y = \frac{\sigma}{\varepsilon} = \frac{5.91 \times 10^{7} \text{ Pa}}{4.67 \times 10^{-4}} = 1.27 \times 10^{11} \text{ Pa} \), which is 127 GPa.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Stress
Stress is a measure of the internal forces that occur within a material when an external load is applied. Imagine stretching a rubber band; as you pull, the band applies opposing forces internally—this is stress. Mathematically, stress (\( \sigma \)) is defined as the force (\( F \)) applied to an object divided by the area (\( A \)) over which that force is distributed. The formula for calculating stress is:
  • \( \sigma = \frac{F}{A} \)
Stress is usually measured in Pascals (Pa), where one Pascal is equivalent to one Newton per square meter. In the context of the problem, the stress on the wire was found by dividing the force exerted by the weight of the load (\( 78.48 \text{ N} \)) by the cross-sectional area of the wire (\( 1.327 \times 10^{-6} \text{ m}^2 \)), providing a stress of \( 5.91 \times 10^{7} \text{ Pa} \).
If you find stress too high, think about how tension affects how various materials can withstand forces and hold structures together.
Strain
Strain is closely related to stress. It describes how much deformation an object experiences under stress. Strain essentially tells us how much a material stretches or compresses when exposed to a force. Unlike stress, strain is a dimensionless quantity because it’s a ratio. It is calculated by dividing the change in length (\( \Delta L \)) by the original length (\( L_0 \)):
  • \( \varepsilon = \frac{\Delta L}{L_0} \)
In the problem, the wire stretched \( 0.0350 \text{ cm} \) due to the load, and its original length was \( 75.0 \text{ cm} \). Thus, the strain experienced by the wire was \( 4.67 \times 10^{-4} \).
Understanding strain helps us predict whether a material will deform permanently or return to its original shape. It’s a critical concept in engineering and materials science when designing anything from buildings to bridges.
Mechanical Properties
Mechanical properties refer to how a material reacts under various types of loads. These properties are crucial because they determine a material's ability to withstand force, pressure, or tension. Some common mechanical properties include:
  • Elasticity: Ability of a material to return to its original shape after removing the load.
  • Tensile Strength: Maximum stress a material can withstand when stretched.
  • Yield Strength: Stress at which a material begins to deform plastically.
In the exercise, we calculated the mechanical properties related to Young's Modulus—a measure of a material's elasticity by calculating stress and strain. Understanding these concepts is essential to ensuring the safe and effective use of materials across various applications. This involves matching the right properties of materials with the specific demands of the task.
Elasticity
Elasticity is an essential concept when studying the behavior of materials. It defines how a material can return to its original form once the applied force causing its deformation is removed. This is a critical factor in many materials because it tells us how "stretchy" the material can be without permanently deforming.
Young's Modulus (\( Y \)) plays a central role in understanding elasticity. It is a measure of how easily a material can stretch or compress. The larger the Young's Modulus, the stiffer the material and the less it will stretch under a given force.
  • For example, in our exercise, the wire had a Young's Modulus of \( 127 \text{ GPa} \)
So, elasticity and Young’s Modulus tell engineers and scientists how much a material can be deformed—and returned to shape—when under stress, which is fundamental when designing and selecting materials for specific uses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 15-kg ball of radius \(4.0 \mathrm{~cm}\) is suspended from a point \(2.94 \mathrm{~m}\) above the floor by an iron wire of unstretched length \(2.85 \mathrm{~m}\). The diameter of the wire is \(0.090 \mathrm{~cm}\), and its Young's modulus is 180 GPa. If the ball is set swinging so that its center passes through the lowest point at \(5.0 \mathrm{~m} / \mathrm{s}\), by how much does the bottom of the ball clear the floor? Discuss any approximations that you make. Call the tension in the wire \(F_{T}\) when the ball is swinging through the lowest point. Since \(F_{T}\) must supply the centripetal force as well as balance the weight, $$ F_{T}=m g+\frac{m v^{2}}{r}=m\left(9.81+\frac{25}{r}\right) $$ all in proper SI units. This is complicated, because \(r\) is the distance from the pivot to the center of the ball when the wire is stretched, and so it is \(\mathrm{r}_{0}+\Delta r\), where \(r_{0}\), the unstretched length of the pendulum, is $$ r_{0}=2.85 \mathrm{~m}+0.040 \mathrm{~m}=2.89 \mathrm{~m} $$ and where \(\Delta r\) is as yet unknown. However, the unstretched distance from the pivot to the bottom of the ball is \(2.85 \mathrm{~m}+\) \(0.080 \mathrm{~m}=2.93 \mathrm{~m}\), and so the maximum possible value for \(\Delta r\) is $$ 2.94 \mathrm{~m}-2.93 \mathrm{~m}=0.01 \mathrm{~m} $$ We will therefore incur no more than a \(1 / 3\) percent error in \(r\) by using \(r=r_{0}=2.89 \mathrm{~m}\). This gives \(F_{T}=277 \mathrm{~N}\). Under this tension, the wire stretches by $$ \Delta L=\frac{F L_{0}}{A Y}=\frac{(277 \mathrm{~N})(2.85 \mathrm{~m})}{\pi\left(4.5 \times 10^{-4} \mathrm{~m}\right)^{2}\left(1.80 \times 10^{11} \mathrm{~Pa}\right)}=6.9 \times 10^{-3} \mathrm{~m} $$ Hence, the ball misses by \(2.94 \mathrm{~m}-(2.85+0.0069+0.080) \mathrm{m}=0.0031 \mathrm{~m}=3.1 \mathrm{~mm}\) To check the approximation we have made, we could use \(r=\) \(2.90 \mathrm{~m}\), its maximum possible value. Then \(\Delta L=6.9 \mathrm{~mm}\), showing that the approximation has caused a negligible error.

A solid cylindrical steel column is \(4.0 \mathrm{~m}\) long and \(9.0 \mathrm{~cm}\) in diameter. What will be its decrease in length when carrying a load of \(80000 \mathrm{~kg}\) ? \(\mathrm{Y}=1.9 \times 10^{11} \mathrm{~Pa}\) First find the $$ \begin{array}{l} \text { Cross-sectional area of column }=\pi r^{2}=\pi(0.045 \mathrm{~m})^{2}=6.36 \times 10^{-3} \mathrm{~m}^{2} \\ \text { Then, from } \mathrm{Y}=(\mathrm{F} / \mathrm{A}) /\left(\Delta L / L^{0}\right), \\ \text { Then. from } Y=(F / A) /\left(\triangle L / L_{0}\right), \\ \qquad \Delta L=\frac{F L_{0}}{A Y}=\frac{\left[\left(8.00 \times 10^{4}\right)(9.81) \mathrm{N}\right](4.0 \mathrm{~m})}{\left(6.36 \times 10^{-3} \mathrm{~m}^{2}\right)\left(1.9 \times 10^{11} \mathrm{~Pa}\right)}=2.6 \times 10^{-3} \mathrm{~m}=2.6 \mathrm{~mm} \end{array} $$

A 60 -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? $$ P=\frac{F}{A}=\frac{(60)(9.81) \mathrm{N}}{\left(5.0 \times 10^{-2} \mathrm{~m}\right)^{2}}=2.4 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2} $$

The bulk modulus of water is \(2.1\) GPa. Compute the volume contraction of \(100 \mathrm{~mL}\) of water when subjected to a pressure of \(1.5 \mathrm{MPa}\) From \(\mathrm{B}=\Delta P /\left(\Delta V / V_{0}\right)\), $$ \Delta V=-\frac{V_{0} \Delta P}{B}=-\frac{(100 \mathrm{~mL})\left(1.5 \times 10^{6} \mathrm{~Pa}\right)}{2.1 \times 10^{9} \mathrm{~Pa}}=-0.071 \mathrm{~mL} $$

The specific gravity of cast iron is \(7.20\). Find its density and the mass of \(60.0 \mathrm{~cm}^{3}\) of it. Make use of sp \(\mathrm{gr}=\frac{\text { Density of substance }}{\text { Density of water }}\) and \(\rho=\frac{m}{V}\) From the first equation, Density of iron \(=(\) sp \(\mathrm{gr})(\) Density of water \()=(7.20)\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)=7200\) and so Mass of \(60.0 \mathrm{~cm}^{3}=\rho V=\left(7200 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(60.0 \times 10^{-6} \mathrm{~m}^{3}\right)=0.432 \mathrm{~kg}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.