/*! 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 17 A wire having a length \(L\) and... [FREE SOLUTION] | 91Ó°ÊÓ

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A wire having a length \(L\) and cross-sectional area \(A\) is suspended at one of its ends from a ceiling. Density and Young's modulus of material of the wire are \(\rho\) and \(Y\), respectively. Find its strain energy due to its own weight in \(\mu \mathrm{J} .\) (Given: \(\left.\frac{\rho^{2} g^{2} A L^{3}}{Y}=12 \times 10^{-6} \mathrm{~J}\right)\)

Short Answer

Expert verified
The strain energy is \(6 \mu \text{J}\).

Step by step solution

01

Understand the Problem

We need to calculate the strain energy stored in the wire due to its own weight. The problem provides values for density \(\rho\), cross-sectional area \(A\), length \(L\), and Young's modulus \(Y\). We need to relate these values to find the strain energy.
02

Recall Relevant Formulas

The strain energy \(U\) stored in a wire due to its own weight can be calculated using the formula:\[U = \frac{1}{2} \times \left( \frac{\rho g L}{Y} \right)^2 \times AL^3\]where \(g\) is the acceleration due to gravity.
03

Substitute Known Values

We can substitute the known expression given in the problem: \(\frac{\rho^{2} g^{2} A L^{3}}{Y} = 12 \times 10^{-6} \text{ J}\). Thus, we have the equivalent:\[U = \frac{1}{2} \cdot \frac{\rho^{2} g^{2} A L^{3}}{Y} = \frac{1}{2} \cdot 12 \times 10^{-6} \text{ J}\]
04

Calculate the Strain Energy

Using the simplified expression above, we obtain that the strain energy is:\[U = \frac{1}{2} \times 12 \times 10^{-6} = 6 \times 10^{-6} \text{ J}\]Convert this to microjoules (\(\mu \text{J}\)) if necessary.

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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, often denoted as \( Y \), is a measure of the stiffness of a material. It is a fundamental concept in material science and engineering. Young's Modulus quantifies the ability of a material to resist deformation. When a material is subjected to a force, it either stretches or compresses. Young's Modulus helps describe how much it will deform under a given load.

The Modulus is defined as the ratio of tensile stress (force per unit area) to tensile strain (proportional deformation in length), which gives us the formula:\[ Y = \frac{\text{stress}}{\text{strain}} \]This formula implies that the higher the Young’s Modulus, the less a material will stretch or compress under a given force.
  • In the context of a wire, Young’s Modulus is crucial as it determines how much the wire will stretch under its own weight.
  • This property is essential when calculating the strain energy in our problem.
Understanding Young's Modulus allows engineers to choose appropriate materials for various applications, ensuring materials will perform correctly under expected load conditions.
Density
Density, represented by the Greek letter \( \rho \), is another critical property in material science. It defines how much mass is contained within a given volume. Mathematically, density is expressed as:\[ \rho = \frac{m}{V} \]Where \( m \) is mass and \( V \) is volume. Density is a significant factor when considering how a material will behave under gravitational forces.

For the wire in the exercise, density determines how heavy the wire will be, contributing to the gravitational force acting on it when suspended.
  • More dense materials will generally exert more weight, affecting the amount of strain energy stored.
  • An essential aspect of solving the problem is understanding how density links with gravity and contributes to the overall calculation of strain energy.
In practical applications, choosing the right density for material ensures both strength and stability without unnecessary weight additions.
Cross-sectional Area
The cross-sectional area, denoted \( A \), is a measure of the size of the section of the wire perpendicular to its length. In simpler terms, if you cut the wire and look straight at its face, the area that you see is the cross-sectional area.

In our problem, the cross-sectional area is crucial because:
  • It affects the amount of stress the wire can handle.
  • The area determines how the weight of the wire is distributed across it, which influences the strain energy.
The formula for stress includes the cross-sectional area:\[ \text{stress} = \frac{\text{force}}{A} \]
This relationship indicates that for a given force, a larger cross-sectional area will result in less stress. Thus, in tension applications, larger cross-sectional areas are beneficial for reducing stretching and potential damage. By understanding this concept, one can better appreciate how different factors like weight and modulus interact to affect a wire's performance.
Gravity
Gravity is a fundamental force of nature that acts on all objects with mass. In the context of our exercise, gravity, represented by the constant \( g \), influences how the wire stretches under its weight. Gravitational force pulls the wire downward, causing tension and storing potential energy within it.

Gravitational force is calculated using the formula:\[ F = m \cdot g \]where \( F \) is the force due to gravity, \( m \) is the mass, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth's surface).
  • The mass in our wire's case comes from its density and volume.
  • Understanding gravitational pull is crucial for calculating the strain energy correctly.
This strong downward force significantly affects how materials, such as wires, perform when hung vertically. It is essential to consider gravity's role in engineering and physics, ensuring structures can safely withstand their intrinsic weight over time.

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Most popular questions from this chapter

Which of the following are correct? (1) The shear modulus of a liquid is infinite. (2) Bulk modulus of a perfectly rigid body is infinity. (3) According to Hooke's law, the ratio of the stress and strain remains constant. (4) None of the above.

A copper bar of length \(L\) and area of cross section \(A\) is placed in a chamber at atmospheric pressure. If the chamber is evacuated, the percentage change in its volume will be (compressibility of copper is \(8 \times 10^{-12} \mathrm{~m}^{2} / \mathrm{N}\) and \(\left.1 \mathrm{~atm}=10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\) (1) \(8 \times 10^{-7} \%\) (2) \(8 \times 10^{-5} \%\) (3) \(1.25 \times 10^{-4} \%\) (4) \(1.25 \times 10^{-5} \%\)

When the tension in a metal wire is \(T_{1}\), its length is \(l_{1}\). When the tension is \(T_{2}\), its length is \(l_{2}\). The natural length of wire is (1) \(\frac{T_{2}}{T_{1}}\left(l_{1}+l_{2}\right)\) (2) \(T_{1} l_{1}+T_{2} l_{2}\) (3) \(\frac{l_{1} T_{2}-l_{2} T_{1}}{T_{2}-T_{1}}\) (4) \(\frac{l_{1} T_{2}+l_{2} T_{1}}{T_{2}+T_{1}}\)

The density of water at the surface of ocean is \(\rho\). If the bulk modulus of water is \(B\), then the density of ocean water at depth, when the pressure is \(\alpha p_{0}\) and \(p_{0}\) is the atmospheric pressure, is (1) \(\frac{\rho B}{B-(\alpha-1) p_{0}}\) (2) \(\frac{\rho B}{B+(\alpha-1) p_{0}}\) (3) \(\frac{\rho B}{B-\alpha_{0}}\) (4) \(\frac{\rho B}{B+\alpha p_{0}}\)

Four rods \(A, B, C\) and \(D\) of the same length and material but of different radii \(r, r \sqrt{2}, r \sqrt{3}\) and \(2 r\), respectively, are held between two rigid walls. The temperature of all rods is increased through the same range. If the rods do not bend, then (1) the stress in the rods \(A, B, C\) and \(D\) is in the ratio \(1: 2: 3: 4\) (2) the forces on them exerted by the wall are in the ratio \(1: 2: 3: 4\) (3) the energy stored in the rods due to elasticity is in the ratio \(1: 2: 3: 4\) (4) it is independent of area like surface tension while friction depends

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