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A piece of mohair taken from an Angora goat has a radius of \(31 \times 10^{-6} \mathrm{m} .\) What is the least number of identical pieces of mohair needed to suspend a \(75-\mathrm{kg}\) person, so the strain experienced by each piece is less than \(0.010 ?\) Assume that the tension is the same in all the pieces.

Short Answer

Expert verified
Around 63 pieces of mohair are needed.

Step by step solution

01

Understand the Problem

We need to find the number of pieces of mohair required to suspend a 75-kg person, ensuring that the strain on each piece is less than 0.010. We know the radius of each piece and the mechanical stress and strain relationship: Stress = Force/Area and Strain = Stress/Y, where Y is Young's modulus.
02

Calculate Force per Piece

The force exerted by the 75-kg person is due to gravity. Using 9.8 m/s² for gravitational acceleration, the total force (weight) is \[ F = mg = 75 \times 9.8 = 735 \text{ N} \] This force needs to be distributed among the pieces.
03

Calculate Stress in a Piece

First, compute the cross-sectional area of a mohair piece, using the radius given:\[ A = \pi r^2 = \pi (31 \times 10^{-6})^2 \text{ m}^2 \] Stress is given by Force/Area:\[ \sigma = \frac{F}{A} \] But in this case, we need to distribute the force among all the pieces.
04

Determine the Strain Limit

Strain \(\epsilon\) is the ratio of Stress to Young's modulus (Y). Given that stress is to be distributed and strain should be less than 0.010, we have:\[ \epsilon = \frac{\sigma}{Y} < 0.010 \] This gives us a maximum allowable stress per piece:\[ \sigma < 0.010 \times Y \]
05

Relate Stress to Force and Pieces

Thus, the stress that one piece supports is:\[ \frac{735}{n \cdot A} < 0.010 \times Y \] Here, \(n\) is the number of pieces. Rearrange to find \(n\):\[ n \cdot 0.010 \times Y \times A > 735 \] Solving for \(n\) will give us the minimum number of pieces needed.
06

Final Calculation

Let's pick a typical value for Young's modulus for mohair, say \(Y = 3.8 \times 10^9 \text{ Pa}\).Substitute \(Y\), \(A\), and the condition from Step 5 into the equation to solve for \(n\).
07

Solution Derivation

Compute\[ A = \pi (31 \times 10^{-6})^2 \approx 3.02 \times 10^{-9} \text{ m}^2 \]Plug into inequality,\[ n > \frac{735}{0.010 \times 3.8 \times 10^9 \times 3.02 \times 10^{-9}} \]Calculate and round \(n\) to the nearest whole number.

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Key Concepts

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

Stress and Strain
In the context of physics and materials science, stress and strain are fundamental concepts used to describe forces and their effects on objects. Stress is defined as the force applied to an object divided by the area over which the force is applied. It is expressed in units of pressure, such as Pascals (Pa).
Strain, on the other hand, is a measure of the deformation of the material in response to stress. It is the ratio of the change in length to the original length of the material. Strain is a dimensionless quantity.
  • Stress (\( \sigma \)) = Force/Area
  • Strain (\( \epsilon \)) = Change in Length/Original Length
Understanding the relationship between stress and strain is crucial. The area of stress versus strain is key in determining if a material can sustain specific forces without deforming beyond its limit. Young's modulus (\( Y \)) is an essential numeric value used here, representing the ratio of stress to strain within the elastic limit of a material. A higher Young’s modulus means a stiffer material.
Gravitational Force
Gravitational force is the attraction that earth exerts on objects, giving them weight. It is directly proportional to the mass of the object and the acceleration due to gravity, which is approximately \( 9.8 \, \text{m/s}^{2} \) on Earth's surface.
The force experienced by an object under gravity is calculated using the formula:
  • \( F = mg \)
where \( m \) is the mass of the object, and \( g \) is the gravitational acceleration. In this case, a 75 kg person experiences a force calculated to be \( 735 \, \text{N} \).
This gravitational force needs to be countered by the supporting material's tensile strength, necessitating accurate calculations like those performed for the mohair strands.
Cross-sectional Area
The cross-sectional area is a critical parameter when evaluating the strength of materials subjected to forces. It refers to the area of the section obtained when cutting through an object perpendicular to its length.
For a circular section, as with a mohair fiber, the cross-sectional area \( A \) can be determined using:
  • \( A = \pi r^2 \)
where \( r \) is the radius of the circle. For mohair of radius \( 31 \times 10^{-6} \, \text{m} \), the area becomes crucial in calculating how much force can be distributed across several fibers.
A greater cross-sectional area means that the material can endure more stress before reaching the same strain level.
Material Properties
Material properties, such as the Young's Modulus, dictate how a material behaves under various stress conditions. It refers to the ability of a material to withstand changes in length when under tension or compression.
Young's Modulus (\( Y \)) is especially crucial when analyzing the stretchability or tensile strength of fibers such as mohair. Its value indicates the stiffness of the material.
Commonly used values for Young’s modulus include:
  • Mohair: \( 3.8 \times 10^9 \, \text{Pa} \)
  • Steel: \( 2.1 \times 10^{11} \, \text{Pa} \)
A higher Young’s modulus equates to a less elastic, more rigid material. This property helps inform the calculations for how many strands of mohair are needed to safely suspend a weight, ensuring the material does not exceed its strain limit of 0.010.

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Most popular questions from this chapter

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of \(5.00 \mathrm{m} / \mathrm{s}\) in \(0.500 \mathrm{s}\). In the process, the spring is stretched by \(0.200 \mathrm{m} .\) The block is then pulled at a constant speed of \(5.00 \mathrm{m} / \mathrm{s}\), during which time the spring is stretched by only \(0.0500 \mathrm{m} .\) Find \((\mathrm{a})\) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighbone can endure without breaking is \(7.0 \times 10^{4} \mathrm{N} .\) The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2},\) its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) The mass of the man is \(65 \mathrm{kg} .\) He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m}\) ) is mounted on the floor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel \(\left(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2}\right) .\) It has a radius of \(0.80 \mathrm{mm}\) and an unstrained length of \(0.76 \mathrm{m}\). The radius of the tuning peg is \(1.8 \mathrm{mm} .\) Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions. Ignore the radius of the wire compared to the radius of the tuning peg.

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