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Chapter 11: Problem 30

A nylon rope used by mountaineers elongates \(1.10 \mathrm{~m}\) under the weight of a \(65.0 \mathrm{~kg}\) climber. If the rope is \(45.0 \mathrm{~m}\) in length and \(7.0 \mathrm{~mm}\) in diameter, what is Young's modulus for nylon?

Short Answer

Expert verified
Young's modulus for nylon is approximately \(6.88 \times 10^{8} \, Pa\).

Step by step solution

01

Calculate the force exerted

The force exerted by the climber on the rope due to gravity can be calculated using the formula \( F = m \cdot g \), where \( m = 65.0 \, kg \) is the mass of the climber and \( g \approx 9.81 \, m/s^2 \) is the acceleration due to gravity. This yields \( F \approx 637.65 \, N \).
02

Compute the cross section area of the rope

The rope can be considered as a cylinder. The cross-sectional area \( A \) of the rope can be computed based on its diameter \( d = 7.0 \, mm = 0.007 \, m \) using the formula \( A = \pi \cdot (d/2)^2 \), which equals \( A \approx 3.8 \times 10^{-5} \, m^2 \).
03

Calculate the stress

Stress is the force applied per unit area. In this case, it can be computed using the formula \( \sigma = F/A \). Substituting \( F \approx 637.65 \, N \) and \( A \approx 3.8 \times 10^{-5} \, m^2 \), we find \( \sigma \approx 1.68 \times 10^{7} \, Pa \).
04

Calculate the strain

Strain is the ratio of the change in length to the original length. In this case, the rope elongates by \( \Delta L = 1.10 \, m \), and the original length is \( L = 45.0 \, m \). Hence, the strain \( \varepsilon \) can be calculated using the formula \( \varepsilon = \Delta L / L \), which is \( \varepsilon \approx 0.0244 \).
05

Find Young's modulus

Young's modulus \( E \) is the ratio of stress to strain. So, we find \( E = \sigma / \varepsilon \). Substituting \( \sigma \approx 1.68 \times 10^{7} \, Pa \) and \( \varepsilon \approx 0.0244 \), we find that Young's modulus for nylon is \( E \approx 6.88 \times 10^{8} \, Pa \).

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Key Concepts

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

Stress and Strain
Stress and strain are fundamental concepts when discussing materials under load. Stress is defined as the force applied to an object divided by the area over which the force is applied. It essentially tells us how much force a material experiences per unit area. Stress can be expressed using the formula:
  • Stress (\(\sigma\)) = Force (\(F\)) / Area (\(A\))
Stress is measured in Pascals (Pa), which corresponds to \(N/m^2\).
Strain, on the other hand, measures the deformation of the material. It is the ratio of change in length to the original length, essentially describing how much a material stretches or compresses in response to stress.
  • Strain (\(\varepsilon\)) = Change in length (\(\Delta L\)) / Original length (\(L\))
Since strain is a ratio, it has no units. Stress and strain are closely related and central to understanding how materials behave under various loads.
Mechanical Properties of Materials
The mechanical properties of materials are crucial in determining how they react to different forces and conditions. Key properties include:
  • Elasticity: The ability of a material to return to its original shape after being deformed by a force.
  • Plasticity: When a material deforms permanently after yielding under stress.
  • Hardness: The resistance of a material against deformation or scratching.
  • Toughness: The ability to absorb energy and deform without breaking.
  • Brittleness: A tendency to break without significant deformation.
Young's modulus, discussed in this exercise, is a measure of the elasticity of a material. It quantifies the relationship between stress (force per unit area) and strain (proportional deformation) within the elastic limit of a material. A high Young's modulus indicates a rigid material that does not easily deform under stress, whereas a low modulus suggests a more flexible material.
Elasticity
Elasticity is a mechanical property that refers to a material's ability to resume its normal shape after being stretched or compressed. It's the property that allows objects to return to their original dimensions when the forces causing the deformation are removed.
For practical purposes, the range within which the material behaves elastically is important. Within this range, the stress-strain relationship is linear, meaning stress is directly proportional to strain. Hooke's Law can describe this linear relationship as:
  • Stress (\(\sigma\)) = Young's Modulus (\(E\)) \(\times\) Strain (\(\varepsilon\))
In other words, Young鈥檚 modulus is the constant of proportionality that expresses stiffness.
Elasticity is key in many applications, such as ensuring that a cable, like the nylon rope used by mountaineers, can absorb the forces it encounters without permanently deforming. Understanding elasticity helps engineers design safer and more efficient structures and materials.

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Most popular questions from this chapter

A steel cable with cross-sectional area \(3.00 \mathrm{~cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8} \mathrm{~Pa}\). Find the maximum upward acceleration that can be given a \(1200 \mathrm{~kg}\) elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

A claw hammer is used to pull a nail out of a board (see Fig. \(\mathrm{P} 11.45\) ). The nail is at an angle of \(60^{\circ}\) to the board, and a force \(\overrightarrow{\boldsymbol{F}}_{1}\) of magnitude \(400 \mathrm{~N}\) applied to the nail is required to pull it from the board. The hammer head contacts the board at point \(A,\) which is \(0.080 \mathrm{~m}\) from where the nail enters the board. A horizontal force \(\vec{F}_{2}\) is applied to the hammer handle at a distance of \(0.300 \mathrm{~m}\) above the board. What magnitude of force \(\overrightarrow{\boldsymbol{F}}_{2}\) is required to apply the required \(400 \mathrm{~N}\) force \(\left(F_{1}\right)\) to the nail? (Ignore the weight of the hammer.)

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

You open a restaurant and hope to entice customers by hanging out a sign (Fig. \(\mathbf{P} 1 \mathbf{1 . 5 3}\) ). The uniform horizontal beam supporting the sign is \(1.50 \mathrm{~m}\) long, has a mass of \(16.0 \mathrm{~kg},\) and is hinged to the wall. The sign itself is uniform with a mass of \(28.0 \mathrm{~kg}\) and overall length of \(1.20 \mathrm{~m}\). The two wires supporting the sign are each \(32.0 \mathrm{~cm}\) long, are \(90.0 \mathrm{~cm}\) apart, and are equally spaced from the middle of the sign. The cable supporting the beam is \(2.00 \mathrm{~m}\) long. (a) What minimum tension must your cable be able to support without having your sign come crashing down? (b) What minimum vertical force must the hinge be able to support without pulling out of the wall?

You are a construction engineer working on the interior design of a retail store in a mall. A 2.00 -m-long uniform bar of mass \(8.50 \mathrm{~kg}\) is to be attached at one end to a wall, by means of a hinge that allows the bar to rotate freely with very little friction. The bar will be held in a horizontal position by a light cable from a point on the bar (a distance \(x\) from the hinge) to a point on the wall above the hinge. The cable makes an angle \(\theta\) with the bar. The architect has proposed four possible ways to connect the cable and asked you to assess them: $$ \begin{array}{lllll} \text { Alternative } & \text { A } & \text { B } & \text { C } & \text { D } \\\ \hline x(\mathrm{~m}) & 2.00 & 1.50 & 0.75 & 0.50 \\ \theta(\text { degrees }) & 30 & 60 & 37 & 75 \end{array} $$ (a) There is concern about the strength of the cable that will be required. Which set of \(x\) and \(\theta\) values in the table produces the smallest tension in the cable? The greatest? (b) There is concern about the breaking strength of the sheetrock wall where the hinge will be attached. Which set of \(x\) and \(\theta\) values produces the smallest horizontal component of the force the bar exerts on the hinge? The largest? (c) There is also concern about the required strength of the hinge and the strength of its attachment to the wall. Which set of \(x\) and \(\theta\) values produces the smallest magnitude of the vertical component of the force the bar exerts on the hinge? The largest? (Hint: Does the direction of the vertical component of the force the hinge exerts on the bar depend on where along the bar the cable is attached?) (d) Is one of the alternatives given in the table preferable? Should any of the alternatives be avoided? Discuss.
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