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

Short Answer

Expert verified
Young's modulus for nylon is approximately \(6.80 \times 10^8 \) N/m².

Step by step solution

01

Define Young's Modulus

Young's modulus, denoted as \( E \), is a measure of the stiffness of a material. It is defined as the ratio of stress (force per unit area) to the strain (relative deformation) that occurs in a material.
02

Calculate the Stress in the Rope

Stress is calculated using the formula: \[ \text{Stress} = \frac{F}{A} \]\ where \( F \) is the force applied, and \( A \) is the cross-sectional area of the rope. The force \( F \) is due to the weight of the climber, and can be calculated as \( F = mg \), where \( m = 65.0 \) kg and \( g = 9.81 \) m/s². Thus, \( F = 65.0 \times 9.81 = 637.65 \) N. The cross-sectional area \( A \) of a circular rope can be calculated as \( A = \pi \times \left( \frac{d}{2} \right)^2 \), where \( d = 7.0 \) mm \( = 0.007 \) m. Therefore, \( A = \pi \times \left( \frac{0.007}{2} \right)^2 \approx 3.85 \times 10^{-5} \) m². So, the stress is \( \frac{637.65}{3.85 \times 10^{-5}} \approx 1.66 \times 10^7 \) N/m².
03

Calculate the Strain in the Rope

Strain is calculated as \( \text{Strain} = \frac{\Delta L}{L} \), where \( \Delta L = 1.10 \) m is the elongation of the rope, and \( L = 45.0 \) m is the original length of the rope. Thus, \( \text{Strain} = \frac{1.10}{45.0} \approx 0.0244 \).
04

Determine Young's Modulus

Young's modulus \( E \) can be determined by the formula \[ E = \frac{\text{Stress}}{\text{Strain}} \]. Using the calculated values, \( E = \frac{1.66 \times 10^7}{0.0244} \approx 6.80 \times 10^8 \) N/m². This value is Young's modulus for nylon in this scenario.

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Key Concepts

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

Stress and Strain
When materials are placed under a load, understanding their response is crucial for predicting their behavior. This is where the concepts of stress and strain come into play. Stress refers to the internal force that is exerted by a material when an external force is applied to it. It is given by the ratio of the force applied to the cross-sectional area over which it is distributed. Mathematically, stress is represented as \( \text{Stress} = \frac{F}{A} \), where \( F \) is the force in newtons and \( A \) is the area in square meters.

Strain, on the other hand, measures how much a material deforms relative to its original length when stress is applied. It is a dimensionless quantity defined as \( \text{Strain} = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length and \( L \) is the original length of the material.

These two concepts are essential in the field of mechanics, allowing engineers to determine how materials will respond to different forces. In our exercise, we used both stress and strain to calculate Young's Modulus, which gives insights into how stiff or flexible a material, such as nylon, can be.
Material Deformation
Material deformation is the change in shape or size of an object due to an applied force or stress. It is an important aspect when analyzing the mechanical properties of materials. Deformation can either be temporary or permanent.
  • Elastic Deformation: This occurs when the material returns to its original shape after the force is removed. The relationship between stress and strain remains linear within this range, which is characterized by Hooke's Law.
  • Plastic Deformation: When the material undergoes a permanent change, it has exceeded its elastic limit. The response to stress becomes nonlinear, and deformation does not revert back when the force is removed.
With nylon, as used in our mountaineering rope example, we primarily see elastic deformation within typical loading scenarios. The understanding of deformation processes is key to ensuring materials are used within their safe operational limits, preventing failure, and ensuring durable application.
Nylon Properties
Nylon, a type of synthetic polymer, is widely used for its favorable properties, particularly in applications like fabric and rope production. Key advantages of nylon stem from its robustness and flexibility.

One critical property is its elasticity, which is measured by Young's Modulus. For nylon, this modulus suggests a good balance between stiffness and flexibility, making it suitable for various applications where stretching and returning to shape are necessary.

Additionally, nylon's durability comes from its resistance to abrasion and wear, which is essential for climbing ropes that face constant friction and pressure. Moreover, nylon can withstand significant tension, shown by how it stretches in our exercise under a climber's weight without breaking.

Overall, the mechanical properties of nylon make it a preferred choice for heavy-duty applications requiring lightweight and reliable materials.

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Most popular questions from this chapter

A 0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.110-kg mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

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

A metal rod that is 4.00 m long and 0.50 cm\(^2\) in crosssectional area is found to stretch 0.20 cm under a tension of 5000 N. What is Young's modulus for this metal?

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 \(\times\) 10\(^{-3}\) cm\(^2\) in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force \(\overrightarrow{F}\) at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force \(\overrightarrow{F}\), (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). \(\textbf{Torques and Tug-of-War.}\) In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0\(^\circ\) to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is \({T_1} =\) 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, \({T_2}\) is not equal to \({T_1}\). His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

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