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Stress on a Mountaineer's Rope. 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 nyon?

Short Answer

Expert verified
Young's modulus is approximately \( 6.67 \times 10^8 \, \text{Pa} \).

Step by step solution

01

Understand the Given Variables

The problem gives us the following values: the elongation of the rope \( \Delta L = 1.10 \text{ m} \), the mass of the climber \( m = 65.0 \text{ kg} \), the initial length of the rope \( L = 45.0 \text{ m} \), and the diameter of the rope's cross-sectional area \( d = 7.0 \text{ mm} \). We need to calculate Young's modulus, \( E \), for the nylon rope.
02

Calculate the Force Exerted by the Climber

The force exerted by the climber, due to gravity, is the weight \( F \) which is calculated using the formula \( F = mg \) where \( g \approx 9.81 \, \text{m/s}^2 \). Hence, \( F = 65.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 637.65 \, \text{N} \).
03

Calculate the Cross-sectional Area of the Rope

The cross-sectional area \( A \) of the rope can be found using the formula for the area of a circle \( A = \pi \left( \frac{d}{2} \right)^2 \). First, convert diameter from mm to m: \( 7.0 \, \text{mm} = 0.007 \, \text{m} \).\( A = \pi \left( \frac{0.007}{2} \right)^2 = \pi (0.0035)^2 \approx 3.85 \times 10^{-5} \, \text{m}^2 \).
04

Use Young's Modulus Formula to Find \( E \)

Young's modulus is calculated using the formula \( E = \frac{FL}{A \Delta L} \). Substitute the known values into the equation:\[ E = \frac{637.65 \, \text{N} \times 45.0 \, \text{m}}{3.85 \times 10^{-5} \, \text{m}^2 \times 1.10 \, \text{m}} \approx 6.67 \times 10^8 \, \text{Pa} \].
05

Conclude the Calculation

The calculated Young's modulus for the nylon rope turns out to be approximately \( 6.67 \times 10^8 \, \text{Pa} \). This signifies the stiffness of the rope, with a higher modulus indicating a stiffer material.

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Key Concepts

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

Stress-Strain Relationship
In the world of physics and engineering, the way a material responds to an applied force is described by the stress-strain relationship. This essential concept helps us understand how objects deform under load.
Stress, identified as \( ext{stress} = rac{F}{A}\), measures the internal forces acting within a material, where \(F\) is the force applied, and \(A\) is the cross-sectional area. It's expressed in pascals (Pa), but sometimes people use MPa or GPa for larger values.
Strain, on the other hand, is a dimensionless number representing deformation and is given by \( ext{strain} = rac{\Delta L}{L}\). Here, \(\Delta L\) is the change in length and \(L\) is the original length. Strain doesn’t have units as it’s a ratio of two lengths.
  • A material showing a linear stress-strain relationship follows Hooke's Law. It's crucial for materials like the nylon rope in our mountaineering example.
  • When stress and strain are proportional, the material exhibits elastic behavior, a favorite in engineering applications.

This relationship is vital to understanding how materials will react in real-world situations, making it a cornerstone for disciplines like engineering physics.
Elasticity
Elasticity is a property of materials that describes their ability to return to their original form after being deformed. It’s like how a rubber band snaps back after being stretched. In the context of the nylon rope used by mountaineers, elasticity determines how much the rope can stretch under the climber’s weight without suffering a permanent deformation.
According to Hooke's Law, the elastic limit is the maximum stress that a material can withstand while still being able to return to its original shape. If the stress surpasses this limit, the material may be permanently deformed. This is known as plastic deformation.
Young's modulus, denoted as \(E\), is a key parameter when we talk about elasticity. It quantifies how much a material can stretch or compress. The formula is \( rac{F}{A} = E rac{\Delta L}{L}\), and it helps determine how stiff a material is. Higher values of \(E\) mean a stiffer material.
  • Flexible materials have lower Young's modulus values. Think about a rubber band.
  • Brittle or hard materials have much higher values, similar to diamonds.

Understanding elasticity helps engineers design materials and structures that can handle everyday forces without catastrophic failure.
Engineering Physics
Engineering physics combines physics principles with engineering practices to solve technical problems. It leverages scientific advancements for practical applications. This discipline is essential for developing new materials, structures, and systems.
One classic example is the design of mountaineer’s gear, such as ropes, which must be strong yet flexible, allowing safe descents. By understanding concepts like stress-strain relationships and elasticity, engineers ensure these ropes perform under the harshest conditions.
Applications of engineering physics extend beyond mountain climbing. It plays a role in the creation of airplanes, bridges, electronics, and countless everyday items.
  • Engineers rely on fundamental physics concepts, like Newton’s laws of motion, thermodynamics, and electromagnetism.
  • Innovations often stem from bridging basic scientific theories with real-world constraints.

This area of study is crucial for pushing the boundaries of what technology can achieve, ensuring safety, efficiency, and innovation across multiple industries.

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Most popular questions from this chapter

Biceps Muscle. A relaxed biceps muscle requires a force of 25.0 \(\mathrm{N}\) for an elongation of 3.0 \(\mathrm{cm}\) ; the same muscle under maximum tension requires a force of 500 \(\mathrm{N}\) for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with blength 0.200 \(\mathrm{m}\) m and cross- sectional area 50.0 \(\mathrm{cm}^{2} .\)

Flying Buttress. (a) A symmetric building has a roof sloping upward at \(35.0^{\circ}\) above the horizontal on each side. If each side of the uniform roof weighs \(10,000 \mathrm{N}\) , find the horizontal force that this roof exerts at the top of the wall, which tends to push out the walls. Which type of building would be more in danger of collapsing: one with tall walls or one with short walls? Explain. (b) As you saw in part (a), tall walls are in danger of collapsing from the weight of the roof. This problem plagued the ancient builders of large structures. A solution used in the great Gothic cathedrals during the 1200 s was the flying buttress, a stone support running between the walls and the ground that helped to hold in the walls. A Gothic church has a uniform roof weighing a total of \(20,000 \mathrm{N}\) and rising at \(40^{\circ}\) above the horizontal at each wall. The walls are 40 \(\mathrm{m}\) tall, and a flying buttress meets each wall 10 \(\mathrm{m}\) below the base of the roof. What horizontal force must this flying buttress apply to the wall?

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

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T\) , where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\) .

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.

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