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Two stretched cables both experience the same stress. The first cable has a radius of \(3.5 \times 10^{-3} \mathrm{m}\) and is subject to a stretching force of 270 \(\mathrm{N}\) . The radius of the second cable is \(5.1 \times 10^{-3} \mathrm{m} .\) Determine the stretching force acting on the second cable.

Short Answer

Expert verified
The stretching force on the second cable is approximately 594 N.

Step by step solution

01

Understanding Stress

Stress is defined as the force applied per unit area. The formula for calculating stress is \( \sigma = \frac{F}{A} \), where \( \sigma \) is the stress, \( F \) is the force applied, and \( A \) is the cross-sectional area.
02

Determine the First Cable's Area and Stress

Calculate the area of the first cable using \( A = \pi r^2 \), where \( r = 3.5 \times 10^{-3} \text{ m} \). Thus, \( A_1 = \pi (3.5 \times 10^{-3})^2 \). Then compute the stress using \( \sigma_1 = \frac{270}{A_1} \).
03

Calculate the Second Cable's Area

Find the cross-sectional area of the second cable using the same formula \( A = \pi r^2 \), where \( r = 5.1 \times 10^{-3} \text{ m} \). Therefore, \( A_2 = \pi (5.1 \times 10^{-3})^2 \).
04

Use Stress Equality to Find the Second Force

Since the stress on both cables is equal (\( \sigma_1 = \sigma_2 \)), set \( \frac{270}{A_1} = \frac{F_2}{A_2} \). Rearranging gives \( F_2 = 270 \times \frac{A_2}{A_1} \).
05

Calculate the Stretching Force on the Second Cable

Substitute the calculated areas into the equation from Step 4 to find \( F_2 \). This results in \( F_2 = 270 \times \frac{\pi (5.1 \times 10^{-3})^2}{\pi (3.5 \times 10^{-3})^2} \). Simplifying, \( F_2 = 270 \times \left( \frac{5.1^2}{3.5^2} \right) \). Compute this to find the value of \( F_2 \).

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Key Concepts

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

Cross-Sectional Area
When talking about cables and their ability to withstand force, the cross-sectional area is very important. It is the area of a slice of the cable perpendicular to its length. Think of it as the size of the surface that directly opposes the force applied. To find this area for a round cable, use the formula: \[ A = \pi r^2 \] Here, \( r \) represents the radius of the cable, and \( \pi \) is approximately 3.14159. This calculation tells us how much space inside the cable interacts to resist any stretching force. Knowing the cross-sectional area allows us to calculate stress, which ultimately helps in determining safety and performance standards for materials.
Force Calculation
Force is a fundamental concept when dealing with stress and mechanics. It refers to any push or pull acting on an object. In this problem, our aim is to find the stretching force acting on the second cable. The stretch or force in each cable can be derived using stress principles. Since the stress is equal in both cables, and stress equals force divided by area (\[ \sigma = \frac{F}{A} \] -- where \( F \) is the force and \( A \) is the area), we can rearrange the equation to find the force. This rearrangement is crucial, especially when we're given stresses but need to calculate either force or area. Employing this principle helps you solve the problem and apply it to other practical scenarios.
Stress Formula
Stress gives us an understanding of how materials handle forces. It defines how much force per unit area is exerted on a material. It’s vital in predicting potential material failure points. The formula for calculating stress is: \[ \sigma = \frac{F}{A} \] Where \( \sigma \) represents stress, \( F \) is the force applied, and \( A \) is the cross-sectional area. This equation highlights a key relationship: to maintain equal stress on two objects, if one’s area increases, the force applied must also change proportionately. This idea is useful in different engineering problems, ensuring uniform material properties across variable dimensions. In our exercise, this formula connects the two cables with differing radii but equal stresses.
Radius and Area Calculation
Understanding the radius and area of a cable is crucial for determining the extent of stress it can handle. The radius is half the diameter and is the starting point for calculating the area. The formula for finding the area based on radius is: \[ A = \pi r^2 \] This polynomial equation shows the area increases with the square of the radius, meaning small changes in the radius greatly impact the area. Knowing the area allows you to determine force and stress for cable design. For example, in our problem, different radii for two cables led to different forces needed to achieve the same stress, showcasing how area calculations are pivotal in engineering applications. These principles apply widely across materials science and structural engineering, guiding the design and safety standards of structures we use and encounter daily.

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Most popular questions from this chapter

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a ram spring \((k=32000 \mathrm{N} / \mathrm{m})\) . The mass of this spring may be ignored The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\)when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{Hz}\) . On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{m} .\)

The length of a simple pendulum is 0.79 \(\mathrm{m}\) and the mass of the particle (the bob) at the end of the cable is 0.24 \(\mathrm{kg}\) . The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.6 \(\mathrm{m}\) and its radius is 65 \(\mathrm{cm} .\) When an object is placed on top of the cylinder, the cylinder compresses by an amount of \(5.7 \times 10^{-7} \mathrm{m} .\) What is the weight of the object?

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 174 \(\mathrm{N} / \mathrm{m}\) . The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2 . The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2 .

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