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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 667.1 N.

Step by step solution

01

Understand the Problem

Both cables experience the same stress, which is the force per unit area. We know the force on the first cable and its radius, as well as the radius of the second cable. We need to find the stretching force on the second cable.
02

Define Stress

Stress is defined as the force acting on an object divided by the cross-sectional area of the cable through which the force acts. Mathematically, stress can be given as \( \sigma = \frac{F}{A} \) where \( F \) is the force and \( A \) is the cross-sectional area of the cable.
03

Calculate the Cross-Sectional Area of the First Cable

The cross-sectional area of a cable with radius \( r \) is given by \( A = \pi r^2 \). For the first cable, the radius is \( 3.5 \times 10^{-3} \) meters. Therefore, the area \( A_1 = \pi (3.5 \times 10^{-3})^2 \). Calculate this value.
04

Calculate the Stress in the First Cable

Using the stress formula \( \sigma = \frac{F}{A} \), where \( F = 270 \) N and \( A_1 \) calculated in Step 3, find the stress in the first cable: \( \sigma_1 = \frac{270}{A_1} \).
05

Calculate the Cross-Sectional Area of the Second Cable

For the second cable with radius \( 5.1 \times 10^{-3} \) meters, the area \( A_2 = \pi (5.1 \times 10^{-3})^2 \). Calculate this value.
06

Calculate the Stretching Force on the Second Cable

Since both cables experience the same stress \( \sigma_1 = \sigma_2 \), use the equation \( \sigma = \frac{F}{A} \) to find the force on the second cable. Rearrange the formula to \( F_2 = \sigma_1 \cdot A_2 \) to solve for \( F_2 \), using the stress found in Step 4 and the area \( A_2 \) from Step 5.

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Key Concepts

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

Cross-Sectional Area
When we talk about the cross-sectional area of a cable, we are referring to the area of the face of the cable that is perpendicular to its length. Imagine slicing a cable like slicing a loaf of bread.
This slice, which would look like a circle, is the cross-sectional area. For a round cable, the formula for calculating this area is quite straightforward. It is given by the formula: - - \[ A = \pi r^2 \] Here, \(A\) is the cross-sectional area, \(\pi\) is a mathematical constant approximately equal to 3.14159, and \(r\) is the radius of the cable. The radius is simply the distance from the center of the circle to the edge.
So, if you know the radius of a cable, like \(3.5 \times 10^{-3} \) meters or \(5.1 \times 10^{-3} \) meters for our cables, you can easily find the area by plugging the radius into this formula. This area is crucial because it helps us understand how much space is available for the force to act upon the cable.
The larger the area, the more the cable can handle forces without stress levels becoming too high.
Stretching Force
The stretching force is a term used to describe the amount of force applied along the length of a material that tends to elongate it. Imagine pulling on both ends of a rubber band; the force you exert by pulling is similar to the stretching force.
In the context of cables, this force is crucial as it determines how much the material can stretch before it starts to permanently deform or fail. The unit of force, the Newton (N), gives us a measure of the strength of the pull that the cable is enduring.
For instance, the first cable experiences a stretching force of 270 N. This is a substantial force, and knowing this helps us calculate the resulting stress each cable experiences. The stretching force on the second cable can be determined using the concept of stress because both cables are under the same stress state.
Stress Formula
Stress is a key concept in material science. It describes how a force is distributed over an area within a material and gives us a sense of the intensity of the internal forces that the material experiences. Stress is mathematically given by the formula: - - \[ \sigma = \frac{F}{A} \] Where \( \sigma \) represents stress, \( F \) is the force applied, and \( A \) is the cross-sectional area.Stress is measured in pascals (Pa), which is equivalent to one Newton per square meter.
This formula allows us to calculate the stress experienced by any material if we know the applied force and the area over which it is applied. Importantly in our case, both cables experience the same stress, meaning that the ratio of force to area is identical for both.
So, when you calculate the stress for the first cable using its known force and area, you can apply that same stress value to determine the force acting upon the second cable by solving for the unknown force. This approach leads us to finding that the second cable must have a different force acting on it to maintain the same stress level due to its different cross-sectional area.

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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 mass less, 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 m/s in 0.500 s. In the process, the spring is stretched by 0.200 m. The block is then pulled at a constant speed of 5.00 m/s, during which time the spring is stretched by only 0.0500 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m} .\) The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{N}\). The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}\). (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See Section 6.9 for a discussion of the work done by a variable force.)

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m}\), released from rest, and subsequently oscillate back and forth with an angular frequency of 11.3 rad/s. What is the speed of the object at the instant when the spring is stretched hy \(0.048 \mathrm{m}\) relative to its \(\mathrm{un}\) strained length?

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