/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 An elastic cable is to be design... [FREE SOLUTION] | 91Ó°ÊÓ

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An elastic cable is to be designed for bungee jumping from a tower 130 ft high. The specifications call for the cable to be 85 ft long when unstretched, and to stretch to a total length of 100 ft when a 600-lb weight is attached to it and dropped from the tower. Determine (a) the required spring constant k of the cable, (b) how close to the ground a 186-lb man will come if he uses this cable to jump from the tower.

Short Answer

Expert verified
(a) \( k = 40 \text{ lbs/ft} \), (b) 40.35 ft from ground.

Step by step solution

01

Understand the Problem

We have a cable which is initially 85 ft long and stretches to 100 ft with a 600-lb weight. This means it stretches an additional 15 ft under a 600-lb load. We need to find the spring constant \( k \) of the cable and determine the distance from the ground for a 186-lb man jumping from the same height of the tower.
02

Calculate the Spring Constant

The spring constant \( k \) is determined using Hooke's Law, which states \( F = kx \), where \( F \) is the force exerted on the spring (600 lbs), \( x \) is the displacement (15 ft). Thus, we have:\[ k = \frac{F}{x} = \frac{600 \text{ lbs}}{15 \text{ ft}} = 40 \text{ lbs/ft} \]. Hence, the spring constant \( k \) is 40 lbs/ft.
03

Calculate the Initial and Maximum Extension for 186-lb Man

With a 186-lb man, using the same formula \( F = kx \), we determine how much the cable will stretch. Solving for \( x \):\[ x = \frac{F}{k} = \frac{186 \text{ lbs}}{40 \text{ lbs/ft}} = 4.65 \text{ ft} \].Therefore, the total stretched length of the cable will be \(85 + 4.65 = 89.65 \text{ ft}\).
04

Calculate the Distance from the Ground

The man jumps from a 130 ft tower. The total length of the cable at maximum stretch will be \(89.65 \text{ ft}\). Therefore, the distance between the man and the ground is:\[ 130 \text{ ft} - 89.65 \text{ ft} = 40.35 \text{ ft} \].Hence, the man comes 40.35 ft close to the ground.

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Key Concepts

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

spring constant
The spring constant, often denoted by the symbol \( k \), is a measure of a spring's stiffness or elasticity. It represents how much force is needed to stretch or compress a spring by a unit length. In the context of bungee jumping, understanding the spring constant is crucial because it tells us how the elastic cable will behave under a load. A higher spring constant means the cable is stiffer, while a lower value indicates it is more elastic and can stretch more. In the problem, we calculated the spring constant to be 40 lbs/ft, meaning every added foot requires 40 lbs of force.
Hooke's Law
Hooke's Law provides the fundamental relationship between the force exerted on a spring and its resulting displacement. The law is expressed by the equation \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the spring's original length.

In the context of bungee jumping, Hooke's Law helps predict how far the cable will stretch when a jumper falls. It is instrumental in ensuring safety, as accurate stretching predictions prevent the jumper from hitting the ground. For instance, in the solution, applying Hooke's Law allowed us to calculate the displacement when a 600-lb weight was attached, showing the spring constant as 40 lbs/ft.
elastic cable dynamics
Elastic cable dynamics refers to how the cable stretches and behaves when subjected to forces during bungee jumping. These dynamics are governed by factors like the spring constant and Hooke’s Law, as well as the material properties of the cable.

Understanding these dynamics helps in designing bungee cables that provide a thrilling yet safe experience. Cables must stretch enough to absorb energy and slow down the jumper safely without excessive rebound. The dynamics ensure that the cable is neither too stiff nor too loose, avoiding both abrupt stops and excessive stretching close to hitting the ground.
force and displacement
Force and displacement are key components when analyzing the physical behavior of a bungee jumping cable. Force, measured in pounds, is the weight or load applied to the cable. Displacement is the change in length from the cable’s original, unstressed length.

Each time a weight, such as a jumper, is applied to the cable, these two factors interplay. In our example, a 600-lb force resulted in a 15 ft displacement. Calculating displacement for different weights is vital to ensure that every jumper remains safe. By understanding this relationship, engineers can design bungee cables that predictably behave under different conditions.

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Most popular questions from this chapter

Two identical billiard balls can move freely on a horizontal table. Ball \(A\) has a velocity \(\mathbf{v}_{0}\) as shown and hits ball \(B,\) which is at rest, at a point \(C\) defined by \(\theta=45^{\circ} .\) Knowing that the coefficient of restitution between the two balls is \(e=0.8\) and assuming no friction, determine the velocity of each ball after impact.

A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

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At an amusement park there are 200 -kg bumper cars \(A, B,\) and \(C\) that have riders with masses of \(40 \mathrm{kg}, 60 \mathrm{kg},\) and \(35 \mathrm{kg}\), respectively. Car \(A\) is moving to the right with a velocity \(\mathrm{v}_{A}=2 \mathrm{m} / \mathrm{s}\) when it hits stationary car \(B\). The coefficient of restitution between each car is \(0.8 .\) Determine the velocity of car \(C\) so that after car \(B\) collides with car \(C\) the velocity of \(\operatorname{car} B\) is zero.

The \(1-\mathrm{kg}\) ball at \(A\) is suspended by an inextensible cord and given an initial horizontal velocity of \(5 \mathrm{m} / \mathrm{s}\). If \(l=0.6 \mathrm{m}\) and \(x_{\mathrm{B}}=0,\) determine \(y_{\mathrm{B}}\) so that the ball will enter the basket.

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