/*! 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 9 A steel cable with a radius of \... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 9

A steel cable with a radius of \(1.5 \mathrm{~cm}\) supports a chair lift at a ski area. If the maximum stress is not to exceed \(10^{8} \mathrm{~N} / \mathrm{m}^{2}\), what is the maximum load the cable can support? (a) \(7 \times 10^{5} \mathrm{~N}\) (b) \(7 \times 10^{6} \mathrm{~N}\) (c) \(7 \times 10^{4} \mathrm{~N}\) (d) \(9 \times 10^{5} \mathrm{~N}\)

Short Answer

Expert verified
(c) \(7 \times 10^{4} \, \text{N}\).

Step by step solution

01

Understand the Problem

We need to find the maximum load a steel cable can support without exceeding the given stress limit of \(10^8 \, \text{N/m}^2\). The stress on the cable when supporting a load is defined as force per unit area.
02

Calculate the Cross-Sectional Area of the Cable

The cable's radius is given as \(1.5 \, \text{cm}\), which is \(0.015 \, \text{m}\) in meters. The cross-sectional area \(A\) of the cable (assumed to be a circle) can be calculated using the formula: \[ A = \pi r^2 = \pi (0.015)^2 \text{m}^2 \] Calculating the area yields: \[ A = \pi \times (0.015)^2 \approx 7.07 \times 10^{-4} \, \text{m}^2 \]
03

Calculate the Maximum Load Using the Stress Formula

The formula for stress \(\sigma\) is given by: \[ \sigma = \frac{F}{A} \] Rearranging this formula to solve for the force \(F\), we get: \[ F = \sigma \times A \] Substitute the given stress \(\sigma = 10^8 \, \text{N/m}^2\) and the area \(A = 7.07 \times 10^{-4} \, \text{m}^2\): \[ F = 10^8 \times 7.07 \times 10^{-4} = 7.07 \times 10^4 \, \text{N} \]
04

Select the Closest Answer from the Choices

The calculated maximum load the cable can support is \(7.07 \times 10^4 \, \text{N}\). Comparing with the options provided, the closest answer is: (c) \(7 \times 10^4 \, \text{N}\).

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Key Concepts

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

Stress Calculation
Stress calculation is a fundamental concept in mechanics that relates force to the area over which it is applied. It helps in understanding how materials behave under load. Stress is denoted by the symbol \( \sigma \). The formula to calculate stress is:
  • \( \sigma = \frac{F}{A} \)
where \( F \) is the force applied (in newtons), and \( A \) is the cross-sectional area (in square meters).

In this particular exercise with the steel cable, understanding the stress helps in determining the cable's capability to bear loads without deforming or failing. By knowing the maximum allowable stress, engineers choose materials and design dimensions that ensure safety and performance under expected load conditions.
Cross-Sectional Area
The concept of cross-sectional area is crucial when determining the strength capacity of a structural element like a cable. For round cables, this area is the area of a circle and is calculated using the formula:
  • \( A = \pi r^2 \)
Here, \( r \) is the radius of the cable. In the given problem, the radius is converted from centimeters to meters as the standard units in stress calculations are meters.

This conversion is important because the unit area determines how effectively a material can distribute the load. By determining this area for the steel cable, we can figure out how much force the cable can handle per unit of area without exceeding the material's stress limit.
Load Capacity
The load capacity refers to the maximum amount of force or weight that a structural element can support. It is determined by combining knowledge of stress and cross-sectional area. In the case of the steel cable, calculating the load it can handle involves rearranging the stress equation:
  • \( F = \sigma \times A \)
This formula shows that the force supported by the cable (load capacity) is the product of the allowable stress limit and the cross-sectional area.

By substituting the given values from the exercise, we calculated that the maximum force the steel cable can support is \(7.07 \times 10^4 \) N. This shows how understanding these primary concepts allows us to predict how much load the cable can safely carry without risk of failure.
Steel Properties
Steel is a widely used material in engineering due to its strength and durability. Its properties like tensile strength, elasticity, and ductility make it ideal for cables and load-bearing structures.
  • Steel's high tensile strength allows it to withstand large stresses before failing.
  • Its ductility provides some flexibility, ensuring that it can bear dynamic loads without sudden failure.
These properties are why steel is chosen in situations like supporting chair lifts, where constant, substantial load stability is crucial.

In understanding these characteristics, engineers can optimize design and safety, ensuring that structures made from steel can handle specified loads repeatedly over time without degradation. These material properties also complement how well the calculations of stress, cross-sectional area, and load capacity inform safe structural design.

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Most popular questions from this chapter

The Young's modulus of a rubber string \(8 \mathrm{~cm}\) long and density \(1.5 \mathrm{~kg} / \mathrm{m}^{3}\) is \(5 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\), is suspended on the ceilling in a room. The increase in length due to its own weight will be (a) \(9.5 \times 10^{-5} \mathrm{~m}\) (b) \(9.6 \times 10^{-11} \mathrm{~m}\) (c) \(9.6 \times 10^{-1} \mathrm{~m}\) (d) \(9.6 \mathrm{~m}\)

A uniform wire, fixed at its upper end, hangs vertically and supports a weight at its lower end. If its radius is \(r\), its length \(L\) and the Young's modulus for the material of the wire is \(E\), the extension is 1\. directly proportional to \(E\) 2\. inversely proportional to \(r\) 3\. directly proportional to \(L\) (a) If only 3 is correct (b) if 1,2 are correct (c) if 2,3 are correct (d) if only \(\mathbb{I}\) is comect

For a given material, the Young's modulus is \(2.4\) times that of modulus of rigidity. Its Poisson's ratio is (a) \(0.1\) (b) \(0.2\) (c) \(0.3\) (d) \(0.4\)

Two wires of the same material and length but diameters in the ratio \(1: 2\) are stretched by the same force. The potential energy per unit volume for the two wires when stretched will be in the ratio (a) \(16: 1\) (b) \(4: 1\) (c) \(2: 1\) (d) \(1: 1\)

A steel wire has length \(2 \mathrm{~m}\), radius \(1 \mathrm{~mm}\) and \(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\). A \(1 \mathrm{~kg}\) sphere is attached to one end of the wire and whirled in a vertical circle with an angular velocity of 2 revolutions per second. When the sphere is at the lowest point of the vertical circle, the elongation of the wire is nearly (Take, \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(1 \overline{\mathrm{mm}}\) (b) \(2 \mathrm{~mm}\) (c) \(0.1 \mathrm{~mm}\) (d) \(0.01 \overline{\mathrm{mm}}\)
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