Problem 18 - College Physics

Question

A steel wire has the following properties: $$ \begin{array}{l} \text { Length }=5.00 \mathrm{~m} \\ \text { Cross-sectional area }=0.040 \mathrm{~cm}^{2} \end{array} $$ Young's modulus \(=2.0 \times 10^{11} \mathrm{~Pa}\) Shear modulus \(=0.84 \times 10^{11} \mathrm{~Pa}\) Proportional limit \(=3.60 \times 10^{8} \mathrm{~Pa}\) Breaking stress \(=11.0 \times 10^{8} \mathrm{~Pa}\) The proportional limit is the maximum stress for which the wire still obeys Hooke's law (see point \(\mathrm{B}\) in Figure 11.12 ). The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

Step-by-Step Solution

Answer

(a) 1440 N, (b) 9 mm, (c) 4400 N.

Step 1: Convert Cross-sectional Area to Meters Squared

The cross-sectional area is given in \([\mathrm{cm}^2]\). Convert it to \([\mathrm{m}^2]\) by using the conversion factor \(1 \, \mathrm{cm}^2 = 10^{-4} \, \mathrm{m}^2\). Thus, \(0.040 \, \mathrm{cm}^2 = 0.040 \, \times 10^{-4} \, \mathrm{m}^2 = 4.0 \, \times 10^{-6} \, \mathrm{m}^2\).

Step 2: Find Maximum Weight Without Exceeding Proportional Limit

The proportional limit is given as stress, and stress is defined as force per unit area (\(\sigma = \frac{F}{A}\)). Plug in the proportional limit for stress and the cross-sectional area to find force: \(3.60 \times 10^{8} = \frac{F}{4.0 \times 10^{-6}}\). Solve for \(F\): \(F = 3.60 \times 10^{8} \, \times 4.0 \times 10^{-6} = 1440 \, \mathrm{N}\). The maximum weight is \(1440 \, \mathrm{N}\).

Step 3: Calculate Wire Stretch Under Proportional Limit Load

The change in length (stretch) can be found using Hooke's Law for solids: \(\Delta L = \frac{FL}{EA}\), where \(E\) is Young's modulus. Using \(F = 1440 \, \mathrm{N}\), \(L = 5.00 \, \mathrm{m}\), \(E = 2.0 \times 10^{11} \, \mathrm{Pa}\), and \(A = 4.0 \times 10^{-6} \, \mathrm{m}^2\): \(\Delta L = \frac{1440 \, \times \, 5.00}{2.0 \times 10^{11} \times 4.0 \times 10^{-6}} = \frac{7200}{8.0 \times 10^{5}} = 0.009 \, \mathrm{m}\). Thus, the wire stretches \(0.009 \, \mathrm{m}\) or \(9 \, \mathrm{mm}\) under the load.

Step 4: Determine Maximum Weight Before Breaking

The maximum weight before breaking is determined using the breaking stress. Set up the equation \(\sigma = \frac{F}{A}\) with the breaking stress. \(11.0 \times 10^{8} = \frac{F}{4.0 \times 10^{-6}}\). Solve for \(F\): \(F = 11.0 \times 10^{8} \, \times 4.0 \times 10^{-6} = 4400 \, \mathrm{N}\). Thus, the maximum weight that can be supported is \(4400 \, \mathrm{N}\).

Key Concepts

Young's modulus

Young's modulus is a measure of the stiffness of a material. It indicates how much a material will deform under tension or compression. Young's modulus is denoted by the symbol \(E\), and it is defined as the ratio of tensile stress to tensile strain. This concept is crucial because it helps predict how materials will behave under various forces.When you apply a force to a material, it stretches or compresses. Young’s modulus gives us a way to quantify this change. The higher the Young's modulus, the stiffer the material, meaning it will deform less for a given load. For example, steel, with a Young's modulus of \(2.0 \times 10^{11} \, \text{Pa}\), is much stiffer than rubber. This means steel will stretch much less than rubber when the same force is applied. This property is especially important in structural engineering, where materials are expected to withstand high levels of stress with minimal deformation.

Hooke's law

Hooke's Law is fundamental for understanding the elastic behavior of materials. It states that the amount of deformation (strain) is directly proportional to the applied force (stress), as long as the material's proportional limit is not exceeded.Formally, Hooke's Law is expressed as \(F = k \cdot x\), where \(F\) is the force applied, \(k\) is the spring constant, which represents the stiffness of the material, and \(x\) is the deformation of the material.In the context of solids like steel wires, Hooke's law can be rephrased using stress and strain, which is given by the formula \(\sigma = E \cdot \epsilon\), where \(\sigma\) is stress, \(E\) is Young's modulus, and \(\epsilon\) is strain.This principle remains valid until the proportional limit is reached. Beyond this point, the material will no longer return to its original shape and may suffer permanent deformation. Thus, Hooke’s law is essential for designing components to ensure they remain within their elastic limits during use.

Tensile stress

Tensile stress is a type of stress that's applied to a material by pulling it. It's defined as the force per unit area applied to a material, trying to stretch it. Mathematically, tensile stress \(\sigma\) is given by the formula \(\sigma = \frac{F}{A}\), where \(F\) is the applied force and \(A\) is the cross-sectional area of the material. This formula shows us that as the force increases, or as the area decreases, the stress increases.Tensile stress is particularly important in applications where materials are subjected to forces that try to elongate them, such as cables in bridges or tendons in buildings. Understanding tensile stress helps ensure that materials are used safely and effectively, without exceeding their limits which could lead to failure.

Proportional limit

The proportional limit marks the boundary between elastic and plastic behavior of a material. It is the highest level of stress a material can withstand while still maintaining a direct proportional relationship between stress and strain, as dictated by Hooke's law.Up to the proportional limit, materials will return to their original shape upon unloading. If stress exceeds this limit, the material will experience plastic, or permanent deformation.Knowing the proportional limit is essential in engineering and materials science, as it helps in designing structures and components that can safely endure the expected stresses in their applications without experiencing irreversible damage. For the steel wire in question, the proportional limit is given as \(3.6 \times 10^{8} \mathrm{~Pa}\). Beyond this point, the wire would no longer behave elastically, risking permanent stretching or even failure.