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\(\bullet\) A steel wire has the following properties: $$\begin{array}{l}{\text { Length }=5.00 \mathrm{m}} \\ {\text { Cross- sectional area }=0.040 \mathrm{cm}^{2}} \\ {\text { Young's modulus = } 2.0 \times 10^{11} \mathrm{Pa}} \\ {\text { Shear modulus = } 0.84 \times 10^{11} \mathrm{Pa}} \\ {\text { Proportional limit }=3.60 \times 10^{8} \mathrm{Pa}} \\\ {\text { Breaking stress }=11.0 \times 10^{8} \mathrm{Pa}}\end{array}$$ 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?

Short Answer

Expert verified
(a) 1440 N, (b) 9 mm, (c) 44000 N.

Step by step solution

01

Determine the Weight for Proportional Limit

To find the maximum weight that can be hung without exceeding the proportional limit, use the formula \( F = A \times \sigma_{PL} \), where \( A \) is the cross-sectional area in \( ext{m}^2 \) and \( \sigma_{PL} \) is the proportional limit stress. First, convert the area from \( ext{cm}^2 \) to \( ext{m}^2 \):\[ A = 0.040 imes 10^{-4} ext{ m}^2 \ = 4.0 \times 10^{-6} \text{ m}^2 \]Then calculate the force:\[ F = (4.0 \times 10^{-6} ext{ m}^2)(3.60 \times 10^{8} ext{ Pa}) = 1440 ext{ N} \]Therefore, the weight that can be supported at the proportional limit is \( 1440 ext{ N} \).
02

Calculate the Stretch under Proportional Limit Load

To find how much the wire stretches, use Hooke's Law for stretching \( \Delta L = \frac{F \cdot L}{A \cdot Y} \), where \( L \) is the original length, \( A \) is the cross-sectional area in \( ext{m}^2 \), and \( Y \) is Young's Modulus:\[ \Delta L = \frac{1440 ext{ N} \cdot 5.00 ext{ m}}{4.0 \times 10^{-6} ext{ m}^2 \cdot 2.0 \times 10^{11} ext{ Pa}} \]\[ \Delta L = 0.009 ext{ m} \text{ or } 9 ext{ mm} \]Therefore, the wire stretches by \( 9 ext{ mm} \).
03

Determine Maximum Weight Before Breaking

To find the maximum weight before the wire breaks, use the formula \( F = A \times \sigma_{b} \), where \( \sigma_{b} \) is the breaking stress:\[ F = (4.0 \times 10^{-6} ext{ m}^2)(11.0 \times 10^{8} ext{ Pa}) = 44000 ext{ N} \]Therefore, the maximum weight the wire can support before breaking is \( 44000 ext{ N} \).

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Key Concepts

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

Young's modulus
Young's modulus is a fundamental concept in the study of the mechanical properties of materials. It measures a material's ability to resist deformation when subjected to stress. In essence, it tells us how stiff or elastic a material is. The higher the Young's modulus, the less it deforms, meaning the material is stiffer. This modulus is expressed in Pascals (Pa).
To better understand it, think of a spring. Young's modulus is like the stiffness of this spring. When you pull or compress it, the spring resists the change in length. Similarly, materials with a high Young's modulus resist deformation. In the context of the steel wire from our exercise, the given Young's modulus of \(2.0 \times 10^{11} \text{ Pa}\) indicates that steel is quite stiff and doesn't easily stretch under tension. This makes steel an excellent material for applications requiring strength and minimal elongation under loads.
  • It is often used to compare the elasticity of different materials.
  • Defined as the ratio of stress to strain in the linear elastic region, Young's modulus helps us calculate how much a material will stretch.
In practical terms, knowing Young's modulus allows us to predict how much a wire, beam, or any material will deform under a specific load, which is crucial for designing structures and mechanical systems.
proportional limit
The proportional limit is an important point on the stress-strain curve of a material. It marks the maximum stress that can be applied to a material while it still obeys Hooke's Law. Beyond this limit, the material will begin to deform plastically, which means it won't return to its original shape after removing the stress.
In the context of the exercise, the steel wire has a proportional limit of \(3.60 \times 10^{8} \text{ Pa}\). Up to this stress level, the material's stress and strain relationship is linear, meaning it can return to its original state when the load is removed.
  • Beyond the proportional limit, permanent deformation begins.
  • It is crucial for ensuring safety and durability in material applications.
Knowing the proportional limit is essential in mechanical and civil engineering to ensure structural integrity, as it helps in determining the maximum load materials can bear without being permanently deformed.
breaking stress
Breaking stress, also known as tensile strength or ultimate strength, refers to the maximum stress that a material can withstand before breaking or failing. In simpler terms, it is the point at which a material will snap or fracture if the stress continues to increase.
In our exercise, the steel wire has a breaking stress of \(11.0 \times 10^{8} \text{ Pa}\). This means if the stress applied to the wire reaches this level, the wire will no longer hold and will break apart. Understanding breaking stress is crucial for safety calculations in engineering and material science, ensuring that structures and components can support the intended loads without failure.
  • Breaking stress determines the maximum load before fracture.
  • The margin between proportional limit and breaking stress is critical for designing resilient structures.
By predicting the breaking stress, engineers and designers can create safer and more efficient structures and materials that perform as needed under a wide range of conditions.
Hooke's Law
Hooke's Law is a fundamental principle that relates the force needed to extend or compress a material to the distance it is stretched. The law states that, for small deformations, the force is proportional to the displacement. In mathematical terms, Hooke’s Law can be expressed as:\[ F = k \cdot x \]where \(F\) is the force applied, \(k\) is the spring constant (a measure of the material's stiffness), and \(x\) is the displacement or change in length.
Hooke's Law applies in the elastic region of the material, where the ratio of stress to strain is constant, as seen in Young's modulus. In the exercise, Hooke's Law is used to calculate the stretch of the wire under a specific load, such as when determining how much a wire will elongate when a weight is added without exceeding its proportional limit.
  • It is applicable only up to the proportional limit of the material.
  • Helps predict how much a material will deform when a certain force is applied.
Understanding Hooke's Law is integral to designing springs, beams, and other elements where elastic properties are crucial, ensuring that they return to their original shape after the force is removed.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) One end of a stretched ideal spring is attached to an airtrack and the other is attached to a glider with a mass of 0.355 \(\mathrm{kg}\) . The glider is released and allowed to oscillate in SHM. If the distance of the glider from the fixed end of the spring varies between 1.80 \(\mathrm{m}\) and \(1.06 \mathrm{m},\) and the period of the oscillation is \(2.15 \mathrm{s},\) find (a) the force constant of the spring, (b) the maximum speed of the glider, and (c) the magnitude of the maximum acceleration of the glider.

\(\bullet\) \(\bullet\) The effect of jogging on the knees. High-impact activities such as jogging can cause considerable damage to the cartilage at the knee joints. Peak loads on each knee can be eight times body weight during jogging. The bones at the knee are separated by cartilage called the medial and lateral meniscus. Although it varies considerably, the force at impact acts over approximately 10 \(\mathrm{cm}^{2}\) of this cartilage. Human cartilage has a Young's modulus of about 24 MPa (although that also varies). (a) By what percent does the peak load impact of jogging compress the knee cartilage of a 75 kg person? (b) What would be the percentage for a lower-impact activity, such as power walking, for which the peak load is about four times body weight?

\(\bullet\) \(\bullet\) Human hair. According to one set of measurements, the tensile strength of hair is 196 \(\mathrm{MPa}\) , which produces a maximum strain of 0.40 in the hair. The thickness of hair varies considerably, but let's use a diameter of 50\(\mu \mathrm{m}\) . (a) What is the magnitude of the force giving this tensile stress? (b) If the length of a strand of the hair is 12 \(\mathrm{cm}\) at its breaking point, what was its unstressed length?

\(\bullet\) \(\bullet\) An object of unknown mass is attached to an ideal spring with force constant 120 \(\mathrm{N} / \mathrm{m}\) and is found to vibrate with a frequency of 6.00 \(\mathrm{Hz}\) . Find (a) the period, (b) the angular frequency, and (c) the mass of this object.

\(\bullet\) The wings of the Blue-throated Hummingbird (Lampornis clemenciae), which inhabits Mexico and the southwestern United States, beat at a rate of up to 900 times per minute. Calculate (a) the period of vibration of the bird's wings, (b) the frequency of the wings' vibration, and (c) the angular frequency of the bird's wingbeats.

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