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Chapter 11: Problem 39

A steel wire with radius \(r_{\text {steel }}\) has a fractional increase in length of \(\left(\Delta l / l_{0}\right)_{\text {steel }}\) when the tension in the wire is increased from zero to \(T_{\text {steel }}\). An aluminum wire has radius \(r_{\text {al }}\) that is twice the radius of the steel wire: \(r_{\mathrm{al}}=2 r_{\text {steel }} .\) In terms of \(T_{\text {steel }},\) what tension in the aluminum wire produces the same fractional change in length as in the steel wire?

Short Answer

Expert verified
The tension in the aluminum wire required to produce the same fractional change in length as in the steel wire is four times the tension in the steel wire, \( T_{\text {al }} = 4 T_{\text {steel }} \).

Step by step solution

01

Set up the Equation for the Steel Wire

We know that tension \( T = F \) and the force \( F \) applies equally over the cross-sectional area \( A \) of the wire, thus the equation for stress \( \sigma \) can be written as \( \sigma_{\text {steel }} = \frac{T_{\text {steel }}}{\pi r_{\text {steel }}^2} \). Also, the strain \( \epsilon \) on the steel wire is given by \( \epsilon_{\text {steel }} = \left(\Delta l / l_{0}\right)_{\text {steel }} \) .
02

Set up the Equation for the Aluminum Wire

Since \( r_{\mathrm{al}}=2 r_{\text {steel }} \), the equation for stress \( \sigma \) for the aluminum wire can be written as \( \sigma_{\text {al }} = \frac{T_{\text {al }}}{\pi (2 r_{\text {steel }})^2} = \frac{T_{\text {al }}}{4 \pi r_{\text {steel }}^2} \). Also, we assume that the strain \( \epsilon \) is the same in both wires, thus \( \epsilon_{\text {al }} = \epsilon_{\text {steel }} \).
03

Equate Stress for Both Wires

We now equate the two expressions for stress from Step 1 and Step 2. This gives us \( \frac{T_{\text {steel }}}{\pi r_{\text {steel }}^2} = \frac{T_{\text {al }}}{4 \pi r_{\text {steel }}^2} \). Solving for \( T_{\text {al }} \) yields \( T_{\text {al }} = 4 T_{\text {steel }} \)
04

Interpret the Result

Thus, the tension in the aluminum wire required to produce the same strain, or fractional change in length, as the steel wire must be four times the tension in the steel wire. This result is independent of the radii of the wires and relies only on the assumption about the equality of the strains.

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Key Concepts

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

Steel Wire Mechanics
Steel wire mechanics explores the behavior of steel wires under different forces. Steel is an incredibly strong material, able to withstand considerable tension.
When you stretch a steel wire, its length changes slightly. This is called strain. The force applied to achieve this elongation is known as tension.
The tension in a wire like steel generates stress, a measure of force distributed over its area. For steel wires, engineers must understand how much tension a wire can handle before deforming beyond acceptable limits.
In practical applications, knowing these limits ensures safe and efficient design, ensuring the wire holds up under expected loads.
Aluminum Wire Tension
Aluminum wires, while also robust, behave differently under tension compared to steel. They're lighter and more flexible, making them ideal for applications where weight is a concern.
When we increase the tension over an aluminum wire, it stretches in a way that's mathematically predictable. The key is to balance how much tension is needed to achieve the desired change in length, without causing harm to the wire.
In this example, we see that to produce the same strain as in the steel wire, the tension in aluminum wire must be adjusted. By understanding tension, you can ensure that aluminum wires perform optimally in a variety of scenarios.
Young's Modulus
Young's Modulus is a crucial concept in materials science. It provides a measure of a material's stiffness or rigidity.
Represented by the symbol "E," Young's Modulus relates stress to strain. Think of it as the material's resistance to deformation when a tension force is applied. It is particularly useful for comparing different materials like steel and aluminum.
For both wires, even if they strain equally, their Young's Moduli differ. This impacts how much tension each material can sustain before deforming unacceptably. In engineering, choosing materials with the right Young's Modulus ensures structures can withstand forces safely and efficiently.
Cross-sectional Area
The cross-sectional area of a wire is crucial for determining how it responds to tension. This can be understood as the side slice or cut through the wire, viewed as a circle for round wires.
The larger the cross-sectional area, the more evenly the tension can distribute across the wire, reducing the likelihood of breaking or deforming under stress.
  • Steel wire: Smaller cross-sectional area means more concentrated stress.
  • Aluminum wire: Larger cross-sectional area, due to doubling the radius, disperses the same tension more.
Understanding this helps engineers predict how wires will behave under different forces and make safer, more effective designs.
Strain Equality in Materials
Strain equality refers to two different materials having the same change in length relative to their original lengths under specific forces. It鈥檚 a crucial aspect when comparing materials like steel and aluminum.
In this context, achieving strain equality denotes applying a force to each material so they stretch the same amount. The formulas used in material science exploit this concept to calculate necessary adjustments in tension.
By understanding these properties, engineers can ensure material components interact harmoniously in a structure, maximizing performance and durability in practical uses.

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Most popular questions from this chapter

A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity threequarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge (Fig. \(\mathbf{P} 1 \mathbf{1 . 5 6}\) ). The drawbridge is \(40.0 \mathrm{~m}\) long and has a mass of \(18,000 \mathrm{~kg} ;\) its center of gravity is at its midpoint. The cement mixer, with driver, has mass \(30,000 \mathrm{~kg}\). When the drawbridge has been raised to an angle of \(30^{\circ}\) above the horizontal, the cable makes an angle of \(70^{\circ}\) with the surface of the bridge. (a) What is the tension \(T\) in the cable when the drawbridge is held in this position? (b) What are the horizontal and vertical components of the force the hinge exerts on the span?

A 3.00-m-long, \(190 \mathrm{~N}\), uniform rod at the \(\mathrm{zoo}\) is held in a horizontal position by two ropes at its ends (Fig. E11.21). The left rope makes an angle of \(150^{\circ}\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A \(90 \mathrm{~N}\) howler monkey (Alouatta seniculus) hangs motionless \(0.50 \mathrm{~m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

A \(350 \mathrm{~N}\), uniform, \(1.50 \mathrm{~m}\) bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of \(500.0 \mathrm{~N}\) without breaking, and cable \(B\) can support up to \(400.0 \mathrm{~N}\). You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

You are to use a long, thin wire to build a pendulum in a science mu- seum. The wire has an unstretched length of \(22.0 \mathrm{~m}\) and a cir- cular cross section of \(\begin{array}{lll}\text { diameter } & 0.860 & \mathrm{~mm}\end{array}\) it is made of an alloy that has a large breaking stress. One end of the wire will be attached to the ceiling, and a \(9.50 \mathrm{~kg}\) metal sphere will be attached to the other end. As the pendulum swings back and forth, the wire's maximum angular displacement from the vertical will be \(36.0^{\circ}\). You must determine the maximum amount the wire will stretch during this motion. So, before you attach the metal sphere, you suspend a test mass (mass \(m\) ) from the wire's lower end. You then measure the increase in length \(\Delta l\) of the wire for several different test masses. Figure \(\mathbf{P} 1 \mathbf{1} .86,\) a graph of \(\Delta l\) versus \(m\) shows the results and the straight line that gives the best fit to the data. The equation for this line is \(\Delta l=(0.422 \mathrm{~mm} / \mathrm{kg}) m\) (a) Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2},\) and use Fig. \(\mathrm{P} 11.86\) to calculate Young's modulus \(Y\) for this wire. (b) You remove the test masses, attach the \(9.50 \mathrm{~kg}\) sphere, and release the sphere from rest, with the wire displaced by \(36.0^{\circ} .\) Calculate the amount the wire will stretch as it swings through the vertical. Ignore air resistance.

You apply a force of magnitude \(F_{\perp}\) to one end of a wire and another force \(F_{\perp}\) in the opposite direction to the other end of the wire. The cross-sectional area of the wire is \(8.00 \mathrm{~mm}^{2}\). You measure the fractional change in the length of the wire, \(\Delta l / l_{0},\) for several values of \(F_{\perp}\) When you plot your data with \(\Delta l / l_{0}\) on the vertical axis and \(F_{\perp}\) (in units of \(\mathrm{N}\) ) on the horizontal axis, the data lie close to a line that has slope \(8.0 \times 10^{-7} \mathrm{~N}^{-1} .\) What is the value of Young's modulus for this wire?
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