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Chapter 12: Problem 30

A steel wire of diameter \(1 \mathrm{mm}\) can support a tension of \(0.2 \mathrm{kN} .\) A cable to support a tension of \(20 \mathrm{kN}\) should have diameter of what order of magnitude?

Short Answer

Expert verified
The diameter of the cable required to support a tension of 20 kN is approximately 10 mm.

Step by step solution

01

Understand the principle of geometric similarity

In a geometrically similar object, if one dimension (length, width, or height) increases by a certain factor, its volume increases by the cube of that factor. Here, we can apply that principle inversely. The tension a wire can bear is proportional to its cross-sectional area (which is a function of the square of the diameter, considering the wire as a cylinder), not its length.
02

Calculate the increase factor of tension

Determine the increased factor of the tension from 0.2 kN to 20 kN. The calculation is done by dividing the targeted tension (20 kN) by the initial tension (0.2 kN), wich gives 100. So the tension increased by a factor of 100.
03

Calculate the increase factor of diameter

As tension increases directly with the cross-sectional area (and thus the square of the diameter) and not the length of the cable, the wire diameter will increase by the square root of tension's increase factor. Hence, calculate the square root of the increased tension factor. The square root of 100 is 10.
04

Compute the diameter

Multiply the initial diameter of the wire (1 mm) by the square root of the increased factor of tension (10). The resulting diameter is 10 mm.

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Key Concepts

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

Tension in Wires
Tension in wires is an essential concept when exploring the strength and stability of materials. When we talk about tension, we refer to the force that is being applied along the length of a wire or cable, stretching it. The ability of a wire to withstand tension is a key factor in many engineering applications.
The tension a wire can handle depends significantly on its material and cross-sectional area.
  • A wire made of stronger material, like steel, can generally support more tension than one made of a weaker material, such as plastic.
  • The wire's diameter also plays a crucial role; a larger diameter means a larger cross-sectional area, leading to a higher capacity to bear tension.
Understanding tension in wires helps in designing safe and efficient structures. Engineers need to calculate the expected forces that structures like bridges or buildings might experience and choose materials and dimensions accordingly to prevent failures.
Cross-sectional Area
The cross-sectional area of a wire is a critical factor that influences its tension-bearing capacity. It refers to the area of the face of the cut surface if you were to slice straight through the wire.
For a cylindrical wire, the cross-sectional area can be calculated using the formula: \[ A = \pi \left(\frac{d}{2}\right)^2 \]where \(d\) is the diameter of the wire.
It's important to grasp why the cross-sectional area is so vital:
  • The larger the cross-sectional area, the more force it can support before breaking.
  • When the diameter of a wire increases, its cross-sectional area increases considerably because the area is proportional to the square of the diameter.
This relationship shows why a small increase in diameter results in a much larger increase in tension capacity, emphasizing the importance of careful design and calculation in practical engineering scenarios.
Square Root Calculation
Understanding the square root calculation is crucial when increasing wire dimensions for additional tension capacity. The square root is a mathematical function that, given a number, finds another number which, when multiplied by itself, gives the original number.
In the context of increasing wire diameter to support more tension, we use the square root calculation to find out by what factor the diameter must increase.
  • If the tension a wire can withstand increases by a factor of 100, we calculate the new diameter by finding the square root of this factor (i.e., 10 in this case).
  • This makes sense since the diameter must increase proportionally to the square root of the factor increase in tension capacity, due to the quadratic relationship between diameter and cross-sectional area.
Understanding square roots helps us effectively scale dimensions in engineering problems, ensuring we optimize material use while maintaining safety and functionality.

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Most popular questions from this chapter

In the What If? section of Example \(12.3,\) let \(x\) represent the distance in meters between the person and the hinge at the left end of the beam. (a) Show that the cable tension in newtons is given by \(T=93.9 x+125 .\) Argue that \(T\) increases as \(x\) increases. (b) Show that the direction angle \(\theta\) of the hinge force is described by \(\tan \theta=\left(\frac{32}{3 x+4}-1\right) \tan 53.0^{\circ}\) How does \(\theta\) change as \(x\) increases? (c) Show that the magnitude of the hinge force is given by $$R=\sqrt{8.82 \times 10^{3} x^{2}-9.65 \times 10^{4} x+4.96 \times 10^{5}}$$ How does \(R\) change as \(x\) increases?

Review problem. A bicycle is traveling downhill at a high speed. Suddenly, the cyclist sees that a bridge ahead has collapsed, so she has to stop. What is the maximum magnitude of acceleration the bicycle can have if it is not to flip over its front wheel-in particular, if its rear wheel is not to leave the ground? The slope makes an angle of \(20.0^{\circ}\) with the horizontal. On level ground, the center of mass of the woman-bicycle system is at a point \(1.05 \mathrm{m}\) above the ground, \(65.0 \mathrm{cm}\) horizontally behind the axle of the front wheel, and \(35.0 \mathrm{cm}\) in front of the rear axle. Assume that the tires do not skid.

A \(200-\mathrm{kg}\) load is hung on a wire having a length of \(4.00 \mathrm{m}\) cross-sectional area \(0.200 \times 10^{-4} \mathrm{m}^{2},\) and Young's modulus \(8.00 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) What is its increase in length?

A steel cable \(3.00 \mathrm{cm}^{2}\) in cross-sectional area has a mass of \(2.40 \mathrm{kg}\) per meter of length. If \(500 \mathrm{m}\) of the cable is hung over a vertical cliff, how much does the cable stretch under its own weight? \(Y_{\text {steel }}=2.00 \times 10^{11} \mathrm{N} / \mathrm{m}^{2}\).

A \(1500-\mathrm{kg}\) automobile has a wheel base (the distance between the axles) of \(3.00 \mathrm{m}\). The center of mass of the automobile is on the center line at a point \(1.20 \mathrm{m}\) behind the front axle. Find the force exerted by the ground on each wheel.
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