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Chapter 7: Problem 890

A wire of diameter \(1 \mathrm{~mm}\) breaks under a tension of \(100 \mathrm{~N}\). Another wire of same material as that of the first one, but of diameter \(2 \mathrm{~mm}\) breaks under a tension of \(\ldots \ldots \ldots\) (A) \(500 \mathrm{~N}\) (B) \(1000 \mathrm{~N}\) (C) \(10,000 \mathrm{~N}\) (D) \(4000 \mathrm{~N}\)

Short Answer

Expert verified
The tension at which the second wire breaks is 400 N. (Answer: D)

Step by step solution

01

Calculate the cross-sectional area of each wire

The cross-sectional area of a wire can be found using the formula for the area of a circle (\(A = 蟺r^2\)), where \(r\) is the radius. Since the diameter is given, we need to find the radius for each wire (half of the diameter) and then calculate the cross-sectional areas. Wire 1 Diameter: 1mm, so Radius: \(0.5\ \mathrm{mm}\) Wire 2 Diameter: 2mm, so Radius: \(1\ \mathrm{mm}\) Cross-sectional Area of Wire 1: \(A1 = 蟺(0.5^2)\ \mathrm{mm}^2\) Cross-sectional Area of Wire 2: \(A2 = 蟺(1^2)\ \mathrm{mm}^2\)
02

Find the relationship between tensions and cross-sectional areas

Since both wires are made of the same material, the ratio of their tensions should be proportional to the ratio of their cross-sectional areas. \(\frac{T2}{T1} = \frac{A2}{A1}\) Where \(T1\) and \(A1\) are the tension and cross-sectional area for Wire 1, and \(T2\) and \(A2\) are the tension and cross-sectional area for Wire 2.
03

Substitute the known values and solve for T2

Now we can substitute the known values for Wire 1 (T1 = 100N) and the calculated cross-sectional areas (A1 and A2) into the equation and solve for T2, the tension for Wire 2. \(\frac{T2}{100\ \mathrm{N}} = \frac{蟺(1^2)\ \mathrm{mm}^2}{蟺(0.5^2)\ \mathrm{mm}^2}\) Simplify the equation by canceling 蟺 in the ratio of areas: \(\frac{T2}{100\ \mathrm{N}} = \frac{1}{0.25}\) Now solve for T2: \(T2 = 100\ \mathrm{N} \times \frac{1}{0.25}\) \(T2 = 100\ \mathrm{N} \times 4\) \(T2 = 400\ \mathrm{N}\) The tension at which the second wire breaks is 400N, which corresponds to option (D) in the exercise.

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Key Concepts

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

Cross-sectional Area
The cross-sectional area of a wire is a key factor in understanding its tensile strength. This area represents the size of the wire's surface that is exposed in a cross-section, crucial for calculations like determining tensile strength. For a circular wire, this is calculated using the formula:
\( A = \pi r^2 \), where \( r \) is the radius of the circle.
Since the diameter is provided in the problem, the radius is found by halving the diameter. For instance:
  • Wire 1 with a diameter of 1 mm has a radius of 0.5 mm.
  • Wire 2 with a diameter of 2 mm has a radius of 1 mm.
The cross-sectional area for each wire is then:
  • Wire 1: \( \pi (0.5)^2 \ \text{mm}^2 \)
  • Wire 2: \( \pi (1)^2 \ \text{mm}^2 \)
Cross-sectional area directly affects how much force a wire can withstand before breaking. Larger areas can endure greater tension due to more material being present to bear the load.
Material Properties
The material properties of a wire determine its behavior under tension. These include its strength, ductility, and ability to withstand stress. For materials in engineering, properties such as tensile strength are critical.
Tensile strength measures how much pulling force a material can withstand before it fails or breaks. This strength varies with material type and is dependent on factors like:
  • Composition: Different materials have different tensile strengths based on their molecular structure.
  • Temperature: Many materials become weaker at higher temperatures.
  • Cross-sectional Area: A larger area can usually withstand more stress due to having more material to absorb the force.
In this exercise, both wires are made from the same material, ensuring that any differences in breaking tension are due solely to dimensional differences, specifically their cross-sectional area.
Ratio Analysis
Ratio analysis is essential in this problem to relate the tension tolerances of two wires made from the same material. Because the wires are identical in every way except diameter, their breaking tensions relate directly to the ratio of their cross-sectional areas.
The relationship can be described by:
\( \frac{T_2}{T_1} = \frac{A_2}{A_1} \),
where \( T_1 \) and \( A_1 \) are the tension and area for the first wire, and \( T_2 \) and \( A_2 \) for the second.
Given \( T_1 = 100 \ \text{N} \), the task is to find \( T_2 \) knowing \( A_2 \) and \( A_1 \).Cancel out common factors like \( \pi \) from the areas:
\( \frac{T_2}{100} = \frac{1}{0.25} \)
This simplifies to \( \frac{T_2}{100} = 4 \), giving:
\( T_2 = 400 \ \text{N} \).
The analysis shows that the larger diameter increases the cross-sectional area, allowing the wire to withstand higher tension before breaking.

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