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Chapter 7: Problem 10

Two wires of the same length and same material but radil in the ratio of \(1: 2\) are stretched by unequal forces to produce equal elongation. The ratio of the two forces is (1) \(1: 1\) (2) \(1: 2\) (3) \(1: 3\) (4) \(1: 4\)

Short Answer

Expert verified
The ratio of the two forces is 1:4.

Step by step solution

01

Formula for Elongation

The elongation of a wire is given by the formula \( \Delta L = \frac{FL}{AY} \), where \( F \) is the force applied, \( L \) is the original length, \( A \) is the cross-sectional area, and \( Y \) is the Young's Modulus of the material. Since the material and length are the same for both wires, \( Y \) and \( L \) remain constant.
02

Relationship of Forces and Areas

For equal elongation, the right side of the equation \( \frac{FL}{AY} \) must be equal for both wires. Since \( L \) and \( Y \) are constant, the ratio based on the cross-sectional area \( A \) and force \( F \) can be written as \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \).
03

Calculate the Areas using the Radii

The cross-sectional area \( A \) of a wire with radius \( r \) is given by \( A = \pi r^2 \). Therefore: \( A_1 = \pi r_1^2 \) and \( A_2 = \pi r_2^2 \). Given the radii ratio \( r_1:r_2 = 1:2 \), then \( A_1:A_2 = r_1^2 : r_2^2 = 1^2 : 2^2 = 1:4 \).
04

Relate the Forces with the Calculated Area Ratio

Since \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \) and \( A_1:A_2 = 1:4 \), rearrange for the force ratio: \( \frac{F_1}{F_2} = \frac{A_1}{A_2} = \frac{1}{4} \). Therefore, the ratio of forces is \( F_1:F_2 = 1:4 \).

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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, denoted as \( Y \), is a fundamental property that describes the stiffness of a material. It is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. This property is crucial in understanding wire elongation, as it helps determine how much a wire will stretch given a specific force. In the problem provided, both wires are made from the same material, meaning they share the same Young's Modulus. This allows us to focus solely on other variables like force and cross-sectional area when considering their elongation properties.
Cross-Sectional Area
The cross-sectional area \( A \) is the area of the face that is perpendicular to the wire's length. For a wire with a circular cross-section, the area can be calculated using the formula \( A = \pi r^2 \), where \( r \) is the radius of the wire. The significance of the cross-sectional area in this exercise is that it influences the wire's capacity to stretch; a smaller area means more stretch for a given force, while a larger area results in less stretch. In our scenario, the wires have radii in the ratio \( 1:2 \), leading to an area ratio of \( 1:4 \), which plays a key role in how the forces are compared.
Force Ratio
The force ratio refers to the comparative force applied to two objects, which affects their elongation when the material properties are consistent. Given the requirement of equal elongation in this problem, the ratio of the forces must inversely relate to the areas. Since the areas have a ratio of \( 1:4 \), it means that the smaller cross-sectional area must bear a larger force to stretch equally with the larger area. This relationship serves as the basis for determining the correct force ratio. As derived, the forces must be applied in a \( 1:4 \) ratio to maintain equal elongation.
Wire Elongation Formula
The wire elongation formula \( \Delta L = \frac{FL}{AY} \) is an equation that describes how much a wire stretches when a force is applied. The variables are: \( F \) for force, \( L \) for the original length of the wire, \( A \) for cross-sectional area, and \( Y \) for Young's Modulus. This formula shows that for a given material and wire length, the elongation changes with force and inverse of the area. In this exercise, since the material and length are consistent, understanding how variations in force and area affect elongation becomes straightforward. Change in each element directly impacts the others in the pursuit of equal elongation between the wires.

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Most popular questions from this chapter

A cube is shifted to a depth of \(100 \mathrm{~m}\) is a lake. The change in volume is \(0.1 \%\). The bulk modulus of the material is nearly (1) \(10 \mathrm{~Pa}\) (2) \(10^{4} \mathrm{~Pa}\) (3) \(10^{7} \mathrm{~Pa}\) (4) \(10^{9} \mathrm{~Pa}\)

Two wires of the same material have lengths in the ratio \(1: 2\) and their radii are in the ratio \(1: \sqrt{2}\). If they are stretched by applying equal forces, the increase in their lengths will be in the ratio (1) \(\sqrt{2}: 2\) (2) \(2: \sqrt{2}\) (3) \(1: 1\) (4) \(1: 2\)

Choose the correct statements from the following: (1) Steel is more elastic than rubber. (2) The stretching of a coil spring is determined by the Young's modulus of the wire of the spring. (3) The frequency of a tuning fork is determined by the shear modulus of the material of the fork. (4) When a material is subjected to a tensile (stretching) stress the restoring forces are caused by interatomic attraction.

A wire can sustain the weight of \(20 \mathrm{~kg}\) before breaking. If the wire is cut into two equal parts, each part can sustain a weight of (1) \(10 \mathrm{~kg}\) (2) \(20 \mathrm{~kg}\) (3) \(40 \mathrm{~kg}\) (4) \(35 \mathrm{~kg}\)

A wire is stretched \(1 \mathrm{~mm}\) by a force of \(1 \mathrm{kN}\). How far would a wire of the same material and length but of four times that diameter be stretched by the same force? (1) \(\frac{1}{2} \mathrm{~mm}\) (2) \(\frac{1}{4} \mathrm{~mm}\) (3) \(\frac{1}{8} \mathrm{~mm}\) (4) \(\frac{1}{16} \mathrm{~mm}\)
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