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A force stretches a wire by \(1.0 \mathrm{mm}\). a. A second wire of the same material has the same cross section and twice the length. How far will it be stretched by the same force? b. A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?

Short Answer

Expert verified
a. The second wire will stretch by 2.0 mm. b. The third wire will stretch by 0.250 mm.

Step by step solution

01

Calculate for the second wire

In the first situation, the length of the second wire is twice the length of the first one. Given that the wire material and the force are the same, then the strain (deformation/original length) produced on the second wire will be the same as on the first one. Since the length of the second wire is double, then the overall deformation (strain times length) will be double as well. So, for the second wire it would be \(2 \times 1.0 \mathrm{mm} = 2.0 \mathrm{mm}\)
02

Calculate for the third wire

As for the second scenario, the diameter of the third wire is twice the diameter of the first one. This means the cross-sectional area (which is \(\pi \times (diameter/2)^2\)) is four times larger. Therefore, given that stress (force/area) needs to be the same for the same material, then the strain will be a quarter of the one in the first wire. That means the deformation will be \(1.0 \mathrm{mm}/4 = 0.250 \mathrm{mm}\).

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Key Concepts

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

Strain and Deformation
Understanding how materials behave under force is vital in various fields of engineering and physics. When a material, such as a wire, is subjected to a force that causes it to stretch, this change in length is known as deformation.

The extent to which the material deforms under this force is quantified by strain, which is a dimensionless ratio. It is calculated by dividing the deformation (the change in length) by the original length of the material. In the given exercise, a wire stretches by 1.0 mm, indicating its deformation. Strain helps us understand that despite the differences in the sizes of the wires, the deformation is proportionate to their respective original lengths.

This concept helps to predict how much a second wire, which is identical in material and cross-sectional area but twice as long, will stretch under the same force. Because strain is the same for similar materials under the same force, the second wire's deformation will also be double, resulting in a 2.0 mm stretch.
Stress-Strain Relationship
The relationship between stress and strain is fundamental in the physics of elasticity. Stress, often measured in pascals (Pa), is the force applied per unit area on a material.

In our textbook exercise, we note that the third wire has the same length but twice the diameter compared to the first wire. Therefore, its cross-sectional area is significantly larger. This is where the stress-strain relationship comes into play. The wire's resistance to deformation is dictated by how the applied stress translates into strain. The stress applied to the third wire is distributed over a larger area, leading to less strain and thus less deformation, only 0.25 mm, which is a quarter of the first wire's deformation.

Understanding this relationship is crucial in materials science and structural engineering, where it's essential to know how materials will respond under various loads to ensure safety and functionality.
Hooke's Law
A core principle that describes elasticity for many materials is Hooke's Law. It states that, within the elastic limit, the strain in a material is directly proportional to the applied stress. The law is often articulated as \( F = kx \), where \( F \) is the force exerted on the object, \( k \) is the spring constant or force constant, which is a measure of the stiffness of the spring or material, and \( x \) is the displacement of the spring or material, i.e., the deformation.

When the force stretches the initial wire mentioned in the exercise by 1.0 mm, we see Hooke's law in action under the assumption that the deformation is within the elastic range. The stretching force is directly proportional to the deformation of the wire. For the second and third wires, although the length and diameter change respectively, the way each responds to the same force can still be described by Hooke's law, as long as the deformations are elastic and the material remains the same.

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