Question

On increasing the length by \(0.5 \mathrm{~mm}\) in a steel wire of length \(2 \mathrm{~m}\) and area of cross-section \(2 \mathrm{~mm}^{2}\) the force required is..... \(\left[\mathrm{Y}\right.\) for steel \(\left.=2.2 \times 10^{11} \mathrm{~N} / \mathrm{m}\right]\) (A) \(1.1 \times 10^{5} \mathrm{~N}\) (B) \(1.1 \times 10^{4} \mathrm{~N}\) (C) \(1.1 \times 10^{3} \mathrm{~N}\) (D) \(1.1 \times 10^{2} \mathrm{~N}\)

Short answer

The required force to increase the length of the steel wire by 0.5 mm is \(2.2 \times 10^2 N\).

Step-by-step solution

Understand the problem

We are given the original length, change in length, area of cross-section and Young's modulus of steel. We need to find the force required to cause the given increase in length.

Write the formula relating stress, strain, and Young's modulus

The formula relating stress, strain, and Young's modulus is: \(Stress = Young's~modulus × Strain\) where \(Stress = \frac{Force}{Area}\) and \(Strain = \frac{Change~in~length}{Original~length}\).

Substitute the given values

Substitute the given values in the formula: \(Force = Stress × Area = (Young's~modulus × Strain) × Area\) \(Force = (2.2 × 10^{11} N/m~ × \frac{0.5 × 10^{-3} m}{2 m}) × 2 × 10^{-6} m^2\)

Calculate the required force

Now, calculate the force: \(Force = (2.2 × 10^{11} × \frac{0.5 × 10^{-3}}{2}) × 2 × 10^{-6}\) \(Force = (2.2 × 10^{11} × 0.25 × 10^{-3}) × 2 × 10^{-6}\) \(Force = 1.1 × 10^{8} × 10^{-3} × 2 × 10^{-6}\) \(Force = 1.1 × 10^{5} × 2 × 10^{-6}\) \(Force = 1.1 × 2 × 10^{5-6}\) \(Force = 2.2 × 10^{5-6}\) \(Force = 2.2 × 10^{-1}\) \(Force = 2.2 × 10^{2} N\) So, the required force to increase the length of the steel wire by 0.5 mm is \(2.2 \times 10^2 N\) which corresponds to the answer choice (D).

Key concepts

Stress and Strain

In the context of materials science and engineering, it's important to understand stress and strain when assessing how materials respond to external forces. Stress represents the force applied to a material divided by its area over which the force acts. It is mathematically given by

• \(Stress = \frac{Force}{Area} \).

Strain, on the other hand, measures the deformation of the material and is defined as the change in length divided by the original length:

• \(Strain = \frac{Change~in~length}{Original~length} \).

In practical terms, when you stretch or compress an object, the stress tells you about the force applied per unit area, and strain tells you how much the object is deformed in response to this force. Understanding these concepts is crucial for analyzing and predicting how materials will behave under different loads and conditions.
Both stress and strain are dimensionless quantities that can describe an object's ability to withstand forces before reaching a critical point like breaking or permanently deforming. Calculating stress and strain allows engineers to ensure that structures are safe and efficient by choosing materials with the appropriate properties.

Force Calculation

Calculating force in materials like steel wires requires linking the concepts of stress, strain, and material properties through formulas. One central formula here, Young's modulus, helps express this connection. Young's modulus, denoted as \( Y \), is a measure of a material's stiffness. It relates stress and strain through the equation:

• \( Stress = Young's~modulus \times Strain \).

From this, the force can be calculated using the rearranged equation:

• \( Force = Stress \times Area = (Young's~modulus \times Strain) \times Area \).

To compute the force that causes a certain deformation in the materials, you begin by calculating strain using the given change in length and the original length. Then, multiply the result by the Young's modulus of the material to find stress. Finally, the force can be determined by multiplying this stress by the cross-sectional area of the material. This method ensures that you compute the exact force required to produce the specified deformation, considering the material's stiffness properties.

Material Properties

Material properties like Young's modulus play a significant role in engineering and physics, especially when understanding how different materials deform under stress. Young's modulus ( \( Y \)) is an intrinsic property that describes the material's elasticity, essentially how stiff or flexible the material is. Different materials have differing Young's moduli, making them either better or worse choices depending on the application.

• Higher Young's modulus: Material is less elastic (stiffer).
• Lower Young's modulus: Material is more elastic (flexible).

Using steel as an example, it has a relatively high Young's modulus ( \( 2.2 \times 10^{11} \, \text{N/m}^2 \)), indicating that steel is a stiff material and does not easily deform under forces. Knowing the Young's modulus helps engineers select appropriate materials for constructs like buildings or bridges, ensuring they can withstand expected loads without excessive deformation. Additionally, these properties help in understanding the durability and longevity of materials under continuous or cyclic loading. Recognizing the intrinsic properties of materials aids in predicting how they will handle stress and strain during their operational life.