Question

A metal rod that is 4.00 m long and 0.50 cm\(^2\) in crosssectional area is found to stretch 0.20 cm under a tension of 5000 N. What is Young's modulus for this metal?

Short answer

Young's modulus for the metal is \(2.00 \times 10^{11} \text{ N/m}^2\).

Step-by-step solution

Understand the Problem

We are asked to find Young's modulus, which is a measure of the stiffness of a material. The formula for Young's modulus is given by \( E = \frac{F}{A} \times \frac{L_0}{\Delta L} \) where \( E \) is Young's modulus, \( F \) is the force applied, \( A \) is the cross-sectional area, \( L_0 \) is the original length, and \( \Delta L \) is the change in length.

Identify Given Values

From the problem statement, we have:- Force \( F = 5000 \text{ N} \)- Original Length \( L_0 = 4.00 \text{ m} \)- Cross-sectional Area \( A = 0.50 \text{ cm}^2 = 0.50 \times 10^{-4} \text{ m}^2 \) (converted to square meters)- Change in Length \( \Delta L = 0.20 \text{ cm} = 0.0020 \text{ m} \) (converted to meters).

Substitute the Values into the Formula

Substitute the values into the formula: \[ E = \frac{5000 \text{ N}}{0.50 \times 10^{-4} \text{ m}^2} \times \frac{4.00 \text{ m}}{0.0020 \text{ m}} \].

Calculate the First Fraction

First, calculate the stress (force per unit area): \[ \sigma = \frac{5000 \text{ N}}{0.50 \times 10^{-4} \text{ m}^2} \] \[ \sigma = 1.00 \times 10^8 \text{ N/m}^2 \].

Calculate the Second Fraction

Next, calculate the strain (change in length per original length): \[ \epsilon = \frac{0.0020 \text{ m}}{4.00 \text{ m}} = 0.0005 \].

Complete the Calculation for Young's Modulus

Now calculate Young's modulus using the values from Steps 4 and 5: \[ E = 1.00 \times 10^8 \text{ N/m}^2 \times \frac{1}{0.0005} \]. Evaluate this to obtain: \[ E = 2.00 \times 10^{11} \text{ N/m}^2 \].

Conclusion

The value of Young's modulus for the metal is \( 2.00 \times 10^{11} \text{ N/m}^2 \). This suggests that the metal is quite stiff, which is typical for metals like steel.

Key concepts

Stress

Stress is a measure of the internal forces that develop within a material when an external force is applied. In simple terms, stress can be thought of as the pressure experienced by the material. It is defined as the force applied divided by the cross-sectional area over which the force acts. Mathematically, you can express stress (\( \sigma \)) using the formula:
\[ \sigma = \frac{F}{A} \]where \( F \) represents the applied force, and \( A \) represents the cross-sectional area.

• Unit: Stress is measured in Pascals (Pa), where 1 Pascal equals 1 Newton per square meter (N/m²).
• Application: Knowing the stress helps predict whether a material will withstand the applied forces or fail.

Stress gives a fundamental insight into the mechanical behavior of materials and is critical in designing structural components.

Strain

Strain is related to the deformation a material undergoes under stress. When a force is applied to an object, it may change its shape or size. Strain quantifies this change and is defined as the ratio of the change in length (\( \Delta L \)) to the original length (\( L_0 \)) of the object. Mathematically, strain (\( \epsilon \)) is given by:
\[ \epsilon = \frac{\Delta L}{L_0} \]Strain is a dimensionless quantity, meaning it has no units. It simply represents how much a material deforms in response to stress.

• Types of Strain: There are mainly two types of strains: tensile strain (stretching) and compressive strain (squeezing).
• Elastic vs Plastic Strain: Elastic strain is reversible, while plastic strain is permanent.

Understanding strain is crucial for assessing how materials deform and behave under various conditions.

Elasticity

Elasticity describes a material's ability to return to its original shape and size after the external forces are removed. Think of it like a spring that stretches when pulled but returns to its original shape when released. The extent to which an object can stretch or compress and still return to its original size determines its elasticity.

• Hooke's Law: This principle states that, within the elastic limit, the amount of stretch or compression (strain) is directly proportional to the applied load (stress). Mathematically, it is expressed as \( \sigma = E \cdot \epsilon \)
• Elastic Limit: The maximum extent to which a material can return to its original shape after the applied stress is removed. Beyond this limit, permanent deformation occurs.

Elasticity is vital for understanding how structures flex under loads and is critical in ensuring materials perform safely under stress.

Mechanical Properties

Mechanical properties indicate how a material reacts to applied forces, including tension, compression, and bending. They define the material's behavior and capabilities under mechanical loads and conditions and can be specific to types of applications, such as construction or manufacturing.

• Young's Modulus: This is a significant mechanical property that measures a material's stiffness, indicating how much it will deform under stress. It's defined as the ratio of stress to strain.
• Tensile Strength: The maximum stress a material can withstand while being stretched before breaking.
• Ductility and Brittleness: Ductility describes a material's ability to deform under tensile stress. Brittleness refers to a lack of ductility, where a material breaks rather than deforms.

Understanding these properties is crucial for selecting materials for specific applications, ensuring both functionality and safety in their end use.

Material Deformation

Material deformation refers to the change in shape, size, or volume of a material due to applied forces. In engineering, it is vital to understand the limits up to which a material can deform.

• Elastic Deformation: Temporary shape change that is recoverable upon the removal of stress.
• Plastic Deformation: Permanent change, where the material doesn't return to its original shape after the stress is removed.
• Factors Influencing Deformation: These include material composition, temperature, and type of applied stress.

Understanding material deformation helps in designing components that can withstand external forces without failing or suffering permanent damage.