Calculating the Force Required to Stretch a Steel Wire
Understanding the mechanics of stretching a steel wire requires a deep dive into fundamental principles of solid mechanics. This piece explores the force required to stretch a steel wire of a cross-sectional area of 1 cm2 to double its original length, using both Hooke's Law and Young's Modulus. We'll also delve into why this question might be seen as a 'trick question' and explore the implications of elastic versus plastic deformation.
Introduction to Stretching a Steel Wire
When a steel wire is stretched, it responds based on the physical properties of the material and the external force applied. Hooke's Law is the cornerstone of this physics, stating that the force required to stretch a material is proportional to the change in length and inversely proportional to the original length. This relationship is mathematically expressed as:
F E * A * ΔL / L
Applying Hooke's Law to Steel Wire
To calculate the force required to stretch a steel wire with a cross-sectional area of 1 cm2 to double its original length, we use Hooke's Law and Young's Modulus. The key parameters are:
Variables and Values
Force (F) - The force required to stretch the wire. Young's Modulus (E) - The mechanical property that measures the material's resistance to being deformed elastically when stress is applied. Cross-sectional Area (A) - 1 cm2 or 0.0001 m2. Change in Length (ΔL) - The wire doubles in length, hence ΔL L. Original Length (L) - The original length of the wire.Plugging in the values:
F 200 × 10 9 ( 0.0001 ) m/22 m N / L × L / m 20000 N
This indicates that the force required is 20,000 Newtons, or 20 kN.
Alternative Calculation Using Linear Elasticity
Another way to calculate this force is through the formula derived from linear elasticity theory:
F A E
where:
Area (A) - 1 cm2 0.0001 m2. Young's Modulus (E) - 200 GPa 200 × 109 N/m2.Substituting the values:
F 0.0001 × 200 × 10 9 N / m2 20000 N
This confirms the force required is indeed 20,000 N, or 20 kN.
The 'Trick Question' Element and Elastic Deformation
While the above calculations are straightforward, one might question this as a 'trick question.' This is because it assumes elastic deformation, which may not be feasible for the materials considered. In typical steel, the yield strength is around 400–1000 MPa. Doubling the length would put the steel well into the plastic deformation region, far beyond the elastic stretch where Young's Modulus is applicable.
Understanding Yield Strength
The stress (σ) in the wire when elongated by double its original length would be:
σ E Δ L / L
Given that ΔL/L 1, we have:
σ 200 × 10 9 N / m2 / m 200 109 N / m2 200 GPa
This stress exceeds the yield strength of typical steel by two orders of magnitude, implying that the steel would enter the plastic deformation region and will not return to its original shape once the force is removed.
Implications of Force and Area Reduction
Even assuming a uniform reduction in cross-sectional area while preserving volume, the area would decrease by 50%. Therefore, the stress would be:
S t r e s s F o r c e F / A
With the new area being 0.5 * 1 cm2 0.00005 m2, the force required would need to be adjusted accordingly.
Conclusion
In conclusion, calculating the force required to stretch a steel wire involves careful consideration of material properties and the deformation region. While the initial calculations using Hooke's Law and Young's Modulus are straightforward, the practical implications of this deformation reveal the complexity of real-world applications. Understanding these principles is crucial for engineers and physicists dealing with the mechanics of materials.