Question Detail
Question
A 2.0 m-long steel rod has a cross-sectional area of 0.30 cm². The rod is a part of a vertical support that holds a heavy 550-kg platform that hangs attached to the rod's lower end. Ignoring the weight of the rod, what is the elongation of the rod under the stress? Take Young's modulus of elasticity of the steel rod as 2.0 × 10¹¹ N/m² and acceleration due to gravity as 9.8 m/s².
Correct Answer
Option A is the correct answer.
Detailed Explanation
Young's modulus Y = (F/A)/(ΔL/L) = FL/(AΔL). Rearranging: ΔL = FL/(AY). Force F = mg = 550 kg × 9.8 m/s² = 5390 N. Length L = 2.0 m. Area A = 0.30 cm² = 0.30 × 10⁻⁴ m² = 3.0 × 10⁻⁵ m². Y = 2.0 × 10¹¹ N/m². ΔL = (5390 × 2.0)/(3.0 × 10⁻⁵ × 2.0 × 10¹¹) = (10780)/(6.0 × 10⁶) = 10780/6,000,000 = 0.001797 m = 1.797 mm ≈ 1.8 mm.
Hint
Use ΔL = FL/(AY). Ensure all units are in SI (meters, kg, seconds).