Question Detail
Question
Suspension cables are used to carry gondolas at ski resorts. Consider a suspension cable that includes an unsupported span of 3020 m. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of 5.6 cm and the maximum tension it can withstand is 3.0 × 10⁶ N. Take the Young's modulus of elasticity of the suspension cables as 210 × 10⁹ N/m².
Correct Answer
Option D is the correct answer.
Detailed Explanation
Using Young's modulus: Y = (F/A) / (ΔL/L). So ΔL = (F × L)/(A × Y). Area A = π(d/2)² = π(0.028)² = 0.00246 m². F = 3.0 × 10⁶ N, L = 3020 m, Y = 2.1 × 10¹¹ N/m². ΔL = (3.0×10⁶ × 3020) / (0.00246 × 2.1×10¹¹) = (9.06×10⁹) / (5.166×10⁸) = 17.54 m ≈ 18 m.
Hint
Use Young's modulus formula: ΔL = (F L)/(A Y).