Question Detail
Question
Suspension cables are used to carry gondolas at ski resorts. Consider a suspension cable that includes an unsupported span of 1020 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 20 × 10⁴ N. Take the Young's modulus of elasticity of the steel as 2 × 10¹¹ N/m².
Correct Answer
Option C is the correct answer.
Detailed Explanation
Young's modulus Y = (F/A)/(ΔL/L) = FL/(AΔL). Rearranging: ΔL = FL/(AY). F = 20 × 10⁴ N = 200,000 N. L = 1020 m. A = πr² = π × (0.028 m)² = π × 0.000784 = 0.002463 m². Y = 2 × 10¹¹ N/m². ΔL = (200,000 × 1020)/(0.002463 × 2 × 10¹¹) = (2.04 × 10⁸)/(4.926 × 10⁸) = 0.414 m ≈ 0.4 m.
Hint
Use ΔL = FL/(AY). Calculate cross-sectional area from diameter.