Question

For safety in climbing, a mountaineer uses a nylon rope that is 60 m long and 0 . 4 cm in diameter. When supporting a 91 kg climber, the rope elongates 1 . 3 m . Find its Young’s modulus. The acceleration of gravity is 9 . 8 m / s 2 . Answer in units of Pa.

Answer 1

Solution:

Length of the nylon rope, L = 60 m

The change length of the rope, ΔL = 1.3 m

Diameter of the rope, d = 0.4 cm

Radius of the rope, r = d/2 = 0.2 cm = 2.0e-3 m

Area of cross section of the rope,

A = πr²

A = 3.1416*(2.0e-3 m)²

A = 1.25664e-5 m²

The force F applied at the bottom of the rope is the weight m*g of the climber.

F = m*g

F = (91 kg)*(9.8 m/s²)

F = 891.8 N

By definition, the Young's modulus E is

E = tensile stress/strain

E = [F/A]/[ΔL/L]

E = [(891.8 N)/(1.25664e-5 m²)]/[(1.3 m)/(60 m)]

E = 3.2754e9 N/m²

E = 3.8*10⁹ Pa                           (since 1 N/m² = 1Pa)

Hence the answer is 3.8*10⁹ Pa