Question

For safety in climbing, a mountaineer uses a nylon rope that is 55 m long and 1.3 cm in diameter. When supporting a(n) 83-kg climber, the rope elongates 1.4 m. Find its Young's modulus.

Answer 1

Answer:

Young's modulus(Y) = 2.4 × 10⁸ (Approx)

Explanation:

Given:

Length of rope (l) = 55 m

Diameter of rope = 1.3 cm

Radius of rope (r) = 1.3 / 2 = 0.65 cm = 0.0065 m

Weight of climber (m) = 83 kg

Rope elongate(Δl) = 1.4 m

Find:

Young's modulus(Y).

Computation:

Force (F) = Weight of climber (m) × g

Force (F) = 83 × 9.8

Force (F) = 813.4 N

Area of rope (A) = πr²

Area of rope (A) = 0.000132665

Young's modulus(Y) =$\frac{(F)(l)}{(\delta l)(A)}$

Young's modulus(Y) = 44,737 / 0.000185731

Young's modulus(Y) = 2.4 × 10⁸ (Approx)

Answer 2

The Young's modulus of the rope is $1.55 \times 10^{5} \;\rm N/m^{2}$.

Given data:

The length of rope is, L = 55 m.

The diameter of rope is, d = 1.3 cm = 0.013 m.

The mass of climber is, m = 83 kg.

The elongation of rope is, $\Delta L = 1.4 \;\rm m$.

The ratio of stress and strain produced in a material is known as Young's Modulus of material . Then,

$E=\dfrac{\sigma }{\epsilon}\\\\E=\dfrac{W/A }{L/ \Delta L}\\\\\\E = \dfrac{W \times \Delta L}{AL}$

Here,

W is the weight of climber and its value is, W = mg.

A is the cross-sectional area of rope and its value is, $A =\dfrac{\pi}{4}d^{2}$.

Solving as,

$E = \dfrac{mg \times \Delta L}{\dfrac{\pi}{4} \times d^{2} \times L}\\\\\\E = \dfrac{83 \times 9.8 \times 1.4}{\dfrac{\pi}{4} \times (0.013)^{2} \times 55}\\\\\\E=1138.76/7.30 \times 10^{-3}\\\\E = 1.55 \times 10^{5} \;\rm N/m^{2}$

Thus, we can conclude that the Young's modulus of the rope is $1.55 \times 10^{5} \;\rm N/m^{2}$.

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