Question

After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter but has stretched by 3.1 cm. For the rope, calculate
(a) the strain,
(b) the stress, and
(c) the Young's modulus.

Answer 1

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, $\Delta L=3.1\ cm$

(a) The change in length per unit original length is called strain. So,

$\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193$

(b) The force acting per unit area is called stress.

$\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa$

(c) The ratio of stress to the strain is called Young's modulus. So,

$Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2$

Hence, this is the required solution.