Question

After a fall, a 75 kg rock climber finds himself dangling from the end of a rope that had been 18 m long and 11 mm in diameter but has stretched by 2.1 cm. For the rope, calculate (a) the strain, (b) the stress, and (c) the Young's modulus.

Answer 1

(A) Strain = 0.0012

(b) Stress
`=77.44* 10^5N/m^2`

(c) Young's modulus
`=6.45* 10^9N/m^2`

Step-by-step explanation:

Mass of the rock m = 75 kg

So weight of the rock
`F=mg=75* 9.8=735N`

Length of the rope l = 18 m

Diameter of the rope d = 11 mm

Change in length of rope
`\Delta l=2.1cm =0.021m`

So radius r = 5.5 mm = 0.0055 m

Cross sectional area
`A=\pi r^2`

`A=3.14* 0.0055^2=9.49* 10^(-5)m^2`

(a) Strain is equal to ratio change in length to original length

So strain
`=(\Delta l)/(l)=(0.021)/(18)=0.0012`

(b) Stress
`=(Weight)/(area)`

`=(735)/(9.49* 10^(-5))=77.44* 10^5N/m^2`

(c) Young's modulus is equal to ratio of stress and strain

So young's modulus
`=(77.44* 10^5)/(0.0012)=6.45* 10^9N/m^2`