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

Answer:

(A) Strain = 0.0012

(b) Stress $=77.44\times 10^5N/m^2$

(c) Young's modulus $=6.45\times 10^9N/m^2$    

Explanation:

Mass of the rock m = 75 kg

So weight of the rock $F=mg=75\times 9.8=735N$

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 $A=\pi r^2$

$A=3.14\times 0.0055^2=9.49\times 10^{-5}m^2$

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

So strain $=\frac{\Delta l}{l}=\frac{0.021}{18}=0.0012$

(b) Stress $=\frac{Weight}{area}$

$=\frac{735}{9.49\times 10^{-5}}=77.44\times 10^5N/m^2$

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

So young's modulus $=\frac{77.44\times 10^5}{0.0012}=6.45\times 10^9N/m^2$