Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials known. Each thread is typically 2.0 μm in diameter and the silk has a Young's modulus of

4.0 ✕ 109 N/m2.

(a) How many strands would be needed to make a rope 11 m long that would stretch only 1.00 cm when supporting a pair of 85-kgmountain climbers?
✕ 106 strands

(b) Assuming that there is no appreciable space between the parallel strands, what would be the diameter of the rope?

a) N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) diameter of single rope with the same effect = 2.415 cm

Explanation:

Hooke's law explains that stress is directly proportional to strain.

Stress ∝ Strain.

Stress = E × Strain

E = constant of proportionality = Young's Modulus = 4.0 ✕ 10⁹ N/m².

Stress = (Load/Total Cross sectional Area)

Load = a pair of 85 kg mountain climbers = 85 × 2 × 9.8 = 1666 N

Total Cross sectional Area = (Number of strands) × (Area of one strand) = A

Strain = (ΔL/L)

ΔL = 1.00 cm = 0.01 m

L = 11 m

Strain = (0.01/11) = 0.0009091

Stress = (Young's Modulus) × (Strain) = (4.0 ✕ 10⁹) × (0.0009091) = 3,636,363.64 N/m²

(Load/ total Area) = 3,636,363.64

Total area = (Load/3,636,363.64) = (1666/3,636,363.64) = 0.00045815 m²

Recall,

Total Cross sectional Area = (Number of strands) × (Area of one strand)

Area of one strand = (πd²/4)

diameter of one strand = 2 μm = (2×10⁻⁶) m

Area of one strand = (πd²/4)

= π × (2×10⁻⁶)² ÷ 4 = (3.142 × 10⁻¹²) m²

Total Cross sectional Area = (Number of strands) × (Area of one strand)

0.00045815 = N × (3.142 × 10⁻¹²)

N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) If it was a single rope, the cross sectional Area would just be equal to the total cross sectional Area obtained in (a)

A = 0.00045815 m²

A = (πD²/4)

where D = diameter of the single rope

0.00045815 = (πD²/4)

D² = (4×0.00045815) ÷ π = 0.0005833347

D = 0.02415 m = 2.415 cm

Hope this Helps!!!

hypothesising or postulating