Respuesta :
Answer:
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!!!
The given Young's modulus of elasticity, indicates the resistance of the caterpillars thread to deformation.
(a) The number of strands needed is approximately 1.45982484 × 10⁸ strands.
(b) The diameter of a single rope of silk to support the two climbers is approximately 2.42 cm.
Reasons:
The given Young's modulus of the caterpillar thread, E = 4.0 × 10⁹ N/m²
The diameter of each thread = 2.0 μm
(a) Length of the rope = 11 m
The stretch of the rope = 1.00 cm
Mass of mountain climbers = 85 kg
Number of mountain climbers = 2
Therefore;
Weight of mountain climbers = Load = 2 × 85 × 9.81 = 1667.7
[tex]Young's \ modulus = \mathbf{\dfrac{Stress, \, \sigma}{Strain, \, \epsilon}}[/tex]
[tex]Stress, \, \sigma = \mathbf{\dfrac{Load}{Area}} = \dfrac{1667.7 \, N}{Area}[/tex]
[tex]Strain,\, \epsilon = \dfrac{\Delta L}{L} = \dfrac{0.01}{11}[/tex]
Therefore;
[tex]4.0 \times 10^9 \, N/m^2 = \dfrac{ \dfrac{1667.7 \, N}{Area}}{ \dfrac{0.01}{11}}[/tex]
Which gives;
Area, A = 4.586175 × 10⁻³ m²
[tex]Number \ of \ strands, \ n = \dfrac{4.586175 \times 10^{-3}}{\left(\dfrac{(2.0 \times 10^{-6})^2 \times \pi }{4} \right) } \approx 1.45982484 \times 10^8[/tex]
The number of strands needed, n ≈ 1.45982484 × 10⁸ strands
(b) Area of the circular rope, A = 4.586175 × 10⁻³ m²
[tex]Area \ of \ a \ circle, \ A = \mathbf{\pi \cdot \dfrac{D^2}{4}}[/tex]
Where;
D = The diameter of the rope
Therefore;
[tex]4.586175 \times 10^{-3} = \mathbf{\pi \cdot \dfrac{D^2}{4}}[/tex]
[tex]D = \sqrt{\dfrac{4.586175 \times 10^{-3} \times 4 }{\pi} } \approx 0.0242[/tex]
The diameter of the rope, D ≈ 0.0242 m = 2.42 cm
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