Question

One end of a nylon rope, of length 4.5 cm and diameter 6mm, id fixed to a free limb. A monkey , weighing 100 N , jumps to catch the free end and stays there. Find the elongation of the rope and the corresponding change in the diameter, Given Young's modules of nylon = 4.8 x 10 to the power of 11 N/m square and Poisson's ratio of nylon = 0.2.?

Answer 1

Hi

Here is your answer,

Here, l = 4.5 m ; D = 6 mm = 6 × 10⁻³ m 

F = Mg = 100N

Y = 4.8 × 10¹¹ N/m² ; Δl = ? ; ΔD = ?

Y = F/ (πD² /4 ) × l/Δl

OR,      Δl = 4Fl/πD²Y = 4 × 100 × 4.5/ 3.14 × (6×10⁻³)² × (4.8 × 10¹¹) 

→ 3.315 × 10⁻⁵ m

σ = ΔD/D/Δl/l (in magnitude) 

OR,       Δ DΔ =σΔl/l × D 

→ 0.2 × (3.315 × 10⁻⁵)/ 4.5 × (6 ×10⁻³)

→ 8.814 × 10⁻⁹ m.



Hope it helps you !