Answer:
(a) the change in length of the silk is 0.001585 cm
(b) the maximum weight that a single thread can support is 17.67 N
Explanation:
Given;
mass of the spider, m = 0.50 g = 0.5 x 10⁻³ kg
length of the silk, L = 12 mm = 0.012 m
diameter of the silk, d = 0.15 mm
radius of the silk, r = d / 2 = 0.075 mm = 0.075 x 10⁻³ m
The cross sectional area of the silk;
A = πr² = π(0.075 x 10⁻³)²
A = 1.767 x 10⁻⁸ m²
The Young's modulus of elasticity of spider-silk is given by;
2.1 Gpa = 2.1 x 10⁹ N/m²
(a)
Apply Young's modulus of elasticity equation to determine the change in length of the silk;
[tex]E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm[/tex]
[tex]x = 0.001585 \ cm[/tex]
(b)
the maximum weight that a single thread can support is given by;
[tex]T_{tensile \ strength} = \frac{F_{max}}{A}[/tex]
The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²
[tex]F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N[/tex]
