Question

Orb spiders make silk with a typical diameter of 0.15 mm. a. A typical large orb spider has a mass of 0.50 g. If this spider suspends itself from a single 12-cm-long strand of silk, by how much will the silk stretch?b. What is the maximum weight that a single thread of this silk could support?

Answer 1

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;

$E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm$

$x = 0.001585 \ cm$

(b)

the maximum weight that a single thread can support is given by;

$T_{tensile \ strength} = \frac{F_{max}}{A}$

The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²

$F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N$