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

(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

Step-by-step 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 = (FL)/(Ax) = (F_gL)/(Ax)\\\\x = (F_gL)/(AE)\\\\x = ((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) = (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`