Question

A typical large orb spider has a mass of 0.50

Answer 1

The tensile strength of spider silk is about 1150 MPa, which is 1.15 x 10^9 N/m². 

So given that the diameter of the spider silk is 0.15 mm or 1.5 x 10^-4 m. The cross sectional area is approximately: 

A = πr² = (3.14)(1.5 x 10^-4 / 2)² = 1.767 x 10^-8 m² 

So this means that the amount of force we can apply before the the silk thread is broken is: 

F = PA = (1.15 x 10^9)(1.767 x 10^-8) = 20.32 N 

F = mg, so m = F/g = (20.32)/(9.81) = 2.07 kg!!!