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!!!
