Question

A nylon string on a tennis racket is under a tension of 275 N. If its diameter is 1.00 mmmm , by how much is it lengthened from its untensioned length of 32.0 cm? No modulus was given.

Answer 1

given
F = 275 N
d = 1 mm = 0.001 m
Lo = 32 cm = 0.32 m
delta_L = ?

radius, r = d/2 = 0.001/2 = 0.0005 m

we know,

modulus of elasticity of nylon, Y = 2.7 Gpa = 2.7*10^9 Pa

we know, modulus of elasticity = stress/starin

Y = (F/A)/(delta_L/Lo)

delta)L/Lo = F/(A*Y)

delta_L = F*Lo/(A*Y)

= F*Lo/(pi*r^2*Y)

= 275*0.32/(pi*0.0005^2*2.7*10^9)

= 0.0415 m (or) 4.15 cm <<<<<<<<<<<--------------Answer