Question

A nylon thread is subjected to a 8.5-N tension force. Knowing that E 5 3.3 GPa and that the length of the thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.

Answer 1

a)diameter of the thread = 0.546mm

b) stress in the thread = 36.3MPa

Step-by-step explanation:

Hooke's law

`E = (Stress)/(Strain) \\`

strain = 1.1% = 0.011

Stress = strain* Young’s modulus

= 3.3 × 10⁹ × 0.011

= 36.3 × 10⁶Pa

stress = 36.3MPa

stress in the thread = 36.3MPa

b) Now,

stress = Force / area

Force = 8.5N

Stress = 36.3 × 10⁶

Area = Force / stress

= 8.5 / 36.3 × 10⁶

= 0.234 × 10⁻⁶m²

Area =

`(\pi )/(4) d^2\\Thus,\\d^2 =0.234 * 10^-^6. (4)/(\pi ) \\d = √(0.298 * 10^-^6 ) \\= 5.46 * 10^-^4m\\=0.546mm`

diameter of the thread = 0.546mm