Question

A force of 68 Newtons is applied to a wire with a diameter of 0.7 mm. What is the tensile stress (in N/m2) in the wire? Do not include units with the answer.

Answer 1

Answer:

7.07 x 10⁸ N/m²

Explanation:

F = Force applied to the wire = 68 N

d = diameter of the wire = 0.7 mm = 0.7 x 10⁻³ m

r = radius of the wire = (0.5) d = (0.5) (0.7 x 10⁻³) = 0.35 x 10⁻³ m

Area of cross-section of wire is given as

A = (0.25) πr²

A = (0.25) (3.14) (0.35 x 10⁻³)²

A = 9.61625 x 10⁻⁸ m²

Tensile stress is given as

$P = \frac{F}{A}$

$P = \frac{68}{9.61625\times 10^{-8}}$

P = 7.07 x 10⁸ N/m²

Answer 2

Given:

Applied Force on wire = 68 N

Diameter of wire, d = 0.7 mm = $0.7\times 10^{-3}$ m

Radius of wire, r = $\frac{d}{2}$ = 0.35 mm = $0.35\times 10^{-3}$ m

Formula used:

Stress =  $\frac{Applied Force}{cross-sectional area}$

Explanation:

Cross-sectional area, A = $\pi r^{2}$  =  $\pi (0.35\times 10^{-3})^{2}$

A = $3.84\times 10^{-7} m^{2}$

Using the formula for stress:

Stress =  $\frac{68}{3.84\times 10^{-7}}$ =  $1.76\times 10^{8}$