Question

How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?

Answer 1

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

• diameter of the steel, d = 4 mm

• the radius of the wire, r = 2mm = 0.002 m

• original length of the wire, L₁

• final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)

• extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁

• the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

$A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2$

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}$

$F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N$

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915