Q11-25E - University Physics with Modern Physics

Question

A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 700N is applied to each end of the wire. What minimum diameter is required for the wire?

Step-by-Step Solution

Answer

Minimum diameter is d = 1.9mm

Step 1: Given information:

Force F = 700N, Elongation $∆I=0.25cm$, length l₀ = 2m

Step 2: Concept/Formula used:

$\mathit{Y}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{0}}\mathbf{F}}{\mathbf{A}\mathbf{\Delta }\mathbf{l}}$

Where, Y is Young’s modulus, I₀ is length of muscle, F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathit{l}$ is elongation.

Step 3: Calculation for diameter:

$\begin{array}{rcl}Y& =& \frac{l_{0}F}{A\Delta l}\\ A& =& \frac{l_{0}F}{Y\Delta l}\\ For steel T& =& 2.0\times {10}^{11}Pa\\ A& =& \frac{\left(2.0m\right)\left(700N\right)}{\left(2.0\times {10}^{11}Pa\right)\left(0.25\times {10}^{-2}m\right)}\\ & =& 2.8\times {10}^{-6}{m}^2\\ A& =& \pi r^2\\ r& =& \sqrt{\frac{A}{\pi }}\\ & =& \sqrt{\frac{2.8\times {10}^{-6}{m}^2}{\pi }}\\ & =& 9.44\times {10}^{-4}m\\ d& =& 2r\\ & =& 2\times 9.44\times {10}^{-4}m\\ & =& 1.9mm\end{array}$