Q25 E - 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 700 N 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 = 1.9 mm .

Step 1: Given information:

Force: F= 700 N , 

Elongation: $∆l=0.25 cm$ ,

length: $l_0=2 m$ .

Step 2: Concept/Formula used:

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

Where, Y is Young’s modulus, ${\mathbf{l}}_{\mathbf{0}}$ is length of muscle, F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathbf{l}$ is elongation.

Step 3: Calculation for diameter:

$$\begin{gathered} Y=\frac{I_{0}F}{∆l} \\ A=\frac{I_{0}F}{∆l} \end{gathered}$$

For steel: $Y=2.0\times {10}^{11} Pa$

$$\begin{gathered} A=\frac{(2.0 m)(700 N)}{(2.0\times {10}^{11} Pa)(0.25\times {10}^{-2} m)} \\ =2.8\times {10}^{-6} m^2 \end{gathered}$$

$$\begin{gathered} A=\pi r^2 \\ r=\sqrt{\frac{A}{\mathrm{\pi }}} \\ =\sqrt{\frac{2.8\times {10}^{-6} m^2}{\mathrm{\pi }}} \\ =9.44\times {10}^{-4} m \end{gathered}$$ 

$$\begin{gathered} d=2r \\ =2\times 9.44\times {10}^{-4} m \\ =1.9 mm \end{gathered}$$