Hooke’s Law for a Wire. A wire of length${\mathit{l}}_{\mathit{0}}$and cross-sectional areasupports a hanging weight W.

1. Show that if the wire obeys Eq. (11.7), it behaves like a spring of force constant$\frac{\mathbf{A}\mathbf{Y}}{{\mathbf{l}}_{\mathbf{0}}}$, where Yis Young’s modulus for the wire material.
2. What would the force constant be for alength of 16-gauge (diameter = 1.291 mm) copper wire? See Table 11.1.
3. What wouldhave to be to stretch the wire in part (b) by 1.25 mm?

Consider the formulas:

$\begin{array}{r}\mathbf{Y}\mathbf{=}\frac{\mathbf{\text{Tensile stress}}}{\mathbf{\text{Tensile strain}}}\\ \mathbf{=}\frac{{\mathbf{F}}_{\mathbf{\perp }}\mathbf{/}\mathbf{A}}{\mathbf{\Delta }\mathbf{I}\mathbf{/}{\mathbf{I}}_{\mathbf{0}}}\\ \mathbf{=}\frac{{\mathbf{F}}_{\mathbf{\perp }}}{\mathbf{\Delta }\mathbf{V}}\frac{{\mathbf{I}}_{\mathbf{0}}}{\mathbf{A}}\end{array}$

Length of wire is $l_0$

Cross sectional area of object is A

Weight of object is W

(a)

Here we have to prove:$k=\frac{YA}{I_0}$

Our goal is to express$\frac{F_{\perp }}{∆l}$because that is the force constant which we can see in formula (11.7)

Now, by formula (11.7) which is given by,

$\mathrm{Hooke}\text{'}\mathrm{s}\mathrm{law}=\frac{\mathrm{Stress}}{\mathrm{Strain}}$

Here, stress is measure of forces applied to body. Strain is measure of how much deformation results from stress.

Now, from equation (1), we have,

$\begin{array}{c}Y=\frac{F_{\perp }}{A}\frac{I_0}{\mathrm{\Delta }I}\\ \frac{YA}{I_0}=\frac{F_{\perp }}{\mathrm{\Delta }I}\\ \mathrm{Now},\mathrm{we}\mathrm{know}\mathrm{that}\frac{{\mathrm{F}}_{\perp }}{∆\mathrm{l}}=\mathrm{k}\mathrm{So},\mathrm{above}\mathrm{equation}\mathrm{becomes}\\ k=\frac{YA}{I_0}\end{array}$

Hence proved.

(b)

From part (a) we have,

$k=\frac{YA}{I_0}$

Now, for copper wire above equation becomes

$k=\frac{Y_{Cu}A}{I_0}$ (2)

Now, $Y_{Cu}=11\times {10}^{10}$

Determine the area:

$\begin{array}{r}A=\pi r^2\\ =\pi \left(\frac{1.29\times {10}^{-3}}{2}\right)\\ =1.31\times {10}^{-6}{\mathrm{m}}^2\end{array}$

Now, by substituting all the numerical values in equation (2).

$\begin{array}{r}k=\frac{\left(11\times {10}^{10}\right)\left(1.31\times {10}^{-6}{\mathrm{m}}^2\right)}{0.75\mathrm{m}}\\ =1.92\times {10}^5\frac{\mathrm{N}}{\mathrm{m}}\end{array}$

Hence, the force constant is$1.92\times {10}^5\frac{\mathrm{N}}{\mathrm{m}}$.

(c)

To find the weight W for given$∆l$is given by,

$\begin{array}{r}k=\frac{W}{∆l}\\ W=k\mathrm{\Delta }l\\ W=\left(1.92\times {10}^5\mathrm{N}/\mathrm{m}\right)\cdot \left(1.25\times {10}^{-3}\mathrm{m}\right)\\ W=2.4\times {10}^2\end{array}$

Hence, Weight is$2.4\times {10}^2\mathrm{N}$ .