Question

A cylindrical aluminum rod, with an initial length of $\mathbf{0}\mathbf{.8000}\mathbf{\text{\hspace{0.17em}m}}$ and radius$\mathbf{1000.0}\mathbf{\text{\hspace{0.17em}μm}}$ , is clamped in place at one end and then stretched by a machine pulling parallel to its length at its other end. Assuming that the rod’s density (mass per unit volume) does not change, find the force magnitude that is required of the machine to decrease the radius to$\mathbf{999}\mathbf{.9}\mathbf{\text{\hspace{0.17em}μm}}$ . (The yield strength is not exceeded.)

Short answer

The magnitude of force applied on the cylindrical rod by machine is $44\mathrm{N}$.

Step-by-step solution

Understanding the given information

The initial length of the rod,$\text{L}=0.8000\text{\hspace{0.17em}m}$

The initial radius of the rod,$\text{r}=1000.0\text{\hspace{0.17em}μm}\left(\frac{{10}^{-6}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}μm}}\right)=1000.0\times {10}^{-6}\text{m}$

The final radius of the rod, $=999.9\text{\hspace{0.17em}μm}\left(\frac{{10}^{-6}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}μm}}\right)=999.9\times {10}^{-6}\text{\hspace{0.17em}m}$

Concept and formula used in the given question

You can use the concept of modulus of elasticity, and expression of density and volume of the rod. The formulas used are given below.

$\begin{array}{c}\mathbf{\rho }\mathbf{=}\frac{\mathbf{M}}{\mathbf{V}}\\ \mathbf{V}\mathbf{=}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{h}\\ \frac{\mathbf{F}}{\mathbf{A}}\mathbf{=}\mathbf{E}\frac{\mathbf{\Delta L}}{\mathbf{L}}\end{array}$

Calculation for the force magnitude that is required of the machine to decrease the radius to  999.9 mm

One end of the rod is clamped and the other end is stretched by the machine in such a way that the rod’s density remains the same after elongation of the rod.

The expression of the density of the rod is

$\begin{array}{l}\mathrm{\rho }=\frac{\mathrm{M}}{\mathrm{V}}\\ \mathrm{V}=\frac{\mathrm{M}}{\mathrm{\rho }}=\mathrm{constant}\\ \mathrm{V}={\mathrm{\pi r}}^2\mathrm{L}=\mathrm{constant}\end{array}$

The expression of the volume of the rod before and after elongation is:

$\begin{array}{c}{\mathrm{\pi r}}^2\mathrm{L}=\mathrm{\pi }{{\mathrm{r}}^{\text{'}}}^2\mathrm{L}\text{'}\\ {\mathrm{r}}^2\mathrm{L}={{\mathrm{r}}^{\text{'}}}^2\mathrm{L}\text{'}\\ {\mathrm{L}}^{\text{'}}=\frac{{\mathrm{r}}^2\mathrm{L}}{{{\mathrm{r}}^{\text{'}}}^2}\end{array}$

Change in length of the rod due to the applied force by the machine is:

$\begin{array}{c}\mathrm{\Delta L}={\mathrm{L}}^{\text{'}}-\mathrm{L}\\ =\frac{{\mathrm{r}}^2\mathrm{L}}{{{\mathrm{r}}^{\text{'}}}^2}-\mathrm{L}\\ =\mathrm{L}\left(\frac{{\mathrm{r}}^2}{{{\mathrm{r}}^{\text{'}}}^2}-1\right)\end{array}$

The expression of Young’s modulus of elasticity is:

$\begin{array}{c}\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{E}\frac{\mathrm{\Delta L}}{\mathrm{L}}\\ \mathrm{F}=\mathrm{AE}\frac{\mathrm{\Delta L}}{\mathrm{L}}\\ \mathrm{F}={\mathrm{\pi r}}^2\mathrm{E}\frac{\mathrm{L}\left(\frac{{\mathrm{r}}^2}{{\mathrm{r}}^{\text{'}2}}-1\right)}{\mathrm{L}}\\ \mathrm{F}=3.14\times {(1000\times {10}^{-6}\mathrm{m})}^2\times 70\times {10}^9{\text{\hspace{0.17em}N/m}}^{\text{2}}\times \left[\frac{{(1000\times {10}^{-6}\mathrm{m})}^2}{{(999.9\times {10}^{-6}\mathrm{m})}^2}-1\right]\\ \mathrm{F}=44\text{\hspace{0.17em}N}\end{array}$