Q11-26E - University Physics with Modern Physics

Question

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right) Steel: 1.1\times {10}^{-4} \\ Copper: 2.1\times {10}^{-4} \\ \left(b\right) Steel:8.3\times {10}^{-5}m \\ Copper: 1.6\times {10}^{-4}m \end{gathered}$$

Step 1: Given information

Tensile force F = 400 N, l₀ = 0.75m

Area$\begin{array}{rcl}A& =& \pi \left(\frac{d^2}{4}\right)\\ & =& \pi \left(\frac{{\left(1.50\times {10}^{-2}m\right)}^2}{4}\right)\\ & =& 1.77\times {10}^{-4}{m}^2\end{array}$ 

Step 2: Concept/Formula used

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

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: Strain Calculation

(a) The strain is $\frac{\Delta l}{l_0}=\frac{F}{YA}$

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

$\begin{array}{rcl}\frac{\Delta l}{l_0}& =& \frac{4000N}{\left(2.0\times {10}^{11}Pa\right)\left(1.77\times {10}^{-4}{m}^2\right)}\\ & =& 1.1\times {10}^{-4}\end{array}$

$\begin{array}{rcl}For copper Y& =& 1.1\times {10}^{11}Pa\\ \frac{\Delta l}{l_0}& =& \frac{4000N}{\left(1.1\times {10}^{11}Pa\right)\left(1.77\times {10}^{-4}{m}^2\right)}\\ & =& 2.1\times {10}^{-4}\end{array}$

Step 4: Elongation Calculation

(b)

For steel:

$\left(1.1\times {10}^{-4}\right)\left(0.75\right)=8.3\times {10}^{-5}m$

For Copper:

$\left(2.1\times {10}^{-4}\right)\left(0.75\right)=1.6\times {10}^{-4}m$