Q26 E - 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

(a) Steel : $1.1\times {10}^{-4}$

      Copper : $2.1\times {10}^{-4}$

(b) Steel : $8.3\times {10}^{-5} m$

      Copper: $1.6\times {10}^{-4} m$

Step 1: Given information

Tensile force: F = 4000 N,

 $l_0 = 0.75 m$

$$\begin{gathered} Area=\left(\frac{d^2}{4}\right) \\ =\left(\frac{{(1.50\times {10}^{-2} m)}^2}{4}\right) \\ =1.77\times {10}^{-4} m^2 \end{gathered}$$

Step 2: Concept/Formula used

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

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

Step 3: (a) Strain Calculation

The strain is : $\frac{∆l}{l_0}=\frac{F}{YA}$

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

$$\begin{gathered} \frac{∆l}{l_0}=\frac{4000 N}{(2.0\times {10}^{11} Pa)(1.77\times {10}^{-4} m^2)} \\ =1.1\times {10}^{-4} \end{gathered}$$

For copper : $Y=1.1\times {10}^{11} Pa$

$$\begin{gathered} \frac{∆l}{l_0}=\frac{4000 N}{(1.1\times {10}^{11} Pa)(1.77\times {10}^{-4} m^2)} \\ =2.1\times {10}^{-4} \end{gathered}$$

Step 4: Elongation Calculation

For steel:

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

For Copper:

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