Q11-29E - University Physics with Modern Physics

Question

In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 ^* 10^-3 cm² in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0kg. For each wire, compute (a) the tensile strain and (b) the elongation.

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right) Upper wire: 3.1\times {10}^{-3} \\ Lower wire: 2.0\times {10}^{-3} \\ \left(b\right) Upper wire: 1.6mm \\ Lower wire: 1.0mm \end{gathered}$$

Step 1: Given information

$Length I_0=0.50.0m, Area A=2.5\times {10}^{-3} c{m}^2, Aluminum sphere mass m_A=6kg, Brass cube mass m_B=10kg$

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: Tension calculation in wires

Tension in the below steel wire

$\begin{array}{rcl}{T}_2& =& m_{B}g\\ & =& \left(10kg\right)\left(9.80\right)\\ & =& 98N\end{array}$

Tension in the above steel wire

$\begin{array}{rcl}{T}_1& =& m_{A}g+T_2\\ & =& \left(6kg\right)\left(9.80\right)+98N\\ & =& 157N\end{array}$

Step 4: Calculation for Tensile strain

(a) Strain in upper wire

 $\begin{array}{rcl}Y& =& \frac{stress}{strain}\\ strain\left({\epsilon }_U\right)& =& \frac{stress}{Y}=\frac{T_1}{AY}\\ {\epsilon }_U& =& \frac{157N}{\left(2.5\times {10}^{-7}{m}^2\right)\left(2\times {10}^{11}Pa\right)}\\ & =& 3.1\times {10}^{-3}\end{array}$

Strain in lower wire

$\begin{array}{rcl}strain\left({\epsilon }_L\right)& =& \frac{stress}{Y}=\frac{T_2}{AY}\\ {\epsilon }_L& =& \frac{98N}{\left(2.5\times {10}^{-7}{m}^2\right)\left(2\times {10}^{11}Pa\right)}\\ & =& 2.0\times {10}^{-3}\end{array}$

Step 5: Calculation for elongation in wires

(b) Elongation in upper wire

$\begin{array}{rcl}\Delta l_U& =& l_0\times {\epsilon }_U\\ & =& \left(0.50\times m\right)\left(3.1\times {10}^{-3}\right)\\ & =& 1.6\times {10}^{-3}m=1.6mm\end{array}$ 

Elongation in Lower wire

$\begin{array}{rcl}\Delta l_L& =& l_0\times {\epsilon }_L\\ & =& \left(0.50\times m\right)\left(2.0\times {10}^{-3}\right)\\ & =& 1.0\times {10}^{-3}m=1.0mm\end{array}$