Q29 E - 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 $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}$ 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 10.0 kg. For each wire, compute (a) the tensile strain and (b) the elongation.

Step-by-Step Solution

Answer

(a) Upper wire: $3.1\times {10}^{-3}$ 

     Lower wire: $2.0\times {10}^{-3}$

(b) Upper wire: $1.6 mm$

      Lower wire: $1.0 mm$

Step 1: Given information

Length: $l_0=0.50 m$,

Area: $A=2.5\times {10}^{-3} c{m}^2$,

Aluminum sphere mass: $m_A=6 kg$,

Brass cube mass: $m_B=10 kg$ .

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: Tension calculation in wires

Tension in the below steel wire 

$$\begin{gathered} T_2=m_{B}g \\ =(10 kg)(9.80) \\ =98 N \end{gathered}$$

Tension in the above steel wire

$$\begin{gathered} T_1=m_{A}g+T_2 \\ =(6 kg)(9.80)+98 N \\ =157 N \end{gathered}$$

Step 4: (a) Calculation for Tensile strain

Strain in upper wire

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

Strain in lower wire

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

Step 5: (b) Calculation for elongation in wires

(b) Elongation in upper wire

$$\begin{gathered} ∆l_u=l_0\times {\epsilon }_u \\ =(0.50 m)(3.1\times {10}^{-3}) \\ =1.6\times {10}^{-3} m \\ =1.6 mm \end{gathered}$$

Elongation in Lower wire

$$\begin{gathered} ∆l_L=l_0\times {\epsilon }_L \\ =(0.50 m)(2.0\times {10}^{-3}) \\ =1.0\times {10}^{-3} m \\ =1.0 mm \end{gathered}$$