A light rod of length 3m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of Aluminium and is of cross-section $6\times {10}^{-3}{m}^2$ and other is of copper of cross section $2\times {10}^{-3}{m}^2$ . Find the position along the rod at which a weight may be hung to produce

i) Equal stresses in both wires and $Y_{A\ell }=7\times {10}^{10}N/m^2$

ii) Equal strain in both wires $Y_{cu}=11\times {10}^{10}N/m^2$

i) Equal stress

$a_1=6\times {10}^{-3}{m}^2$

 Here $a_2=2\times {10}^{-3}{m}^2$

$$\begin{gathered} \Rightarrow \frac{T_1}{a_1}=\frac{T_2}{a_2} \\ \Rightarrow \frac{T_1}{T_2}=\frac{a_1}{a_2}=\frac{6\times {10}^{-3}}{2\times {10}^{-3}}=3 \\ \therefore T_1=3T_2\mathrm{................}\left(1\right) \end{gathered}$$

In equilibrium, moment of all the forces acting on the rod about a point is zero.

Let the centre of gravity acts at a distance of x from string 1 and 3-x from string 2 

$$\begin{gathered} T_{1}x=T_2(3-x)----\left(2\right) \\ from \left(1\right) and \left(2\right) \\ 3T_{2}x=T_2(3-x) \\ 3x=3-x \\ 4x=3 \\ x=\frac{3}{4}=0.75m \end{gathered}$$ 

ii) Equal strain, $e_2=e_2$

$$\begin{gathered} \frac{T_1}{a_1{y}_1}=\frac{T_2}{a_2{y}_2}, \\ \frac{T_1}{T_2}=\frac{a_1{y}_1}{a_2{y}_2}=\frac{6\times {10}^{-3}\times 7\times {10}^{10}}{2\times {10}^{-3}\times 11\times {10}^{10}}\Rightarrow \frac{T_1}{T_2}=\frac{21}{11}\mathrm{....................}\left(3\right) \end{gathered}$$

From (2) and (3) we get

 $$\begin{gathered} T_{1}x=T_2(3-x) \\ \Rightarrow \frac{T_1}{T_2}=\frac{(3-x)}{x} \\ \Rightarrow \frac{21}{11}=\frac{(3-x)}{x} \\ \Rightarrow 21x=33-11x \\ \Rightarrow 32x=33 \\ \Rightarrow x=\frac{33}{32}=1.03m \end{gathered}$$