A light rod of length L is suspended from a support horizontally by means of two vertical wires A and B of equal length as shown in Fig. The cross sectional area of A is half that of B and the Young's modulus of A is twice that of B. A weight W is hung as shown. The value of x so that W produces equal stress in wires A and B is-----$\frac{L}{3}$

Let T_A and T_B be the tensions in wires A and B respectively. If a_A and a_B are the respective cross-sectional  areas, then

Stress in wire $\mathrm{A}=\frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{a}}_{\mathrm{A}}}$

Stress in wire $\mathrm{B}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{a}}_{\mathrm{B}}}$

The stress in wires A and B will be equal if

$\frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{a}}_{\mathrm{A}}}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{a}}_{\mathrm{B}}}\text{ or }\frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{B}}}=\frac{{\mathrm{a}}_{\mathrm{A}}}{{\mathrm{a}}_{\mathrm{B}}}=\frac{1}{2}\text{ (given). }$

Since the system is in equilibrium, the moments of forces T_A and T_B about C will be equal (see Fig), i.e.

$\begin{array}{l}{\mathrm{T}}_{\mathrm{A}}\times \mathrm{x}={\mathrm{T}}_{\mathrm{B}}\times (\mathrm{L}-\mathrm{x})\\ \frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{T}}_{\mathrm{B}}}=\frac{\mathrm{L}-\mathrm{x}}{\mathrm{x}}\text{ or }\frac{1}{2}=\frac{\mathrm{L}-\mathrm{x}}{\mathrm{x}}\end{array}$

which gives $\mathrm{x}=\frac{2\mathrm{L}}{3}$.