Question

A $2m$ long light metal rod $AB$ is suspended from the ceiling horizontally by means of two vertical wires of equal length, tied to its ends. One wire is of brass and has cross-section of $0.2\times {10}^{-4}{m}^2$ and the other is of steel with $0.1\times {10}^4{m}^2$ cross-section. In order to have equal stresses in the two wires, a weight should be hung from the rod from end $A$ a distance of

• A. $66.6cm$
• B. $133cm$
• C. $44.4cm$
• D. $155.6cm$

Answer 1

Answer: (A)

$66.6cm$

$\begin{array}{cc}& \text{Since stress is equal}\to \\ & \begin{array}{cc}\frac{T_1}{0.2\times {10}^{-4}}& =\frac{T_2}{0.1\times {10}^{-4}}\\ \Rightarrow T_1& =2T_2⟶\left(1\right)\end{array}\end{array}$

$\begin{array}{c}\text{Torque about}c\text{should be}0.\sum \tau =0\\ T_{1}x=T_2(2-x)\\ \Rightarrow T_{1}x=2T_2-x{T}_2\\ \Rightarrow x(T_1+T_2)=2T_2\\ \Rightarrow x=\frac{2T_2}{3T_2}=\frac{2}{3}m=66.6cm\end{array}$