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 stress in the two wires, a weight is hang from the rod. The position of the weight along the rod from end A should be.

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

Answer 1

The correct option is A

66.6 cm

Let $T_1$ and $T_2$ be the tensions in brass and steel wire respectively. Let weight ‘w’ be hanged from a point which is at a distance x from A.
Then $T_{1}x=T_2(2-x)-\left(1\right)$
$T_1+T_2=w-\left(2\right)$
As stresses in the wires are same
$\frac{T_1}{T_2}=\frac{A_1}{A_2}=\frac{0.2\times {10}^{-4}}{0.1\times {10}^{-4}}=2\Rightarrow T_1=2T_2-\left(3\right)$
From (2) and (3)
$T_1=\frac{2w}{3}and{T}_2=\frac{w}{3}$
From (1)$\therefore \frac{2w}{3}x=\frac{w}{3}(2-x)\Rightarrow 3x=2\Rightarrow x=0.666m=66.6cm$