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A light rod of length $200\,cm$ 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 steel and is of cross-section $0.1\,cm^2$ and the other of brass of cross-section $0.2\,cm^2$ . Along the rod at which distance a weight may be hung to produce equal stresses in both the wires?
$\frac {4}{3}\,m$ from steel wire
$\frac {4}{3}\,m$ from brass wire
$1\,m$ from steel wire
$\frac {1}{4}\,m$ from brass wire
Solution
As stresses are equal, $\frac{\mathrm{T}_{1}}{\mathrm{A}_{1}}=\frac{\mathrm{T}_{2}}{\mathrm{A}_{2}}$
$\text { i.e., } \quad \frac{T_{1}}{T_{2}}=\frac{A_{1}}{A_{2}}=\frac{0.1}{0.2} \quad \text { or } T_{2}=2 T_{1} \quad \ldots(1)$
Now for translatory equilibrium of the rod,
$\mathrm{T}_{1}+\mathrm{T}_{2}=\mathrm{W}$ $…(2)$
From $( 1)$ and $( 2)$$: T_{1}=\frac{W}{3} ; T_{2}=\frac{2 W}{3}$
Now if $x$ is the distance of weight $W$ from steel wire, then for rotational equilibrium of rod,
$\mathrm{T}_{1} \mathrm{x}=\mathrm{T}_{2}(2-\mathrm{x})$
$\mathrm{Or}$ $\frac{W}{3} x=\frac{2 W}{3}(2-x)$
$x=\frac{4}{3} m$
