A light rod of length 200,cm is suspended fro | vedclass

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(A) Let $T_1$ be the tension in the steel wire and $T_2$ be the tension in the brass wire. Let $A_1 = 0.1\,cm^2$ and $A_2 = 0.2\,cm^2$ be their respec

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$\frac{4}{3}\,m$ from steel wire

Vedclass · Answer

(A) Let $T_1$ be the tension in the steel wire and $T_2$ be the tension in the brass wire. Let $A_1 = 0.1\,cm^2$ and $A_2 = 0.2\,cm^2$ be their respective cross-sectional areas.Since the stresses are equal,$\frac{T_1}{A_1} = \frac{T_2}{A_2}$.Substituting the values,$\frac{T_1}{0.1} = \frac{T_2}{0.2}$,which gives $T_2 = 2T_1$.For the translatory equilibrium of the rod,$T_1 + T_2 = W$,where $W$ is the weight hung.Substituting $T_2 = 2T_1$,we get $T_1 + 2T_1 = W$,so $T_1 = \frac{W}{3}$ and $T_2 = \frac{2W}{3}$.Let $x$ be the distance of the weight $W$ from the steel wire. For rotational equilibrium about the point where the weight is hung,the torques must balance: $T_1 x = T_2 (L - x)$,where $L = 200\,cm = 2\,m$.Substituting the values,$\frac{W}{3} x = \frac{2W}{3} (2 - x)$.Dividing by $\frac{W}{3}$,we get $x = 2(2 - x) = 4 - 2x$.Thus,$3x = 4$,which gives $x = \frac{4}{3}\,m$ from the steel wire.

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