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A rod of length $l$ and negligible mass is suspended at its two ends by two wires of steel (wire $A$ ) and aluminium (wire $B$ ) of equal lengths (figure).
The cross-sectional areas of wires $A$ and $B$ are $1.0 \mathrm{~mm}^2$ and $2.0 \mathrm{~mm}^2$, respectively.
$\left(Y_{\mathrm{Al}}=70 \times 10^9 \mathrm{Nm}^{-2}\right.$ and $\left.Y_{\text {steel }}=200 \times 10^9 \mathrm{Nm}^{-2}\right)$
Options:
The cross-sectional areas of wires $A$ and $B$ are $1.0 \mathrm{~mm}^2$ and $2.0 \mathrm{~mm}^2$, respectively.
$\left(Y_{\mathrm{Al}}=70 \times 10^9 \mathrm{Nm}^{-2}\right.$ and $\left.Y_{\text {steel }}=200 \times 10^9 \mathrm{Nm}^{-2}\right)$
Solution:
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Verified Answer
The correct answers are:
Mass $m$ should be suspended close to $B$ to have equal stresses in both the wires
,
Mass $m$ should be suspended close to wire $A$ to equal strain in both wires
Mass $m$ should be suspended close to $B$ to have equal stresses in both the wires
,
Mass $m$ should be suspended close to wire $A$ to equal strain in both wires
Consider the given diagram, and a rod is balanced by weight $(m g)$, i.e. (rod remain horizontal)
Let the mass is placed at $x$ from the end $B$
Let $T_A$ and $T_B$ be the tensions in wire $A$ and wire $B$ respectively.
For the rotational equilibrium of the system, $\Sigma \tau=0 \quad($ Total torque $=0$ )
$T_B(x)-T_A(l-x)=0$ (taking moment about point A) $\frac{T_B}{T_A}=\frac{l-x}{x}$
Stress in wire $A=S_A=\frac{T_A}{a_A}$
Stress in wire $B=S_B=\frac{T_B}{a_B}$
where $a_A$ and $a_B$ are cross-sectional areas of wire $A$ and $B$ respectively.
$$
a_B=2 a_A \quad \text { (given) }
$$
As the stress on steel $\left(S_A\right)=$ Stress on $A 1\left(S_B\right)$
So, $S_A=S_B$
$$
\begin{aligned}
&\frac{T_A}{a_A}=\frac{T_B}{a_B} \Rightarrow \frac{T_B}{T_A}=\frac{a_B}{a_A}=2 \\
&\frac{l-x}{x}=\frac{2}{1} \Rightarrow \frac{l}{x}-1=2 \Rightarrow x=\frac{l}{3} \quad(\text { from } B)
\end{aligned}
$$
distance of $m$ from $A$
$$
\Rightarrow \quad l-x=l-l / 3=\frac{2 l}{3}
$$
Hence, mass $m$ should be placed closer to $B$.
(b) Strain : Strain in both wire is equal because rod remain horizontal in balanced condition.
For equal strain,
$(\text { strain })_A=(\text { strain })_B$
$$
\frac{Y_A}{S_A}=\frac{Y_B}{S_B}
$$
(where $Y_A$ and $Y_B$ are Young modulii)
$$
\begin{aligned}
&\frac{Y_{\text {stecl }}}{T_A / a_A}=\frac{Y_{\mathrm{Al}}}{T_B / a_B} \\
&\frac{Y_{\text {steel }}}{Y_{\mathrm{Al}}}=\frac{T_A}{T_B} \times \frac{a_B}{a_A}=\left(\frac{x}{1-x}\right)\left(\frac{2 a_A}{a_A}\right) \\
&\frac{200 \times 10^9}{70 \times 10^9}=\frac{2 x}{1-x} \Rightarrow \frac{20}{7}=\frac{2 x}{1-x} \\
&\Rightarrow 14 x=20 l-20 x \Rightarrow 34 x=20 l \\
&\text { or } 17 x=10 l \Rightarrow x=\frac{10 l}{17} \quad \text { (from } B \text { ) } \\
&(l-x) \text { from } A=\left[l-\frac{10 l}{17}\right]=\frac{7 l}{17} \quad \text { (from } A \text { ) }
\end{aligned}
$$
Hence for equal strain, mass $m$ should be closer to wire $A$.
Let the mass is placed at $x$ from the end $B$
Let $T_A$ and $T_B$ be the tensions in wire $A$ and wire $B$ respectively.
For the rotational equilibrium of the system, $\Sigma \tau=0 \quad($ Total torque $=0$ )
$T_B(x)-T_A(l-x)=0$ (taking moment about point A) $\frac{T_B}{T_A}=\frac{l-x}{x}$
Stress in wire $A=S_A=\frac{T_A}{a_A}$
Stress in wire $B=S_B=\frac{T_B}{a_B}$
where $a_A$ and $a_B$ are cross-sectional areas of wire $A$ and $B$ respectively.
$$
a_B=2 a_A \quad \text { (given) }
$$
As the stress on steel $\left(S_A\right)=$ Stress on $A 1\left(S_B\right)$
So, $S_A=S_B$
$$
\begin{aligned}
&\frac{T_A}{a_A}=\frac{T_B}{a_B} \Rightarrow \frac{T_B}{T_A}=\frac{a_B}{a_A}=2 \\
&\frac{l-x}{x}=\frac{2}{1} \Rightarrow \frac{l}{x}-1=2 \Rightarrow x=\frac{l}{3} \quad(\text { from } B)
\end{aligned}
$$
distance of $m$ from $A$
$$
\Rightarrow \quad l-x=l-l / 3=\frac{2 l}{3}
$$
Hence, mass $m$ should be placed closer to $B$.
(b) Strain : Strain in both wire is equal because rod remain horizontal in balanced condition.
For equal strain,
$(\text { strain })_A=(\text { strain })_B$
$$
\frac{Y_A}{S_A}=\frac{Y_B}{S_B}
$$
(where $Y_A$ and $Y_B$ are Young modulii)
$$
\begin{aligned}
&\frac{Y_{\text {stecl }}}{T_A / a_A}=\frac{Y_{\mathrm{Al}}}{T_B / a_B} \\
&\frac{Y_{\text {steel }}}{Y_{\mathrm{Al}}}=\frac{T_A}{T_B} \times \frac{a_B}{a_A}=\left(\frac{x}{1-x}\right)\left(\frac{2 a_A}{a_A}\right) \\
&\frac{200 \times 10^9}{70 \times 10^9}=\frac{2 x}{1-x} \Rightarrow \frac{20}{7}=\frac{2 x}{1-x} \\
&\Rightarrow 14 x=20 l-20 x \Rightarrow 34 x=20 l \\
&\text { or } 17 x=10 l \Rightarrow x=\frac{10 l}{17} \quad \text { (from } B \text { ) } \\
&(l-x) \text { from } A=\left[l-\frac{10 l}{17}\right]=\frac{7 l}{17} \quad \text { (from } A \text { ) }
\end{aligned}
$$
Hence for equal strain, mass $m$ should be closer to wire $A$.
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