A copper wire of length $2.2 \;m$ and a steel wire of length $1.6\; m ,$ both of diameter $3.0 \;mm ,$ are connected end to end. When stretched by a load, the net elongation is found to be $0.70 \;mm$. Obtain the load applied in $N$.
A copper wire of length $2.2 \;m$ and a steel wire of length $1.6\; m ,$ both of diameter $3.0 \;mm ,$ are connected end to end. When stretched by a load, the net elongation is found to be $0.70 \;mm$. Obtain the load applied in $N$.
The copper and steel wires are under a tensile stress because they have the same tension (equal to the load $W$ ) and the same area of cross-section $A$. we have stress $=$ strain $\times$ Young's modulus. Therefore
$W / A=Y_{c} \times\left(\Delta L_{c} / L_{c}\right)=Y_{s} \times\left(\Delta L_{s} / L_{s}\right)$
where the subscripts $c$ and s refer to copper and stainless steel respectively. Or,
$\Delta L_{c} / \Delta L_{s} =\left(Y_{s} / Y_{c}\right) \times\left(L_{c} / L_{s}\right)$
$\text { Given } L_{c} =2.2 m , L_{s}=1.6 m$
$Y_{c}=1.1 \times 10^{11} N . m ^{-2},$ and
$Y_{c}=2.0 \times 10^{11} N \cdot m ^{-2}$
$\Delta L_{c} / \Delta L_{s}=\left(2.0 \times 10^{11} / 1.1 \times 10^{11}\right) \times(2.2 / 1.6)=2.5$
The total elongation is given to be
$\Delta L_{c}+\Delta L_{s}=7.0 \times 10^{-4} m$
Solving the above equations,
$\Delta L_{c}=5.0 \times 10^{-4} m , \quad \text { and } \quad \Delta L_{s}=2.0 \times 10^{-4} m$
Therefore $W=\left(A \times Y_{c} \times \Delta L_{J}\right) / L_{C}$
$=\pi\left(1.5 \times 10^{-3}\right)^{2} \times\left[\left(5.0 \times 10^{-4} \times 1.1 \times 10^{11}\right) / 2.2\right]$
$=1.8 \times 10^{2}\, N$
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