Question

Two wires of diameter $0.25cm$, one made of steel and the other made of brass are loaded as shown in $\text{figure}$. The unloaded length of steel wire is $1.5m$ and that of brass wire is $1.0m$. Compute the elongations of the steel and the brass wires.

Answer 1

Find total force exerted on the steel wire and brass wire.

Total force exerted on the steel wire,
$F_1=$ gravitational force due to both the blocks.
$=(4+6)g$
$=10\times 9.8$
$=98N$

Total force exerted on brass wire,
$F_2=$ gravitational force due to $6kg$ block
$=6g$
$=6\times 9.8$
$=\mathrm{5.8.8}N$
Find area of cross section of steel wire and brass wire.

Given, diameter of both wires, $d=0.25cm$
So, radius of the wires,$r=\frac{d}{2}$
$r=\frac{0.25}{2}$
$=0.125cm$
$=1.25\times {10}^{-3}m$
Area of cross section of wire
$A_1=A_2=A=\pi r^2$
$=3.14(1.25\times {10}^{-3}{)}^2$
$=4.9\times {10}^{-6}{m}^2$

Find elongation of the steel wire.

Let the elongation of the steel wire be $\Delta L_1$.
Given, Length of steel wire, $L_1=1.5m$
Young’s modulus for steel wire, $Y_1=2\times {10}^{11}N/m^2$
As we know that Young’s modulus,

$Y_1=\frac{\frac{F_1}{A}}{\frac{\Delta L_1}{L_1}}$​
$\Rightarrow Y_1=\frac{F_1}{A}\times \frac{L_1}{\Delta L_1}$
$\Rightarrow \Delta L_1=\frac{F_1\times L_1}{Y_1\times A}$​​​​​​

Using given values, we get

$\Delta L_1=\frac{98\times 1.5}{2\times {10}^{11}\times 4.9\times {10}^{-6}}$

$=1.5\times {10}^{-4}m$
Find elongation of the brass wire.

Let elongation for brass wire be $\Delta L_2$
Given, length of brass wire, $L_2=1m$
Young’s modulus for brass wire, $Y_2=0.91\times {10}^{11}N/m^2$

So, elongation $\Delta L_2=\frac{F_2\times L_2}{Y_2\times A}$

$\Delta L_2=\frac{58.8\times 1}{0.91\times {10}^{11}\times 4.9\times {10}^{-6}}$

$\Delta L_2=1.3\times {10}^{-4}m$
$\text{Final Answer}:1.5\times {10}^{-4}m\text{(steel)},1.3\times {10}^{-4}m\text{(brass)}$