Question

A steel wire of original length 1 m and cross-sectional area 4.00 mm² is clamped at the two ends so that it lies horizontally and without tensions. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?
Y of the steel = 2.0 × 10¹¹ N m⁻². Take g = 10 m s⁻².

Answer 1

Given:
Original length of steel wire (L) = 1 m
Area of cross-section (A) = 4.00 mm² = 4 × 10⁻² cm²
Load = 2.16 kg
Young's modulus of steel (Y) = 2 × 10¹¹ N/m²
Acceleration due to gravity (g) = 10 m s⁻²

Let T be the tension in the string after the load is suspended and θ be the angle made by the string with the vertical, as shown in the figure:


$\cos \theta =\frac{x}{\sqrt{x^2+l^2}}=\frac{x}{l}{\left\{1+\frac{x^2}{l^2}\right\}}^{-1/2}$
Expanding the above equation using the binomial theorem:
$$\begin{gathered} \cos \theta =\frac{x}{l}\left\{1-\frac{1}{2}\frac{x^2}{l^2}\right\}\left(\mathrm{neglecting}\mathrm{the}\mathrm{higher}\mathrm{order}\mathrm{terms}\right) \\ \mathrm{Since}x<<l,\frac{x^2}{l^2}\mathrm{can}\mathrm{be}\mathrm{neglected}. \\ \Rightarrow \cos \theta =\frac{x}{l} \end{gathered}$$
Increase in length:
ΔL = (AC + CB) − AB
AC = (l² + x²)^1/2
ΔL = 2 (l² + x²)^1/2 − 2l

We know that:
$$\begin{gathered} Y=\frac{F}{A}\frac{L}{∆L} \\ \Rightarrow 2\times {10}^{12}=\frac{T\times 100}{\left(4\times {10}^{-2}\right)\times \left[2{\left({50}^2+x^2\right)}^{1/2}-100\right]} \end{gathered}$$

From the free body diagram:
$$\begin{gathered} 2T\cos \theta =mg \\ 2T\left(\frac{x}{50}\right)=2.16\times {10}^3\times 980 \\ \Rightarrow \frac{2\times \left(2\times {10}^{12}\right)\times \left(4\times {10}^{-2}\right)\times \left[2\left({50}^2+x^2{\displaystyle \frac{1}{2}}\right)-100\right]x}{100\times 50}=\left(2.16\right)\times {10}^3\times 980 \end{gathered}$$

On solving the above equation, we get x = 1.5 cm.
Hence, the required vertical depression is 1.5 cm.