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:

\[\cos\theta = \frac{x}{l}\left\{ 1 - \frac{1}{2}\frac{x^2}{l^2} \right\} \left( \text{ neglecting the higher order terms } \right)\]

\[\text{ Since x} < < \text{l}, \frac{x^2}{\text{l}^2} \text{can be neglected }. \]

\[ \Rightarrow \cos\theta = \frac{x}{\text{l}}\] 

Increase in length:
ΔL = (AC + CB) − AB
AC = (l² + x²)^1/2
ΔL = 2 (l² + x²)^1/2 − 2l

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

From the free body diagram:

\[2T\cos\theta =\text{ mg}\]

\[2T\left( \frac{x}{50} \right) = 2 . 16 \times {10}^3 \times 980\]

\[ \Rightarrow \frac{2 \times \left( 2 \times {10}^{12} \right) \times \left( 4 \times {10}^{- 2} \right) \times \left[ 2\left( {50}^2 + x^2 \frac{1}{2} \right) - 100 \right]x}{100 \times 50} = \left( 2 . 16 \right) \times {10}^3 \times 980\]

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