Question

A wire of cross-sectional area $A$, modulus of elasticity $2 \times 10^{11} \, \text{Nm}^{-2}$, and length $2 \, \text{m}$ is stretched between two vertical rigid supports.
When a mass of $2 \, \text{kg}$ is suspended at the middle, it sags lower from its original position making angle $\theta = \frac{1}{100}$ radian on the points of support.  
The value of $A$ is ____ $\times 10^{-4} \, \text{m}^2$ (consider $x \ll l$).  
(Given: $g = 10 \, \text{m/s}^2$)

Answer 1

Correct Answer: 1

The objective is to determine the cross-sectional area \(A\) of a wire supporting a mass, considering its sag. Input parameters include the wire's original length, Young's modulus, the mass's magnitude, and the angle the wire makes with the horizontal at the points of support.

Principles Utilized:

The solution is based on the integration of the following physics concepts:

1. Mechanical Equilibrium: The system is in a state of static equilibrium. The sum of the vertical forces acting on the suspended mass is zero. This implies that the upward vertical components of the tension in both segments of the wire must counteract the downward gravitational force (weight) of the mass.
2. Elasticity (Young's Modulus): Applied tension within the wire induces stretching. Young's modulus (\(Y\)) quantifies this elastic behavior, relating stress (force per unit area) to strain (relative deformation) via the equation: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta L / L_0} \] Here, \(T\) represents tension, \(A\) is the cross-sectional area, \(\Delta L\) denotes the elongation, and \(L_0\) signifies the initial length.
3. Small Angle Approximation: For small angles \(\theta\) expressed in radians, the trigonometric functions \(\sin\theta\) and \(\tan\theta\) can be approximated by \(\theta\). This simplification is instrumental in linking the geometric configuration of the sag to the strain experienced by the wire.

Solution Procedure:

Step 1: Force Analysis and Tension Determination.

The suspended mass, with a value of \(m = 2 \, \text{kg}\), is in equilibrium. Its weight, \(W = mg\), acts vertically downward and is counterbalanced by the combined vertical components of the tension \(T\) acting on the two segments of the wire. The angle each wire segment makes with the horizontal is denoted by \(\theta\).

The equation for vertical equilibrium is:

\[2T \sin\theta = mg\]

Given that the angle \(\theta = \frac{1}{100}\) radian is small, the approximation \(\sin\theta \approx \theta\) is applicable.

\[2T\theta = mg\]

Solving for the tension \(T\):

\[T = \frac{mg}{2\theta} = \frac{2 \, \text{kg} \times 10 \, \text{m/s}^2}{2 \times \frac{1}{100}} = \frac{20}{0.02} = 1000 \, \text{N}\]

Step 2: Geometric Analysis for Strain Calculation.

The initial length of the wire is \(L_0 = 2 \, \text{m}\). Let the half-length be \(L = L_0/2 = 1 \, \text{m}\). Upon suspension of the mass, the wire sags by a vertical distance \(x\), resulting in a new length \(L'\) for each half. A right-angled triangle is formed by the sides \(L\), \(x\), and the hypotenuse \(L'\).

\[L' = \sqrt{L^2 + x^2} = L\sqrt{1 + \left(\frac{x}{L}\right)^2}\]

For a small angle \(\theta\), the approximation \(\tan\theta \approx \theta = \frac{x}{L}\) holds true.

Applying the binomial approximation \((1+u)^n \approx 1+nu\) for small values of \(u\), the length \(L'\) can be approximated as:

\[L' \approx L\left(1 + \frac{1}{2}\left(\frac{x}{L}\right)^2\right) = L + \frac{x^2}{2L}\]

The total elongation of the wire is \(\Delta L = 2L' - 2L = 2(L' - L) = \frac{x^2}{L}\). The strain is then calculated as \(\frac{\Delta L}{L_0} = \frac{x^2/L}{2L} = \frac{x^2}{2L^2}\).

Since \(\frac{x}{L} = \theta\), the strain can be expressed in terms of \(\theta\):

\[\text{Strain} = \frac{\theta^2}{2} = \frac{(1/100)^2}{2} = \frac{1}{2 \times 10^4} = 0.5 \times 10^{-4}\]

Step 3: Cross-sectional Area Calculation Using Young's Modulus.

The relationship defined by Young's Modulus is \(Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\text{Strain}}\).

Rearranging the formula to solve for the area \(A\):

\[A = \frac{T}{Y \times \text{Strain}}\]

The known values are:

• \(T = 1000 \, \text{N}\)
• \(Y = 2 \times 10^{11} \, \text{N/m}^2\)
• \(\text{Strain} = 0.5 \times 10^{-4}\)

Substituting these values into the equation for \(A\):

\[A = \frac{1000}{(2 \times 10^{11}) \times (0.5 \times 10^{-4})} = \frac{10^3}{1 \times 10^7} = 10^{-4} \, \text{m}^2\]

Step 4: Formatting the Final Answer.

The question requires the value of A to be presented in the format _____ \(\times 10^{-4} \, \text{m}^2\).

Our computed area is \(A = 1 \times 10^{-4} \, \text{m}^2\).

Therefore, the numeral to be inserted into the blank space is 1.