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 problem asks for the cross-sectional area \(A\) of a wire that sags under the weight of a suspended mass. We are given the wire's initial length, its Young's modulus, the suspended mass, and the angle the wire makes with the horizontal at the supports.

Concept Used:

The solution involves combining two main physics principles:

1. Mechanical Equilibrium: The system is in static equilibrium. The vertical components of the tension in the two segments of the wire must balance the gravitational force (weight) of the suspended mass.
2. Elasticity (Young's Modulus): The tension in the wire causes it to stretch. The relationship between stress, strain, and Young's modulus (\(Y\)) is given by: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta L / L_0} \] where \(T\) is the tension, \(A\) is the cross-sectional area, \(\Delta L\) is the elongation, and \(L_0\) is the original length.
3. Small Angle Approximation: For a small angle \(\theta\) (in radians), \(\sin\theta \approx \tan\theta \approx \theta\). This approximation is used to relate the geometry of the sag to the strain.

Step-by-Step Solution:

Step 1: Analyze the forces and find the tension (\(T\)).

The suspended mass \(m = 2 \, \text{kg}\) is in equilibrium. Its weight \(W = mg\) acts downwards and is balanced by the vertical components of the tension \(T\) from the two halves of the wire. Let \(\theta\) be the angle each half of the wire makes with the horizontal.

The equilibrium equation in the vertical direction is:

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

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

\[ 2T\theta = mg \]

We can now solve 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: Analyze the geometry to find the strain.

The initial length of the wire is \(L_0 = 2 \, \text{m}\). Let's denote the half-length as \(L = L_0/2 = 1 \, \text{m}\). When the mass is suspended, the wire sags by a distance \(x\) and each half has a new length \(L'\). From the diagram, we have a right-angled triangle with sides \(L\), \(x\), and hypotenuse \(L'\).

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

For a small angle \(\theta\), we have \(\tan\theta \approx \theta = \frac{x}{L}\).

Using the binomial approximation \((1+u)^n \approx 1+nu\) for small \(u\), we can find \(L'\):

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

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

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

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

Step 3: Use Young's Modulus to calculate the area (\(A\)).

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

Rearranging for the area \(A\):

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

We have:

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

Substituting these values:

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

Step 4: Express the answer in the required format.

The problem asks for the value of A in the form _____ \(\times 10^{-4} \, \text{m}^2\).

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

Therefore, the value to be filled in the blank is 1.

Answer 2

In the vertical direction:
\[2T \sin \theta = 20\]
Using the small angle approximation \( \sin \theta \approx \theta \):
\[\theta = \frac{1}{100} \implies T = \frac{10}{\theta} = 1000 \, \text{N}\]
The change in length \( \Delta L \) is given by:
\[\Delta L = 2\sqrt{x^2 + L^2} - 2L\]
\[\Delta L \approx 2L \left( \frac{x^2}{2L^2} \right) = \frac{x^2}{L}\]
Modulus of elasticity \( E \) is defined as:
\[E = \frac{\text{stress}}{\text{strain}}\]
Substitute \( E = 2 \times 10^{11} \, \text{Nm}^{-2} \):
\[2 \times 10^{11} = \frac{10^3}{A} \times \frac{x^2}{L} \times 2L\]
Solve for \( A \):
\[A = 1 \times 10^{-4} \, \text{m}^2\]