A rubber string 10m long is suspended from a rigid support at its one end. Calculate the extension in the string due to its own weight. The density of rubber is

Extension of Rubber String Due to Its Own Weight

A rubber string 10m long is suspended from a rigid support at its one end. Calculate the extension in the string due to its own weight. The density of rubber is \(1.5 \times 10^3 \, \text{kg/m}^3\) and Young's modulus for the rubber is \(5 \times 10^6 \, \text{N/m}^2\). Take \(g = 10 \, \text{m/s}^2\).

Solution:

To calculate the extension of the rubber string due to its own weight, we use the formula for Young's modulus:

\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{F}{A}}{\frac{\Delta L}{L}} \]

Rearranging for the extension \( \Delta L \):

\[ \Delta L = \frac{F L}{A Y} \]

The weight of the string (Force \( F \)) is given by:

\[ F = \text{mass} \times g \]

The mass of the string is:

\[ \text{mass} = \rho \times \text{volume} = \rho \times A \times L \]

Thus, the weight \( F \) is:

\[ F = \rho \times A \times L \times g \]

Substituting into the formula for extension:

\[ \Delta L = \frac{\rho \times A \times L \times g \times L}{A \times Y} \]

Simplifying the expression:

\[ \Delta L = \frac{\rho \times g \times L^2}{Y} \]

Now, we substitute the given values:

• Density, \( \rho = 1.5 \times 10^3 \, \text{kg/m}^3 \)
• Length, \( L = 10 \, \text{m} \)
• Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)
• Young's modulus, \( Y = 5 \times 10^6 \, \text{N/m}^2 \)

Substituting these values into the equation:

\[ \Delta L = \frac{(1.5 \times 10^3) \times 10 \times (10)^2}{5 \times 10^6} \]

\[ \Delta L = \frac{1.5 \times 10^3 \times 100 \times 10}{5 \times 10^6} \]

\[ \Delta L = \frac{1.5 \times 10^6}{5 \times 10^6} = 0.3 \, \text{m} \]

Therefore, the extension in the rubber string due to its own weight is:

\[ \Delta L = 0.3 \, \text{m} \]