Question

A rubber cord \(10 \mathrm{~m}\) long is suspended vertically. How much does it stretch under its own weight. [Density of rubber is \(1500\left(\mathrm{~kg} / \mathrm{m}^{3}\right), \mathrm{Y}=5 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\), \(\left.\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (A) \(15 \times 10^{-4} \mathrm{~m}\) (B) \(7.5 \times 10^{-4} \mathrm{~m}\) (C) \(12 \times 10^{-4} \mathrm{~m}\) (D) \(25 \times 10^{-4} \mathrm{~m}\)

Short answer

The stretch of the rubber cord under its own weight is \(3 \times 10^{-3} \mathrm{~m}\).

Step-by-step solution

Find the Volume of the Rubber Cord

To find the volume of the rubber cord, we need to assume a uniform cylindrical shape, with length L and cross-sectional area A. As we are only given the length of the cord (10 meters), but not the thickness, we cannot determine the exact volume with the given data. We can find the expression for the mass of the rubber cord in terms of unknown cross-sectional area A, using the given density of rubber (ρ = 1500 kg/m³). Thus, the volume (V) of the cord can be represented as: \[ V = AL \]

Find the Mass of the Rubber Cord

Next, we'll find the mass (m) of the rubber cord using the previously calculated volume expression (A * L) and the given density (ρ = 1500 kg/m³); this is done as follows: \[ m = \rho V = 1500(AL) \]

Calculate the Weight of the Rubber Cord

Now, we'll calculate the weight (W) of the rubber cord using the mass (m) we found in the previous step and the given acceleration due to gravity (g = 10 m/s²): \[ W = mg = 15000(AL) \]

Find the Stress on the Rubber Cord

Stress is defined as the force (or the weight in this case) divided by the cross-sectional area (A): \[ \sigma = \frac{W}{A} = \frac{15000(AL)}{A} = 15000L \]

Calculate the Strain on the Rubber Cord

Next, we'll calculate the strain on the rubber cord using Young's modulus formula. Young's modulus (Y) is given as \(5\times10^8 \mathrm{~N}\) / \(\mathrm{m}^{2}\) thus: \[ Y = \frac{\sigma}{\epsilon} \] where ε is the strain. Solving for ε, we get: \[ \epsilon = \frac{\sigma}{Y} = \frac{15000L}{5\times10^8} \]

Calculate the Stretch of the Rubber Cord

Finally, we'll calculate the stretch (ΔL) of the rubber cord using the strain (ε) and the initial length (L) of the cord as follows: \[ \Delta L = \epsilon L = \frac{15000L^2}{5\times10^8}\] Now, plugging in the given length L = 10m we get: \[ \Delta L = \frac{15000(10^2)}{5\times10^8} = \frac{1500000}{5\times10^8} = \frac{3}{1000} \] So the stretch of the rubber cord is: \[ \Delta L = 0.003 \mathrm{~m} = 3 \times 10^{-3} \mathrm{~m} \] None of the provided options are correct.

Key concepts

Young's Modulus

Young's Modulus is a fundamental property in physics that describes the ability of a material to withstand changes in length when under lengthwise tension or compression. It is a measure of the stiffness of a solid material. - Young's Modulus, denoted by the symbol \(Y\), is defined mathematically as the ratio of stress (force per unit area) to strain (deformation in length per original length).- The formula for Young's Modulus is given by: \[ Y = \frac{\text{stress}}{\text{strain}} \] This ratio is critical because it tells us how much a material will stretch or compress under a given amount of force. Materials with a high Young's Modulus are relatively stiff and do not deform easily, whereas materials with a lower Young's Modulus are more elastic and deform more readily. For our rubber cord example, having a Young's Modulus of \(5 \times 10^8 \, \text{N/m}^2\) indicates that rubber, although flexible, has a defined limit to how much it can stretch under tension.

Density of Rubber

Density, often represented by the Greek letter \(\rho\), is the mass per unit volume of a substance. It's an essential property in physics because it helps determine the amount of mass in a given volume of a material.- For rubber, the density given is \(1500 \, \text{kg/m}^3\). This means that for every cubic meter of rubber, it contains 1500 kilograms of mass.- In practical terms, understanding the density of rubber is crucial when calculating the mass of objects like rubber cords since it helps establish how much the objects weigh.The density is part of what allows us to find the weight of the rubber cord when combined with the volume (derived from length and cross-sectional area). This weight is directly used to calculate the stress and, eventually, the stretch of the rubber cord when subjected to gravitational force.

Stress and Strain

Stress and strain are two core concepts in the study of elasticity. They help quantify how materials react under various forces.- **Stress** is defined as the force exerted per unit area within materials. It’s measured in \( \,\text{N/m}^2 \), or Pascals (Pa). Stress can be induced by tension, compression, or shear.- For the rubber cord, the stress is the weight of the cord per its cross-sectional area: \[ \text{Stress} = \frac{W}{A} = 15000L \] Where \(W\) is the weight calculated previously. - **Strain**, on the other hand, is the measure of deformation representing the displacement between particles in the material body. Strain is dimensionless as it’s a ratio of lengths: \[ \text{Strain} = \frac{\Delta L}{L} \]Where \(\Delta L\) is the change in length and \(L\) is the original length.Understanding stress and strain helps predict how much a material like rubber will extend or compress when under certain physical conditions.

Gravitational Force

Gravitational force is the attractive force that objects exert on each other due to their masses. It’s a fundamental force of nature that affects all matter.- Gravitational force is calculated using Newton's law of gravitation. When considering objects near the Earth’s surface, it simplifies to the weight \(W\) of an object: \[ W = mg \] Where \(m\) is mass and \(g\) is the gravitational acceleration given as \(10 \, \text{m/s}^2\) in this context.- In the case of the rubber cord, this gravitational force is what causes the stretch. The entire length of the cord provides gravitational pull, making it susceptible to stretching.Understanding gravitational force is crucial to calculating how much a material stretches under its own weight, and is a key part of solving elasticity problems in physics.