Question

A 20 -kg chandelier is suspended from a ceiling by four vertical steel wires. Each wire has an unloaded length of \(1 \mathrm{~m}\) and a diameter of \(2 \mathrm{~mm},\) and each bears an equal load. When the chandelier is hung, how far do the wires stretch?

Short answer

Question: Calculate the elongation of the steel wires when a 20 kg chandelier is suspended from a ceiling using four vertical steel wires with a diameter of 2 mm and length of 1 meter each. Answer: \(\Delta L = \epsilon \times L_0 = \frac{49.05\,\text{N}}{\pi\, (1\,\text{mm})^2 \cdot 200\, \mathrm{GPa}} \times 1\,\text{m}\)

Step-by-step solution

Calculate the force exerted by the chandelier on the wires

First, we will find the force exerted by the chandelier on the wires. Since the chandelier weighs 20 kg and each wire bears an equal load, the force acting on each wire would be equal to one-fourth of the total force due to gravity acting on the chandelier. The force due to gravity can be calculated using the mass of the chandelier and the acceleration due to gravity (9.81 \(m/s^2\)). \(F_{total} = m \times g = 20\,\text{kg} \times 9.81\, m/s^2 = 196.2\,\)N Now, we will find the force acting on each wire: \(F_{wire} = \frac{F_{total}}{4} = \frac{196.2\,\text{N}}{4} = 49.05\,\text{N}\)

Calculate the stress on the wires

Next, we will calculate the stress on the wires. Stress is the force per unit area and can be calculated as: \(\sigma = \frac{F}{A}\) First, we need to find the cross-sectional area of the wires. Since they have a diameter of 2 mm, we can use the formula for the area of a circle to find the area: \(A = \pi r^2 = \pi (\frac{1}{2}\cdot 2\,\text{mm})^2 = \pi (1\,\text{mm})^2\) Now, we can calculate the stress on the wires: \(\sigma = \frac{F_{wire}}{A} = \frac{49.05\,\text{N}}{\pi\, (1\,\text{mm})^2}\)

Calculate the wires' elongation using the stress-strain relationship

The final step is to use the stress-strain relationship to find the elongation of the wires. The stress-strain relationship, also known as Hooke's Law, is given by: \(\sigma = E \times \epsilon\) Where \(E\) is the modulus of elasticity (also known as Young's modulus) and \(\epsilon\) is the strain (elongation per unit length). For steel, the modulus of elasticity is approximately \(200\, \mathrm{GPa}\). Divide the stress by the modulus of elasticity to find the strain: \(\epsilon = \frac{\sigma}{E} = \frac{49.05\,\text{N}}{\pi\, (1\,\text{mm})^2 \cdot 200\, \mathrm{GPa}}\) Finally, we will find the elongation of the wires using the strain: \(\Delta L = \epsilon \times L_0 = \frac{49.05\,\text{N}}{\pi\, (1\,\text{mm})^2 \cdot 200\, \mathrm{GPa}} \times 1\,\text{m}\) This gives us the elongation of the wires when the chandelier is hung.

Key concepts

Force Due to Gravity

When we hang an object like a chandelier, it's the force due to gravity that is pulling it down towards the center of the Earth. This force is also known as weight and can be calculated using the object's mass and the acceleration due to gravity, which is approximately 9.81 meters per second squared on Earth's surface. Imagine you're holding a 1 kg apple; gravity pulls on it with a force of 9.81 Newtons. Similarly, our 20 kg chandelier experiences a gravitational pull of 196.2 Newtons. Since the chandelier is held up by four wires, this force is evenly divided among them, causing each wire to experience a load of 49.05 Newtons.

Understanding this concept is crucial in physics problem solving as it lays the foundation for analyzing the effects of forces on objects and structures.

Stress and Strain

The terms 'stress' and 'strain' are fundamental in the study of materials in physics and engineering. Stress is a measure of the internal forces that particles within a material exert on each other and is defined as force per unit area. It's a bit like the pressure you feel on your fingertips if you press them against a balloon - the tighter you press, the more stress you're applying. When we calculated the stress on the wires holding the chandelier, we divided the force each wire carries by the wire's cross-sectional area.

Strain, on the other hand, describes how much an object is stretched or compressed relative to its original length. It's a dimensionless number, a ratio - think of it as the percentage by which the length of the wire has been changed. To compute strain, we compare the amount of stretch (which we will calculate) to the wire's original length.

Modulus of Elasticity

The modulus of elasticity, often called Young's modulus, is a material property that measures its stiffness or resistance to being deformed elastically (i.e., non-permanently) when a force is applied. The higher the modulus, the stiffer the material. Imagine a rubber band versus a steel rod; it's much easier to stretch the rubber band than the steel rod because the modulus of elasticity of rubber is much lower than that of steel.

In the context of the chandelier problem, the modulus of elasticity of steel tells us how much the wires will stretch under the stress we calculated. It's crucial to note that these calculations assume the deformations are within the elastic limit. That means once the load is removed, the wires will return to their original length without any permanent deformation.

Hooke's Law

Hooke's Law is a principle of physics that relates the force needed to extend or compress a spring to the distance it is stretched or compressed. Simply put, it says that the stretch (or strain) in a wire is proportional to the applied stress, as long as the material remains in its elastic limit. Think of a slinky – you pull on it, and it stretches; when you release it, it bounces back to its original shape. That's Hooke's Law in action. In our chandelier scenario, we apply Hooke's Law to find out just how much each steel wire will stretch. The formula for the stress-strain relationship, \(\sigma = E \times \epsilon\), helps us determine the strain, which, when multiplied by the original length of each wire, gives the elongation due to the weight of the chandelier.

Mastering Hooke's Law is not just important for solving textbook problems; it's also crucial for designing and understanding the behavior of springs, which play integral roles in various mechanical systems.