Question

A straight wire clamped vertically at its lower end stands vertically if it is short, but bends under its own weight if it is long. It can be shown that the greatest length for vertical equilibrium is \(l,\) where \(k l^{3 / 2}\) is the first zero of \(J_{-1 / 3}\) and $$k=\frac{4}{3 r^{2}} \sqrt{\frac{\rho g}{\pi Y}},$$ \(r=\) radius of the wire, \(\rho=\) linear density, \(g=\) acceleration of gravity, \(Y=\) Young's modulus. Find \(l\) for a steel wire of radius \(1 \mathrm{mm}\); for a lead wire of the same radius.

Short answer

The greatest length for equilibrium is approximately 0.051 m for steel and 0.012 m for lead.

Step-by-step solution

Identify the given data

Given is the formula for the greatest length for vertical equilibrium, where \(k l^{3 / 2}\) is the first zero of \(J_{-1 / 3}\). The value of \(k\) is given by \(k=\frac{4}{3 r^{2}} \sqrt{\frac{\rho g}{\pi Y}}\). The radii, linear densities, accelerations due to gravity, and Young's moduli for materials need to be identified for calculation.

Define constants for steel and lead

Set the radius for both wires: \(r = 1 \, \mathrm{mm} = 1 \times 10^{-3} \, \mathrm{m}\). For steel:- \(\rho_{\text{steel}} = 7800 \, \mathrm{kg/m^3}\)- \(Y_{\text{steel}} = 2.0 \times 10^{11} \, \mathrm{Pa}\)For lead:- \(\rho_{\text{lead}} = 11340 \, \mathrm{kg/m^3}\)- \(Y_{\text{lead}} = 1.6 \times 10^{10} \, \mathrm{Pa}\)Acceleration due to gravity, \(g = 9.81 \, \mathrm{m/s^2}\).

Calculate constant \(k\)

Use the formula \(k = \frac{4}{3 r^{2}} \sqrt{\frac{\rho g}{\pi Y}}\) for each material. Substitute the respective values.For steel:\[ k_{\text{steel}} = \frac{4}{3 (1 \times 10^{-3})^{2}} \sqrt{\frac{7800 \times 9.81}{\pi \times 2.0 \times 10^{11}}} \]For lead:\[ k_{\text{lead}} = \frac{4}{3 (1 \times 10^{-3})^{2}} \sqrt{\frac{11340 \times 9.81}{\pi \times 1.6 \times 10^{10}}} \]

Evaluate the calculations for \(k\)

For steel:\[ k_{\text{steel}} = \frac{4}{3 \times 1 \times 10^{-6}} \times \sqrt{\frac{76438}{6.283 \times 10^{11}}} \]\[ k_{\text{steel}} = \frac{4}{3 \times 10^{-6}} \times \sqrt{1.217 \times 10^{-7}} \]\[ k_{\text{steel}} = 1.333 \times 10^{6} \times 3.488 \times 10^{-4} \]\[ k_{\text{steel}} \approx 465 \]For lead:\[ k_{\text{lead}} = \frac{4}{3 \times 10^{-6}} \times \sqrt{70702.2}{6.283 \times 1.6 \times 10^{10}} \]\[ k_{\text{lead}} = 1.333 \times 10^{6} \times 2.65 \times 10^{-3} \]\[ k_{\text{lead}} \approx 3539 \]

Determine greatest length for equilibrium

From the problem statement, \(k l^{3 / 2}\) is the first zero of \(J_{-1 / 3}\). Let's assume the first zero of \(J_{-1/3}\) is around 2.8558 (a known value).For steel:\[ l_{\text{steel}} = \left( \frac{2.8558}{465} \right)^{2/3} \]\[ l_{\text{steel}} \approx 0.051 \, \mathrm{m} \]For lead:\[ l_{\text{lead}} = \left( \frac{2.8558}{3539} \right)^{2/3} \]\[ l_{\text{lead}} \approx 0.012 \, \mathrm{m} \]

Key concepts

Young's modulus

Young's modulus, often denoted by the symbol \(Y\), is a measure of the stiffness of a material. It is a fundamental mechanical property of materials that quantifies the relationship between tensile stress (force per unit area) and tensile strain (proportional deformation in length). Young's modulus is expressed in Pascals (Pa) and is defined by the formula: \[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} \] It tells us how much a material will deform under a given load. A high Young's modulus means a material is very stiff, and a low Young's modulus means it is more flexible. In the exercise, we used the Young's modulus for steel (\(2.0 \times 10^{11}\text{Pa}\)) and lead (\(1.6 \times 10^{10}\text{Pa}\)). This helped us determine how much each material bends under its own weight when subjected to gravity.

Linear density

Linear density, symbolized as \(\rho\), is the mass per unit length of a material. It is primarily used in the context of long, slender objects like wires and strings. The formula for linear density is: \[ \rho = \frac{\text{mass}}{\text{length}} \] Linear density is typically measured in \( \text{kg/m} \). In our exercise, we used the linear densities of steel (\(7800 \text{kg/m}^3\)) and lead (\(11340 \text{kg/m}^3\)). Linear density is a crucial factor in determining how the wire behaves under its own weight because it affects how much force is acting downwards due to gravity.

Bending of beams

The bending of beams or wires under their own weight is a key topic in mechanics of materials. This concept involves analyzing how a beam deflects when subjected to external forces, including its own weight. The equation used in the exercise, \[ k = \frac{4}{3 r^{2}} \sqrt{\frac{\rho g}{\pi Y}} \], relates the material properties and geometry (radius \(r\), linear density \( \rho \), gravitational acceleration \(g\), and Young's modulus \(Y\)) to the constant \(k\). This constant helps determine the critical length \(l\) at which the wire bends significantly under its own weight. A material with higher linear density or lower Young's modulus tends to bend more easily.

Special functions in physics

Special functions often arise in the solutions of complex physical problems. In this exercise, we encountered Bessel functions, specifically \( J_{-1 / 3} \). Bessel functions are solutions to Bessel's differential equation and frequently appear in problems involving cylindrical or spherical symmetry. In the context of our problem, the first zero of the Bessel function \( J_{-1/3} \), which is approximately 2.8558, helped us calculate the greatest length \( l \) for which the wire remains in vertical equilibrium. Understanding special functions, like Bessel functions, is crucial for solving advanced physics and engineering problems, involving wave propagation, heat transfer, and structural mechanics.