Question

5. A 20 kg chandelier is suspended from the ceiling by four vertical steel wires. Each wire has an unloaded length of 1 m and a diameter 2 mm and each bears an equal load. When the chandelier is hung, the stretch in each wire is approximately : \[ \left(Y_{\text {steel }}=2 \times 10^{11} N / m ^{2}, g=10 m / s ^{2}\right) \] (1) 0.08 mm (2) 0.04 mm (3) 0.02 mm (4) 0.01 mm

Answer 1

When a 20 kg chandelier is suspended from the ceiling using four vertical steel wires, we need to determine the force exerted by the chandelier on the wires. Since each of the four wires bears the same 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/s²).

F_total​ = m × g = 20kg × 9.81m/s² = 196.2 N

Now, we will find the force acting on each wire:

F_wire​ = \(\frac {F_{Total}}4 = \frac {196.2N}{4}\)​ = 49.05N

Next, we will calculate the stress on the wires. Stress is the force per unit area and can be calculated as: 

σ = \(\frac FA\)

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 = πr² = π(\(\frac 12\)​⋅2mm)² = π(1 mm²)

Now, we can calculate the stress on the wires:

\(\sigma = \frac { F_{wire}}A = \frac {49.05}{\pi (1 \ mm^2)}\)

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:

σ = E × ϵ 

Where E is the modulus of elasticity (also known as Young's modulus) and ϵ is the strain (elongation per unit length). For steel, the modulus of elasticity is approximately 200GPa. Divide the stress by the modulus of elasticity to find the strain:

\(\epsilon = \frac {\sigma} E = \frac {49.05 N}{\pi ( mm^2).200GPa}\)

Finally, we will find the elongation of the wires using the strain:

\(\Delta L = \epsilon \times L_0 = \frac {49.05N}{\pi (1\ mm^2 ).200GPa} \times1 m\)

Therefore, the elongation of the steel wires is \(\Delta L = \frac {49.05N}{\pi (1\ mm^2 ).200GPa} \times1 m = 7.82 \times 10^{-5} m\) when a 20 kg chandelier is suspended from the ceiling using four vertical steel wires with a diameter of 2 mm and a length of 1 meter each.