A 100 kg chandelier is hanging 2 meters from the ceiling of a cathedral. The chandelier is 
attached to a steel cable whose elastic modulus is 200 MPA. Determine the minimum 
diameter of the cable so that no more than 5 mm of deformation occurs in the cable

To determine the minimum diameter of the steel cable so that the deformation does not exceed 5 mm, we'll use the formula for elongation due to a tensile load in a material, which is given by:

\[
\Delta L = \frac{F \cdot L}{A \cdot E}
\]

Where:
- \(\Delta L\) is the elongation (deformation) in meters.
- \(F\) is the force applied (the weight of the chandelier) in newtons (N).
- \(L\) is the original length of the cable in meters.
- \(A\) is the cross-sectional area of the cable in square meters.
- \(E\) is the elastic modulus (Young's modulus) of the material in pascals (Pa).

### Step 1: Calculate the force \(F\)
The force \(F\) is the weight of the chandelier, which can be calculated using:

\[
F = m \cdot g
\]

Where:
- \(m = 100 \, \text{kg}\) (mass of the chandelier)
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity)

\[
F = 100 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 981 \, \text{N}
\]

### Step 2: Use the deformation equation

We rearrange the deformation equation to solve for the cross-sectional area \(A\):

\[
A = \frac{F \cdot L}{E \cdot \Delta L}
\]

Given:
- \(F = 981 \, \text{N}\)
- \(L = 2 \, \text{m}\)
- \(E = 200 \, \text{MPa} = 200 \times 10^6 \, \text{Pa}\)
- \(\Delta L = 5 \, \text{mm} = 0.005 \, \text{m}\)

Substituting the values:

\[
A = \frac{981 \, \text{N} \cdot 2 \, \text{m}}{200 \times 10^6 \, \text{Pa} \cdot 0.005 \, \text{m}}
\]

\[
A = \frac{1962 \, \text{N} \cdot \text{m}}{1 \times 10^6 \, \text{N/m}^2}
\]

\[
A = 1.962 \times 10^{-3} \, \text{m}^2 = 1962 \, \text{mm}^2
\]

### Step 3: Calculate the diameter of the cable

The cross-sectional area \(A\) of the cable is related to the diameter \(d\) by the formula:

\[
A = \frac{\pi d^2}{4}
\]

Solving for the diameter \(d\):

\[
d^2 = \frac{4A}{\pi}
\]

\[
d = \sqrt{\frac{4 \times 1962 \, \text{mm}^2}{\pi}} \approx \sqrt{2497.43 \, \text{mm}^2}
\]

\[
d \approx 49.97 \, \text{mm}
\]

### Conclusion

The minimum diameter of the cable should be approximately **50 mm** to ensure that the deformation does not exceed 5 mm.

Do you want details or have any questions?

Here are some related questions:
1. How would the cable diameter change if the chandelier weighed 200 kg?
2. What would be the required diameter if the cable length were doubled?
3. How does the elastic modulus affect the cable's diameter?
4. What is the effect of a different material with a higher elastic modulus?
5. How would a decrease in allowed deformation impact the cable diameter?

**Tip:** Always check the units when working with different physical quantities to ensure consistency.

Learn how to calculate the minimum diameter of a steel cable attached to a chandelier to limit deformation to 5mm. This problem involves mechanics and elasticity concepts, including Hooke's Law and calculations of elongation due to tensile load. Suitable for university-level engineering and physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation