Solution to Question 2
Problem Analysis:
We are comparing the extension of an elastic cord in two configurations:
- Hanging vertically from one end.
- Hanging between two nails (A and B) forming a catenary.
Since the cord obeys Hooke’s Law, the extension $\Delta L$ is directly proportional to the average tension in the cord: $\Delta L \propto \int T \, dl$.
Case 1: Vertical Hanging
The tension $T$ varies linearly from $Mg$ at the top to $0$ at the bottom.
$$ T_{\text{avg}} = \frac{Mg + 0}{2} = 0.5 Mg $$Case 2: Catenary (with $\theta = 45^\circ$)
Let the tension at the ends be $T_{\text{end}}$. The vertical component supports half the weight:
$$ T_{\text{end}} \sin 45^\circ = \frac{Mg}{2} \implies T_{\text{end}} = \frac{Mg}{2 \sin 45^\circ} = \frac{Mg}{\sqrt{2}} \approx 0.707 Mg $$The tension is minimum at the lowest point ($T_{\text{min}}$), which equals the horizontal component:
$$ T_{\text{min}} = T_{\text{end}} \cos 45^\circ = \frac{Mg}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 0.5 Mg $$In this case, the tension varies from $0.5 Mg$ (at the bottom) to $0.707 Mg$ (at the ends).
Conclusion:
Comparing the two cases:
- Vertical Case Tension Range: $[0, Mg]$ with average $0.5 Mg$.
- Catenary Case Tension Range: $[0.5 Mg, 0.71 Mg]$.
While the vertical case has a higher maximum tension, the catenary case maintains a significant tension throughout the entire length (never dropping to zero). Consequently, the integrated average tension is higher in the catenary configuration.
Answer: In the second case (catenary), the cord is stretched longer.