Question

A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.

Answer 1

Given

• Rigid bar of mass \( 15 \,\text{kg} \) supported symmetrically by three wires.
• Each wire has length \( L = 2.0 \,\text{m} \).
• Outer two wires: copper; middle wire: iron.
• Each wire is to have the same tension \( T \).
• Young’s modulus (typical values): \( Y_{\text{Cu}} \approx 1.2 \times 10^{11} \,\text{Pa} \), \( Y_{\text{Fe}} \approx 1.9 \times 10^{11} \,\text{Pa} \).

1. Condition for equal tension

All three wires have the same original length and support the rigid bar symmetrically. If each wire has the same tension, then each must undergo the same extension, so each has the same strain.

For a wire:

\( Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{T/A}{\Delta L / L} \Rightarrow Y \propto \dfrac{1}{A} \) (since \( T, L, \Delta L \) are same)

Thus, for different materials under the same tension and strain:

\( Y \propto \dfrac{1}{A} \Rightarrow A \propto \dfrac{1}{Y} \)

2. Express area in terms of diameter

For a circular wire of diameter \( d \):

\( A = \dfrac{\pi d^{2}}{4} \Rightarrow A \propto d^{2} \)

So

\( d^{2} \propto \dfrac{1}{Y} \Rightarrow d \propto \dfrac{1}{\sqrt{Y}} \)

3. Ratio of diameters

Let \( d_{\text{Cu}} \) and \( d_{\text{Fe}} \) be the diameters of copper and iron wires, respectively. Then:

\( \dfrac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\dfrac{Y_{\text{Fe}}}{Y_{\text{Cu}}}} \)

Substitute values:

\( \dfrac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\dfrac{1.9 \times 10^{11}}{1.2 \times 10^{11}}} = \sqrt{\dfrac{19}{12}} \approx \sqrt{1.583} \approx 1.26 \)

Required ratio of diameters (copper : iron) \( d_{\text{Cu}} : d_{\text{Fe}} \approx 1.26 : 1 \) (often rounded to \( \approx 1.25 : 1 \)).