Question

A uniform wire of length \(l\) of weight \(w\) is suspended from the roof with a weight \(W\) at the other end. The stress in the wire at \( \frac{l}{3} \) distance from the top is} \[ \left(\frac{W}{A} + \frac{2}{\gamma}\frac{w}{A}\right) \] where \(A\) is the cross sectional area of the wire. The value of \(\gamma\) is ____.}

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
Stress at any point in a suspended wire is defined as the total weight acting below that point divided by the cross-sectional area. The total weight includes the load attached at the bottom and the weight of the portion of the wire hanging below the chosen point.
Step 2: Key Formula or Approach:
1. Weight of wire below point \(P\): \(W_{\text{wire\_below}} = w \times \frac{l_{\text{below}}}{l}\).
2. Total Tension at \(P\): \(T = W + W_{\text{wire\_below}}\).
3. Stress: \(\sigma = \frac{T}{A}\).
Step 3: Detailed Explanation:
The point is at a distance \(l/3\) from the top.
Therefore, the length of the wire hanging below this point is:
\[ l_{\text{below}} = l - \frac{l}{3} = \frac{2l}{3} \]
Since the wire is uniform, the weight of this lower portion is:
\[ W_{\text{wire\_below}} = \frac{2}{3} w \]
The total force (tension) at the point \(l/3\) from the top is the sum of the load \(W\) and the weight of the wire below it:
\[ F = W + \frac{2}{3} w \]
Stress at this point is:
\[ \sigma = \frac{F}{A} = \frac{W + \frac{2}{3} w}{A} = \frac{W}{A} + \frac{2}{3} \frac{w}{A} \]
Comparing this with the given formula \(\frac{W}{A} + \frac{2}{\gamma} \frac{w}{A}\):
\[ \frac{2}{\gamma} = \frac{2}{3} \implies \gamma = 3 \]
Step 4: Final Answer:
The value of \(\gamma\) is 3.