Stress in Suspended Wire Under Gravity and External Mass

To find the value of $\gamma$, we analyze the forces acting on a cross-section of the wire located at a distance of $\frac{l}{3}$ from the top support.

1. Length of the wire below the point: Since the cross-section is at a distance of $\frac{l}{3}$ from the top, the length of the wire remaining below this section is: $l' = l - \frac{l}{3} = \frac{2}{3}l$

2. Weight of the remaining portion of the wire: Given that the total weight of the wire of length $l$ is $w$, the weight per unit length is $\frac{w}{l}$. Thus, the weight of the lower segment of length $l'$ is: $w' = \left(\frac{w}{l}\right) \cdot \left(\frac{2}{3}l\right) = \frac{2}{3}w$

3. Total downward force (Tension) at the cross-section: The total tension $T$ at this point supports both the external mass of weight $W$ hanging at the lower end and the weight of the lower portion of the wire $w'$: $T = W + w' = W + \frac{2}{3}w$

4. Stress at the cross-section: The stress in the wire at this position is defined as the force per unit cross-sectional area $A$: $\text{Stress} = \frac{T}{A} = \frac{W + \frac{2}{3}w}{A} = \frac{W}{A} + \frac{2}{3}\frac{w}{A}$

5. Comparing with the given expression: The problem specifies the stress as: $\text{Stress} = \frac{W}{A} + \frac{2}{\gamma}\frac{w}{A}$

   Comparing the two expressions: $\frac{2}{\gamma} = \frac{2}{3} \implies \gamma = 3$

Final Answer: The value of $\gamma$ is 3.