Question

If stress at $x = \ell/3$ from bottom is $\frac{W}{A} + \frac{2}{\gamma} \frac{w}{A}$ then find $\gamma$ :

Answer 1

The question asks to find \( \gamma \) in the formula for the stress at \( x = \frac{\ell}{3} \) from the bottom of the rod, given by:

Stress = \( \frac{W}{A} + \frac{2}{\gamma} \frac{w}{A} \)

Step 1: Define the forces at height \( x \) from the bottom.
The total force (tension) at any point \( x \) is the sum of the load at the bottom (\( W \)) and the weight of the portion of the rod below that point. Let \( w \) be the weight of the total length \( \ell \). The weight of a section of length \( x \) is \( w' = \frac{w}{\ell} \cdot x \).

Step 2: Expression for Stress.
The stress is given by: \[ \text{Stress} = \frac{\text{Total Force}}{\text{Area}} = \frac{W + w'}{A} = \frac{W + \frac{wx}{\ell}}{A} = \frac{W}{A} + \frac{wx}{A\ell}. \]

Step 3: Calculate Stress at \( x = \frac{\ell}{3} \).
Substitute \( x = \frac{\ell}{3} \) into the equation: \[ \text{Stress} = \frac{W}{A} + \frac{w(\ell/3)}{A\ell} = \frac{W}{A} + \frac{w}{3A}. \]

Step 4: Compare with the given expression.
The given expression is: \[ \frac{W}{A} + \frac{2}{\gamma} \frac{w}{A}. \] Equating the second terms: \[ \frac{1}{3} = \frac{2}{\gamma} \implies \gamma = 6. \]

Final Answer: \( \gamma = 6 \).