Question

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

Answer 1

Step 1: Identify 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: Write the expression for Stress.
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 $x = \ell/3$.
Substitute $x = \ell/3$ into the equation:
Stress = $\frac{W}{A} + \frac{w(\ell/3)}{A\ell} = \frac{W}{A} + \frac{w}{3A}$.

Step 4: Compare with the given expression to find $\gamma$.
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: $\gamma = 6$.