Question

A uniform elastic rod of cross-sectional area $A,$ natural length $L$ and young’s modulus $Y$ is placed on a smooth horizontal surface. Now, two horizontal forces of magnitude $F$ and $3F$ directed along the length of rod and in opposite direction act at two of its ends as shown. After the rod has acquired steady state, the extension of the rod is $\frac{\alpha F}{YA}L$ Find $\alpha$

Answer 1

Step 1: Draw a labelled diagram.
Step 2: Find the value of $\alpha$
Formula used: $Y=\frac{F/A}{\Delta l/l}$
Tension in the rod at a distance 𝑥 from its
right edge is
$T=F(3-2\frac{x}{L})$

$\therefore$ Net extension in the rod,

$={\int }_0^L\frac{T}{YA}dx=\frac{2F}{YA}L$

Final Answer: (2)