Question

A rod of mass $M$ and cross-section area $A$ is subjected to two forces $f$ and $2f$ at both the ends as shown in the figure. If Young's modulus of its material is $y$ and its length is $L$, then the total elongation of the rod is $\frac{\alpha fL}{\beta Ay}$. Find $(\alpha +\beta )$.

Answer 1

Formula used: $Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$

Tension in the rod at a distance 𝑥 from its
right edge is

$T=F(2-\frac{x}{L})$

Therefore,
Net extension in the rod,

$\delta ={\int }_0^L\frac{T}{yA}dx=\frac{3FL}{2yA}$

Final Answer: $\left(5\right)$