Question

A uniform steel rod of density $\rho$, cross - sectional area $A$ and length $L$ is suspended so that it hangs vertically. The stress at the middle point of the rod is given by $\frac{\rho gL}{N}$. Then, the value of $N$ is

Answer 1

Middle point of the rod has to support the lower half portion of the rod. The weight of the lower half -portion of the rod will cause stress at the middle point of the rod.


Mass of half portion of the rod is given as,
$m=\text{volume}\times \text{density}$
$m=\rho A\frac{L}{2}$
$\because [\text{Volume}=\text{Area}\times \text{Length}]$
Weight of lower half-portion acting vertically downwards is,
$F=mg=\frac{\rho ALg}{2}$
$\Rightarrow \text{Stress}=\frac{F}{A}=\frac{\rho ALg}{2A}$
$\Rightarrow \text{Stress}=\frac{\rho Lg}{2}$
On comparing with the given value of stress,
$\text{Stress}=\frac{\rho gL}{N}$
$\therefore N=2$