Question

A vertical hanging bar of length $l$ and mass \[m\] per unit length carries a load of mass $M$ at the lower end, its upper end clamped at a rigid support. The tensile stress a distance \[x\] from support is ($A$- area of cross-section of the bar)
A) $\dfrac{{Mg + mg(l - x)}}{A}$
B) $\dfrac{{Mg}}{A}$
C) $\dfrac{{Mg + mgl}}{A}$
D) $\dfrac{{(M + m)gx}}{{Al}}$

Answer 1

Hint: To answer this question we will calculate the mass of the hanging bar and its sum with the load mass at any arbitrary point on the rod. Only the mass of the rod and the load below an arbitrary point will contribute to the tensile stress at that point.

Formula used: In this solution, we will use the following formula:
Tensile stress: $S = \dfrac{F}{A}$ where $F$ is the force acting on a point on the rod and $A$ is the cross-sectional area of the rod

Complete step by step answer:
In the setup given to us, a vertical hanging bar of length $l$ and mass \[m\] per unit length carries a load at its lower end. The force that will be acting on this rod will be due to the weight of the rod itself and due to the weight of the load mass.
For any point $x$ from the support, only the mass below the point will contribute to the tensile stress at that point. The mass of the rod below the point will be calculated as follows. Since the mass of the rod is uniformly distributed, the weight of the rod below the point will be due to a length of $l - x$.
Then the mass of the rod below the point $x$ will be the product of the density of the rod which will be $m$ and the length of the rod below the point i.e. $l - x$ and the gravitational acceleration as: $m(l - x)g$
So, the net force will be due to the weight of the rod below the point and the load mass as well which lies completely below the point. So, the net force will be
$F = \dfrac{{mg(l - x) + Mg}}{A}$

Hence the correct choice is option (A).

Note: Here we should not be confused by the mass density of the rod which has been given to us as $m$. It is not the mass of the rod but the mass density so to get its mass we need to multiply the density by length. As a consistency check, we can check that the stress on the rod will be maximum at the support i.e. $x = 0$ and minimum at the end of the rod i.e. at $x = l$ which is satisfied only in option (A).