Question

A vertical hanging bar of length $l$ and mass $m$ per unit length carries a load of mass M at 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 bar)
A) ${\displaystyle \frac{Mg+mg(1-x)}{A}}$
B) ${\displaystyle \frac{Mg}{A}}$
C) ${\displaystyle \frac{Mg+mgl}{A}}$
D) ${\displaystyle \frac{(M+M)gx}{Al}}$

Answer 1

Hint: In order to find the correct solution of the given question, we need to find the value of total force acting at a distance $x$from the support. After that we need to use the formula for the tensile stress which relates the tensile stress with total force and area on which the force is acting. Then we can finally conclude with the correct solution of the given question.

Complete step by step solution:
First of all we need to find the total forces acting on the system.
The force acting on the body of mass$m$at distance of $$(l-x)$$ can be written as, $F_1=m(l-x)g$
and the force acting on the body of mass $M$ can be written as, $F_2=Mg$
Now, we need to find the total force acting on the system.
So, we need to add both the forces acting on the given system.
Therefore, we can write it as,
$F=F_1+F_2$
$\Rightarrow F=m(l-x)g+Mg$……………….. (i)
Now, we know that the tensile stress of a body is the ratio of the force acting on the body per unit area.
Mathematically, we can write it as, $\sigma ={\displaystyle \frac{F}{A}}$
From equation (i), we can get,
$\Rightarrow \sigma ={\displaystyle \frac{m(l-x)g+Mg}{A}}$

Hence, option (A), i.e. ${\displaystyle \frac{Mg+m(l-x)g}{A}}$ is the correct choice of the given question.

Note: We define tensile stress as the force applied to a cross-sectional area. Mathematically, it is represented as,$\sigma ={\displaystyle \frac{F}{A}}$. We should not confuse tensile stress with tensile strength. The tensile strength of a body is the amount of force required to break a cross-sectional area. Mathematically, it is represented by, $S={\displaystyle \frac{P}{A}}$.