Question

A $1.05m$ having negligible mass is supported at its ends by two wires of steel (wire $A$) and aluminium (wire $B$) of equal lengths as shown in Fig. The cross-sectional areas of wires $A$ and $B$ are $1.0m{m}^2$ and $2.0m{m}^2$, respectively. At what point along the rod should a mass $m$ be suspended in order to produce
(a) Equal stress in $A$ and $B$ and
(b) Equal strains in $A$ and $B$?

Answer 1

Cross-sectional area of wire A, $a_1=1.0m{m}^2=1.0\times {10}^{-6}{m}^2$

Cross-sectional area of wire B, $a_2=2.0m{m}^2=2.0\times {10}^{-6}{m}^2$

Young’s modulus for steel, $Y_1=2\times {10}^{11}N{m}^{-2}$

Young’s modulus for aluminium, $Y_2=7.0\times {10}^{10}N{m}^{-2}$

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area = F / a
If the two wires have equal stresses, then:
$F_1/a_1=F_2/a_2$
Where,
$F_1=$ Force exerted on the steel wire
$F_2=$ Force exerted on the aluminum wire
$F_1/F_2=a_1/a_2=1/2$ ..... (i)
The situation is shown in the figure.

Taking torque about the point of suspension, we have:
$F_{1}y=F_2(1.05-y)$

$F_1/F_2=(1.05-y)/y$ ..... (ii)
Using equations (i) and (ii), we can write:
$\frac{(1.05-y)}{y}=\frac{1}{2}$
$2(1.05-y)=y$
$y=0.7m$
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young's modulus = Stress / Strain

Strain = Stress / Young's modulus $=\frac{\left(F/a\right)}{Y}$
If the strain in the two wires is equal, then:
$\frac{\left(F_1/a_1\right)}{Y_1}=\frac{\left(F_2/a_2\right)}{Y_2}$
$\frac{F_1}{F_2}=\frac{a_1{Y}_1}{a_2{Y}_2}$

$\frac{a_1}{a_2}=\frac{1}{2}$

$\frac{F_1}{F_2}=\left(\frac{1}{2}\right)(2\times {10}^{11}/7\times {10}^{10})=\frac{10}{7}$ ..... (iii)
Taking torque about the point where mass m, is suspended at a distance y1 from the side where wire A attached, we get:
$F_1{y}_1=F_2(1.05–y_1)$

$\frac{F_1}{F_2}=\frac{(1.05-y_1)}{y_1}$ ..... (iv)

Using equations (iii) and (iv), we get:
$\frac{(1.05-y_1)}{y_1}=\frac{10}{7}$
$7(1.05-y_1)=10y_1$
$y_1=0.432m$
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.