Question

A massless rod of length 1 m is suspended by two vertical strings of same length. Area of cross-section of left string is 'A' and its young's modulus is $\frac{3}{2}Y$, whereas area of cross-section of right string is 2A and its young's modulus is Y. A block of mass 'm' is placed at 'x' cm from left string. If strains in both strings is same, then 'x' is . (Answer upto two digits after the decimal point)

Answer 1

$\text{Strain}=\frac{\text{stress}}{Y}=\frac{T}{AY}$

${\text{Strain}}_L={\text{Strain}}_R$

$\frac{T_L}{A_L{Y}_L}=\frac{T_R}{A_R{Y}_R}\Rightarrow \frac{T_L}{A\times \frac{3}{2}Y}=\frac{T_R}{2A\times Y}$

$4T_L=3T_R$

Taking moments about the point where $m$ is placed, we get:
$T_L$x = $T_R(100-$x)
$\Rightarrow \frac{3}{4}$ x = $(100-$x) $\Rightarrow 7$ x = 400

x $=\frac{400}{7}=57.14$