Question

A light rod of length $2.00m$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section ${10}^{-3}{m}^2$ and the other is of brass of cross-section $2\times {10}^{-3}{m}^2$ . Find out the position along the rod at which a weight may be hung to produce.(Youngs modulus for steel is 2x10${}^{11}$N /m${}^2$ and for brass is 10${}^{11}$N / m${}^2$ )
a) equal stress in both wires
b) equal strains on both wires

• A. $1.33m,1m$
• B. $1m,1.33m$
• C. $1.5m,1.33m$
• D. $1.33m,1.5m$

Answer 1

Answer: (A)

$1.33m,1m$

For equal stress

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\frac{F_1}{F_2}=\frac{A_1}{A_2}=\frac{{10}^{-3}{m}^2}{2\times {10}^{-3}{m}^2}=\frac{1}{2}$

$2F_1=F_2$

For balance of rod

$W=F_1+F_2$

$W=\frac{3F_2}{2}$

$F_2=\frac{2}{3}W$

Now equating torque

$Wx=F_2\times 2$

$x=\frac{2}{3}\times 2=\frac{4}{3}=1.33m$

For equal strain

$\frac{△l_1}{l}=\frac{△l_2}{l}$

or

$\frac{{\sigma }_1}{Y_1}=\frac{{\sigma }_2}{Y_2}$

or

$\frac{F_1}{{10}^{-3}\times 2\times {10}^{11}}=\frac{F_2}{2\times {10}^{-3}\times {10}^{11}}$

Thus we get $F_1=F_2$.

So, weight will be hanging mid - way 1m