Tag: mechanical properties of solids

Questions Related to mechanical properties of solids

A light rod of length $2.00 m$ 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 $2\times10^{-3}m^{2}$ . 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

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $1.33 m, 1m$

  2. $1m, 1.33 m$

  3. $1.5 m, 1.33 m$

  4. $1.33m, 1.5 m$

Show answer
Correct Option: A
Explanation:
For  equal  stress
$ \dfrac{F _{1}}{A _{1}} = \dfrac{F _{2}}{A _{2}}$
$ \dfrac{F _{1}}{F _{2}} = \dfrac{A _{1}}{A _{2}} = \dfrac{10^{-3} m^{2}}{2 \times 10^{-3}m^{2}} = \dfrac{1}{2}$
$ 2F _{1} = F _{2}$
For  balance  of  rod
$ W = F _{1} + F _{2}$
$W = \dfrac{3 F _{2}}{2}$
$ F _{2} = \dfrac{2}{3} W$
Now equating torque
$Wx = F _{2} \times 2$
$x = \dfrac{2}{3} \times 2 = \dfrac{4}{3} = 1.33m$
For equal strain
$ \dfrac{\triangle l _{1}}{l} = \dfrac{\triangle l _{2}}{l}$
or
$\dfrac{\sigma _1}{Y _1}=\dfrac{\sigma _2}{Y _2}$
or
$\dfrac{F _1}{10^{-3}\times 2\times 10^{11}}=\dfrac{F _2}{2\times 10^{-3}\times 10^{11}}$
Thus we get $F _1=F _2$.
So, weight  will  be  hanging  mid - way 1m