A light rod of length 2 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 xx 10^(-3) m^2` . Find out the position along the rod at which a weight may be hung to produce, i) equal stress in both wires ii) equal strains in both wires
Young's modulus of brass `= 1 xx 10^11 N//m^2`
Young's modulus of steel `= 2 xx 10^11 N//m^2`

Explanation

Suppose `a_1` and `a_2` are the cross-sectional areas, and `Y_1` and `Y_2` are the Young.s moduli of steel and brass wire respectively. Let `T_1` and `T_2` are tensions in the steel and brass wires respectively.
Let x is distance of the position of the hanging weight from the steel wire.
i) First case : For equal stress in both wires, we have
`(T_1)/(a_1) = (T_2)/(a_2)`
`(T_1)/(10^(-3)) = (T_2)/(2 xx 10^(-3)) " or " T_2 = 2T_1 ` ....(i)
As the whole system is in equilibrium, so `sum vec tau = 0.` Taking moment of all the forces acting on the rod about C, we have
`T_1 x = T_2 (2-x) =0` ....(ii)
Solving eqn (i) and (ii)
we get `x = 4/3 m`
ii) second case :
For equal strain in both the wires `e_1 = e_2`
`(T_1l)/(a_1Y_1) = (T_2l)/(a_2Y_2)`
`(T_1)/(10^(-3) xx 2 xx 10^11) = (T_2)/(2 xx 10^(-3) xx 10^11)`
`T_1 = T_2` ...(iii)
From equation (ii) and (iii) , we get x = 1m