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^2and 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 wiresYoung's modulus of brass = 1 xx 10^11 N//m^2Young's modulus of steel = 2 xx 10^11 N//m^2

A light rod of length 2 m is suspended from the ceiling horizontally b

1.	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^2and 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 wiresYoung's modulus of brass = 1 xx 10^11 N//m^2Young's modulus of steel = 2 xx 10^11 N//m^2
Answer» Solution :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