A light rod of length 2m 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 sectional area 0.1 sq. cm and the other is of brass of cross-sectional area 0.2 sq. cm find out the position along te rod at which a weight may be hung to produce equal stresses in both wires. Y for steel `=20xx10^(11)` dyne `cm^(-2)` and Y for brass `=10xx10^(11)` dyne `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