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 Steel the wires is made of steel and is of cross-section 10^(-3)m^(2) and the other wire is of brass of cross-section 2xx 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 =2xx10^(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 Steel the wires is made of steel and is of cross-section 10^(-3)m^(2) and the other wire is of brass of cross-section 2xx 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 =2xx10^(11)N//m^(2)
Answer» Solution :SUPPOSE a, and a, 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) or (T(1))/(10^(-3))=(T_(2))/(2xx10^(-3)) (or) T_(2)=2T_(1)...(i)` As the whole system is in equilibrium, so `sum bar 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 equation(i) and (ii) we get `x=(4)/(3)m` (ii) Second case : For equal strains in both the wires`e_(1)=e_(2)` `(T_(1)l)/(a_(1)y_(1))=(T_(2)l)/a_(2)Y_(2)) or (T_(1))/(10^(-3)xx2xx10^(12))=(T_(2))/(2xx10^(-3)xx10^(11) (or) T_(1)=T_(2).... (ii)` From equation (ii) and (III) , we get x=1m