A rod of length 1.05 m having negliaible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in fig. The cross-sectional area of wire A and B are `1 mm^(2)` and 2` mm^(2)`, respectively . At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires. Given,
`Y_(steel) = 2 xx 10^(11) Nm^(-2) and Y-(alumi n i um) = 7.0 xx 10^(10)N^(-2)`

Explanation

For steel wire `A, l_(1) = l, A_(1) = 1 mm^(2) , Y_(1) = 2 xx 10^(11)Nm^(-2)`
For aluminium wire B, `l_(2) = l , A_(2) = 2 mm^(2) , Y_(2) = 7 xx 10^(10) Nm^(-2)`
(a) Let mass m be suspeneded from the rod at distance x from the end where wire A is connected. let `F_(1)` and `F_(2)` be the tensions in two wires and there is equal stress in two wires, then
`(F_1)/(A_1) = (F_2)/(A_2) or (F_1)/(F_2) = (A_1)/(A_2) = 1/2 ` ..(i)
Taking moment of forces about point of suspension of mass from the rod, we have
`F_(1) x = F_(2) (1.05 -x) or (1.05 = x)/(x) = (F_1)/(F_2) = 1/2`
or ` 2.10 -2x = x or x=0.70 m = 70 cm`.
(b) Let mass m be suspeneded from the rod at distance x from the end where wire A is connected. let `F_(1)` and `F_(2)` be the tensions in two wires and there is equal stress in two wires, ie. ,
`(F_1)/(A_(1)Y_(1)) = (F_2)/(A_(2)Y_(2))` or `(F_1)/(F_2) = (A_(1)Y_(2))/(A_(2)Y_(2)) = 1/2 xx (2 xx 10^(11))/(7 xx 10^(10)) = 10/7`
As the rod is stationary, os `F_(1) x = F_(2) (1.05-x) or (1.05 = x)/(x) = (F_1)/(F_2) =10/7`
or, `10x = 7.35 - 7x or x=0.4324m = 43.2cm`.