A rod `PQ` of length `1.05m` having negligible mass is supported at its ends by two wires one of stell (wire `A`), and the other of aluminium (wire `B`) of equal lengths as shown in fig. The cross-sectional areas of wires `A` and `B` are `1.0mm^(2)` and `2.0mm^(2)` respectively. At what point along the rod a load `W` be suspended in order to produce
(a) equal stress, (b) equal strains in both steel an aluminium.
`(Y_("steel") = 200 GPa, Y_("aluminium") = 70 GPa)`

Explanation

(a) For wire `A, A_(1) = 1mm^(2)` and for wire `B, A_(2) = 2mm^(2)`. If `F_(1)` and `F_(2)` are the tensions in the wire `A` and `B` to produce equal stresses in both as shown in Fig, then
`(F_(1))/(A_(1)) = (F_(2))/(A_(2))` or `(F_(1))/(F_(2)) = (A_(1))/(A_(2)) = (1 mm^(2))/(2 mm^(2)) = (1)/(2)`

If mass `m` is at a distance `x` from the end `P` of the rod `PQ`, then for rotational equilibrium of the rod about `O`,
`F_(1) x = F_(2) (1.05-x)` (or) `(F_(1))/(F_(2)) = (1.05-x)/(x)`
`(1.05-x)/(x) = (1)/(2) rArr 2.1-2x = x rArr x = 0.7m`
(b) If `F_(1)` and `F_(2)` are the tensions in the wires `A` and `B` respectively to produce equal strain in both wire then
`(F_(1))/(A_(1)Y_(1)) = (F_(2))/(A_(2)Y_(2)) rArr (F_(1))/(F_(2)) = (A_(1)Y_(1))/(A_(2)Y_(2)) = ((1mm^(2))/(2mm^(2)))((200G Pa)/(70G Pa)) = (10)/(7)`
If mass `m` is now placed at a distance `x^(1)` from the end `P` of the rod `PQ` for the rotational equilibrium of the rod about `O`.

`F_(1) x' = F_(2) (1.05-x')` or `(F_(1))/(F_(2)) = (1.05-x')/(x')`
`rArr 7.35-7x' = 10x' rArr x' = (7.35)/(17) = 0.43m`