A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wire of equal length tied to its ends. One of the wires is made of steel and is of crosss-section `0.1cm^(2)` and the other brass of cross-section `0.2cm^(2)`. Along the rod at whcih distance may a weight be hung to produce (a) equal stresses in both the wires (b) equal strains in both the wires ? Y or brass and steel are `10 xx 10^(11) and 20xx 10^(11)` dyne/ cm respectively.

Explanation

(a) As stresses are equal,
`(T_(1))/(A_(1))=(T_(2))/(A_(2))i.e., (T_(1))/(T_(2))=(A_(1))/(A_(2))=(0.1)/(0.2)orT_(2)=2T_(1)..........(i)`
Now for translatory equilibrium of the rod, `T_(1) + T_(2) =W`
Which in the light of Eqn. (i) gives `T_(1) = ((W)/(3)) and T_(2) = ((2W)/(3)) ......... (ii)`

Now if x is the distance of weight W from steel, wire, for rotational equlibrium of rod,
`T_(1)x = T_(2) (2 -x) or (W//3)x = (2W//3) (2 - x)`.
i.e., x = (4/3) m
(b) As strains are equal,
`(T_(1))/(A_(1)Y_(1))=(T)/(A_(2)Y_(2))["as train"=("stress")/(Y)]`
`T_(1) = T_(2) = ((W)/(2)) .......... (iv)`
`T_(1) x =T_(2) (2 -x)or ((W)/(2)) x = ((W)/(2)) (2 -x)`
i.e., x = 1 m