A light rod of length 2m is suspended horizontally by means of 2 vertical wires of equal lengths tied to its ends. One of the wire is made of steel & is of cross section `A_(1) = 0.1 cm^(2)` & other of brass & is of cross section `A_(2) = 0.2 cm^(2)`, find out the position along the rod at which a weight must be suspended to produce (i) equal stresses in both wires, (ii) equal strains in both wires for steel, `y = 20 xx 10^(10)Nm^(-2)` & for brass `y = 10 xx 10^(10) Nm^(-2)`.

Explanation

The situation in diagram
Let & B be a rod of length 2m suppose a weight W is hung at C at a distance x from A. Let `T_(1)` & `T_(2)` be the tension in the steel & brass rods.
(i) stress in steel wire `= (T_(1))/(A_(1))`
stress in brass wire = `(T_(2))/(A_(2))`
as both the stresses are equal, so
`(T_(1))/(A_(1)) = (T_(2))/(A_(2))` or `(T_(1))/(T_(2)) = (A_(1))/(A_(2))`

`= (0.1)/(0.2) = (1)/(2)`
Now moments about C are equals as the system is in equilibrium
`T_(1) = T_(2)(2-x)` or `(T_(1))/(T_(2)) = (2-x)/(x)`
`(1)/(2) = (2-x)/(x)`
`x = 4-2x`
`3x = 4` or `x = (4)/(3) = 1.33 m`
(ii) Now `y = ("stress")/("strain")`
`:. "strain" = ("stress")/(y)`
strain in steel wire `= (T_(1)//A_(1))/(y)`
strain in brass wire `= (T_(2)//A_(2))/(y)`
Now, `(T_(1))/(A_(1)y_(1)) = (T_(2))/(A_(2)y_(2)) : (T_(1))/(T_(2)) = (A_(1)y_(1))/(A_(2)y_(2))`
`= (0.1 cm^(2) xx 20 xx 10^(10) Nm^(-2))/(0.2m^(2) xx 10 xx 10^(10) Nm^(-2)) = 1`
again, `T_(1)x = T_(2)(2-x)`
`1 = (2-x)/(x)`
`x = 2 - x`
`2x = 2` or x = 1m