A light rod of length 200cm is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is cross-section `0.1cm^(2)` and the other of brass of cross-section `0.2cm^(2)`. Along the rod at which 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 `10xx10^(11)` and `20xx10^(11)"dyne"//cm^(2)` 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)` or `T_(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 equilibrium 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_(2))/(A_(2)Y_(2))`[as strain `=("stress")/(Y)]`
So, `(T_(1))/(T_(2))=(A_(1)Y_(1))/(A_(2)Y_(2))=(0.1xx20xx10^(11))/(0.2xx10xx10^(11))=1`, i.e., `T_(1)=T_(2)`..........(iii)
So, for translatory equilibrium of rod, `T_(1)+T_(2)=W` in the light of Eqn. (iii) yields.
`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=1m`