A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in figure. The cross-sectional areas of wires A and B are 1.0 `mm^(2)` and 2.0 `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.

Explanation

Let system 1 for streel and 2 for aluminium,
Length of steel wire `A, l _(1) = l`
Area of cross section of steel wire `A _(1) = 1 mm ^(2)`
Young.s modulus of steel wire `Y_(1) = 2 xx 10 ^(11) Nm ^(-2)`

Length of aluminium `l _(2) =l`
Area of cross-section
`A _(2) =2mm ^(2)`
Young.s modulus `Y_(2) = 7 xx 10 ^(10) Nm ^(-2)`
(a) Let mass m be suspended from the rod at distance x from the end.
Let `F_(1) and F_(2)` be the tension in two wires and there is equal stress in two wires then,
`therefore (F _(1))/(A _(1)) = (F_(2))/(A _(2))`
`therefore (F _(1))/(F_(2)) = (A_(1))/(A_(2))`
`therefore (F _(1))/(F_(2)) = 1/2 " "...(1)`
Taking moment of forces about the point of suspension of mass from the rod.
`F_(1) x =F_(2) (1.05 -x)`
`therefore (F_(1))/(F _(2))= (1.05 -x)/(x)`
`therefore 1/2 = (1.05 -x)/(x) " "[because` From equation (1)]
`therefore x =2.1 -2x`
`therefore 3x =2.1`
`therefore x =0.7 m or x = 70 cm`
(b) Let mass m be suspended from the rod at distance x from the end. There is equal strain in two wires,
`Y = (F //A)/( epsi _(l ))` where `epsi _(l) =` longitudinal strain
`therefore epsi _(l) = (F)/(AY)`
`therefore` For steel wire `: (epsi _(l)) _(1) = (F_(1))/( A _(1) Y_(1))`
For aluminium wire `: (epsi _(l)) _(1) = (F_(1))/( A _(1) Y_(1))`
For aluminium wire `: (epsi _(l)) _(2) = (F_(2))/(A_(2) Y_(2))`
but `(epsi l ) _(1) = (epsi _(l)) _(2)`
`therefore (F._(1))/(A _(1) Y _(1)) =(F._(2))/(A _(2) Y_(2))`
`therefore (F ._(1))/(F ._(2)) = (A_(1)Y _(1))/( A _(2) Y _(2))`
`= 1/2 xx (2 xx 10 ^(11))/( 7 xx 10 ^(10))`
`(F._(1))/(F ._(2)) = (10)/(7)" "...(2)`
Taking moment of force about the point of suspension of mass from the rod,
`F._(1) x = F ._(2) (1.05 -x)`
`therefore (F._(1))/(F._(2)) = (1.05 -x)/(x)`
`(10)/(7) = (1.05 -x)/(x) " "[because `From equation (2)]
`therefore 10 x =7.35 -7x`
`therefore 17 x -7.35`
`therefore x = (7.35)/(17) =0.43235 m or ~~43.2 cm`