A rod of length 1.05 m having negliaible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in fig. The cross-sectional area of wire A and B are `1 mm^(2)` and 2` 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. Given,
`Y_(steel) = 2 xx 10^(11) Nm^(-2) and Y-(alumi n i um) = 7.0 xx 10^(10)N^(-2)`

Explanation

Answer: (a) 0.7 m from the steel-wire end
(b) 0.432 m from the steel-wire end
Cross-sectional area of wire A, `a_(1)=1.0"mm"^(2)=1.0xx10^(-6)"m"^(2)`
Cross-sectional area of wire B,`a_(2)=2.0"mm"^(2)=2.0xx10^(-6)"m"^(2)`
Young’s modulus for steel, `Y_(1)=2xx10^(11)"Nm"^(-2)`
Young’s modulus for aluminium, `Y_(2)=7.0xx10^(10)"Nm"^(-2)`
(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire= `("Force")/("Area")=(F)/(a)`
If the two wires have equal stresses, then:
`(F_(1))/(a_(1))=(F_(2))/(a_(2))`
Where,
`F_(1)` = Force exerted on the steel wire
`F_(2)` = Force exerted on the aluminum wire
`(F_(1))/(F_(2))=(a_(1))/(a_(2))=(1)/(2)" "...(i)`
The situation is shown in the following figure.

Taking torque about the point of suspension, we have:
`F_(1)y=F_(2)(1.05-y)`
`(F_(1))/(F_(2))=((1.05-y))/(y)" "...(ii)`
Using equations (i) and (ii), we can write:
`((1.05-y))/(y)=(1)/(2)`
`2(1.05-y)=y`
`2.1-2y=y`
`3y=2.1`
`thereforey=0.7"m"`
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.
(b) Young's modulus = `("Stress")/("Strain")`
Strain = `("Stress")/("Young's modulus")=((F)/(a))/(Y)`
If the strain in the two wires is equal, then:
`((F_(1))/(a_(1)))/(Y_(1))=((F_(2))/(a_(2)))/(Y_(2))`
`(F_(1))/(F_(2))=(a_(1))/(a_(2))(Y_(1))/(Y_(2))=(1)/(2)xx(2xx10^(11))/(7xx10^(10))=(10)/(7)" "...(iii)`
Taking torque about the point where mass m, is suspended at a distance y1 from the side where wire A attached, we get:
`(F_(1))/(F_(2))=((1.05-y_(1)))/(y_(1))" "...(iii)`
Using equations (iii) and (iv), we get:
`((1.05-y_(1)))/(y_(1))=(10)/(7)`
`7(1.05-y_(1))=10y_(1)`
`17y_(1)=7.35`
`thereforey_(1)=0.432` m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.