A Copper wire of length 2.2m and a steel wire of length 1.6m, both of diameter 3.0mm are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied . Young's modulus of copper is `1.1 xx 10^(11)Nm^(-2)` and Young's modulus of steel is `2.0 xx 10^(11)Nm^(-2)`.

Explanation

The copper and steel wires are under a tensile stress because they have the same area of cross-section `A`. From Eq. (9.7) we have stress = strain `xx` Young's modulus. Therefore
`" "W//A=Y_(c)xx(DeltaL_(c)//L_(c))=Y_(s)xx(DeltaL_(s)//L_(s))`
where the subscripts `c` and `s` refer to copper and stainless steel respectively. Or,
`" "DeltaL_(c)//DeltaL_(s)=(Y_(s)//Y_(c))xx(L_(c)//L_(s))`
Given `L_(c)` = 2.2 m, `L_(s)` = 1.6 m,
From Table 9.1 `Y_(c)=1.1xx10^(11)"N.m"^(-2)`, and
`" "Y_(s)=2.0xx10^(11)"N.m"^(-2)`
`DeltaL_(c)//DeltaL_(s)=(2.0xx10^(11)//1.1xx10^(11))xx(2.0xx1.6)=2.5`
The total elongation is given to be
`" "DeltaL_(c)+DeltaL_(s)=7.0xx10^(-4)"m"`
Solving the above equations,
`DeltaL_(c)=5.0xx10^(-4)"m",and DeltaL_(s)=20.0xx10^(-4)"m"`.
Therefore
`W=(AxxY_(c)xxDeltaL_(c))//L_(c)`
`" "pi(1.5xx10^(-3))^(2)xx[(5.0xx10^(-4)xx1.1xx10^(11))//2.2]`
`" "=1.8xx10^(2)"N"`