A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load , the net elongation is found to be `0.70` mm. Obtain the load applied.

Explanation

The copper and steel wires are under a tensile stress because they have the same tension (equal to load W) and 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(2.2//1.6)=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)=2.0xx10^(-4),.`
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`