Two wires one of Steel and other of aluminium each of 2 m long and diameter 20 mm are joined end to end and a composite wire of 4 m is formed. What is the tension in the wire producef on a total extension of 0.9 millimetres?

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Solution

L = 4m ; A = πr2 = 3.14 x (1x10-3)2 = 3.14 x 10-9m2 ; F = ?
dlsteel+dlAl = 0.9 x 10-3m -- (Equ. 1)
Young's Modulus (Y) = F.L/A.dl
dlsteel = F.L/A.Ysteel
dlAl = F.L/A.YAl
Substituting in Equ 1
FL/A.Ysteel + F.L/A.YAl = 0.9 x 10-3m
FL/A x (1/Ysteel + 1/YAl ) = 0.9 x 10-3m
F = 0.9 x 10-3 x 3.14 x 10-9/4(1/Ysteel + 1/YAl )
F = 0.9 x 10-3 x 3.14 x 10-9/4[(YAl + Ysteel)/(Ysteel x YAl)]
F = (0.9 x 3.14 x 2 x 7 x 1011 x 1011 x 10-12) / 4 x 1011(2 + 7)
F = 9.891 x 1010 / 1011 x 9
F = 9.891 x 10-1 / 9
F = 0.109 N

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