Question

Q-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 0.70mm.Obtain the load applied.

Answer 1

Length of copper wire=2.2 cm
Length of steal wire=1.6 cm
Diameter cl=3 mm=3x10*-3m
cross section are=/piR*2=/pi(d/2)*2
3.14x2.25x10*-6m*2
7.1x10*-6m*2
Net elongated =0.70mm
=7x10*-4m
Let load applied is F
y=2.0x10*11N/m*2
T=1.3x10*11N/m*2
Ys=stress/strain=F/A/(ls/Ls)
Ys=(F/A)/(lc/Lc)
Ys/Yc=(F/A)/(ls/Ls)x(lc/Lc)/(F/A)
=(Ls/Lc)*(lc/ls)
2.0x10*11/1.3x10*11
(2.2)/(1.16)x(lc/ls)
lc/ls=2x1.6/1.3x2.2=1.12
lc=1.12ls
1.12ls+ls=7x10*-4m
(1.12ls)=7x10*4m
ls=3.30x10*-4m
Ys=Fxls/Axls
F=Ys x A x ls/LS
=2x10*11x7.16x10*-6x3.30x10*-4/1.6
=2x7.1x33/1.6
=292.87N

Answer 2

The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A.Stress = strain × Young’s modulus.  

ThereforeW/A = Yc × (ΔLc/Lc) = Ys × (ΔLs/Ls)

 whereThe subscripts c and s refer to copper and stainless steel respectively.  

Or,ΔLc /ΔLs = (Ys/Yc) × (Lc /Ls)Given Lc = 2.2 m, Ls = 1.6 m, Yc = 1.1 × 1011 Nm–2, and Ys = 2.0 × 1011 Nm–2.ΔLc/ΔLs = (2.0 × 1011/1.1 × 1011) × (2.2/1.6) = 2.5.

The total elongation is given to be

ΔLc + ΔLs = 7.0 × 10-4 mSolving the above equations,ΔLc = 5.0 × 10-4 m, and ΔLs = 2.0 × 10-4 m.

ThereforeW = (A × Yc × ΔLc)/Lc= π (1.5 × 10-3)2 × [(5.0 × 10-4 × 1.1 × 1011)/2.2]= 1.8 × 102 N