Question

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.


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Answer 1

Answer: 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

so the answer is 1.8 × 102 N

Explanation :

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Answer 2

Answer:

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

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