A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a composite wire of length 3 m. Find the resulting change in the length of this composite wire if an object with a mass of 5 kg is hung vertically from one of its ends. (Neglect any effects the masses of the two wires have on the changes in their lengths.). Use 2 x 1011 N/m2 and 0.7 x 1011 N/m2 for the Young's modulus for steel and aluminum respectively.

Question

Jase Washington

Physics

17 May, 10:04

A steel wire of length 1.5 m and diameter 1 mm is joined to an aluminum wire of identical dimensions to make a composite wire of length 3 m. Find the resulting change in the length of this composite wire if an object with a mass of 5 kg is hung vertically from one of its ends. (Neglect any effects the masses of the two wires have on the changes in their lengths.). Use 2 x 1011 N/m2 and 0.7 x 1011 N/m2 for the Young's modulus for steel and aluminum respectively.

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Simone Moses · Answer 1

1.805 mm

Explanation:

Extension in the steel wire = WL_{steel}/AE_{steel}

Extension in the aluminium wire = WL_{Al}/AE_{Al}

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

we have:

W = mg

W = 5 * 9.8

W = 49 N

Area A = π/4 * (0.001) ²

= 7.85398 * 10 ⁻⁷ m²

Total extension = W/A * (L_{steel}/E_{steel} + L_{Al}/E_{Al})

Total extension = 49 / 7.85398 * 10 ⁻⁷ ((1.5 / 200*10⁹) + 1.5 / 70*10⁹))

Total extension = 0.0018048

Total extension = 1.805 mm

Thus, the total extension = the resulting change in the length of this composite wire = 1.805 mm