Question

Find the minimum diameter of an l = 16.7 m long steel wire that will stretch no more than 9.53 mm when a mass of 361 kg is hung on the lower end. (Hint: The Young's modulus of steel is 200.0 GPa.)

Answer 1

Young's Elasticity, E = stress / strain = (F/A) / (ΔL/L) = (F* L) / (A* ∆L)

A = F * L / (E * ∆L)= 361kg * 9.8m/s² * 16.7m / (2 * 10^11N/m² * 0.00953m)

A = 3.099 * 10^-5 m²

and since area A = π * d²/4, we have

π * d²/4 = 3.099 * 10^-5 m²

solving for diameter,

d = 0.00628 m = 6.28 mm

the minimum diameter of the steel wire is 6.28 mm.

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