A steel wire has a cross sectional area 5×10^-5m^2.calculate the amount of tensile force to be applied to double it's length.Ysteel=2×10^11Nm^-2.
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Elongation (e) = L
Y = FL / Ae
F = YAe / L
= (2 * 10^11 * 5 * 10^-5 * L) / L
= 10^7 N
= 10 MN [where MN is mega Newton]
Tensile Force to be applied is 10 meganewton.
Y = FL / Ae
F = YAe / L
= (2 * 10^11 * 5 * 10^-5 * L) / L
= 10^7 N
= 10 MN [where MN is mega Newton]
Tensile Force to be applied is 10 meganewton.
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