A steel wire of 80m length 1mm2 cross sectional area is freely hanging from a tower. What will be its elongation due to its self weight? Take specific weight of the steel as 78.6KN/m2 and modulus of elasticity as 200GPa
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Answer:
The force required will be 400 N
Explanation :
In the first case
Length L = 1 m
Area A = 1 mm² = 10⁻⁶ m²
Force F = 200 N
Change in length , L' = 1 mm = 10⁻³ m
we know that young's modulus is given by
Y = FL/AL'
= 200 x 1/(10⁻⁶ x 10⁻³)
= 2 x 10¹¹
Similarly in the second case,
L = 10 m
Area A = 1 mm² = 10⁻⁶ m²
New length = 1002 cm = 10.02 m
Change in length , L' = 10.02 - 10 = 0.02 m = 2 x 10⁻² m
Hence,
Y = FL/AL'
=> F = YAL'/L
= (2 x 10¹¹ x 10⁻⁶ x 2 x 10⁻²)/10
= 400 N
Hence the force required will be 400 N
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A Steel wire of 80m length 1mm2 cross sectional area is freely hanging from a tower. what will be it's elongation due to its self weight? Take specific weight of the steel as 78.6N/m2 and modulus of elasticity as 200Gpa
ANS:-- 1.3mm
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