Force P is 11304 N and normal stress is 400 N/mm²
Explanation:
Given-
Length, l = 9 m = 9000 mm
Diameter, d = 6 mm
Radius, r = 3 mm
Stretched length, Δl= 18 mm
Modulus of elasticity, E = 200 GPa = 200 X 10³MPa
Force, P = ?
According to Hooke's law,
Stress is directly proportional to strain.
So,
σ ∝ ε
σ = E ε
Where, E is the modulus of elasticity
We know,
ε = Δl / l
So,
σ = E X Δl/l
σ =
[tex]200 X 10^3 * \frac{18}{9000} \\\\ = 400N/mm^2[/tex]
We know,
σ = P/A
And A = π (r)²
σ = P / π (r)²
[tex]400 N/mm^2 = \frac{P}{3.14 X (3)^2} \\\\400 = \frac{P}{28.26} \\\\P = 11304N[/tex]
Therefore, Force P is 11304 N and normal stress is 400 N/mm²
The magnitude of the force is 11.3KN and the normal stress is 400 MPa
Given that length (L) = 9 m, diameter (d) = 6 mm = 6 * 10⁻³ m, extension (δ) = 18 mm = 18 * 10⁻³ m, E = 200 GPa = 200 * 10⁹ Pa
The area of the wire (A) is:
[tex]A=\pi*\frac{diameter^2}{4}=\pi*\frac{(6*10^{-4})^2}{4} =28*10^{-6}\ m^2[/tex]
[tex]\delta=\frac{PL}{AE} \\\\P=\frac{AE\delta}{L}=\frac{28*10^{-6}*200*10^9*18*10^{-3}}{9}=11300N\\\\\\Normal\ stress(\sigma)=\frac{P}{A} =\frac{11300}{28*10^{-6}} =400*10^6\ Pa[/tex]
The magnitude of the force is 11.3KN and the normal stress is 400 MPa
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