A 2.2m of steel wire hangs vertically down from a support. A 1.5m length of brass wire of the same diameter is attached to the lower end of the steel wire.when a 10 kg mass is attached to the lower end of the brass wire a total extension of 0.05 m is produced. by how much does each wire extend? (young's modulus for steel = 2.0 x 10^11pa, young's modulus for brass = 9.2 x 10^10 pa)
To find out how much each wire extends, we can use Hooke's Law, which states that the extension of a spring or wire is directly proportional to the force applied to it.
The extension of a wire is given by the equation:
extension = (Force * Length) / (Young's modulus * Cross-sectional area)
For the steel wire:
Let's denote the extension of the steel wire as E₁.
Length of the steel wire = 2.2 m
Young's modulus for steel (E₁) = 2.0 x 10^11 Pa
Cross-sectional area for the steel wire = (π * r^2) -- (1)
For the brass wire:
Let's denote the extension of the brass wire as E₂.
Length of the brass wire = 1.5 m
Young's modulus for brass (E₂) = 9.2 x 10^10 Pa
Cross-sectional area for the brass wire = (π * r^2) -- (2)
Given:
Total extension = 0.05 m
Force = mass * acceleration due to gravity = 10 kg * 9.8 m/s^2 = 98 N
Since the wires are attached in series (one wire is attached to the other), the total extension is divided between the two wires:
Total extension = E₁ + E₂
Now, since the extension is proportional to the force applied, we can write:
E₁ / E₂ = (Force₁ / Force₂) = (Length₁ / Length₂) * (Young's modulus₂ / Young's modulus₁)
Substituting the given values:
E₁ / E₂ = (2.2 / 1.5) * (9.2 x 10^10 / 2.0 x 10^11)
Now, solve for E₁ and E₂:
E₁ + E₂ = 0.05 (from the total extension equation)
Now, you can solve these simultaneous equations to find out the extensions E₁ and E₂.