What do we know?
Radius of Steel Wire = 1.25 X 10^{-3}m
Radius of Brass Wire = 1.25 X 10^{-3}m
Length of Steel Wire = 1.5 m
Length of Brass Wire = 1.0 m
Force on Steel Wire = 60N + 40N = 100N
Force on Brass Wire = 60N
Young's Modulus for steel is 2.0 X 10^{11} Nm^{-2}
Extension of brass wire = Extension of Steel wire
Plan to solve:
We know Stress = [^{[Force]} / _{[Cross Sectional Area]}]
Young's Modulus = [^{[Stress]} / _{[Strain]}]
Strain = [^{[ΔL]} / _{[L]}]
So we can get ΔL for the steel wire if we calculate Stress of Steel Wire and use the Young's modulus to find strain.
Strain = [^{[ΔL]} / _{[L]}]
Then we calculate the Stress and Strain of the Brass wire and from this we can find the Young's Modulus of the Brass wire.
Answer:
Steel Wire
Stress of Steel Wire= [(Force) / (Cross Sectional Area)]
Stress of Steel Wire= [(100N) / (πr^{2})]
Stress of Steel Wire= [(100N) / [(π) {1.25 X 10^{-3} m}^{2})]
Stress of Steel Wire= 2.04 X 10^{7 }Nm^{-2}
Modulus of Steel Wire= [^{[Stress]} / _{[Strain]}]
Strain of Steel Wire= [^{[Stress]} / _{[Modulus]}]
Strain of Steel Wire= 2.04 X 10^{7 }Nm^{-2} / 2.0 X 10^{11} Nm^{-2}
Strain of Steel Wire= 1.02 X 10^{-4 }Nm^{-2}
[^{[ΔL]} / _{[L]}] = 1.02 X 10^{-4 }Nm^{-2}
[ΔL] of Steel Wire = 1.02 X 10^{-4 }Nm^{-2} / [1.5m]
[ΔL] of Steel Wire = 1.53 X 10^{-4 }m
[ΔL] of Brass Wire = 1.53 X 10^{-4 }m
Brass
Stress of Brass Wire = [(Force) / (Cross Sectional Area)]
Stress of Brass Wire = [(60N) / (πr^{2})]
Stress of Brass Wire = [(60N) / [(π) {1.25 X 10^{-2} m}^{2})]
Stress of Brass Wire = 1.22 X 10^{5 }Nm^{-2}
Strain of Brass Wire = [^{[ΔL]} / _{[L]}]
Strain of Brass Wire = (1.53 X 10^{-4 }m) / 1.0 m
Strain of Brass Wire = (1.53 X 10^{-4})
Modulus of Brass Wire = [^{[Stress]} / _{[Strain]}]
Modulus of Brass Wire = [(1.22 X 10^{7 }Nm^{-2}) / (1.53 X 10^{-4})]
Modulus of Brass Wire = 8.02 X 10^{10} Nm^{-2} |