Physis Tension Question?
Two pillars, separated by 2m, have a cord strung between them that consists of two different wires attached to each other, a 5mm diameter nylon wire and a 1mm diameter copper wire. At the initial temperature of 25°C, there is 100N of tension in the wire.
a) If the temperature drops to 5°C, by how much does the joining point of the two wires move?
b) What is the final tension in the wires?
For nylon, we have Y = 3GPa and α = 5 × 10-5 K-1
For copper, we have Y = 117GPa and α = 1.6 × 10-5 K-1
- NCSLv 710 months agoFavorite Answer
Let's assume that each wire is 1 m long.
Y = σ / ε = (F/A) / (Δℓ/ℓ) = Fℓ / AΔℓ
so Δℓ = Fℓ / AY
for thermal expansion, Δℓ = ℓ*α*ΔT
Δℓ₁ + Δℓ₂ = ℓ₁*α₁*ΔT + ℓ₂*α₂*ΔT = F₁ℓ₁ / A₁Y₁ + F₂ℓ₂ / A₂Y₂
for nylon ₁ + and copper ₂
1m * 5e-5/K * -20K + 1m * 1.6e-5/K * -20K =
F₁*1m / (π/4)(0.005m)²(3e9Pa) + F₂*1m / (π/4))(0.001m)²(117e9Pa)
And since by Newton's Third
F₁ = F₂,
let's call it F, and then
-1.32e-3 m = F * 2.786e-5m/N
F = -47.4 N
which I take to mean that the tension has INCREASED by 47.4 N
So let's look at the left side of the splice.
The change in length of the nylon is a combination of the increased tension (stretching) and the decreased temperature (contracting).
Δx = F₁ℓ₁ / A₁Y₁ - ℓ₁*α₁*ΔT
Δx = 47.4N*1m / (π/4)(0.005m)²(3e9Pa) - 1m*5e-5/K*20K
Δx = -1.95e-4 m
so the splice moves 0.195 mm to the left.
Let's check that by looking at the copper:
Δx = 47.4N*1m / (π/4)(0.001m)²(117e9Pa) - 1m*1.6e-5*20K
Δx = 1.96e-4 m
so the splice moves 0.195 mm to the left (stretching the copper)
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