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# 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

### 1 Answer

- 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

Then

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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