promotion image of download ymail app
Promoted
Anonymous
Anonymous asked in Science & MathematicsPhysics · 6 months ago

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

Relevance
  • NCS
    Lv 7
    6 months ago
    Favorite 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)

    If you find this helpful, please award Best Answer!

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.