An 85.0-N backpack is hung from the middle of an aluminum wire, as the drawing shows. The temperature of the wire then drops by 20.0 C. Find the tension in the wire at the lower temperature. Assume that the distance between the supports does not change, and ignore any thermal stress.

T = 995.95 N

Explanation:

Using the equation to find the tension in the wire

2 * T * sin (φ) = m * g

T = m∙g / (2∙sin (φ))

cos (φ) = d/l ⇒ d = l∙cos (φ)

we know the angle of thew "hot" wire as φ₀ = 3°

d = l₀∙cos (φ₀)

α = 23*10⁻⁶K⁻¹ thermal expansion coefficient of aluminum

So the length changes to

l₁ = l₀ - ∆l = l₀∙ (1 - α∙∆T)

Since distance d does not change, the angle decreases such that:

d = l₁∙cos (φ₁) = l₀ * (1 - α * ∆T) * cos (φ₁)

l₀∙cos (φ₀) = l₀ * (1 - α * ∆T) * cos (φ₁)

φ₁ = cos⁻¹ (cos (φ₀) / (1 - α * ∆T))

φ₁ = cos⁻¹ (cos (3°) / (1 - 23*10⁻⁶K⁻¹ * 20.0 C))

φ₁ = 2.4457°

T = 85.0 N / (2 * sin (2.4457°))

T = 995.95 N