Question

In a pickup basketball game, your friend cracked one of his teeth in a collision with another player while attempting to make a basket. To correct the problem, his dentist placed a steel band of initial internal diameter \(4.40 \mathrm{~mm},\) and a cross-sectional area of width \(3.50 \mathrm{~mm},\) and thickness \(0.450 \mathrm{~mm}\) on the tooth. Before placing the band on the tooth, he heated the band to \(70.0^{\circ} \mathrm{C}\). What will be the tension in the band once it cools down to the temperature in your friend's mouth \(\left(36.8^{\circ} \mathrm{C}\right) ?\) The steel used for the band has a linear expansion coefficient of \(\alpha=13.0 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and a Young's modulus of \(Y=200 . \cdot 10^{9} \mathrm{~N} / \mathrm{m}^{2}\).

Short answer

Answer: The tension in the steel dental band after cooling down is approximately -136 N, which indicates contraction due to the temperature decrease.

Step-by-step solution

Calculate the change in diameter due to temperature change

We can use the linear expansion formula to determine how much the diameter of the steel band changes due to the decrease in temperature: ∆D = α * D_initial * ∆T We are given the following values: - α = 13.0 × 10⁻⁶ °C⁻¹ - D_initial = 4.40 mm - ∆T = T_final - T_initial = 36.8 °C - 70.0 °C = -33.2 °C Now we can plug in these values and compute ∆D: ∆D = (13.0 × 10⁻⁶ °C⁻¹) × (4.40 × 10⁻³ m) × (-33.2 °C) ≈ -1.9 × 10⁻⁶ m

Calculate the strain caused by the change in diameter

By definition, strain (ε) is the fractional change in dimension caused by the applied stress or force. In our case, it's the change in diameter (∆D) divided by the initial diameter (D_initial): ε = ∆D / D_initial ≈ (-1.9 × 10⁻⁶ m) / (4.40 × 10⁻³ m) ≈ -4.3 × 10⁻⁴

Calculate the tension in the steel band

The mechanical stress (σ) experienced by the steel band is related to the strain (ε) through Young's modulus (Y): σ = Y × ε We are given Young's modulus (Y) as 200 × 10⁹ N/m². Plugging in the values for Y and ε, we can compute the stress (σ): σ = (200 × 10⁹ N/m²) × (-4.3 × 10⁻⁴) ≈ -8.6 × 10⁴ N/m² Now, we know that stress is force per unit area, which means that tension (T) is the stress (σ) multiplied by the cross-sectional area (A) of the steel band. The cross-sectional area can be computed as the product of the width and thickness of the band, both given: A = 3.50 mm × 0.450 mm = 1.58 × 10⁻⁶ m². Now we can calculate the tension (T) in the steel band: T = σ × A ≈ (-8.6 × 10⁴ N/m²) × (1.58 × 10⁻⁶ m²) ≈ -136 N After cooling down to the temperature of your friend's mouth, the steel dental band experiences a negative tension of approximately -136 N, meaning that it contracts due to the temperature decrease.