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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 \(4.4 \mathrm{~mm},\) and a crosssectional area of width \(3.5 \mathrm{~mm},\) and thickness \(0.45 \mathrm{~mm}\) on the tooth. Before placing the band on the tooth, he heated the band to \(70 .{ }^{\circ} \mathrm{C}\). What will be the tension in the band once it cools down to the temperature in your friend's mouth \(\left(37^{\circ} \mathrm{C}\right) ?\)

Short Answer

Expert verified
Answer: The tension in the steel band is approximately \(368.65\,\mathrm{N}\).

Step by step solution

01

Calculate the change in temperature

We need to determine the change in temperature experienced by the steel band when it cools from \(70^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\). The change in temperature, \(\Delta T\), is given by: \(\Delta T = T_{final} - T_{initial}\) Here, \(T_{initial} = 70^{\circ} \mathrm{C}\) and \(T_{final} = 37^{\circ} \mathrm{C}\). So, we get: \(\Delta T = 37 - 70 = -33 ^{\circ} \mathrm{C}\)
02

Calculate the change in diameter due to temperature change

Now, we will use the linear expansion formula to determine the change in diameter due to the temperature change. The formula for linear expansion is: \(\Delta L = L_{initial} \cdot \alpha \cdot \Delta T\) Here, \(\Delta L\) is the change in length (or diameter in this case), \(L_{initial} = 4.4\,\mathrm{mm}\) is the initial diameter, \(\alpha\) is the coefficient of linear expansion for steel (approximately \(12 \cdot 10^{-6} \mathrm{ \, K^{-1}}\)), and \(\Delta T\) is the change in temperature. Plugging in the values, we have: \(\Delta L = 4.4\,\mathrm{mm} \cdot 12 \cdot 10^{-6} \mathrm{ \, K^{-1}} \cdot (-33)^{\circ} \mathrm{C}\) \(\Delta L \approx -0.00175\, \mathrm{mm}\)
03

Calculate the final diameter of the steel band

Now that we know the change in diameter, we can find the final diameter \(D_{final}\) by adding the change in diameter, \(\Delta L\), to the initial diameter, \(D_{initial}\). \(D_{final} = D_{initial} + \Delta L = 4.4\,\mathrm{mm} - 0.00175\,\mathrm{mm} \approx 4.39825\,\mathrm{mm}\)
04

Calculate the Young's modulus for steel

To find the tension in the band, we will need to use Hooke's Law. Hooke's Law relates stress, strain, and Young's modulus (Y) of the material. For steel, the average value of Young's modulus is approximately \(2 \times 10^{11} \mathrm{\, N/m^2}\).
05

Apply Hooke's Law to calculate the tension in the band

Hooke's Law states: \(\sigma = Y \cdot \epsilon\) We can find the stress in the band (\(\sigma\)) as the tension (\(F\)) divided by the cross-sectional area of the band (\(A\)). \(\sigma = \frac{F}{A}\) Strain (\(\epsilon\)) is defined as the change in diameter divided by the initial diameter: \(\epsilon = \frac{\Delta L}{D_{initial}}\) We can rearrange the Hooke's Law equation to find tension: \(F = \sigma \cdot A = Y \cdot \epsilon \cdot A\) Using our values, we get: \(F = (2 \times 10^{11} \mathrm{\, N/m^2})\left(\frac{-0.00175\,\mathrm{mm}}{4.4\,\mathrm{mm}}\right)(3.5\,\mathrm{mm} \times 0.45\,\mathrm{mm})\) After plugging in these values and performing the calculation, we obtain: \(F \approx -368.65 \,\mathrm{N}\) The negative sign indicates that the force is compressive, which is expected as the band is trying to contract. In terms of tension, the magnitude of the force is 368.65 N, and that's the answer. So, the tension in the steel band when it cools down to the temperature of your friend's mouth is approximately \(368.65\,\mathrm{N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Expansion
Linear expansion refers to the phenomenon where materials change in length in response to a change in temperature. When a substance is heated, its particles move more rapidly and tend to take up more space, leading to an expansion. Conversely, cooling a substance slows down the particle movement, causing it to contract.

Considering the practical example of a steel band heated before being placed on a tooth, the change in its diameter upon cooling is a direct result of linear expansion. It's essential to understand that linear expansion affects solids in one dimension, primarily length, and is a critical consideration in construction, manufacturing, and even dental care as seen in this example.
Coefficient of Linear Expansion
The coefficient of linear expansion is a specific value that quantifies how much a material will expand or contract per degree change in temperature. This coefficient, denoted as \( \alpha \), is a property inherent to the material and can vary widely from one substance to another.

In our example, steel's coefficient of linear expansion is used to calculate how much the steel band will shrink when it cools. By knowing \( \alpha \) and the change in temperature, we can predict the precise dimensional change a material will undergo, allowing for accurate design and engineering of components that must endure temperature fluctuations.
Young's Modulus
Young's modulus is a measure of the stiffness of a solid material. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in a material. Young's modulus is a fundamental property that indicates the tensile (or compressive) elasticity, which is the ability of a material to undergo elongation (or compression) along an axis when opposing forces are applied.

Regarding our steel band example, the Young's modulus of steel is used to relate the tension in the band to the strain produced by the contraction due to cooling. A high Young's modulus indicates that steel is a highly stiff material, which means it resists deformation well, ensuring that once the band contracts, it will apply a significant force to remain in its new size.
Hooke's Law
Hooke's Law is a principle that states that the force needed to extend or compress a spring by some distance is proportional to that distance. This law can also apply to any elastic material, like our steel band. The formula for Hooke's Law is \( F = -kx \), where \( F \) is the force, \( k \) is the spring constant (or stiffness), and \( x \) is the distance of extension or compression.

In the context of the problem, Hooke's Law provides the framework to understand how the steel band's tension is calculated. It enables us to determine how the internal forces within the band react to the external disturbance caused by the temperature change, and thus, we can calculate the tension in the band once it is cooled down to body temperature.

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Most popular questions from this chapter

When a 50.0 -m-long metal pipe is heated from \(10.0^{\circ} \mathrm{C}\) to \(40.0^{\circ} \mathrm{C}\), it lengthens by \(2.85 \mathrm{~cm}\). a) Determine the linear expansion coefficient. b) What type of metal is the pipe made of?

a) Suppose a bimetallic strip is constructed of copper and steel strips of thickness \(1.0 \mathrm{~mm}\) and length \(25 \mathrm{~mm},\) and the temperature of the strip is reduced by \(5.0 \mathrm{~K}\). Determine the radius of curvature of the cooled strip (the radius of curvature of the interface between the two strips). b) If the strip is \(25 \mathrm{~mm}\) long, how far is the maximum deviation of the strip from the straight orientation?

In order to create a tight fit between two metal parts, machinists sometimes make the interior part larger than the hole into which it will fit and then either cool the interior part or heat the exterior part until they fogether. Suppose an aluminum rod with diameter \(D_{1}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right)\) is to be fit into a hole in a brass plate that has a diameter \(D_{2}=10.000 \mathrm{~mm}\) (at \(\left.2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\right) .\) The machinists can cool the rod to \(77.0 \mathrm{~K}\) by immersing it in liquid nitrogen. What is the largest possible diameter that the rod can have at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C}\) and just fit into the hole if the rod is cooled to \(77.0 \mathrm{~K}\) and the brass plate is left at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} ?\) The linear expansion coefficients for aluminum and brass are \(22 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and \(19 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\), respectively.

The background temperature of the universe is a) \(6000 \mathrm{~K}\). b) \(288 \mathrm{~K}\). c) \(3 \mathrm{~K}\). d) \(2.73 \mathrm{~K}\). e) \(0 \mathrm{~K}\).

You are designing a precision mercury thermometer based on the thermal expansion of mercury \(\left(\beta=1.8 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\right)\) which causes the mercury to expand up a thin capillary as the temperature increases. The equation for the change in volume of the mercury as a function of temperature is \(\Delta V=\beta V_{0} \Delta T\) where \(V_{0}\) is the initial volume of the mercury and \(\Delta V\) is the change in volume due to a change in temperature, \(\Delta T .\) In response to a temperature change of \(1.0^{\circ} \mathrm{C}\), the column of mercury in your precision thermometer should move a distance \(D=1.0 \mathrm{~cm}\) up a cylindrical capillary of radius \(r=0.10 \mathrm{~mm} .\) Determine the initial volume of mercury that allows this change. Then find the radius of a spherical bulb that contains this volume of mercury.

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