An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by $\Delta T$. Find change in the angle ABC. [Coeff. of linear expansion for Cu is $\alpha_1$, Coeff. of linear expansion for Al is $\alpha_2$]

> [!TLDR] > When an equilateral triangle made of different materials expands due to a temperature increase, the unequal expansion of its sides changes its angles. The change in angle ABC is $\boxed{\frac{2}{\sqrt{3}}(\alpha_2 - \alpha_1)\Delta T}$. The core idea here is that different materials expand by different amounts when heated. This differential expansion will change the relative lengths of the sides of the triangle, which in turn alters its angles. We will use the principle of linear thermal expansion to find the new lengths of the rods and then apply the Law of Cosines to determine the new angle. Since thermal expansion coefficients are very small, we will use approximations for small changes in length and angle. 1. **Initial Geometry of the Equilateral Triangle** An equilateral triangle has all three sides equal in length and all three interior angles equal to $60^\circ$. Let the initial length of each rod be $L$. So, $AB = BC = AC = L$. The initial angle $\angle ABC = \theta = 60^\circ$. 2. **Thermal Expansion of the Rods** When the temperature of each rod increases by $\Delta T$, its length changes. The change in length $\Delta L$ for a rod of initial length $L$, coefficient of linear expansion $\alpha$, and temperature change $\Delta T$ is given by $\Delta L = L \alpha \Delta T$. The new length $L'$ is $L' = L + \Delta L = L(1 + \alpha \Delta T)$. * Rods AB and BC are made of Copper (Cu), with a coefficient of linear expansion $\alpha_1$. Their new lengths will be: $AB' = L(1 + \alpha_1 \Delta T)$ $BC' = L(1 + \alpha_1 \Delta T)$ * Rod AC is made of Aluminum (Al), with a coefficient of linear expansion $\alpha_2$. Its new length will be: $AC' = L(1 + \alpha_2 \Delta T)$ 3. **Applying the Law of Cosines to the Expanded Triangle** After expansion, the triangle $A'B'C'$ is generally no longer equilateral because the sides have expanded by different amounts. We need to find

An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that

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\Delta T

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Concept understanding — Thermal Expansion Coefficient

Thermal Expansion Coefficient: From Intuition to Precision

The Intuition: What Happens When Things Get Hot?

Think about a metal railway track on a summer day. The track is laid in sections with small gaps between them. On a hot afternoon, those gaps get smaller — sometimes the track even buckles. Why? Because the metal expands when heated.

Or consider a mercury thermometer. The liquid mercury sits in a bulb at the bottom. When your body warms the bulb, the mercury expands and rises up the narrow tube. The hotter you are, the higher it climbs.

This is thermal expansion: most materials get bigger when heated and smaller when cooled. The atoms inside vibrate more vigorously as temperature rises, pushing their neighbours slightly farther apart. The entire object grows in every direction.

But different materials expand by different amounts. A steel rod and an aluminium rod of the same length, heated by the same amount, will not end up the same length. Aluminium expands more. So we need a number that tells us how much a given material expands per degree of temperature change. That number is the thermal expansion coefficient.

The Precise Statement: Defining the Coefficient

There are actually three coefficients, depending on whether we care about length, area, or volume. For a first meeting, we focus on the most common one: the linear thermal expansion coefficient, denoted by the Greek letter $\alpha$ (alpha).

\alpha = \frac{1}{L_0} \cdot \frac{\Delta L}{\Delta T}

Where:

$L_0$ is the original length of the object (at some starting temperature)
$\Delta L$ is the change in length (final length minus original length)
$\Delta T$ is the change in temperature (final temperature minus initial temperature)

What this formula says in plain English: The coefficient $\alpha$ is the fractional change in length per degree of temperature change. If $\alpha = 2.5 \times 10^{-5} \, \text{per} \, ^\circ\text{C}$ , it means that for every $1^\circ\text{C}$ rise in temperature, the material expands by $0.0025\%$ of its original length.

How to Use It: The Working Formula

From the definition, we can rearrange to get the practical formula:

\Delta L = \alpha \, L_0 \, \Delta T

So the final length $L$ after a temperature change is:

L = L_0 + \Delta L = L_0 (1 + \alpha \, \Delta T)

Tip

For small temperature changes (say, less than $100^\circ\text{C}$ ), this linear approximation is excellent. For very large changes, the coefficient itself may change slightly with temperature, but at the introductory level we treat $\alpha$ as constant.

A Concrete Example

A steel bridge girder is $50.00 \, \text{m}$ long at $20^\circ\text{C}$ . The linear expansion coefficient of steel is $\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$ . How much longer is it on a $40^\circ\text{C}$ day?

Step 1: Identify the quantities.

$L_0 = 50.00 \, \text{m}$
$\Delta T = 40 - 20 = 20^\circ\text{C}$
$\alpha = 1.2 \times 10^{-5} \, /^\circ\text{C}$

Step 2: Apply the formula.

\Delta L = \alpha \, L_0 \, \Delta T = (1.2 \times 10^{-5})(50.00)(20)

Step 3: Calculate.

\Delta L = 1.2 \times 10^{-5} \times 1000 = 0.012 \, \text{m} = 1.2 \, \text{cm}

So the girder expands by $1.2 \, \text{cm}$ . That is why bridges have expansion joints — without them, the structure would buckle.

Two Important Cousins: Area and Volume Expansion

For a thin sheet (like a metal plate), we care about area expansion. The area expansion coefficient is approximately $2\alpha$ . For a solid object, the volume expansion coefficient is approximately $3\alpha$ . These come from the same idea: if every linear dimension grows by a factor $(1 + \alpha \Delta T)$ , then area grows by $(1 + \alpha \Delta T)^2 \approx 1 + 2\alpha \Delta T$ , and volume by $(1 + \alpha \Delta T)^3 \approx 1 + 3\alpha \Delta T$ .

Watch out

These approximations ( $2\alpha$ and $3\alpha$ ) are valid only when $\alpha \Delta T$ is small compared to 1. For most solids and modest temperature changes, this is true. For gases, the expansion is much larger and a different treatment is needed.

What the Coefficient Tells Us About Materials

| Material | $\alpha$ (per $^\circ\text{C}$ ) | Behaviour |

|----------|--------------------------------|-----------| …

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The key idea is that the thermal expansion of the rods changes their lengths, which in turn alters the angles of the triangle.

Let the initial length of each rod be $L$ . When the temperature increases by $\Delta T$ , the new lengths of the rods are:
- $L'_{AB} = L(1 + \alpha_1 \Delta T)$
- $L'_{BC} = L(1 + \alpha_1 \Delta T)$
- $L'_{CA} = L(1 + \alpha_2 \Delta T)$
Let the new angle ABC be $\theta'$ . Applying the cosine rule to the new triangle A'B'C':

$(L'_{CA})^2 = (L'_{AB})^2 + (L'_{BC})^2 - 2 (L'_{AB}) (L'_{BC}) \cos \theta'$

Substituting the new lengths and dividing by $L^2$ :

$(1 + \alpha_2 \Delta T)^2 = 2(1 + \alpha_1 \Delta T)^2 - 2(1 + \alpha_1 \Delta T)^2 \cos \theta'$

This simplifies to:

$\cos \theta' = 1 - \frac{(1 + \alpha_2 \Delta T)^2}{2(1 + \alpha_1 \Delta T)^2}$
Since $\alpha \Delta T$ is very small, we use the approximation $(1+x)^n \approx 1+nx$ and $(1+x)^{-1} \approx 1-x$ :

$\cos \theta' \approx 1 - \frac{1 + 2\alpha_2 \Delta T}{2(1 + 2\alpha_1 \Delta T)} \approx 1 - \frac{1}{2}(1 + 2\alpha_2 \Delta T)(1 - 2\alpha_1 \Delta T)$

Neglecting terms with $(\Delta T)^2$ :

$\cos \theta' \approx 1 - \frac{1}{2}(1 + 2\alpha_2 \Delta T - 2\alpha_1 \Delta T) = \frac{1}{2} - (\alpha_2 - \alpha_1) \Delta T$ …

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When an equilateral triangle made of different materials expands due to a temperature increase, the unequal expansion of its sides changes its angles. The change in angle ABC is $\boxed{\frac{2}{\sqrt{3}}(\alpha_2 - \alpha_1)\Delta T}$ .

The core idea here is that different materials expand by different amounts when heated. This differential expansion will change the relative lengths of the sides of the triangle, which in turn alters its angles. We will use the principle of linear thermal expansion to find the new lengths of the rods and then apply the Law of Cosines to determine the new angle. Since thermal expansion coefficients are very small, we will use approximations for small changes in length and angle.

Initial Geometry of the Equilateral Triangle

An equilateral triangle has all three sides equal in length and all three interior angles equal to $60^\circ$ .

Let the initial length of each rod be $L$ .

So, $AB = BC = AC = L$ .

The initial angle $\angle ABC = \theta = 60^\circ$ .
Thermal Expansion of the Rods

When the temperature of each rod increases by $\Delta T$ , its length changes. The change in length $\Delta L$ for a rod of initial length $L$ , coefficient of linear expansion $\alpha$ , and temperature change $\Delta T$ is given by $\Delta L = L \alpha \Delta T$ . The new length $L'$ is $L' = L + \Delta L = L(1 + \alpha \Delta T)$ .
- Rods AB and BC are made of Copper (Cu), with a coefficient of linear expansion $\alpha_1$ .
  
  Their new lengths will be:
  
  $AB' = L(1 + \alpha_1 \Delta T)$
  
  $BC' = L(1 + \alpha_1 \Delta T)$
- Rod AC is made of Aluminum (Al), with a coefficient of linear expansion $\alpha_2$ .
  
  Its new length will be:
  
  $AC' = L(1 + \alpha_2 \Delta T)$
Applying the Law of Cosines to the Expanded Triangle

After expansion, the triangle $A'B'C'$ is generally no longer equilateral because the sides have expanded by different amounts. We need to find the new angle $\angle A'B'C'$ , which we will denote as $\theta'$ .

The Law of Cosines relates the lengths of the sides of a triangle to one of its angles:
$c^2 = a^2 + b^2 - 2ab \cos C$
Applying this to triangle $A'B'C'$ for the angle $\theta'$ at vertex B:

$(AC')^2 = (AB')^2 + (BC')^2 - 2(AB')(BC') \cos\theta'$

Substitute the expanded lengths from Step 2:

$[L(1 + \alpha_2 \Delta T)]^2 = [L(1 + \alpha_1 \Delta T)]^2 + [L(1 + \alpha_1 \Delta T)]^2 - 2[L(1 + \alpha_1 \Delta T)][L(1 + \alpha_1 \Delta T)] \cos\theta'$

Divide the entire equation by $L^2$ :

$(1 + \alpha_2 \Delta T)^2 = 2(1 + \alpha_1 \Delta T)^2 - 2(1 + \alpha_1 \Delta T)^2 \cos\theta'$
Using Approximations for Small Changes

The terms $\alpha_1 \Delta T$ and $\alpha_2 \Delta T$ are typically very small (e.g., $10^{-5}$ to $10^{-3}$ ). This allows us to use the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$ .

Applying this approximation to the squared terms:

$(1 + \alpha_2 \Delta T)^2 \approx 1 + 2\alpha_2 \Delta T$

$(1 + \alpha_1 \Delta T)^2 \approx 1 + 2\alpha_1 \Delta T$

Substitute these into the Law of Cosines equation:

$1 + 2\alpha_2 \Delta T \approx 2(1 + 2\alpha_1 \Delta T) - 2(1 + 2\alpha_1 \Delta T) \cos\theta'$

Now, rearrange the equation to solve for $\cos\theta'$ :

$2(1 + 2\alpha_1 \Delta T) \cos\theta' \approx 2(1 + 2\alpha_1 \Delta T) - (1 + 2\alpha_2 \Delta T)$

$2(1 + 2\alpha_1 \Delta T) \cos\theta' \approx 1 + 4\alpha_1 \Delta T - 2\alpha_2 \Delta T$

$\cos\theta' \approx \frac{1 + (4\alpha_1 - 2\alpha_2)\Delta T}{2(1 + 2\alpha_1 \Delta T)}$

$\cos\theta' \approx \frac{1}{2} [1 + (4\alpha_1 - 2\alpha_2)\Delta T] (1 + 2\alpha_1 \Delta T)^{-1}$

Again, using the approximation $(1+x)^{-1} \approx 1-x$ for small $x$ :

$\cos\theta' \approx \frac{1}{2} [1 + (4\alpha_1 - 2\alpha_2)\Delta T] (1 - 2\alpha_1 \Delta T)$

Expand the product, neglecting terms involving $(\Delta T)^2$ because they are negligibly small: …

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