Three particles of masses `1kg`, `2kg` and `3kg` are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge `1m`. Find the distance of their centre of mass from A.

Explanation

To find the distance of the center of mass from point A for three particles of masses 1 kg, 2 kg, and 3 kg placed at the corners of an equilateral triangle ABC with an edge length of 1 m, we can follow these steps:

### Step 1: Define the coordinates of the particles
- Let the coordinates of the particles be:
- A (1 kg) at (0, 0)
- B (2 kg) at (1, 0)
- C (3 kg) at \((\frac{1}{2}, \frac{\sqrt{3}}{2})\)

### Step 2: Calculate the total mass
- The total mass \( M \) of the system is:
\[ M = m_A + m_B + m_C = 1 \, \text{kg} + 2 \, \text{kg} + 3 \, \text{kg} = 6 \, \text{kg} \]

### Step 3: Calculate the center of mass coordinates
- The center of mass \( R_{cm} \) can be calculated using the formula:
\[ R_{cm} = \frac{1}{M} \sum m_i r_i \]
where \( m_i \) is the mass and \( r_i \) is the position vector of each particle.

- Calculate the x-coordinate of the center of mass:
\[ x_{cm} = \frac{1}{M} (m_A \cdot x_A + m_B \cdot x_B + m_C \cdot x_C) \]
\[ x_{cm} = \frac{1}{6} (1 \cdot 0 + 2 \cdot 1 + 3 \cdot \frac{1}{2}) = \frac{1}{6} (0 + 2 + \frac{3}{2}) = \frac{1}{6} \cdot \frac{7}{2} = \frac{7}{12} \]

- Calculate the y-coordinate of the center of mass:
\[ y_{cm} = \frac{1}{M} (m_A \cdot y_A + m_B \cdot y_B + m_C \cdot y_C) \]
\[ y_{cm} = \frac{1}{6} (1 \cdot 0 + 2 \cdot 0 + 3 \cdot \frac{\sqrt{3}}{2}) = \frac{1}{6} (0 + 0 + \frac{3\sqrt{3}}{2}) = \frac{3\sqrt{3}}{12} = \frac{\sqrt{3}}{4} \]

### Step 4: Write the position of the center of mass
- The center of mass \( R_{cm} \) is located at:
\[ R_{cm} = \left( \frac{7}{12}, \frac{\sqrt{3}}{4} \right) \]

### Step 5: Calculate the distance from point A
- The distance \( d \) from point A to the center of mass can be calculated using the distance formula:
\[ d = \sqrt{(x_{cm} - x_A)^2 + (y_{cm} - y_A)^2} \]
\[ d = \sqrt{\left( \frac{7}{12} - 0 \right)^2 + \left( \frac{\sqrt{3}}{4} - 0 \right)^2} \]
\[ d = \sqrt{\left( \frac{7}{12} \right)^2 + \left( \frac{\sqrt{3}}{4} \right)^2} \]
\[ d = \sqrt{\frac{49}{144} + \frac{3}{16}} \]
- Convert \(\frac{3}{16}\) to a fraction with a denominator of 144:
\[ \frac{3}{16} = \frac{27}{144} \]
\[ d = \sqrt{\frac{49 + 27}{144}} = \sqrt{\frac{76}{144}} = \frac{\sqrt{76}}{12} = \frac{2\sqrt{19}}{12} = \frac{\sqrt{19}}{6} \]

### Final Answer
The distance of the center of mass from point A is:
\[ \frac{\sqrt{19}}{6} \, \text{meters} \]

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