Question

Three point masses of 2g, 3g and 4g are placed at the vertices of an equilateral triangle of side 1meter. Find the centre of mass of the system.
​

Answer 1

Solution :

As per the given data ,

• m₁ = 2 g
• m₂ = 3 g
• m₃ = 4 g
• side = 1m

The given triangle is an equilateral triangle. Let the three vertices of the triangle be A, B, and C

Hence,

α = β = γ = 60⁰

Coordinates

• A = (0,0 ) [ origin ]
• B = (1,0 )
• C = ( 1/2 , √3/2)

Now,

$\implies\sf{X_{com} = \dfrac{m_1x_1 + m_2x_2+ m_3x_3}{m_1+m_2+m_3}}$

Now let's substitute the values in the above equation,

$\implies\sf{X_{com} = \dfrac{2\times 0 + 3 \times 1 + 4 \times \dfrac{1}{2}}{2+3+4}}$

$\implies\boxed{\sf{X_{com} = \dfrac{5}{9}} }$

Similarly,

$\implies\sf{Y_{com} = \dfrac{m_1y_1 + m_2y_2+ m_3y_3}{m_1+m_2+m_3}}$

$\implies\sf{Y_{com} = \dfrac{2\times 0 + 3 \times 0+ 4 \times \dfrac{\sqrt{3} }{2}}{2+3+4}}$

$\implies\boxed{\sf{Y_{com} = \dfrac{2\sqrt{3} }{9}}}$

The coordinates of COM of the system is $\sf{\bigg(\dfrac{5}{9} ,\dfrac{2\sqrt{3} }{9}\bigg )}$.

Answer 2

Answer:

Solution :

As per the given data ,

m₁ = 2 g

m₂ = 3 g

m₃ = 4 g

side = 1m

The given triangle is an equilateral triangle. Let the three vertices of the triangle be A, B, and C

Hence,

α = β = γ = 60⁰

Coordinates

A = (0,0 ) [ origin ]

B = (1,0 )

C = ( 1/2 , √3/2)

Now,

\begin{gathered}\implies\sf{X_{com} = \dfrac{m_1x_1 + m_2x_2+ m_3x_3}{m_1+m_2+m_3}}\\ \\\end{gathered}

⟹X

com

=

m

1

+m

2

+m

3

m

1

x

1

+m

2

x

2

+m

3

x

3

Now let's substitute the values in the above equation,

\begin{gathered}\implies\sf{X_{com} = \dfrac{2\times 0 + 3 \times 1 + 4 \times \dfrac{1}{2}}{2+3+4}}\\ \\\end{gathered}

⟹X

com

=

2+3+4

2×0+3×1+4×

2

1

\begin{gathered}\implies\boxed{\sf{X_{com} = \dfrac{5}{9}} }\\ \\\end{gathered}

⟹

X

com

=

9

5

Similarly,

\begin{gathered}\implies\sf{Y_{com} = \dfrac{m_1y_1 + m_2y_2+ m_3y_3}{m_1+m_2+m_3}}\\ \\\end{gathered}

⟹Y

com

=

m

1

+m

2

+m

3

m

1

y

1

+m

2

y

2

+m

3

y

3

\begin{gathered}\implies\sf{Y_{com} = \dfrac{2\times 0 + 3 \times 0+ 4 \times \dfrac{\sqrt{3} }{2}}{2+3+4}}\\ \\\end{gathered}

⟹Y

com

=

2+3+4

2×0+3×0+4×

2

3

\begin{gathered}\implies\boxed{\sf{Y_{com} = \dfrac{2\sqrt{3} }{9}}}\\ \\\end{gathered}

⟹

Y

com

=

9

2

3

The coordinates of COM of the system is \sf{\bigg(\dfrac{5}{9} ,\dfrac{2\sqrt{3} }{9}\bigg )}(

9

5

,

9

2

3

) .