Question

Calculate the coordinate of center of a mass of a triangular lamina , whose vertices are (0,0) , (4,0) , (0,4).

Answer 1

The coordinate of center of mass is $(\frac{4}{3} ,\frac{4}{3} )$

Explanation:

Center of mass of triangle is the centroid which is point of intersection of medians.

The vertices of the triangle are (0,0),(4,0) and (0,4)

The coordinate of centroid  is $(\frac{x_{1},y_{1} ,z_{1} }{3},\frac{x_{2},y_{2},z_{2} }{3} )$

$(\frac{0+4+0}{3} ,\frac{0+0+4}{3} )\\\=(\frac{4}{3} ,\frac{4}{3} )$

Answer 2

The coordinate of centre of a mass of a triangular lamina is $\bold{(\frac{4}{3},\frac{4}{3})}$.

Given:

Vertex 1 = (0,0)

Vertex 2 = (4,0)

Vertex 3 = (0,4)

To find:

Coordinate of centre of a mass = ?

Formula:

The centre of the mass is given by the formula:

$\bold{G = \frac{(m_1x_1+m_2x_2+m_3x_3)}{(m_1+m_2+m_3)}, \frac{(m_1y_1+m_2y_2+m_3y_3)}{(m_1+m_2+m_3)}}$

Solution:

$\bold{m_1=0; \ x_1=0; \ m_2=4; \ x_2=0; \ m_3=0; \ x_3=4}$

On substituting the known values,

$\bold{\Rightarrow G= {\frac{(m \times 0 + m \times 4 + m \times 0)}{(m + m + m)}, \frac{(m \times 0 + m \times m + 0 \times 4)}{(m + m + m)}}}$

$\bold{\Rightarrow G = (\frac{4m}{3m},\frac{4m}{3m})}$

$\bold{\therefore G = (\frac{4}{3},\frac{4}{3})}$