Question

Three particles of mass 1kg, 2kg and 3kg are placed at the corners A, B and C respectively of an equilateral triangle ABC of edge 1m. How do I find the distance of their centre of mass from A?

Answer 1

Step-by-step explanation:

centre of mass is the equal distance from each. so

x=m1x1+m2x2+m3x3/m1+m2+m3

1*1+1*2+1*3/1+1+1

6/2=3

Answer 2

The distance of their centre of mass from A is $\frac{\sqrt{19}}{6} \ m$

Solution:

As given, three particles of mass 1 kg, 2 kg and 3 kg are placed at the corners A, B and C respectively of an equilateral triangle.

For A, $m_1 =1 \ kg, \ x_1 =0 \ m, \ y_1 =0 \ m$

For B, $m_2 =2 \ kg, \ x_2 =1 \ m, \ y_2 =0 \ m$

For C, $m_3 =3 \ kg, \ x_3= \left(\frac{1}{2}\right) \ m, \ y_3 = \frac{\sqrt{3}}{2} \ m$

In respect to point A the center of mass need to be find.

$x=\frac{m_1 \times 1+m_2 \times 2+m_3\times 3}{m_1+m_2+m_3}=\frac{(1 \times 0)+(2 \times 1)+\left(3 \times\left(\frac{2}{2}\right)\right)}{1+2+3}=\left(\frac{7}{12}\right) \ m$

$\mathrm{y}=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3}=\frac{(1 \times 0)+(2 \times 0)+\left(3 \times \frac{\sqrt{3}}{2}\right)}{1+2+3}=\frac{3 \sqrt{3}}{12} \mathrm{m}$

So distance $d=\sqrt{x^{2}+y^{2}}=\frac{\sqrt{19}}{6} \ m$