Question

Three particles of masses 1 kg, 2 kg and 3 kg
are placed at the vertices A, B and C of an
equilateral triangle ABC. If A and B lie at (0,0)
and (1,0) m, the co-ordinates of their centre
of mass are
​

Answer 1

hope it helps !!!!!!!!!!

Answer 2

Correct Question: Three particles of masses 1 kg, 2 kg and 3 kg are placed at the vertices A, B and C of an equilateral triangle ABC. If A and B lie at (0,0) and (1,0) m, and C at (3, 2)m, then co-ordinates of their centre of mass are?

Answer:

Co-ordinates of Centre of Mass, Rcm = ($\frac{11}{6}$, $1$)

Explanation:

Given;-

    Masses, m = 1kg, 2kg, 3kg

Let the given mass be placed at A, B and C respectively.

 Co-ordinates of A= (0, 0)

 Co-ordinates of B= (1, 0)

 Co-ordinates of C= (3, 2)

Now, finding the co-ordinates of the centre of mass.

We know that;-

          (Rcm)x = $\frac{m1x1 + m2x2 + m3x3}{m1 + m2 + m3}$

[Note, Rcm = Centre of mass]

Putting the given values in the above formula, we get;-

          (Rcm)x = $\frac{1(0) + 2(1) + 3(3)}{1+2+3}$

          (Rcm)x = $\frac{2 + 9 }{6}$

          (Rcm)x = $\frac{11}{6}$

Also,  (Rcm)y = $\frac{m1y1 + m2y2 + m3y3}{m1 + m2 +m3}$

         (Rcm)y = $\frac{1(0)+ 2(0) + 3(2)}{1+2+3}$

         (Rcm)y = $\frac{6}{6}$

        (Rcm)y = $1$

Hence, the co-ordinates of centre of mass are ($\frac{11}{6}$, $1$).

Hope it helps! ;-))