Question

Particles of masses m1 = 2g, m2 = 2g, m3 = 1g and m4 = 1g are
placed at the corners of a square of side L, as shown. Find the centre
of mass of the system with respect to m .
​

Answer 1

Answer:

The answer will be (L/2, L/3)

Explanation:

Let the masses are plotted in a x-y plane where x is the horizontal axis and y is the vertical axis

Now the mass m1 =2g is plotted at the origin i.e( 0,0) position

mass m4= 1g is plotted at the point L cm vertically up  from the origin i.e(0,L) position

Again m2= 2g is plotted at the point L cm horizontally   from the origin i.e(L,0) position

and m3= 1g  is plotted at the point L cm vertically up  from the m2 position i.e(L,L) position

X= m1x1+m2x2+m3x3+m4x4/(m1+m2+m3+m4)

 =  2 x 0 + 2 x L+ 1 x L+ 1 x0/(2+2+1+1)

 = L/2

Y = m1y1+m2y2+m3y3+m4y4/(m1+m2+m3+m4)

   = 2 x 0 +2 x 0 +1 x L+ 1 x L/(2+2+1+1)

   = L/3

The center of mass with respect to m is (L/2,L/3)

Answer 2

Answer:

X= m1x1+m2x2+m3x3+m4x4/(m1+m2+m3+m4)

=  2 x 0 + 2 x L+ 1 x L+ 1 x0/(2+2+1+1)

= L/2

Y = m1y1+m2y2+m3y3+m4y4/(m1+m2+m3+m4)

  = 2 x 0 +2 x 0 +1 x L+ 1 x L/(2+2+1+1)

  = L/3

The center of mass with respect to m is (L/2,L/3)