Question

If the centre of mass of 3 particles of masses 10,20 and 30g is at a point (1,-2,3),then where should a fourth particle of mass 40g be placed , so that the combined centre of mass of the system is at(1,1,1)?

Answer 1

We can simply treat those three bodies like they are a single point body of mass 60g located at (1,-2,3)

now to find the answer

m2 =40 g , to find r2

CM x = (m1 x1 + m2 x2)/m1+m2

1= (60 *1 + 40 *x2 )/100

x2=1

CM y=(m1 y1 + m2 y2)/m1+m2

1 = (60 * (-2) + 40*y2 )/100

y2= 5.5

CM z = (m1 z1 + m2 z2)/m1+m2

1= ( 60 *3 + 40 *z2 ) /100

z2 = -2

So your 40g mass should be placed at (1,5.5,-2).

Hope it helps

So the

Answer 2

10x1+20 x2+30x3÷10+20+30=1

=10X1+20X2+30X3=60..........1

10X1+20X2+30X3+40X4=0..........2

1-2=

-40X4=60

X4=-1.5

Similarly, Y4 =-1.5,Z4= -1.5