Consider the following distribution of objects: a 3.00-kg object with its center of gravity at (0, 0) m, a 4.20-kg object at (0, 6.00) m, and a 1.40-kg object at (1.00, 0) m. Where should a fourth object of mass 7.00 kg be placed so that the center of gravity of the four-object arrangement will be at (0, 0)

Explanation:

Given the following masses and coordinates

M1 = 3kg at x1 = (0,0)m

M2 = 1.4kg at x2 =(1,0)m

M3 = 4.2kg at x3 = (0, 6)m

M4 = 7kg at x4 = (x, y)m

General center of mass

Xcm=(0,0)

Center of mass is given as

Xcm = 1/M • Σ Mi•xi

M = m1+m2+m3+m4

M = 3+1.4+4.2+7

M = 15.6kg

(0,0) = 1/15.6•Σ Mi•xi

Cross multiply by 15.6

Then,

(0,0) = Σ Mi•xi

(0,0)=M1•x1 + M2•x2 + M3•x3+M4•x4

(0,0)=3(0,0) + 1.4(1,0)+ 4.2(0,6) + 7(x, y)

(0,0) = (0,0) + (1.4,0) + (0,25.56)+ (7x, 7y)

(0,0) = (0+1.4+0+7x, 0+0+25.56+7y)

(0,0) = (1.4+7x, 25.56+7y)

Comparing coefficient

1.4+7x = 0

7x = -1.4

x = -1.4/7

x = -0.2m

Also,

25.56 + 7y = 0

7y = - 25.56

y = -25.56/7

y = -3.65m

Then, the position of fourth mass is

(x, y) = (-0.2, -3.65) m

insects!

explanation:

well clustered together.

points close on a scatter diagram