Question

Three masses are positioned at the following coordinates 3.0 kg at (4.0 m, -5.0 m); 6.0 kg at (-7.0 m, 3.0 m); and 1.0 kg (2.0 m, 2.0 m). Calculate the center of mass of the system.

Answer 1

Mass of the first object = M₁ = 3 kg

Mass of the second object = M₂ = 6 kg

Mass of the third object = M₃ = 1 kg

Position of the first object = (X₁ , Y₁) = (4 m , -5 m)

Position of the second object = (X₂ , Y₂) = (-7 m , 3 m)

Position of the third object = (X₃ , Y₃) = (2 m , 2 m)

Position of the center of mass of the system = (X_cm . Y_cm)

$X_{cm} = \frac{M_{1}X_{1} + M_{2}X_{2} + M_{3}X_{3}}{M_{1} + M_{2} + M_{3}}$

$X_{cm} = \frac{(3)(4) + (6)(-7) + (1)(2)}{3 + 6 + 1}$

X_cm = -2.8 m

$Y_{cm} = \frac{M_{1}Y_{1} + M_{2}Y_{2} + M_{3}Y_{3}}{M_{1} + M_{2} + M_{3}}$

$Y_{cm} = \frac{(3)(-5) + (6)(3) + (1)(2)}{3 + 6 + 1}$

Y_cm = 0.5 m

Position of the center of mass of the system = (-2.8 m , 0.5 m)