Question

A 4.04 kg particle has the xy coordinates (-1.24 m, 0.498 m), and a 3.86 kg particle has the xy coordinates (0.399 m, -0.641 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 2.09 kg particle such that the center of mass of the three-particle system has the coordinates (-0.840 m, -0.910 m)?

Answer 1

given

m₁ = 4.04 kg at (-1.24 m, 0.498 m) = ( x₁ , y₁ )

m₂ = 3.86 kg at (0.399 m, -0.641 m) = ( x₂ , y₂ )

m₃ = 2.09 kg at (-0.840 m, -0.910 m) = ( x_cm , y_cm )

a )

x_cm = m₁ x₁ + m₂ x₂ + m₃ x₃ / ( m₁ + m₂ + m₃ )

-0.84 = 4.04 X ( -1.24 ) + 3.86 X (0.399 ) + 2.09 X x₃ / ( 4.04 + 3.86 + 2.09 )

-0.84 = 4.04 X ( -1.24 ) + 3.86 X (0.399 ) + 2.09 X x₃ / ( 9.99 )

- 8.39 = - 5 + 1.54 + 2.09 X x₃

2.09 X x₃ = - 11.85

x₃ = - 5.66 m

b )

same way for

y_cm = m₁y₁ + m₂ y₂ + m₃ y₃ / ( m₁ + m₂ + m₃ )

-0.91 = 4.04 X ( 0.498 ) + 3.86 X (-0.641 ) + 2.09 X y₃ / ( 4.04 + 3.86 + 2.09 )

-0.91 = 4.04 X ( 0.498 ) + 3.86 X (-0.641 ) + 2.09 X y₃ / ( 9.99 )

- 9.09 = 2.011 - 2.41 + 2.09 X y₃

2.09 X y₃ = - 8.691

y₃ = - 4.158 m

finally ( x₃ , y₃ ) = ( - 5.66 , - 4.158 )