Question

A 4.04 kg particle has the xy coordinates (-1.24 m, 0.498 m), and a 3.86 kg...

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)?

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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) = ( xcm , ycm )

a )

xcm = 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

ycm = 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 )

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