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A 2.70 kg particle has the xy coordinates (-1.22 m, 0.504 m), and a 4.64 kg...

A 2.70 kg particle has the xy coordinates (-1.22 m, 0.504 m), and a 4.64 kg particle has the xy coordinates (0.580 m, -0.526 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 3.98 kg particle such that the center of mass of the three-particle system has the coordinates (-0.0990 m, -0.170 m)?

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Answer #1

Grunen x 4 m,- 2.70 kg at (-1.22 m, 0.504m) m2 = 4.64kg at (osom, -0.526m) Let m3= 3.98 kg at (43, 83) Co-ordinates of Center-0.17=2-7Xo. Sou +4.64X-0.526+3.984 2-7+4.64+3.90 -0.17 =1-3608-2-44064+3.98% 11.32 -0-17X17-32 = -|-67904+3,90 4, - 1.9244 =

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