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AnkhUNC
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[SOLVED] Center of Mass problem
A 3.12 kg particle has the xy coordinates (-1.76 m, 0.652 m), and a 2.89 kg particle has the xy coordinates (0.380 m, -0.636 m). Both lie on a horizontal plane. At what (a)x and (b)y coordinates must you place a 3.74 kg particle such that the center of mass of the three-particle system has the coordinates (-0.565 m, -0.601 m)?
I know this problem is extremely easy but I'm just having trouble getting it. I solve for x(com) and y(com) =
-.565 = (3.12(-1.76)+2.89(.380)+3.74(x))/3.12+2.89+3.74
-.601 = (3.12(.652)+2.89(-.636)+3.74(y))/3.12+2.89+3.74
This is correct right? So how do I pull out to solve for x and y?
Homework Statement
A 3.12 kg particle has the xy coordinates (-1.76 m, 0.652 m), and a 2.89 kg particle has the xy coordinates (0.380 m, -0.636 m). Both lie on a horizontal plane. At what (a)x and (b)y coordinates must you place a 3.74 kg particle such that the center of mass of the three-particle system has the coordinates (-0.565 m, -0.601 m)?
Homework Equations
The Attempt at a Solution
I know this problem is extremely easy but I'm just having trouble getting it. I solve for x(com) and y(com) =
-.565 = (3.12(-1.76)+2.89(.380)+3.74(x))/3.12+2.89+3.74
-.601 = (3.12(.652)+2.89(-.636)+3.74(y))/3.12+2.89+3.74
This is correct right? So how do I pull out to solve for x and y?