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Unformatted text preview: the triangle? CoM of each side: Side a: CoM = a = ( (x A +x B ) , (y A +y B ) ) Side b: CoM = b = ( (x B +x C ) , (y B +y C ) ) Side c: CoM = c = ( (x C +x A ) , (y C +y A ) ) CoM of triangle (average of 3 centers of masses above): CoM = ( 1 / 3 [(x A +x B )+(x B +x C )+(x C +x A )] , 1 / 3 [(y A +y B )+ (y B +y C )+(y C +y A )] ) 3. If mass is uniformly distributed over the interior of ABC, where is the center of mass? For a triangle of uniform interior mass, the center of mass is also located at the centroid of the triangle CoM = 1 / 3 (A+B+C) CoM = ( 1 / 3 (x A +x B +x C ) , 1 / 3 (y A +y B +y C ) )...
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 Spring '00
 GOODE
 Math

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