351_Physics ProblemsTechnical Physics

351_Physics ProblemsTechnical Physics - Chapter 12 P12.6...

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Chapter 12 353 P12.6 Let σ represent the mass-per-face area. A vertical strip at position x , with width dx and height x 3 00 9 2 . a f has mass dm x dx = σ 3 00 9 2 . a f . The total mass is M dm x dx M x x dx M x x x x = = = F H G I K J + = F H G I K J + L N M O Q P = z z z = σ σ σ σ 3 9 9 6 9 9 3 6 2 9 2 0 3 00 2 0 3 00 3 2 0 3 00 a f e j . . . x dx 0 3.00 m x y 1.00 m y = ( x 3 .00) 2 /9 FIG. P12.6 The x -coordinate of the center of gravity is x xdm M x x dx x x x dx x x x CG m 9.00 m = = = + = + L N M O Q P = = z z z 1 9 3 9 6 9 1 9 4 6 3 9 2 6 75 0 750 2 0 3 00 3 2 0 3 00 4 3 2 0 3 00 σ σ σ σ a f e j . . . . . P12.7 Let the fourth mass (8.00 kg) be placed at ( x , y ), then x m x m x CG m = = + + = − = 0 3 00 4 00 12 0 12 0 8 00 1 50 4 4 . . . . . . a fa f a f Similarly, y y CG = = + + 0 3 00 4 00 8 00 12 0 8 00 . . . . . a fa f b g y = 1 50 . m P12.8 In a uniform gravitational field, the center of mass and center of gravity of an object coincide. Thus, the center of gravity of the triangle is located at x =
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