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 300 9 2 . af has mass dm xd x = 9 2 . . The total mass is Md m x Mx x d x M xx x x == = F H G I K J −+ = F H G I K J L N M O Q P = zz z = 3 9 9 69 93 6 2 9 2 0 2 0 32 0 a f ej . . . 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 dx x x xdx xxx CG m 9.00 m = + =− + L N M O Q P z 1 9 3 9 1 94 6 3 9 2 675 0 750 2 0 0 432 0 .. . . . P12.7 Let the fourth mass (8.00 kg) be placed at ( x , y ), then x mx m x CG m + + = − 0 300 400 12 0 12 0 800 150 4 4 . . . . a f Similarly, y y CG + + 0 12 0 8 00 . bg y . 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 = 667 m , y = 233 m (see the Example on the center of mass of a triangle in Chapter 9).
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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