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Chapter-09 - 377 Chapter 9 1 We use Eq 9-5 to solve for 3 3...

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Unformatted text preview: 377 Chapter 9 1. We use Eq. 9-5 to solve for 3 3 ( , ). x y (a) The x coordinates of the system’s center of mass is: ( 29 ( 29 ( 29 3 1 1 2 2 3 3 com 1 2 3 (2.00 kg)( 1.20 m) 4.00 kg 0.600 m 3.00 kg 2.00 kg 4.00 kg 3.00 kg 0.500 m. x m x m x m x x m m m- + + + + = = + + + + = - Solving the equation yields x 3 = –1.50 m. (b) The y coordinates of the system’s center of mass is: ( 29 ( 29 ( 29 3 1 1 2 2 3 3 com 1 2 3 (2.00 kg)(0.500 m) 4.00 kg 0.750 m 3.00 kg 2.00 kg 4.00 kg 3.00 kg 0.700 m. y m y m y m y y m m m +- + + + = = + + + + = - Solving the equation yields y 3 = –1.43 m. 2. Our notation is as follows: x 1 = 0 and y 1 = 0 are the coordinates of the m 1 = 3.0 kg particle; x 2 = 2.0 m and y 2 = 1.0 m are the coordinates of the m 2 = 4.0 kg particle; and, x 3 = 1.0 m and y 3 = 2.0 m are the coordinates of the m 3 = 8.0 kg particle. (a) The x coordinate of the center of mass is ( 29 ( 29 ( 29 ( 29 1 1 2 2 3 3 com 1 2 3 4.0 kg 2.0 m 8.0 kg 1.0 m 1.1 m. 3.0 kg 4.0 kg 8.0 kg m x m x m x x m m m + + + + = = = + + + + (b) The y coordinate of the center of mass is ( 29 ( 29 ( 29 ( 29 1 1 2 2 3 3 com 1 2 3 4.0 kg 1.0 m 8.0 kg 2.0 m 1.3 m. 3.0 kg 4.0 kg 8.0 kg m y m y m y y m m m + + + + = = = + + + + (c) As the mass of m 3 , the topmost particle, is increased, the center of mass shifts toward that particle. As we approach the limit where m 3 is infinitely more massive than the others, the center of mass becomes infinitesimally close to the position of m 3 . CHAPTER 9 378 3. Since the plate is uniform, we can split it up into three rectangular pieces, with the mass of each piece being proportional to its area and its center of mass being at its geometric center. We’ll refer to the large 35 cm × 10 cm piece (shown to the left of the y axis in Fig. 9-38) as section 1; it has 63.6% of the total area and its center of mass is at ( x 1 ,y 1 ) = (- 5.0 cm, - 2.5 cm). The top 20 cm × 5 cm piece (section 2, in the first quadrant) has 18.2% of the total area; its center of mass is at ( x 2 , y 2 ) = (10 cm, 12.5 cm). The bottom 10 cm x 10 cm piece (section 3) also has 18.2% of the total area; its center of mass is at ( x 3 , y 3 ) = (5 cm, - 15 cm). (a) The x coordinate of the center of mass for the plate is x com = (0.636) x 1 + (0.182) x 2 + (0.182) x 3 = – 0.45 cm . (b) The y coordinate of the center of mass for the plate is y com = (0.636) y 1 + (0.182) y 2 + (0.182) y 3 = – 2.0 cm . 4. We will refer to the arrangement as a “table.” We locate the coordinate origin at the left end of the tabletop (as shown in Fig. 9-37). With + x rightward and + y upward, then the center of mass of the right leg is at ( x,y ) = (+ L , – L /2), the center of mass of the left leg is at ( x,y ) = (0, – L /2), and the center of mass of the tabletop is at ( x,y ) = ( L /2, 0)....
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Chapter-09 - 377 Chapter 9 1 We use Eq 9-5 to solve for 3 3...

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