08_InstSolManual_PDF_Part13 - y cm 5 1 23.0 cm 21 8.60 kg 2...

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8.48. Set Up: Solve: 8.49. Set Up: Use coordinates with the origin at the 20 N weight and with toward the 10 N weight and toward the 40 N weight. Since weight and mass are proportional, we can use weight in place of mass in calculating the coordinates of the center of mass. Solve: (a) The center of mass is 0.40 m above and 0.70 m to the right of the 20 N weight. (b) The center of mass of the four rods taken together is at the center of the square. Including their weight moves the center of mass of the complete system closer to the center of the square, so moves the center of mass calculated in (a) up and to the left. Reflect: The object is not symmetric and its center of mass is not at its geometric center. 8.50. Set Up: The leg in each position is sketched in Figure 8.50a and b. Use the coordinates shown. The mass of each part of the leg may be taken as concentrated at the center of that part. Figure 8.50 Solve: (a) The center of mass of the leg is 40.4 cm below the hip.
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Unformatted text preview: y cm 5 1 23.0 cm 21 8.60 kg 2 1 1 69.0 cm 21 5.25 kg 2 8.60 kg 1 5.25 kg 5 40.4 cm x cm 5 0, x y 8.60kg 23.0cm 23.0cm 23.0cm 23.0cm 5.25kg cm cm (b) x y 23.0cm 23.0cm 23.0cm 23.0cm 8.60kg 5.25kg (a) cm cm y cm 5 1 20 N 21 2 1 1 10 N 21 1.00 m 2 1 1 30 N 21 1.00 m 2 1 1 40 N 21 2 20 N 1 10 N 1 30 N 1 40 N 5 0.40 m x cm 5 1 20 N 21 2 1 1 10 N 21 2 1 1 30 N 21 1.00 m 2 1 1 40 N 21 1.00 m 2 20 N 1 10 N 1 30 N 1 40 N 5 0.70 m 1 x 1 y y cm 5 1 0.300 kg 21 0.300 m 2 1 1 0.400 kg 21 2 0.400 m 2 1 1 0.200 kg 21 0.600 m 2 0.300 kg 1 0.400 kg 1 0.200 kg 5 0.0556 m y cm 5 m A y A 1 m B y B 1 m C y C m A 1 m B 1 m C x cm 5 1 0.300 kg 21 0.200 m 2 1 1 0.400 kg 21 0.100 m 2 1 1 0.200 kg 21 2 0.300 m 2 0.300 kg 1 0.400 kg 1 0.200 kg 5 0.0444 m x cm 5 m A x A 1 m B x B 1 m C x C m A 1 m B 1 m C m C 5 0.200 kg m B 5 0.400 kg, m A 5 0.300 kg, Momentum 8-13...
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