**Unformatted text preview: **Figure 10.52 (b) gives (c) gives gives so Since we calculate to be positive, we correctly assumed that it was downward when we drew the free-body diagram. (d) The factor divides out of the equation in part (b), so the force T stays the same as she raises her arm. g t 5 cos u E v E v 5 T 2 w arm 2 w 5 569 N 2 1 2.16 kg 21 9.80 m / s 2 2 2 1 7.50 kg 21 9.80 m / s 2 2 5 474 N T 2 E v 2 w arm 2 w 5 0, g F y 5 E h 5 0. g F x 5 T 5 16.0 w arm 1 38.0 w 5.5 5 16.0 1 2.16 kg 21 9.80 m / s 2 2 1 38.0 1 7.50 kg 21 9.80 m / s 2 2 5.5 5 569 N T 1 5.5 cm 21 cos u 2 2 w arm 1 16.0 cm 21 cos u 2 2 w 1 38.0 cm 21 cos u 2 5 g t 5 w w arm y x T 10.5cm 22.0cm pivot E v E h u 5.5cm E h . E v E S T S m 5 7.50 kg. w 5 mg , m arm 5 1 2 1 0.0600 21 72 kg 2 5 2.16 kg. 1 x 1 y f L h 5 U h 5 140 N. g F x 5 U h 5 1 0.50 m 1.00 m 2 1 280 N 2 5 140 N. 2 w 1 0.50 m 2 1 U h 1 1.00 m 2 5 g t 5 0.50m 1.00m 0.50m U h L h L v U v w cg 0.50m 1 U v 5 L v 5 w / 2 5 140 N. L h L v , U h U v , 10-18 Chapter 10...

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