10_InstSolManual_PDF_Part28

10_InstSolManual_PDF_Part28 - 60 F h T sin60 T cos60 F v n...

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Solve: (a) The center of mass is 24.0 cm to the right and 6.7 cm above the shoulder joint. (b) gives (c) gives gives and 10.73. Set Up: The free-body diagram for the foot is given in Figure 10.73. Take the pivot at the ankle joint and let counterclockwise torques be positive. Figure 10.73 Solve: (a) If we consider the entire person, the upward force n exerted by the floor must equal the weight w of the person; (b) gives and (c) gives and to the left. gives and downward. 10.74. Set Up: The free-body diagram for each marble is given in Figure 10.74. Each marble has weight and are the forces exerted by the container at points A , B , and C . n is the normal force one marble exerts on the other. and u5 30°. sin u5 R / 2 R F C F B F A , w 5 mg 5 0.735 N. F 5 " F h 2 1 F v 2 5 2980 N F v 5 T sin 60° 1 n 5 2001 N 1 750 N 5 2750 N, T sin 60° 2 F v 1 n 5 0 g F y 5 0 F h 5 T cos 60° 5 1160 N, T cos 60° 2 F h 5 0 g F x 5 0 T 5 3.08 n 5 2310 N n 1 16.0 m 21 cos 30° 2 2 T 1 4.5 m 2 5 0 g t5 0 n 5 750 N 30 ° 30
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Unformatted text preview: 60 F h T sin60 T cos60 F v n F y x Pivot T 16.0cm 4.5cm F h 5 W cos 55 5 28.3 N W cos 55 2 F h 5 g F x 5 F v 5 w u 1 w f 2 W sin 55 5 22.8 N 1 21.1 N 2 40.5 N 5 3.4 N F v 2 w u 2 w f 1 W sin 55 5 g F y 5 W 5 1 22.8 N 21 13.0 cm 2 1 1 21.1 N 21 35.75 cm 2 1 45.5 cm 2 sin 55 2 1 27.9 cm 2 cos 55 5 49.4 N 2 w f 1 26.0 cm 1 9.75 cm 2 2 w u 1 13.0 cm 2 5 1 W sin 55 21 26.0 cm 1 19.5 cm 2 2 1 W cos 55 21 27.9 cm 2 g t 5 y cm 5 w 1 17.0 cm 21 sin 55 2 w u 1 w f 5 1 21.1 N 21 13.9 cm 2 22.8 N 1 21.1 N 5 6.7 cm x cm 5 1 22.8 N 21 13.0 cm 2 1 1 21.1 N 21 35.8 cm 2 22.8 N 1 21.1 N 5 24.0 cm x cm 5 w u 1 13.0 cm 2 1 w f 1 26.0 cm 1 3 17.0 cm 43 cos 55 42 w u 1 w f 10-28 Chapter 10...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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