10_InstSolManual_PDF_Part33

# 10_InstSolManual_PDF_Part33 - in Figure 10.83b gives so...

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10.83. Set Up: Figure 10.83a shows the distances and angles. and The distances and are and The free-body diagram for the person is given in Figure 10.83b. is the weight of his feet and legs and is the weight of his trunk. and are the total normal and friction forces exerted on his feet and and are those forces on his hands. The free-body diagram for his legs is given in Figure 10.83c. F is the force exerted on his legs by his hip joints. Figure 10.83 Solve: (a) Consider the force diagram of Figure 10.83b. with the pivot at his feet and counterclockwise torques positive gives so there is a normal force of 175 N at each hand. so so there is a normal force of 200 N at each foot. (b) Consider the force diagram of Figure 10.83c. with the pivot at his hips and counterclockwise torques positive gives There is a friction force of 92 N at each foot.
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Unformatted text preview: in Figure 10.83b gives so there is a friction force of 92 N at each hand. Reflect: In this position the normal forces at his feet and at his hands don’t differ very much. f h 5 f f , g F x 5 f f 5 1 400 N 21 50.0 cm 2 2 1 277 N 21 22.8 cm 2 74.9 cm 5 183 N f f 1 74.9 cm 2 1 w l 1 22.8 cm 2 2 n f 1 50.0 cm 2 5 g t 5 750 N 2 350 N 5 400 N, n f 5 w l 1 w t 2 n h 5 n f 1 n h 2 w l 2 w t 5 n h 5 350 N, n h 1 162 cm 2 2 1 277 N 21 27.2 cm 2 2 1 473 N 21 103.8 cm 2 5 g t 5 90° u f 90 cm 135 cm x 1 x 2 162 cm (a) (b) (c) n h n f 41 cm 49 cm 65 cm 70 cm u u f 76.6 cm 27.2 cm 58.2 cm w l w t n f f f w l F f f f h Axis f h n h f f n f w t 5 473 N w l 5 277 N x 2 5 1 135 cm 2 cos f 5 112 cm. x 1 5 1 90 cm 2 cos u 5 50.0 cm x 2 x 1 f 5 33.7°. u 5 56.3° u 1 f 5 90°. Dynamics of Rotational Motion 10-33...
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