10_InstSolManual_PDF_Part17

# 10_InstSolManual_PDF_Part17 - Dynamics of Rotational Motion...

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(b) gives (c) gives and gives and 10.49. Set Up: The free-body diagram for the leg is given in Figure 10.49. Take the pivot at the hip joint and let counterclockwise torque be positive. There are also forces on the leg exerted by the hip joint but these forces produce no torque and aren’t shown. Figure 10.49 Solve: (a) gives and for (b) For The tension is much greater when he just starts to raise his leg off the ground. (c) as The person could not raise his leg. If the leg is horizontal so is zero, the moment arm for T is zero and T produces no torque to rotate the leg against the torque due to its weight. 10.50. Set Up: Take the pivot at the left-hand end of the branch and let counterclockwise torque be positive. The trunk exerts a force on the branch at its left-hand end. The wood fiber exerts a force at a point from the left-hand end of the branch.
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Unformatted text preview: Solve: (a) The free-body diagram is given in Figure 10.50. Figure 10.50 (b) gives T 5 1 1.75 m 21 60 kg 21 9.80 m / s 2 2 1 0.162 m 2 sin 45° 5 9000 N T sin 45° 1 0.162 m 2 2 w 1 1.75 m 2 5 g t 5 1.75m 0.162m T cos45° T sin45° T F v F h w Pivot 1 45° 1 4 1 0.65 m 2 5 0.162 m T S F S u u S 0. T S ` T 5 7400 N. u 5 5°, T 5 4.4 1 15 kg 21 9.80 m / s 2 2 tan 60° 5 370 N u 5 60°, T 5 4.4 w cos u sin u 5 4.4 w tan u T 1 10 cm 21 sin u 2 2 w 1 44 cm 21 cos u 2 5 g t 5 u w 10cm T 34cm Pivot 1 H v 5 w b 1 w 2 T cos 60.0° 5 2560 N H v 1 T cos 60.0° 2 w b 2 w 5 g F y 5 H h 5 T sin 60.0° 5 5960 N H h 2 T sin 60.0° 5 g F x 5 T 5 1 2500 N 21 2.00 m 2 1 1 3500 N 21 2.50 m 2 2.00 m 5 6880 N 1 T cos 60.0° 21 4.0 m 2 2 w b 1 2.0 m 2 2 w 1 2.5 m 2 5 g t 5 Dynamics of Rotational Motion 10-17...
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