10_InstSolManual_PDF_Part14

10_InstSolManual_PDF_Part14 - 2 f n w 2600 N 5 7600 N F v 5...

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Figure 10.43 Solve: (a) and all produce zero torque. gives and (b) gives and gives and Reflect: For an axis at the right-hand end of the beam, only w and produce torque. The torque due to w is coun- terclockwise so the torque due to must be clockwise. To produce a counterclockwise torque, must be upward, in agreement with our result from 10.44. Set Up: The free-body diagram for the boom is given in Figure 11.44. Let the length of the boom be L . Figure 10.44 Solve: (a) gives and (b) gives and gives and 10.45. Set Up: There is a normal force at the wall. At the floor there is a normal force n and a friction force Let the length of the ladder be L . The center of gravity of the ladder is a distance from each end. Let be upward and let be horizontal, toward the wall. Let the pivot be at the foot of the ladder and let counterclockwise torques be positive. 1 x 1 y L / 2 f
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Unformatted text preview: 2 f . n w 2600 N 5 7600 N F v 5 5000 N 1 F v 2 w 2 w load 5 g F y 5 F h 5 3410 N. F h 2 T 5 g F x 5 T 5 w load cos 60.0° 1 w 1 0.35 cos 60.0° 2 sin 60.0° 5 1 5000 N 2 cos 60.0° 1 1 2600 N 21 0.35 cos 60.0° 2 sin 60.0° 5 3.41 3 10 3 N T 1 L sin 60.0° 2 2 w load 1 L cos 60.0° 2 2 w 1 0.35 L cos 60.0° 2 5 g t 5 x y w Axis T w load F h F v 0.35 L 0.65 60 ° 1 g F y 5 0. H v H v H v H v 5 w 1 w load 2 T sin u 5 150 N 1 300 N 2 1 625 N 21 0.600 2 5 75 N. T sin u 5 H v 2 w 2 w load 1 g F y 5 H h 5 1 625 N 21 0.800 2 5 500 N. H h 5 T cos u 5 g F x 5 T 5 1 150 N 21 2.00 m 2 1 1 300 N 21 4.00 m 2 1 4.00 m 21 0.600 2 5 625 N. T sin u 1 4.00 m 2 5 2 w 1 2.00 m 2 2 w load 1 4.00 m 2 1 g t 5 T x 5 T cos u H h H v , 1 H v 2.00m w x y H h Axis u T cos u T sin u T 2.00m 10-14 Chapter 10...
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