10_InstSolManual_PDF_Part16

10_InstSolManual_PDF_Part16 - that opposes the torque due...

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10.47. Set Up: The distance along the beam from the hinge to where the cable is attached is 3.0 m. The angle that the cable makes with the beam is given by so The center of gravity of the beam is 4.5 m from the hinge. Use coordinates with upward and to the right. Take the pivot at the hinge and let counterclockwise torque be positive. Express the hinge force as compo- nents and Assume is downward and that is to the right. If one of these components is actually in the opposite direction we will get a negative value for it. Set the tension in the cable equal to its maximum possible value, Solve: (a) The free-body diagram is given in Figure 10.47. T has been resolved into its x and y components. Figure 10.47 (b) gives (c) gives and gives and Reflect: and all have zero moment arms for a pivot at the hinge and therefore produce zero torque. If we consider a pivot at the point where the cable is attached we can see that must be downward to produce a torque
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Unformatted text preview: that opposes the torque due to w . 10.48. Set Up: The center of gravity of the beam is 2.0 m from the hinge. Use coordinates with upward and to the right. Take the pivot at the hinge and let counterclockwise torque be positive. Solve: (a) The free-body diagram for the beam is shown in Figure 10.48. The tension T in the wire has been replaced by its x and y components. Figure 10.48 y w b w T cos60 T sin60 x H v H h Pivot Hinge 2.0m 0.5m 1.5m 1 60 1 x 1 y H v H h H v T cos f , H v 5 1 1.00 kN 21 sin 53.1 2 2 530 N 5 270 N T sin f 2 H v 2 w 5 g F y 5 H h 5 1 1.00 kN 21 cos 53.1 2 5 600 N H h 2 T cos f 5 g F x 5 w 5 1 T sin f 21 3.0 m 2 4.5 m 5 1 1.00 kN 21 sin 53.1 21 3.0 m 2 4.5 m 5 530 N 1 T sin f 21 3.0 m 2 2 w 1 4.5 m 2 5 g t 5 T cos f T sin f Hinge Pivot y x 3.0m 4.5m H h H v T w f 1 T 5 1.00 kN. H h H v H h . H v 1 x 1 y f 5 53.1. sin f 5 4.0 m 5.0 m f 10-16 Chapter 10...
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