10_InstSolManual_PDF_Part18

10_InstSolManual_PDF_Part18 - Figure 10.52 (b) gives (c)...

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10.51. Set Up: The free-body diagram for the door is given in Figure 10.51. and are the vertical and horizontal components of the forces exerted on the door by the upper and lower hinges, respectively. Since each hinge supports half the total weight of the door, Figure 10.51 Solve: with the axis at the lower hinge gives and says The upper hinge exerts a horizontal force of 140 N away from the door and the lower hinge exerts a horizontal force of 140 N toward the door. Reflect: Our calculation and Newton’s third law shows that the door pulls outward on the upper hinge; our experi- ence with doors agrees with this. 10.52. Set Up: Let the forearm be at an angle below the horizontal. Take the pivot at the elbow joint and let counterclockwise torques be positive. Let be upward and let be to the right. Each forearm has mass The weight held in each hand is with is the force the biceps muscle exerts on the forearm. is the force exerted by the elbow and has components and Solve: (a) The free-body diagram for the forearm is given in Figure 10.52.
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Unformatted text preview: Figure 10.52 (b) gives (c) gives gives so Since we calculate to be positive, we correctly assumed that it was downward when we drew the free-body diagram. (d) The factor divides out of the equation in part (b), so the force T stays the same as she raises her arm. g t 5 cos u E v E v 5 T 2 w arm 2 w 5 569 N 2 1 2.16 kg 21 9.80 m / s 2 2 2 1 7.50 kg 21 9.80 m / s 2 2 5 474 N T 2 E v 2 w arm 2 w 5 0, g F y 5 E h 5 0. g F x 5 T 5 16.0 w arm 1 38.0 w 5.5 5 16.0 1 2.16 kg 21 9.80 m / s 2 2 1 38.0 1 7.50 kg 21 9.80 m / s 2 2 5.5 5 569 N T 1 5.5 cm 21 cos u 2 2 w arm 1 16.0 cm 21 cos u 2 2 w 1 38.0 cm 21 cos u 2 5 g t 5 w w arm y x T 10.5cm 22.0cm pivot E v E h u 5.5cm E h . E v E S T S m 5 7.50 kg. w 5 mg , m arm 5 1 2 1 0.0600 21 72 kg 2 5 2.16 kg. 1 x 1 y f L h 5 U h 5 140 N. g F x 5 U h 5 1 0.50 m 1.00 m 2 1 280 N 2 5 140 N. 2 w 1 0.50 m 2 1 U h 1 1.00 m 2 5 g t 5 0.50m 1.00m 0.50m U h L h L v U v w cg 0.50m 1 U v 5 L v 5 w / 2 5 140 N. L h L v , U h U v , 10-18 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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