hw10_sol - 6-10 Draw a free—body diagram for the diving...

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Unformatted text preview: 6-10 Draw a free—body diagram for the diving board shown in Fig. P6-10. SOLUTION The action of the pin at support A is represented by force components —. _ Ax and Ay. Support B exerts a normal force 3 on the board. The diver exerts a normal force 3 on ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-19 Draw a free—body diagram for I {a} bar BE shown in Fig. P6—19. (b1 bar DF shown in Fig. P6—19. SOLUTION (a) The link at the left end of bar BE exerts a force g on the bar that is directed along the axis of the link. The action of the pin support at point C of bar BE is represented by forces fix and Cy. The pin at E exerts a force E on bar BE that is normal to the surface of the slot in bar DE. The action of the pin support at the right end of bar DF is represented by force components Fx and F3. The pin at E exerts a force E on bar DF that is normal to the surface of the slot in the bar. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6—34 A beam is loaded and supported as shown in Fig. P6-34. The beam has a uniform cross section and a mass of 180 kg. Determine the reaction at support A. SOLUTION A free—body diagram of the beam is shown at the right. The reaction at support A is represented by force components Xx and Ky and a I moment H . The weight 9 = mg of the beam acts through the center of gravity G of the beam and is directed toward the center of the earth. mg = 180(9.807) = 1765.3 N a 1.765 kN A - W — 2 y Ay - 1.7653 — 2 = 0 3.7653 kN E 3.77 kN = 3.77 3 kN = 3.77 kN T MA — 3 - 1.7653(2) — 2(4) = 0 14.531 kN-m E 14.53 kN'm 14.53 E kN'm = 14.53 kN-m C ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-37 A structural member is loaded and supported as $0” shown in Fig. P6—37. The member has a uniform cross section and weighs 208 lb. Determine the reactions at supports A and B. SOLUTION A free—body diagram of the beam is shown at the right. The reaction at support A is represented by force components Xx and Ky. The reaction at B is a vertical force 3. The weight w of the beam acts through the center of gravity G of the beam and is directed toward the center of the earth. /( ,\ . ‘ /. ls Lid1 + Lad2 + Lad3 6(3) + 4(6) + 3(7.5) G = 4.962 ft A + 200 - 300 = 0 X 100 lb = ~Ay(2) - 208(2.962) — 200(2) - 100(7) - 300(2) = 0 -1158.0 lb = -1158 lb A: + 2 /(100)2 + (-1158)2 = 1162.3 1b a 1162 lb y tan-1 - -65.03° z -85.1° X = 100 T - 1158 A 1b 1162 lb s 85.10 + C EN = 8(2) — 208(4.962) - 200(2) - 100(9) - 300(2) = 0 1466.0 lb = 1466 lb A B: B = 1466 3 lb = 1466 lb T ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6—38 A beam is loaded and supported as shown in Fig. P6-38. The beam has a uniform cross section and a mass of 20 kg. Determine the reaction at support A and the tension T in the cable. SOLUTION The action of the pin at support A is represented by force components — Ax and Ay. The cable is continuous over the pulley; therefore, the force in the cable is constant. At points B and C the cable exerts tensile forces T on the beam that are tangent to the cable. The weight w of the beam acts through the center of gravity G of the beam and is directed toward the center of the earth. W = mg = 20(9.807) = 196.14 N + C EMA = T sin 24°(1 000) — 196.14(1.440) — 750(1.750) + T sin 70° (2880) = 512.3 N a 512 N A + T cos 24° - T cos 700 X A + 512.3 cos 24° - 512.3 cos 70° = o X -292.8 N 3 -293 N = Ay + T sin 24° - w - 750 + T sin 70° = Ay + 512.3 sin 24° -196.14 — 750 + 512.3 sin 700 = 0 256.4 N a 256 N = /A: + A = /(-292.8)2 + (256.4)2 = 389.2 N z 389 N 56.4 -292.8 = -293 i + 256 3 1b = 389 lb 5 41.2° = 138.79° a 138.80 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-40* An angle bracket is loaded and supported as shown in Fig. P6-40. Determine the reactions at supports A and B. 350 N Fig. P640 SOLUTION ,1 From a free-body diagram for the angle bracket: + C EMA = B(0.200) 500(0.100) — 350(0.220) = 0 = 635 N B § = -635 j N = 636 N «— =A +500—B X A + 500 - 635 = X 135 N = A - 350 = 0 y 350 N y = /(135)2 + (350)2 = 375.1 lb a 375 lb '1 0 tan —§§ = 68.9 3 lb = 375 lb a 68.9° ENGINEERING MECHANICS - STATICS, 2nd. Ed. 6-45 A rope and pulley system is used to support a body as shown in Fig. P6—45. Each pulley is free to rotate and the ropes are continuous over the pulleys. Determine the force p required to hold the body in equilibrium if the weight w of the body is 400 lb. SOLUTION From a free—body diagram for pulley A: + T 2F = 2T - 400 = 0 y ‘1 T‘ = 200 lb From a free—body diagram for pulley B: + T 2F = 2T - 200 = 0 y 2 T2 = 100 lb From a free—body diagram for pulley C: +T2Fy=2P-100=O P = 50 lb Ans. W. F. RILEY AND L. D. STURGES ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES The wrecker truck of Fig. P6—63 has a weight of 15,000 lb and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a, tangential component Bx while the force exerted on the front wheel consists of a normal force Ay only. Determine the maximum pull P the wrecker can exert when 6 : 300 if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backwards (the front wheels remain in contact with the ground). SOLUTION From a free-body diagram for the wrecker truck: For impending tipping: + C 2MB = W(8) — p sin 30° (10) — P cos 30° (5) = 15,000(8) - p sin 30° (10) - P cos 30° (5) = 0 12,862 lb 8 12.86 kip = By — W - P cos 300 By - 15,000 - 12,862 cos 30° = 0 26,139 lb 2 26.1 kip —3x + p sin 30° -Bx + 12,862 sin 30° 6431 lb 3 6.43 kip Bx(max) = 0.88V = 0.8(26,139) = 20,911 lb > 6431 lb Therefore: P 12.86 kip 45'} ...
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This homework help was uploaded on 03/26/2008 for the course CVEN 221 taught by Professor - during the Spring '08 term at Texas A&M.

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hw10_sol - 6-10 Draw a free—body diagram for the diving...

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