hw 7 - - "Hqu '7 Melvin!) ENGINEERING MECHANICS -...

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Unformatted text preview: - "Hqu '7 Melvin!) ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-83 The plate shown in Fig. P6-83 is supported in a horizontal position by two hinges and a cable. The hinges have been properly aligned; therefore, they exert only force reactions on the plate. Assume that the hinge at B resists any force along the axis of the hinge pins. Determine the reactions at supports A and B and the tension in the cable. SOLUTION From a free-body diagram for the plate: moment equilibrium: x F) (20 i) x (ij + A22) + (—5 i + 20 3) x (—0.3660TCD 3 + 0.5000TCDE) (10 i + 10 j) x (—100 E) + (25 i + 20 3) x (-300 E) = (F n W) + (F A/B x x) + (PC/B x Tcn) + (F G/B E/B (101*CD - 7000) i + (2.500TCD - 20Az + 3500) j + (4.330'rCD + 20A ) E = O Y From which: 3 4L = 700 lb TCD = —606.2‘2 + 350.0 j/lb 512.5 lb 8 513 lb E -606 i + 350 3 lb -151.55 1b a —151.6 lb 3 = -151.6 3 + 513 E lb force equilibrium: 2F = K + B + TCD - W — P = Ay j + A2 2 + Ex 5 + By 3 + 132 E - 606.2 3 + 350.0 E — 100 E — 300 E = B i + (-151.55 + By - 606.2) 3 + (512.5 + 32 — 50) E = o X which: B By = 757.75 5 758 lb X B = -462.5 s -463 1b 3 = 758 “ - 463 R 1b Ans. Z ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-87* The plate shown in Fig. P6-87 weighs 150 lb and is supported in a Janina—1L horizontal position by two hinges and a cable. The hinges have been properly aligned; therefore, they exert only force reactions on the plate. Assume that the hinge at B resists any force along the axis of the hinge pins. Determine the reactions at supports A and B and the tension in the cable. SOLUTION T a = T 20 1 - 22 J + 18 E I CD/C c 2 . Z Z {(20) + (-22) + (18) V = 0.5754TC 1 - 0.6330TC J + 0.5179TC 2 From a free-body diagram for the plate: For moment equilibrium: (PC/B x Tc) + (FA/B x K) + (FGIB x W) A [(8 i + 22 j) x (0.5754TC i - 0.6330TC J + 0.51791C 2)] A [(24 i) x (Ay J + Az E)] + [(12 i + 11 j) x (—150 2)] (11.3938TC - 1650) i + (-4.1432TC - 24Az + 1800) j + (-17.7228TC + 24Ay) E = 6 ~ (aiming—r. 1m 'thMwJu-mgmnu ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-87 (Continued) Solving yields: = 144.82 lb a 144.8 lb Tc ' TC = 144.32(0.5754 i - 0.6330 3 + 0.5179 E) = 33.33 i — 91.67 3 + 75.00 E lb = 106.98 lb = 50.00 lb K = 106.93 3 + 50.00 E 1b 2 107.0 3 + 50.0 E lb A = /(106.93)2 + (50.00)2 = 113.09 lb a 113.1 lb For force equilibrium: 2? = K + B + Tc + W = 106.93 3 + 50.00 E + 3x 1 + 3y 3 + 32 E + 33.33 i - 91.67 3 + 75.00 E — 150 E = (3x + 33.33) i + (By + 106.93 - 91.67) 3 + (32 + 50.00 + 75.00 - 150.00) E = U + B j + 3 E y 2 = -33.33 i — 15.31 3 + 25.00 E lb A a —33.3 i - 15.31 J + 25.0 E lb 3 = /(-33.33)2 + (-15.31)2 + (25.00)2 = 33.34 lb a 33.3 lb ENGINEERING MECHANICS - STATICS, 2nd. Ed. 6—89 A beam is supported by a ball-and-socket joint and two cables as shown in Fig. P6-89. reaction at support A (the Determine the ball-and—socket joint) and the tensions in the two cables. SOLUTION B I 0 i - 11 ‘ + 6 E T “i—J—J / 02 + (-11)2 + (6)2 -0.8779TB j + 0.4789TB E [ —7 i - 11 i + 3 E J _TC /(-7)2 + (-11)2 + (3)2 = 052321c 1 - 0.8222TC J + 0.22421C E From a free-body diagram for the bar: For moment equilibrium: (FD/A x TB) + (F [(11 3) x (—0.37791 D/A B x TC) + (F D/A x P1) + (F j + 0.47391B E)] [(11 3) x (-0.5232Tc i - 0.8222Tc j + 0.2242Tc [(11 3) x (300 3)] + [(6 3) (5.2679TB + 2.4662TC - 4300) i + (5.75521C - 3300) E = 0 x (-300 E)] 484 E'IA W. F. RILEY AND L. D. x P ) E)] STURGES -Aflmm...m‘ no.an M . .,~.‘_1.‘_v.\..1-nmw,i.‘;.. ..- hum-wwwwunm, : 1 ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-89 (Continued) Solving yields: TB = 642.7 lb 3 643 lb TC = 573.4 lb a 573 lb Ta = 642.7(-0.8779 j + 0.4789 2) = —564.2 3 + 307.8 2 lb TC = 573.4(-0.5232 i — 0.8222 3 + 0.2242 R) = -300.0 E — 471.4 3 + 128.6 E lb For force equilibrium: 1 .,.......1-w...¢._.-..._._._- A“... _..._-_._..‘ WMEJMAW.“ )ZF'=1!A+TB+TC+151+152 (RA - 300.0 + 300.0) T X AV - 564.2 - 471.4) J (R + 307.8 + 128.6 - 800) E = 0 A2 A Ay‘] 1035.6 3 + 363.6 R 1b +RAZE A z 1036 J + 364 2 lb RA = /(1035.6)2 + (363.6)2 = 1097.6 lb a 1098 lb WWGA»‘mem.nwmm.m.u.m_¢n. Nah“- ..NA 1. ...1.._._._._. -3...“ 3.... ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6—90 The plate shown in Fig. P6—90 has a mass of 75 kg. The brackets at supports A and B exert only force reactions on. the plate. Each of the brackets can resist a force along the axis of pins in one direction only. Determine the reactions at supports A and B and the tension in the cable. SOLUTION T = T a = T -1.00 i - 1.40 cos 30° 3 + g1.2o + 1.40 sin 30°) E C C D/C C 2 2 2 /(—1.00) + (-1.40.cos 30°) + (1.20 + 1.40 sin 30°) = -o.4056Tc i - 0.4917Tc 3 + 0.7706TC E W = n5 = 75(9.81)(—E) = -735.75 E N 2 —735.8 E N I From a free-body diagram ' for the plate: For moment equilibrium: (re/B x TC) + (r A/B x A) + (I‘G/B x W) [(1 i + 1.2 E) x (—0.4056TC i - 0.4917Tc j + 0.7706TC 2)] [(2 i) x (AV 3 + A2 2)] [(1 i + 0.70 cos 30° 3 - 0.70 sin 30° E) x (-735.3 2)] (0.5900TC - 446.0) 3 + (-1.2573TC - 2Az + 735.8) 3 + (—0.4917TC + 2A ) E = 6 Y 486 n «wflamwm‘ s. an.) Mei tan...“ in“ n H ~1‘a_¢>~u—H<v;p ( 1 i i i ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-90 (Continued) Solving yields: = 755.9 N 5 756 N Tc Tc = 755.9(-0.4056 T - 0.4917 3 + 0.7706 R) = -306.6 f — 371.7 3 + 582.5 E N 185.84 N = -107.30 N A K = 185.84 3 - 107.30 E N a 185.8 J - 107.3 E N A = {(185.84)2 + (“107.30)2 = 214.59 N 3 215 N For force equilibrium: 2? = K +-§ + TC + W = 185.84 3 — 107.30 E + Bx i + By 3 + Bz E - 306.6 f - 371.7 3 + 582.5 E - 735.8 E = (Bx - 306.6) E + (By + 185.84 — 371.7) 3 + (32 - 107.30 + 582.5 — 735.8) E = 6 y j + Bz E + 185.86 3 + 260.6 E N a 307 i + 185.9 3 + 261 E N B = /(306.6)? + (185.86)2 + (260.6)2 = 443.2 N z 443 N ...
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This note was uploaded on 07/16/2009 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

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hw 7 - - "Hqu '7 Melvin!) ENGINEERING MECHANICS -...

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