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hw10_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: 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 yum the tension in the cable. SOLUTION From a free—body diagram for the plate: moment equilibrium: x W) + (F (FA/B x N) (20 i) x (ij + A22) + (-5 i + 20 3) x (-0.8660TCD j + 0.5000TCDE) (10 i + 10 j) x (-100 E) + (25 i + 20 j) x (-300 E) x K) + (PC/B x TCD) + (F (3/8 E/B (IOTCD - 7000) T + (2.500TCD - 20Az + 8500) 3 + (4.330TCD + 20A ) E = 6 Y From which: = 700 lb TCD = —606.2 E + 350.0 3 lb 512.5 lb 3 513 lb a —606 i + 350 3 lb _. -151.55 lb 2 —151.6 lb A = —151.6 3 + 513 E lb force equilibrium: K+B+TCD—W—P AV 3 + AZ 2 + Bx i + By 3 + Bz E - 606.2 3 + 350.0 E - 100 E — 300 E B i + (-151.55 + By - 606.2) 3 + (512.5 + 32 — 50) fi = 0 X which: Bx = 0 By = 757.75 8 758 lb B = -462.5 z -463 1b 3 = 753 3 - 463 E'lb Z W. F. RILEY AND L. D. STURGES ENGINEERING MECHANICS - STATICS, 2nd. Ed. 6-85* The plate shown in Fig. P6—85 weighs 200 lb and is supported in a horizontal position by a hinge and a cable. Determine the reactions at the hinge and the tension in the cable. SOLUTION Tc = TCeC/B T [ -12 i — 24 i + 20 E C (-12)2 + (—24)2 + (20)2 A = —0.3586TC 1 - 0.7171TC J + 0.5976TC E From a free—body diagram for the plate: SE = C + (F A A G/A x W) + (rB/A x Tc) = MAX 1 + MA: E + [(14 i) x (—200 E)] + [(26 i + 11 3) x (—0.3596TC i — 0.71711C 3 + 0.5976TC E)] = (M + 6.5736TC) 1 + (2800 — 15.5376TC) J + (MAz — 14.7000TC) E = fi Ax Solving yields: i + MA E = —1185 i + 2649 E in.-lb ' Ans. A Ax z T = 180.21 lb a 180.2 lb T = 180.21(—0.3586 i - 0.7171 3 + 0.5976 E) = -64.62 i — 129.22 3 + 107.69 E 1b 2? = K + TC + W = A i + AY j + Az E — 64.62 i - 129.22 3 + 107.69 E - 200 E = 6 X A = (Ax - 64.62) i + (Ay — 129.22) J + (Az — 92.31) E = U K = 64.62 i + 129.22": + 92.31 E 3 64.6 i + 129.2 3 + 92.3 E lb - - Ans. A = /(64.62)2 + (129.22)2 + (92.31)2 171.45 lb 3 171.5 lb ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-87* The plate shown in Fig. 96—87 weighs 150 lb and 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. F¢.Pas7 SOLUTION c c 01c 2 2 2 {(20) + (-22) + (18) = 0.5754TC i — 0.6330TC j + 0.5179Tc E _ . A _ 20 i - 22 “ + 18 t T T e — Tc From a free—body diagram for the plate: For moment equilibrium: (relB x TC) + (I'MB x A) + (rG/B x W) [(3 i + 22 j) x (0.5754TC i - 0.6330TC 3 + 0.5179TC E)] [(24 i) x (AV 3 + A2 E)] + [(12 i + 11 3) x (—150 E)] (11.3938TC — 1650) i + (-4.1432TC - 24Az + 1800) 3 + (-17.72stC + 24Ay) E = a ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-87 (Continued) Solving yields: = 144.82 lb § 144.8 lb TC TC = 144.82(0.5754 i - 0.6330 3 + 0.5179 E) 83.33 i - 91.67 3 + 75.00 E lb 106.98 lb $0.00 lb ‘ = 106.98 3 + 50.00 E lb 9 107.0 3 + 50.0 E lb A = {(106.98)2 + (50.00)2 = 118.09 lb 3 118.1 lb For force equilibrium: 2? =.K + B + TC + W = 106.98 3 + 50.00 E + 8x i + By 3 + 82 E + 83.33 i — 91.67 3 + 75.00 E - 150 E = (8x + 83.33) i + (By + 106.98 - 91.67) 3 + (82 + 50.00 + 75.00 - 150.00) E = 6 +ByJ+BzE 1 — 15.31 3 + 25.00 E 1b a -83.3 i — 15.31 3 + 25.0 E 1b B = /(-83.33)2 + (-15.31)2 + (25.00)2 = 88.34 1b a 88.3 lb ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-91* A bar is supported by a ball-and—socket joint and two cables as shown in Fig. P6—91. Determine the reaction at support A (the ball- and-socket joint) and the tensions in the two cables. SOLUTION _ —22 i + 24 i + 16 E B /(-22)2 + (24)2 + (16)2 A = -0.6064TB i + 0.6616TB J + 0.4411TB E C C T a = Tc[ -56 1.— 14 1 + 24 E J {(-56)2 + (-14)2 + (24)2 = -0.8958TC i - 0.2239TC 3 + 0.3839TC E From a free-body diagram for the bar: For moment equilibrium: (F x P) B/A x Ta) + (rC/A x Tc) +‘(rE/A [(24 3~+ 16 E) x (-0.6064TB i + 0.6616TB j + 0.4411TB E)] [(-14 j + 24 E) x (-0.3958Tc i - 0.2239TC j + 0.3839TC 2)] [(38 i) x (—500 E)1 (-9.7024TB — 21.49ngC + 19,000) 3 + (14.5536TB — 12.5412Tc) E = 6 ENGINEERING MECHANICS * STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-91 (Continued) Solving yields: 548.31 lb a 548 lb 636.31 lb ¥ 636 lb TB 548.3l(—0.6064 i + 0.6616 3 + 0.4411 E) = -332.49 i + 362.76 3 + 241.86 E lb 636.31(-0.8958 i - 0.2239 3 + 0.3839 E) = —670.00 3 ~ 142.47 3 + 244.28 2 lb For force equilibrium: 2? = RA + TB + TC + fi = (RAX - 332.49 — 570.00) 1 + (RAY + 362.76 - 142.47) J + (RA: + 241.86 + 244.28 - 500) 2 = 6 J + RA2 fi ...
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