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HW11_sol

# HW11_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 the tension in the cable. SGLUTION From a free-body diagram for the plate: moment equilibrium: = (rmB x A) + (rc/B x TCD) + (re/a x W) + (rE,B x P) (20 i) x (ij + Azﬁ) + (—5 i + 20 3) x (—0.8660T (10 i + 10 3) x (-100 ﬁ) + (25 i + 20 3) x (—300 E) CD J + 0.5000TCDE) A (10TCD - 7000) 1 + (2.500TCD - 20Az + 3500) J + (4.330TCD + 20Ay) E = 6 From which: = 700 lb TCD = -606.2 E + 350.0 3 lb 512.5 lb 8 513 lb ﬁ -606 T + 350 3 lb -151.55 lb 3 -151.6 lb A = -151.6 3 + 513 E lb force equilibrium: SE = K + E + TCD — W _ P = A; 3 + A, E + B, T + By 3.+ Bz E - 606.2 3 + 350.0 E - 100 E - 300 E Bx i + (-151.55 + By - 606.2) 3 + (512.5 + Bz — 50) E = 0 which: B = 0 ' By = 757.75 a 753 lb X A B = -462.5 a —463 1b E = 758 J - 453 E lb 2 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 6-90 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 .W the cable. ‘>Vi*\(/” Fig. P6-90 SQLUTION . —1 00 i - 1.40 cos 30° “ + (1.20 + 1.40 sin 30°) E V(-1.00)2 + (-1.40 cos 30°)2 + (1.20 + 1.40 sin 30°)2 A = -o.4056TC 1 — 0.4917Tr + 0.7705TC 2 :TC W = m3 = 75(9.81)(—E) = -735.75 E N a -735.8 E N '2 I From a free-body diagram ," for the plate: For moment equilibrium: Xi = (F B C/B x W) [(1 i + 1.2 E) x (—0.4056TC i - 0.4917TC j + 0.7706TC E)] x TC) + (FA/B x K) + (PG/B [(2 i) x (Ay j + A2 2)] A [(1 i + 0.70 cos 30° J - 0.70 sin 30° E) x (-735.8 E)] (0.5900TC - 446.0)-i + (-1.2573TC - 2Az + 735.8) 3 + (—0.4917TC + ZAy) E = O 486 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6-90 (Continued) Solving yields: = 755.9 N 2 756 N TC TC = 755.9(—0.4056 i — 0.4917 3 + 0.7706 E) = -306.6 E - 371.7 3 + 582.5 E N 185.84 N — —107.30 N A 185.84 J - 107.30 E N 2 185.8 3 — 107.3 E N A = /(185.84)2 + (—107.30)2 = 214.59 N z 215 N For force equilibrium: 2? = K + E + TC + W = 185 84 j - 107.30 E + Bx i + By 3 + Bz E — 306.6 1 - 371.7 3 + 582.5 E — 735.8 E = (8x - 306.6) 1 + (By + 185.84 - 371.7) 3 + (82 — 107 30 + 582 5 - 735.8) E = 6 xi+Byj+BzE = 306.6 1 + 185.86 3 + 260.6 E N g 307 i + 185.9 3 + 261 E N B = {(306.6)2 + (185.86)2 + (260.65)2 = 443.2 N 3 443 N 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 _ A _ -22 i + 24 i + 16 E ‘I‘B-TBeB—TEI 2 2 2 (—22) + (24) + (16) A = -0.6064TB 1 + 0.6616TB J + 0.4411TB E 'z. I l I _ A _ -56 i — 14 i + 24 E T —T e —TC‘: /(—56)2 + (-14)2 + (24)2 __ A_ A / _ 0.8958TC 1 0.2239TC J + 0.3839TC E I From a free-body diagram for the bar: For moment equilibrium: EH = (F A B/ A x TB) + (FC/A x TC) + (FE/A x P) = [(24 j + 16 E) x (—0.6064TB i + 0.6616TB 3 + 0.4411TB E)] + [(—14 3,+ 24 E) x (-0.8958TC i - 0.2239TC j + 0.3839TC E)] + [(38 i) x (-500 E)) = (—9.7024TB — 21.4992TC + 19,000) 3 +‘(14.5536TB - 12.5412TC) E = O ‘?€3£5 ENGINEERING MECHANICS - STATICS, 2nd. Ed. 6-91 (Continued) Solving yields: W. F. RILEY AND L. D. = 548.31 lb 3 548 lb 636.31 lb ¥ 636 lb T = 548.31(-O.6064 T + 0.6616 3 + 0.4411 E) B = —332.49 E + 362.76 3 + 241.86 2 lb = 636.31l—0.8958 i - 0.2239 3 + 0.3839 2) C = —570.00 E - 142.47 3 + 244.28 2 1b ' For force equilibrium: 2? RA +_TB + TC + ﬁ = (RA - 332.49 - 570.00) 1 X + (R + 362.76 — 142.47) J AY A A + (RA + 241.86 + 244.28 — 500) E = 6 Z A =RAX1+RAYJ+RAZE = 902.49 E — 220.29 3 + 13.86 2 lb 2 902 i — 220 3 + 13.86 E lb RA = “902.4912 + (—220.29)2 + (13.86)2 929.09 lb 2 929 lb STURGES ...
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HW11_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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