HWS 11 - H093 \\ (WE 935. so u . ENGINEERING MECHANICS -...

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Unformatted text preview: H093 \\ (WE 935. so u . ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6~82* The shaft with two levers shown in Fig. P6-82 is used-to change the direction of a force. Determine the force fi required for equilibrium and the reactions at supports A and B. The support at A is a ball bearing and the support at B is a thrust hearing. The bearings exert only force reactions on the shaft. 200 mm - SOLUTION From a free—body diagram for the system: EMy = 9(200) — 750(200) = 0 fi = 750 N I = —750 E N Ans. For moment equilibrium: A $33 = [(-650 J) x (Ax i + AZ 2)] + [(—200 j + 200 E) x (—750 i)] + [(200 i — 450 3) x (n750 2)] = (-650AZ + 337,500) 3 + (650Ax - 150,000) E = 6 . E = 230.8 1 + 0 j + 519.2 E N s 231 i + 519 E N Ans. J A = /(230.8)2 + (519.2)2 For force equilibrium: 568.2 N % 568 N z? = 230.8 E + 519.2 E + 13x 3 + B j + B E - 750 i e 750 R = (B — 519.2) E + By 3 + (92 — 230.8) E = 6 X B = B i + B j + B E x y z 0 519.2 1 + j + 230.9 E N s 519 i + 231 E N Ans. B = H5192)2 + (230.8)2 = 568.2.0 N 3 568 N ENGINEERING MECHANICS — STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 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. mum Determine the reactions at the hinge and the tension in the cable. SOLUTION C TCeC/B . —12 i — 24 j + 20 E TC (—12)2 + (—24)2 + (20)2 A = -0.3586TC 1 - 0.7171TC J + 0.5976TC E From a free—body diagram for the plate: SEA = EA + (FGIA x W) + (FE/A x Tc) = MAX 1 + MA: E + [(14 i) x (—200 E)] + [(26 1 + 11 3) x (—0.3566TC 1 - 0.7171TC j + 0.5976TC E)] = (MAx + 6.5736TC) 1 + (2800 — 15.5376Tc) J + (MAZ - 14.7000TC) E = 0. Solving yields; C = M i + M E = -1185 i + 2649 E in.-lb Ans. n A Ax A2 TC = 180.21 lb 3 180.2 lb Ans. TC = 180.21(—0.3586 i - 0.7171 3 + 0.5976 E) = -64.62 i — 129.22 3 + 107.69 E lb EF=K+TC+W = Ax i + Ay j + Az E — 64.62 i — 129.22 3 + 107.69 E — 200 E = 0 = (Ax - 64.62) i + (Ay - 129.22) 3 + (Az — 92.31) E = 6 K = 4.62 i + 129.22 3 + 92.31 E 6 a 64.6 i + 129.2 3 + 92.3 E lb Ans. A = -/(64.62)2 + (129.22)2 + (92.31)2 477 171.45 lb % 171.5 lb ENGINEERING MECHANICS _ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 1 6=91* A bar is supported by a F 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. ,- 4/ P} 3mm : SOLUTION '_ A _ —22 i + 24 j + 16 R T — T eB - TB[ /(-22)2 + (24)2 + (16)2 A = —O.6064TB i + 0.6616TB 3 + 0.4411TB E 1; l I l _ A _ -56 i — 14 j + 24 E T — T e - TC[ 4“§““‘# /(—55)2 + (-14)2 + (24)2 ,/ I A A = -0.8958TC 1 - 0.22391:C J I l + 0.3839T E ’ , 1 C I I l I From a free-body diagram for the bar: For moment equilibrium: SE = (FE/A x TB) + (F x fi) A x TC) + (F C/A E/A = [(24 j + 16 E) x (—0.6064TB i + 0.6616TB j + 0.4411TB 2)] + [(—14 j + 24 E) x (—0.8958TC i — 0.2239TC j + 0.33ngC 2)] + [(38 i) x (—500 fi)] (-9.7024TB — 21.4992TC + 19,000) 3 + (14.5536TB - 12.5412TC) E = 6 Wig)“ .._«_ . , . ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6a91 (Continued) Solving yields: TB = 548.31 lb E 548 lb Ans. TC = 636.31 lb s 636 lb Ans. T3 = 548.31(—0.6064 i + 0.6616 3 + 0.4411 2) = ~332.49 i + 362.76 3 + 241.86 E lb Tr = 636.31(—0.8958 i — 0.2239 3 + 0.3839 R) = —570.00 T — 142.47 3 + 244.28 E lb For force equilibrium: 2F = §A 4 TB 4 TC + P = (RAx — 332.49 — 570.00) 1 + (8AY + 362.76 — 142.47) 1 + (8A2 + 241.86 + 244.28 — 500) E = 6 E = R i + R j + R E A Ax Ay A2 = 902.49 3 — 220.29 3 + 13.86 E lb II? 902 1 ~ 220 j + 13.86 R 1b Ans. R = (902.49)2 + (—220.29)2 + (13.86)2 = 929.09 lb 4 929 lb “2.44.6..- . ..... mm .44... 3.1.4,». “y. N. m . ....... .-..... m... . 4 3.. .... ...... .., _. .....,. . 4 .. .. .... .08... ... . .4... 4 .. m M... . 1 V... 487 ENGINEERING MECHANICS — STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 6=101* The shaft shown in Fig. P6—101 is part of a drive system in a factory. Friction between the belts and pulleys prevents slipping of the belts. Determine the torque T required for equilibrium and the reactions at supports A and B. The support at A is a journal bearing and the support at B is a thrust bearing. The bearings exert only force reactions on the shaft. 6 in. SOLUTION From a free—body diagram for the shaft: ZM y = T + 50(3) - 150(3) + 50(3) ~ 200(3) 2 0 T = 750 j in.'lb Ans. For moment equilibrium: SEE = [(:35 j) x (Ax i + AZ E)] + [(—14 j + 3 E) x (—200 1)] + [(-14 j — 3 E) x (—50 3)] + [(3 1 ~ 28 j) x (150 E)1 + [(-3 i — 28 j) x (50 E11 + 750 j = («3542 — 5500) i + (354x — 3500) E = 6 K = 97.22 i — 155.56 E lb 2 97.2 i — 155.5 E 1b Ans. A = ./(97.22)2 + (—155.56)2 = 183.44 lb 2 183.4 1b i For force equilibrium: 2? = 97.22 i — 155.55 E + BR 1 + By 3 + BZ E — 200 i — 50 i + 150 E + 50 E = (Bx — 152.78) 1 + By 3 + (B2 + 44.44) E = 6 E = 3x i + By 3 + 82 E = 152.78 1 + 0 j m 44.44 E 1b e 152.8 E — 44.4 E lb Ans. H B = #(152.78)2 + (—44.44)2 ,. - mwmmmmwewmmmww\ 159.11 lb 5 159.1 lb mm “mlgfggim-mm ...
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HWS 11 - H093 \\ (WE 935. so u . ENGINEERING MECHANICS -...

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