Hwk 13 Soln

Hwk 13 Soln - Hwkfl‘l?) ‘ ' n ENGINEERING MECHANICS...

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Unformatted text preview: Hwkfl‘l?) ‘ ' n ENGINEERING MECHANICS ‘ STATICS. 2nd. Ed. H. F. RILEY AND L; D. STURGES 6—73‘ Determine the reaction 51) Hi at support A of the Pipe system shown in Fig. P6—73 when the force applied to the pipe wrench is 50 lb. SOLUTION From a free-body diagran for the pipe system: For force equilibrium: 2? = R + P A -7 i + 23 3 + 10 E R = F AP BIA x P = (-7 i 4 23 3 + 10 i) x (—50 E) = —1150 i — 350 3 in.'lb For moment equilibrium 23 = CA M1” :01“ — 1150) i + (11Ay - 350) 3 + MAI E z 0 f! u A MA, 1 + MAY 3 + HA! 2 = 1150 1 + 350 3 1n.-lb Ans. (M )2 + 1n 12 = /(1150)2 + 1350)2 = 1202 in.-lb Ax AY Bar AB is used to support an BSD—lb load as shown in Fig. P6—79. End A of the bar is supported with a ball and socket joint. End B of the bar is supported with two cables. Deteruine the components of the reaction at support A and the tensions in the two cables. 21. In mui fig. 1'647‘) S-i—aA—i‘cs SumI~79 6-77‘ The bent bar shown in Fig. P6-77 is supported with three brackets that exert only force reactions on the bar. Determine the reactions at supports A, B, and C. Fig. P6-77 SOLUTION )aoih 1' From a free—body diagram for the bar: moment equilibrium: - (EB/A x E) + (EC/A x C) + (FD/A x D) + (FE/A x E) [(—3 1 + e E) x (Bx i + By 3)} [(-8 i + 9 3 + 14 E) x (cx i + C1 E1] [(-8 i) x 175 3)] + [(-8 i + 14 E) x (50 i — 100 211 (-GB + 9c i i + (GB + 14c + as — 100) 3 y z x x z + (—813y - 94:x — 600) E = O For force equilibrium: 2F=K+B+C+D+E = 4 3 + A E + B i + B 3 + c i + c E + 75 3 + so Y - 100 E Y z x y x z = (B + c 1 so; i + (A + a + 75) 3 + (A + c — 100) i = O x x y y 1 1 Solving yields A = 450 lb Y A = 450 lb 2 450 3 + 450 E lb 400 i < 350 2 lb - —450 i — 525 3 lb ’Zfi SOLUTION From a free—body diagram for the bar: For moment equilibrium: u (16 1 + 26 i) x (—TBC 3) x (32 3 + 26 E) x (—180 i) + (16 i + 32 31 x 1~350 E) A = (26 TBC — 27.200) 1 + (~26 TED + 13,600) 3 + (—16 TBC + 32 TED) E = O From which: 18c 1046.2 lb a 1045 lb T —1046 3 lb TED 523.1 lb 3 523 lb —523 1 lb For force equilibrium: 2F = X + TBC + Tan + 9 = Axl + 4y3 + Ali - TBC 3 — T201 - 850 E (Ax ' 523.!) i + (A — 1046.2) 3 + (AZ v 850) R : 0 Y which: 523.1 lb 2 523 lb 1046.2 lb 3 1046 lb 850 lb 523 i + 1046 3 + 950 2 lb G-Bl‘ A shaft is loaded through a pulley and a lever (see Fig. PG—Bll that are fixed to the shaft. Friction between the belt and pulley prevents slipping of the belt. Determine the force F 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 shafL 501! “1 IS“ lh Fig. P6411 SOLUTION From s free-body disgran for the system: {My = P(14) 4 150(6) - 500(6) = 0 P = 150 lb i : —150 E lb Ans. For moment equilibrium: {HE = [(—10 3) x (Ax i + A2 E11 + [(-30 3 v 6 E) x (-500 3)] + [(-30 3 — 6 E) x (—150 3)} + ((14 i + 14 3) x 1200 3 - 150 E1] = (—10.1z - 2100) i + (131x - 16,700) E = 0 927.8 1 — 116.67 E lb a 923 3 - 116.7 E lb A = H9218)2 + (—116.67)2 = 935.1 lb 3 935 lb For force equilibrium: Ans. 2? = 927.3 1 - 116.67 E + Bx i + By 3 + BZ E - 500 i — 150 3 + 200 3 — 150 E = (Bx 1 277.0) 3 + (By + 200) 3 + (a2 — 266.67) E = O 3 + 61 E = -277.3 3 - 200 3 + 266.67 E lb a —278 3 — 200 3 + 267 E lb B = /(—277.8)2 + (—200)2 + (256.67)2 = 1 Y Ans. 433.9 lb 3 434 lb 6~87* The plate shown in Fig. P6—B7 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. SOLUTION TC=Te =C (20)2 + (—22)2 + 113)2 = 0.5754TC 1 — 0.6330Tc J + 0.5179TC E Pro. a free-body diagram for the plate: For moment equilibrium: ER = (F B C/B x X) + (F x 9) + _ XTC) (rA/B 0/9 [(8 i + 22 3) x (0.5754TC 1 ~ 0.6330Tc J n 1 0.51791c E11 + [(24 I) x (Ay 3 + AZ E)! + {112 i + 11 3) x (v150 E11 (11.3936TC - 1650) i + (-4.1432TC — 24Az + 1800) 3 + (—17.7228T + 24A ) E = U C v Solving yields: a n 144.82 1b I [44.8 lb TC 144.82(0.5754 i — 0.6330 3 + 0.5179 R) M 33.33 3 - 91.67 3 + 75.00 E lb A1 = 106.98 lb A1 = 50.00 lb K = 106.98 3 + 50.00 E lb 2 107.0 3 + 50.0 E lb A = /(106.96)2 + (50.00)2 = For force equilibrium: 116.09 lb I l18.l lb 2F = K + B + TC + D = 106.98 3 + 50.00 E + B 3 + B 3 + B y z + 83.33 i - 91.67 3 + 75 00 E — 150 E = (Bx + 03.33) i + (By + 106.98 — 91.67) 3 + (01 + 50.00 + 75.00 — 150.00) E = O B = B i + B 3 + 0 E x y z = —03.33 3 ~ 15 31 3 + 25 00 E lb 2 -03.3 i - 15.31 3 + 25.0 E lb B = /(-83.33)2 + (-15.31)2 + 125.0012 = 88.34 lb x 99.3 lb Ans. Ans. E Ans. ...
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This note was uploaded on 10/13/2009 for the course ME 274 taught by Professor Valry during the Spring '09 term at Iowa State.

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Hwk 13 Soln - Hwkfl‘l?) ‘ ' n ENGINEERING MECHANICS...

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