{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hwk 13 Soln

# Hwk 13 Soln - Hwkﬂ‘l ‘ n ENGINEERING MECHANICS ‘...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}