{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HWS 11 - H093(WE 935 so u ENGINEERING MECHANICS STATICS 2nd...

This preview shows pages 1–5. Sign up to view the full content.

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

View Full Document

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: 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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

HWS 11 - H093(WE 935 so u ENGINEERING MECHANICS STATICS 2nd...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online