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assignment 18 solutions

# assignment 18 solutions - 0 III.ul In PROBLEM 6.77 For the...

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Unformatted text preview: 0 III. .ul In PROBLEM 6.77 For the frame and loading shown, determine the force acting on member ABC (a) at B, (b) at C. SOLUTION Note that ED is a two-force member FBD member ABC: . (a) ( EMC = 0: (l6 in.)[iF},D] _ (24 in.)(80 lb) = 0 M M A ‘ 5 872 lb Fm): 1501b, FBD = 150.0 lb 7 369° 4 3 (b) 1 2147,, = o: Cy — §(150 lb) = o, q, = 901b —» 25:0: Cr—%(1501b)+801b=0, Cx=401b C=QR§Ih /660°‘ PROBLEM 6.80 A circular ring of radius 8 in. is pinned at A and is supported by rod BC, which is ﬁtted with a collar at C that can be moved along the ring. For the position when 0 = 35°, determine (a) the force in rod BC, (b) the reaction at A. SOLUTION FBD Ring: (a) Q EMA = o; (8 in.)(FBCcos35°) ~ (8 in.)(6 1b) = 0 FBC = 7.32461b, FBC = 7.32 lb C 4 (b) —’ EEC = O: AJr —(7.3246 lb)cos35° = 0 A_\;=61b T >313, = 0: A), + (7.3246 lb)sin35° — 6 lb = 0 11y = 1.79876 lb A = 6.26 lb 416.69o { r'PROBLEM 6.84 Deteimine the components of the reactions atA and E when a 24-1b force directed verticallydownward is applied (a) at B, (b) at D. SOLUTION (a) FBDAC: (b) FBD CE: 24/4, 251 Notcz. CE is a' two-force member FCE ( EMA: 0: (8 1113(3) + (2.1n)[FT‘-‘ZE) — (6 in.)(24:lb)» 0 II FOE =1.4.4J§-1b, so Ex": 14.40 lb ~ 1 E, = 14.40 lb I 4 ~32}; =0: —AX + 14.4 1b = 0 AX =14.40'lb «— 4 (2113, = 0: A), — 24 lb +14.40'1b’= 0 Av = 9.60 lb 1 < Note AC is a two-force member Q ZME = 0: ,(3 m')[J_1_'TE‘C + Wag— (1 in.)(24 1b) =0 FAG -= 1.6Jl—7 1b, Ax = 6.402113 +----‘< A}, =1-.6001b 1 .4' 4 \. —». 2F'-=O:x E —~+ 1. 1711) =0, 11,26,400) —»--.. x x Jﬁ( 6J— ) ‘ L (213-: 0: - .15; +.\/ﬁ (1.6Jﬁ1b) .324 lb = 0 E5, =' 22.4 -lb ( { ...
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assignment 18 solutions - 0 III.ul In PROBLEM 6.77 For the...

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