Ch16_Sec7_Ex-110-115-126

Ch16_Sec7_Ex-110-115-126 - r e “f'zfiefi'v‘E-tfit a.

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Unformatted text preview: r e “f'zfiefi'v‘E-tfit a. chemiyriflwa‘ivil l 3 16-11“, At a certain instant, link AB has a clockwise angulair elocity of to,” = 2 rad/s and a counterclockwise angular eceleration of a“; = 2 rad/s2. Determine the aocelel'a ion of link BC at the instant shown. Velocity Analyst's : The angular velocity of links BC and CD can be obtained by “5ng the method of instantaneous center of zero velocity. 'nie instantaneous center of: o velocity of link BC at the instant shown is located at the intersection of extcn linet; drawn perpendicular from VA, and vc. Here, 1), e m, em) = 12.0 in.ls . of = warn, = 6mm. rm : 5cos 36.87° = 4 in. bud trcuc = Ssin 36.87° = 3 in. The angular velocity oflink BC is given by 12.0 a)“. = L" = T = 3.00 rad/s ’5116 i Thus. Jngular velocity of link CD is given by i 1’c = mec'cuc ‘ 6(0c0 = 1 row = 1.50 rad/s Andalusian Equation .- The acceleration of pointB can be obtained by analyzing the angular motion of linkAB about pointA. Here. r” = {iii} in. '3 = an XI'AB ‘afiufu = (2k) x (6i) -2’(6i) = {—24.0i+ 12.01} in. /s2 The acceleration of point C can be obtained by analyzing the angular motion of link CD about boini D. Here. rm = {6]} in. ‘c = “co x "co ' minaret) = (acpk) x (51) — 1.501(6)) = (—6ac0l — 13.51) in. Is2 Link Bic is subjected to general plane motion. Applying Eq. l6 — 18 with r,,c = («ti—3]} in. we have t “a = .6 “Inc X 1'm: ‘ mgcrn/c 42m + l2.0j = ~6acDi—13.5j+ aukx (—4i - 3j) — 3.002 (—4i — 3t) —2-4.0i + 12.01 = (sun —6ac,, + 36.0) i + ( 13.5 ~4at,c)j Equatiltgi and j component. we have —24.0 = 3a" — 6am + 36.0 12.0 = 13.5 — 4a,”. ‘1 SolvingiEqs.[i] and [2] yields a“ = 0.375 nit/sz a = 101875 nict/sZ Thus. CD at ={—6(10.1875)i—13.5j} ian: = {—61.125i— 13.55} inls2 The magnitude and directional angle of the acceleration of point C are given by ac=t/(—61.125)2+(-13.5)2=62.6 in./s2 Ans 5 -tan" 13' -125° 61.125" ' Ans 415 _.< l ogfife’ nti > L 3 16-115. The hoop is cast on the rough surface such that it has an angular velocity on = 4 rad/s and an angular acceleration a = 5 rad/52. Also, its center has a velocity v0 = 5 m/s and a deceleration a0 = 2 m/s’. Determine the acceleration of point A at this instant. aA=|o+ ~=Ehf M = [03] 84/0 1: $03)] + [5(5): 3)] + [438] m = 4.83 m/s' An: :9 u 8 I drug + a X fun .A = -2i— 4)'(o.31) + (-Sls) x (0-3.!) 1‘ s (—0.5: - 4.31) mls‘ 4‘ = 4.33 mls‘ All :9 g a 84.l°7Am 16-126. At a given instant, the gear has the angular motion shofwn. Determine the accelerations of points A and B on the link and the link’s angular acceleration at this instant, ' For the gear ‘; \ vA = armc = ab = 6 ian- so = —[2(3)i= (fissuinjs‘ rm = {—21} in. a= (12k) ndls' IA = .0 + axrué-OVI'A/o = —36i + (mix (42]) -‘ (<5)1 (—21) = {—12l+72j}‘ an? at = y/(-12)‘+7:l2 = 73.0 nus! arc—3:80.? 2: For link AB Thele arm. somgi=0. i.e.. I. = Gui GA. f-dfigk r.“ = ill = '4 + an Xi'an “affirm a.i= (—12l+ 721);~(-aum)x(scos6m+ 83in60‘])-0 (it) a. =—12i+83in60°(18)= 113 nus: —» Am (+ T) 0= 72 Jamison" a“ a 18 rad/:1) Au ...
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This note was uploaded on 07/09/2011 for the course EGN 3311 taught by Professor Hudyma during the Fall '08 term at UNF.

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Ch16_Sec7_Ex-110-115-126 - r e “f'zfiefi'v‘E-tfit a.

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