homework8 - Problem 17.72 When the mechanism in Problem...

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Unformatted text preview: Problem 17.72 When the mechanism in Problem l7.7l is in the position shown here. use instantaneous centers to determine the horizontal velocity of B. Solution: The strategy is to determine the intersection of lines perpendicular to the motions at A and B. The velocity of A is parallel to the bar AB. A line perpendicular to the motion at A will be parallel to the bar 0/1. From the dimensions given in Problem 17.7 I. the length of bar AB is i343 : V63 + I22 : [3.42 in. Consider the triangle (JAB. The interior angle at B is fl:tan" :24.l‘. his and the interior angle at 0 is H : 9i]u _ fl : 65“)”. The unit vector parallel to the handle on is 20A 2 icosi! +j sint-I. and a point on the line is LOA 2 LOAeUA. where LOA is the magnitude of the distance of the point from the origin. A line perpendicular to the motion at B is parallel to the y axis. At the intersection of the two lines 7.4!? L()A cos it : , cosfi front which 1.9,; = 3b in. The coordinates of the instantaneous center are (14.7. 32.9) (in). Check From geometry. the triangle (MB and the triangle formed by the intersecting lines and the base are similar. and thus the interior angles are known for the larger triangle. From the law of sines Lon fl I'os _ I‘Aa sin 900 sinfi sinflcosfi = 36 in.. and the coordinates follow immediately from LOA = LgAELHL diet-k. The vector distance from (J to A is rm” : fit'icostl +j sinH] : 2.45m + 5.478j (in.). The angular velocity ofthc bar AB is detennined front the known linear velocity at A. i j It vazwaaxrayo: 0 0 —i 2.450 5.477 0 : 5.48i — 2.45] (im's). The vector from the instantaneous center to point A is P‘s/r 2 [as m I'(' 2 680,1 7 + 32.36“ : illlfii 7 2139i (in) The velocity at point A is l J k VA 2rd,”; x er : 0 U was ‘I2.25 fi27.39 0 : (“397.395 7 l2.25j) trust. ‘— Equate the two expressions for the velocity at point A and separate components. 5.48i : 27.39am. —2.45j : 712.25wagj {one of these conditions is superfluous) and solve to obtain m“- = 0.2 radls. coun- terelockwise. [Chet-k: The distance ()A is 6 in. The magnitude of the velocity at A is mantel : (Ute) : 6 inis. The distance to the instantaneous cen— ter from 0 is «I413 + 32.92 : 36 in.. and from C to A is (36— 6) : 30 in. from which 3001,13 : o infs. from which may = (1.2 rad/st t‘t‘te't'kJ. The vector from the instantaneous center to point B is ran" = n; a rr : H.717 (l4.7i+ 32.86j : 732.86j) tin.) The velocity at point B is i R 1‘5 : “AB x rgyr : 0 U 0.2 26,57i (in/s) U k32.86 U Problem 17.75 Bar AB rotates at 6 rad/s in the clock- wise direction. Use instantaneous centers to determine the angular velocity of bar BC. Solution: Choose a coordinate system with origin at A and _v axis vertical. Let C’ denote the instantaneous center. The instantaneous center for bar AB is the point A. by definition. since A is the point of zero velocity. The vector AB is r3” = 41+ 4j B is (in). The velocity at i j k W? : kins X rap; = 0 0 76 = 24i 7 24] (in/5). 4 4 U The unit vector parallel to AB is also the unit vector perpendicular to the velocity at B. l . 9A3 I —(5 +3)- J5 The vector location of a point on a line perpendicular to the velocity at B is LAB : fume“. where L41, is the magnitude of the distance froth point A to the point on the line. The vector location of a point on a perpendicular to the velocity at C is L(- : (l4i + yj) where _v is the yicoordinate of the point referenced to an origin at A‘ When the two lines intersect, L it: Mi. J} L and_v:—"‘E=i4 Ji from which 1.43 = 19.8 in.. and the coordinates of the instmitaneous center are (H. l4) (in). (Check: The line AC’ is the hypotenuse ofa right triangle with a base of 14 in. and interior angles of 45". from which the coordinates of L" are (l4. 14) in check]. The angular velocity of bar BC is determined from the known velocity at B. The vector from the instantaneous center to point B is ry/c : n; ‘ l'r : 4i + 4] 7 Mi — Hj = —I0i lflj. The velocity of point B is i j k V]; = wgf X rgyf : 0 0 Lung —l0 em 0 = {03(‘(l0i — Equate the two expressions for the velocity: 24 : lllwmu - from which Problem 17.86 The disk rolls on the circular surface with a constant clockwise angular velocity of 1 rad/s. What are the accelerations of points A. and B? Strategy: Begin by determining the acceleration of the center of the disk. Notice that the center moves in a cir- cular path and the magnitude of its velocity is censtant. Solution: V3 2 0 V0 = V5 +wk X [0/3 = (—1k) X (0.41) V0 = 0.4 i III/S Point 0 moves in a circle at constant speed. The acceleration of 0 is a0 = —vg/(R + r)j = (—0.16)/(l.2 + 0.4)j a0 = —0.1j (m/sz). a5 = a0 — wer/o = -—0.1j— (1)2(—0.4)j a3 = 0.3] (m/sz). 2A = a0 — wer/o = —o.11- (1)2(04): a4 = —0.Sj (m/sz). 1.2m ’roblem 17.92 "'6 = 45° and the sleeve P is moving 0 the right with a constant velocity of '2 m/s, whul are he angular accelerations of the bars 0Q and PQ'? Solution: W 2 an : 0. VJ. : 2i. a], : U rm“ : |,2 cusifiai + I.25in 45 'j m rmu : liens-1537 l.2§in45“j I11 v0 : wugk X r9,” = (1).,Qk x ((13in + 0.848]! 1'91 : 70.34-3an (1] HQ“ : 0.348mng (2) VP = V9 + ruPQk x (0.848i r 0.848jj 2 : vw + 0.848%,“ {3) 0 : ug‘. +0.848tupg (43 Solving cqns. (H44). mug : —l.l'.'9 rad/s. (UPQ : LIT‘) rad/s v1" = l mls uQ‘ : —1 m/s a9 = aw x r9,” — migrw“ (IQ‘ : 70348009 — 0.848(ugu (5) am. : 0.3430,..9 — 0.348ij {6) Also. 11,, :0 = HQ +u,,Qk X I'WQ - mfiurwg n = aw + ‘).348UI.,Q — 0.843%{9 m u 2 “a, + u.x4m,,Q + umswfig (3) Solving cqns. (5)7(8), we gel HQ. : 0.419} : 0 any : L39 nulls: (CIUCkWiSC) aw : 1.39 radlsJ (counter clockwimd) Problem 17.104 At the instant shown, bar AB has no angular velocity but has a counterclockwise angular acceleration of 10 rad/s2. Determine the acceleration of point E. Solution: The vector locations of A, B, C and D are: rA = 0, r3 = 400} (mm), rc = 700i (mm), to = l100i (mm). r5 = 1800i (mm) The vectors rB/A = 1‘3 - IA 2 400j (mm). mm = n: —— r3 = 700i - 400j (mm), mm = rc — to = —400i (mm) (a) Get the angular velocities ch, wcp. The velocity of point B is zero. The velocity of C in terms of the velocity of B is i j k vc = VB +w5c X 1173 = 0 0 (use 700 ~400 0 = +400w3cl + 700mm: j (mm/s). The velocity of C in terms of the velocity of point D i J k Vc = wco X mm = 0 0 ma) —400 0 0 = ~400wcpj (mm/s). Equate the expressions for vc and separate components: 400mm: = 0, 700mm: 2 ~400wcp. Solve: (03C = 0 rad/s, LOCI) 2: 0 rad/s. v- (b) Get the angular accelerations. The acceleration of point B is i j k BB :00”; X I'B/A - wfiBrg/A = 0 0 10 0 400 0 = —-4000i (mm/s2). The acceleration of point C in terms of the acceleration of point B: ac = as + (IBC X I'C/B — wich/B i j k : *4000i + 0 0 (15.: . 700 —400 0 ac = —4000l + 400a3cl + 700agcj (mm/s2). w.» , |.—- 700mm——|—~ 400 A|—— 700mm mm The acceleration of point C in terms of the acceleration of point D: i J k ac = OICD X I'C/D ~ wé‘DrC/D = 0 0 010) —400 0 0 = —400acoj (mm/s2). Equate the expressions and separate components: —4000 + 4000135 2 0, 7000(3C = ~400acp. Solve: agc = 10 rad/s2, den = —-l7.5 rad/s2, The acceleration of point E in terms of the acceleration of point D is i j k as =0lcp X rE/D = 0 0 ~17.5 7000 0 = —12250j (mm/$2) (clockwise) Problem 17.122 In the system shown in Prob- lem 17.121, the velocity of the pin C relative to the slot is 21 in./s upward and is decreasing at 42 in./s2. What are the angular velocity and acceleration of bar AC? Solution: See the solution of Problem 17.121. Solving Eqs. ('3), (4), (7), and (8) with 11cm : 21 in/s and new. = ~42 in/sz, we obtain wAc = 3 rad/s, one = —6 rad/s2. ...
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homework8 - Problem 17.72 When the mechanism in Problem...

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