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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 sintI. 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 dietk.
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. [Chetk: 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 : liens1537 l.2§in45“j I11
v0 : wugk X r9,” = (1).,Qk x ((13in + 0.848]! 1'91 : 70.343an (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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This note was uploaded on 03/20/2008 for the course EM 311 M taught by Professor N/a during the Spring '08 term at University of Texas.
 Spring '08
 N/A
 Dynamics

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