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Unformatted text preview: PROBLEM 15.40 Rod AB moves over a small wheel at C while end A moves to the right
with a constant velocity of 25 in./s. At the instant shown, determine
(a) the angular velocity of the rod, (b) the velocity of end B of the rod. SOLUTION Slope angle ofrod. tan6 : % = 0.7, 6 = 35°
A—C = 10 = 12.2066 in. '65 = 20 — E : 7.7934 in.
0056 Velocity analysis. VA : 25 in./s., VC = VC 4: 6
VGA = ACwABV 6
VC = VA + vC/A Draw corresponding vector diagram. vC/A : VA sinB = 25 sin35° = 14.34 in./s (a) a) _ VC/A _ 14.34
”‘3 E 12.2066 21.175rad/s wAB 21.175rad/s) 4
VC 2 vA c056 : 25 c056 = 20.479 in./s
vm : 63(1)” = (7.7934)(1.175)
: 9.1551in./s
vmj has same direction as vC/A.
VB : VC + VB/C Draw corresponding vector diagram. tanw : vB/C : 9.1551 (0 = 24 090
vC 20.479’ ‘
(b) v3 : VC = 30—47—9— = 22.4 in./s : 1.869 ft/s cos<p cosZ4.09°
q) + 0 = 59.10
v3 = 1.869 ft/s 431 59.104 , _ _ PROBLEM 15.44
39' 1n Prob. 15.43, determine (a) the velocity of point 0, (b) the point of the ‘ , V ::ovf_ disk with zero velocity.
A/ 0L ‘ WM
24m,
SOLUTION
In units of in./s, VIM = wk x rB/A = wk x (24i + 24j) 2 A24wi + 24a)j
vC/A : wk >< rC/A : wk x 481 : 4860j
VB = VA + VB/A
—296i + (1(9)), j = (vA)xi — 282j — 24m + 24m Components. i: —296 = (VA )X — 24a) (1) j: (V31) : —282 + 2460 (2) vC = VA + VOA
—56i + (vC)y = (VA)xi — 282j + 48a)j Components. i: ~56 2 (VA )x (3) j: (VaV = —282 + 48a) (4)
From (3), (VAL = —56in./s, vA = —56i  282j
From (1), a) = ELK—Si) : 10 rad/s, 0) = (10.00 rad/s)k
(a) v0 = VA + v0M 2 VA + a) x r0“ : 10k >< (24i) = —56i — 282j + 240j = —56i , 42j
v0 = —(56.0in./s)i—(42.01n./s)j<
(b) 02v0+mx(xi+yj)
0 : —56i — 42j+10k x xi + yj) = —56i — 42j+10xj—10yj Components. i: 0 = —56 — 10y, y = —5.60 in. 4 j: 02—42+10x, x=4.20in.4 PROBLEM 15.59 1 In the eccentricity shown, a disk of 40mmradius revolves about shaft 0
that is located 10 mm from the center A of the disk. The distance between
L the center A of the disk and the pin at B is 160 mm. Knowing that the angular velocity of the disk is 900 rpm clockwise, determine the velocity 40 mm
of the block when (9 = 30°.
160mm l0 mm SOLUTION Geometry. o “D B (0A)sin6 = (AB)sin/3
_' , (0A)sin€ 10 sin30°
‘ sm = —— = —, =1.79°
A ﬂ AB 160 [3
Shaﬁ‘ and eccentric disk. (Rotation about 0), 600A = 900 rpm 2 301: rad/s 3 vA = (0A) (00A 2 (10)(30n) = 3001tmm/s7 Rod AB. (Plane motion = Translation with A + Rotation about A.) VB 2 VA +vB/A [va— ]=[VA 7 6001+[VA/B \ ,6] Draw velocity vector diagram. 90° — ﬂ = 88.21°
45' (p=180°—60°—88.21° =31.79° Law of Sines. VB VA simp _ sin(90° — ﬂ) vAsinq2 _(300n)sin31.79° sin(90° — ,8) sin 88.21°
= 497 mm/s PROBLEM 15.77 A double pulley is attached to a slider block by a pin at A. The 1.5in.
radius inner pulley is rigidly attached to the 3in.—radius outer pulley.
Knowing that each of the two cords is pulled at a constant speed as
shown, determine (a) the instantaneous center of rotation of the double
pulley, (b) the velocity of the slider block, (0) the number of inches of
cord wrapped or unwrapped on each pulley per second. VD = lOin./s§, VB = Sin/Si w=vD+vB 210+8z4rad/s
BD 4.5 CD=XP—:B=2.Sin.
a) 4 CA : 3.0 — 2.5 2 0.5 in. (a) C lies 0.500 in. to the right of A. 4 F ’4 (b) VA : 0.50) = (0.5)(4) = 2 in./s VA 2 2.00 in./sl { (c) VD — VA : 12 in./sl Cord DE is unwrapped at 12.00 in./s. 4 VB—VA =6in./sl Cord BF is unwrapped at 6.00 in./s. { J PROBLEM 15.86 /\QP At the instant shown, the angular velocity of bar DE is 8 rad/s
r" counterclockwise. Determine (a) the angular velocity of bar BD, (b) the angular velocity of bar AB, (c) the velocity of the midpoint of bar BD. Hon mm %\/;‘j‘30“ «,6.
2m mm B
N < ’ L
A 5.51% 600 mm Elf}... SOLUTION Bar DE. VD : ewm : (0.6)(8) = 4.8 m/s
VD = 4.8 m/s w
BarAB. VB 2 acoAB = 0.2 (0143 v” = 0.2043 5. 30° Locate the instantaneous center (point C) of bar BD by noting that velocity
directions at points B and D are known. Draw BC perpendicular to vb, and DC perpendicular to VD.
Let I 2 BD = 0.6 m. Law of Sines for triangle CBD. b d l 0.6 . : _ = . = . = 1.2 m
s1n120° srn30° sm30° sm30°
b = 1.03923 m, d = 0.6 m
(a) (030 = 3/2 : 4—8 = 8 rad/s 0’30 = 8.00 rad/s } 4
d 0.6
(b) v5 = waD = (1.03923)(8) : 8.3138 m/s
wAB = 13‘ = 8.3138 2 41.6 rad/s (0,13 : 41.6 rad/5‘; 4
a 0.2 ‘
(c) Law of cosines for triangle CMD.
2 2 1 2 1
m = d + — — 2d—coleOo
2 2
= 0.62 + (0.3)2 — (2)(0.6)(0.3)cos120°
m = 0.793725 m PROBLEM 15.86 CONTINUED Law of Sines. _' ' ° 0.3 ' 120°
Slnﬂ:51n120 , sinﬂ=(——)SI—n——, ,8219.1°
% m 0.79375 Velocity ofM. VM = "’1ng : (0.793725)(8) = 6.35 m/s VM 2 6.35 m/sl‘nl9.1°{ PROBLEM 15.92 At the instant shown, the velocity of collar A is 1.4 ft/s to the right and
T the velocity of collar B is 3.6 ft/s to the left. Determine (a) the angular 72in #103 in!»
’0 l I)
‘ ,1 f . 1.4m. , , ,
l «V as; 3% L velocny of bar AD, (b) the angular ve1001ty of bar BD, ((7) the veloc1ty of
14 «l in B .
L . p01nt D.
142%.; 7, ,.
SOLUTION
Method 1 I Assume V 1) has the direction indicated by the angle ,6 as shown. Draw
CD1 perpendicular to VD. Then, point C is the instantaneous center of
rod AD and point I is the instantaneous center of rod BD. Geometry. AD = 10.82 +14.42 = 18in.
8 BD = 7.22 + 5.42 = 9 in.
sin0=1—0—'8—=0.6, sin¢:E:0.8
18 9
c 18 18 d 9 9 sing _ sin(90° + ﬂ) cosﬂ singp sin(90° —,6) 008,6 ¥ l8sin9 _
cosﬂ cos/3 a 214.4 —10.81an,3 b : 5.4 + 7.2tanﬂ C Kinematics. VA : 1.4 ft/s = 16.8 in./s, v8 = 3.6 ft/s = 43.2 in./s v v v v
0.40:4: 0, wBDZ—BZ—U
a c b d
_CV4 de
vD—w—_—
a b WA [9 _183in6' cosﬂ . 16.8
de cosﬂ 9sinq) 43.2 b = 0.583331) a: 14.4 —10.8tan,B = 0.58333(5.4 + 7.2tanﬂ)
11.25 =15tanﬂ, tanﬂ = 10.75, ﬂ = 36.870 a = 14.4 — (10.8)(0.75) = 6.3 in. PROBLEM 15.92 CONTINUED b 2 5.4 + (7.2)(0.75) = 10.8 in. (a) (0/10 : g = g = 2.6667 rad/s CHAD = 2.67 rad/5‘; 4
(b) wBD = V?” = 2%:— : 4 rad/s (080 = 400 rad/s ; 4 (c) VD = chD = (13.5)(2.6667) = 36 in./s, VI) = 3.00 ft/s ‘5. 531° 4 Method 2 Consider the motion using a frame of reference that is translating with
collar A. For motion relative to this frame. VA = 1.4 ft/s : 16.8 in./s 4... VB 2 3.6 ft/s = 43.2 in./s «~— Ht! VA/A = 0, VB“ : 60 in./s W
tang = E, 6 = 36.87°
14.4
A AD 2 14'4 = 18in.
0056 VD/A = 1850.40 :5 9 Locate the instanteous center (point C) for the relative motion of bar BD
by noting that the relative velocity directions at points B and D are
known. Draw BC perpendicular to VB/A and DC perpendicular to VD/A. 7.2 s1n0
BC = (CD)cos19 + 5.4 = 15 in. CD = = 12in. M=Loz4raWs 0’” _ CB 15
VD/A = (CD)(OBD = (12)(4) = 48 in./s VD/A 48 \
a) =—=7 a) =2.67rad/s 4
(a) .41) AD 18 AU ,2
(12) mm) 2 4.00 rad/s I: 4
(c) vD : v4 + vW = [16.8 in./s «a» ] + [48 in./s 3 6]
= 36 in./S L 53.130 VD = 3 ﬁ/S l 53.l° 4 ...
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 Summer '06
 

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