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Unformatted text preview: ENGINEERING MECHANICS  Dynamics & L.D. Sturges 1444* One end of the 2m long bar AB shown is
attached to a collar which slides on a smooth
vertical shaft. The other end of the bar slides
on a smooth horizontal surface. At the instant
shown, 9 = 300 and collar A is moving downward with a speed of 0.4 m/s. Determine the velocity V73 of end B and the angular velocity wAB of bar A8 at this instant. Solution The collar A has no x—component of A
velocity while the end of the rod B
has no ycomponent of velocity.
Therefore, the relative velocity
equation ZUJAB
‘ VB = VA + vB/A 8
becomes ) A A A O A
v 1 = —0.4 _] + 20A8(cos 300 1 + sin 30 J) Separating this equation .into its x— and ycomponents O : (A)
x VB 2 AB cos 30 y. _Oo4 . 20) 511) 30 and solving gives (I)
AB 0.400 rad/s C . ....................... ........... Ans. V 0.693 m/s—) ....... .......... ..............Ans. B ENGINEERING MECHANICS  Dynamics 1449* Movement of the solenoid plunger
causes a gear to rotate. At the instant shown, the angular velocity of the gear is mo = 4 rad/s counterclockwise. Determine
the angular velocity wAB of the rod AB and
the velocity VA of the plunger at this
instant. Solution At the instant shown, (0:0:
CD 0 4 rad/s c sin ¢ = 5/13 . cos ¢ = 12/13 The gear teeth prevent slip between the
gear and the control rod AB. Therefore,
the velocity of the contacting points of
the two rigid bodies is the same and using
the relative velocity equation to write the
velocity of C in terms of the motion of the
two separate bodies VC= VD + vC/D = VA + vC/A gives 6 + 5(4)(cos ¢ 3 + sin ¢ 3) W.F. Riley & L.D. Sturges IZCUAQ ¢ = VA i + 12wA8(—sin ¢ 3  cos ¢ 3) 0%58 Separating this equation into its x— and ycomponents x: 20(12/13) v  12wAB(5/13) A y: 20(5/13) —12wAB(12/13) and solving gives a.) =—.
AB 0 694 rad/s = 0.694 rad/s C ...... V = 21.7 in./s 21.7 in./s 6— . ..... ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1475* One end of the l8in. long bar AB is attached to a slider that
moves in a vertical slot. The other end is attached to a wheel which :rolls on a horizontal surface. At the instant
Eshown, the center of the wheel is moving to the gright with a speed of 5 in/s. For this instant 3. Locate C, the instantaneous center
of zero velocity of the bar AB. b. Determine the angular velocity (DAB
of the bar AB. c. Determine the velocity 63 of the slider 8. Solution Since slider B moves in a vertical slot,
the instantaneous center of zero velocity
must lie on the horizontal line through 8.
Also, since the end of the rod A moves in
a straight line parallel to the horizontal
surface, the instantaneous center of zero
ivelocity must lie on the vertical line through A. Then, & VA=VC+VA/C
o
5=0+(18cos3O)AB
(0118:0321 rad/Sc ....... . ...... . ........... Ans.
and
VB=VC+VB/C
o
f =0+(185in30)AB = 2.89 in./s T ............ . ....... .. ............. Ans. ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 14—90 In the four—bar linkage shown, the
control link CD has a constant clockwise
angular velocity of 120 rev/min. For the
instant shown, in which B is directly above D a. Locate the instantaneous center of zero
velocity of the bar BC. b. Determine the angular velocities wAB
0) . and BC of both bars c. Determine the velocity GB of the pin B. Solution 4n rad/s D. Bar CD is in ll At the instant shown, wCD = 120 rev/min
fixed—axis rotation about point D; the velocity of point C is
perpendicular to CD and V0 = 5(4H) = 20H in./s Bar AB is also in fixedaxis rotation about point A; the velocity of point
B is perpendicular to AB. Therefore, the instantaneous center of zero
velocity of bar BC lies at the intersection of the lines AB (which is
perpendicular to the velocity 33) and CD (which is perpendicular to the velocity VG). Then 1 o
6 = tan 4/3 = 53.130 ¢ = 1800 — 120° — 9 = 6.8700 C 4’,
DE _ 6 = AE
sin 120o sin 6.8700 sin 53.13o°
(Ar = 40.129 in.
DE = 43.441 in.
and
= t + w
Vc (DE 5’30
wBC = zen/48.441 = 1.297 rad/s D ......... ..... ...... Ans.
= C + w
VB (AE 12)BC°
= 67.6 in./s i 30 ........................... ... Ans.
= 12w
AB
w = 5.63 rad/s D ................................. .. Ans. AB W
ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1496* One end of the 2m long bar AB shown is
‘attached to a collar which slides on a smooth
vertical shaft. The other end of the bar slides
mia smooth horizontal surface. At the instant
ﬂwwn, 9 = 30° and collar A is moving downward
with a constant speed of 0.4 m/s. Determine Um acceleration EB of end B and the angular acceleration “A8 of bar AB at this instant. Solution
From the relative velocity equation between A and B, VB = VA + VB/A
A A 0 A O A
=  . + w + '
VB 1 O 4 J 2 AB‘COS 30 1 Sin 30 J)
the x and ycomponents /\
u)
0 A8
x: VB = ZQAB cos 30 30F OKAG
O
y: 0 = 0.4 + 2w sin 30
AB B luAB
give V
wAB = 0.4 rad/s C 1
am“
VB = 0.69282 m/s + Then, from the relative acceleration equation between A and B, ‘=§+§
38 A B/A a ’1‘ _. o A o A
+ + ‘
B O ZaAB(cos 30 1 Sin 30 J) 2 o A A
+ 2(o.4) (sin 30 1 + cos 30° J) The x— and y— components x: a3 = 2aAB cos 30 — 0.1600
0
: = ' + .
y o 2aAB Sln 30 0 27713
give
2 2
AB = —o.27713 rad/s = 0.277 rad/s D ............... Ans.
2 2
a = —0.640 m/s = 0.640 m/s 6— . ........... ... Ans. :ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 14—101* Movement of the solenoid plunger
causes a gear to rotate. At the instant
shown, the angular velocity of the gear
is w = 4 rad/s counterclockwise and the 0
speed of the plunger is decreasing at a 2
rate of 0.25 ft/s . Determine the angular acceleration (to of the gear and the angular acceleration “A3 of the rod AB at this instant . Solution
At the instant shown, C
(o — w — 4 d/ C .
CD — o — m S 5 [2.
sin 43 = 5/13 cos ¢ = 12/13 A
and the relative velocity equation gives 'D ‘3
ﬂ = —. + —9 = _. + _.
Vc VD VC/D VA vC/A
6 + 5(4)(—cos 4) I + sin 4) 3) 1
_ A _ A A 5w.
 VA 1 + 12wAB(Sin ¢ 1 — cos d) J) 50" I ,l
:  =  w ‘F\\<::;
x 20(12/13) VA 12 ”(s/la) C
y: 20(5/13) = 12w (12/13)
AB ¢ wo
= — _ = . a
(DAB 0 6944 rad/s 0 69444 rad/s C D 0
VA = 21.667 in./s = 21.667 in./s <— Then, if the speed of A is decreasing, 8A = 0.25 ft/s —) = 0.25 1 ft/s and the relative acceleration equation gives . = —o —o = * " 2. ac an + ac/D aA + aC/A lawAB 0 + (5/12)ao(—c:s <13 ’1‘ + sin 4’ 3) B C was
+ (5/12)(4) (sin 43 ’1‘ — cos (b j) c"A8 = 0.25 ’1‘ + (12/12)a (sin dz ’1‘ — cos d: 3)
AB 2 A A mum A
+ (12/12)(0.6944) (cos 4) 1 — sin 42 J) ¢ X: O.38462(Io  2.56410 = 0.25 — 0.384621'1AB + 0.44510
: . a  . =  . a  .
y 0 16026 0 6 15365 0 92308 AB 0 18546
2
“AB = 6.76 rad/s C .............................. Ans.
2 2
or = 1.711 rad/s = 1.711 rad/s D ............... Ans. O ...
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 Spring '09
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