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Since point Ccan only move in the vertical direction, aCx D0, which implies
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Dynamics 2e 1331
Problem 6.130
For the slider-crank mechanism shown, let
RD1:9 in:
,
LD6:1 in:
, and
HD1:2 in:
Assuming that, at the instant shown,
D31ı
,
!AB D4850 rpm
, and
˛AB D P!AB D 280 rad=s2
, determine the angular acceleration of the
connecting rod and the acceleration of point C.
Relating the acceleration of Bto that of A, we have
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For the acceleration of C, we have
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Dynamics 2e 1333
Problem 6.131
For the slider-crank mechanism shown, let
RD1:9 in:
,
LD6:1 in:
, and
HD1:2 in:
Determine the general expression of the acceleration of the piston
C
as a function of L,R,,!AB DP
, and ˛AB DR
.
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permission of McGraw-Hill, is prohibited.
For the acceleration of C, we have
Since point Ccan only move in the vertical direction, aCx D0, which implies
where we have used
sin
and
cos
from Eq. (3) and
!BC
from Eq. (4). Knowing that
aCx D0
,
EaC
in
Eq. (5) becomes
Lcos ;
or, rearranging,
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1335
Problem 6.132
In the four-bar linkage system shown, let the circular guide with
center at
O
be fixed and such that, for
D0ı
, the bars
AB
and
BC
are vertical and horizontal, respectively. In addition,
let
RD0:6
m,
LD1
m, and
HD1:25
m. When
D37ı
,
ˇD25:07ı
, and
D78:71ı
, assume collar
C
is sliding clockwise
with a speed
7m=s
. Assuming that, at the instant in question, the
speed is increasing and that
jEaCj D 93 m=s2
, determine the angular
accelerations of the bars AB and BC .
D˛OC O
kR.cos O{Csin O| / v2
R2R.cos O{Csin O| /
C
C
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where we have used the fact that !OC D vC=R. Since jEaCjis known, we form jEaCj2D EaCEaCas
C
C
C
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Cv2
tan.ˇ / sin.ˇ / #CHRv2
sin2.ˇ / );
and
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permission of McGraw-Hill, is prohibited.
Problem 6.133
Complete the acceleration analysis of the slider-crank mechanism
using differentiation of constraints that was outlined beginning on
p. 474. That is, determine the acceleration of the piston
C
and the
angular acceleration of the connecting rod as a function of the given
quantities
,
!AB
,
R
, and
L
. Assume that
!AB
is constant, and use
the component system shown for your answers.
DR2P
pL2R2cos2R4P
L2R2cos23=2 RP
2sin
DR!2
AB "Rcos 2
pL2R2cos2R3cos2sin2
L2R2cos23=2 sin #
where we have used the trig identity
cos2sin2Dcos 2
and the fact that
P
D!AB Dconstant
.
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1339
To determine
E˛BC
, we start by noting that
!BC DP
and that
˛BC DR
. Therefore, differentiating
Lsin DRcos with respect to time, we obtain
sin
sin
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