PROBLEM 16.133
For the car of Problem 16.132, determine the smallest constant
acceleration that the driver can maintain if the door is to close
and latch, knowing that as the door hits the frame its angular
velocity must be at least 2 rad/s for the latching mechanism to
operate.
PROBLEM 16.133 (Continued)
/2
00
/2
20
0
2
(0.4124 )cos
10.41234 | sin |
2
f
f
A
A
d ad
a
ωπ
ω
π
ωω θ θ
ωθ
=
=
∫∫
PROBLEM 16.134
The hatchback of a car is positioned as shown to
help determine the appropriate size for a
damping mechanism AB. The weight of the door
is 40 lbs, and its mass moment of inertia about
the center of gravity G is 15 lb·ft·s2. The
linkage DEFH controls the motion of the hatch
and is shown in more detail in Fig. b. Assume
that the mass of the links DE, EF, and FH are
negligible compared to the mass of the door.
With AB removed, determine (a) the initial
angular acceleration of the 40 lb door as it is
released from rest, (b) the force on link FH.
SOLUTION
Given:
2
2
40 lb 1.242 slugs
32.2 ft/s
15 lb ft s
0
DE EF FH
m
I
ωωω
= =
=
= = =
PROBLEM 16.134 (Continued)
Kinematics:
(4)
Equating i components: (6)
Equating i components: (8)
Rearrange (11): (12)
Substitute (7)→(14): (15)
Substitute (13)(15)→(3): (16)
FH
=aa 2
FH
ω
−/H /H
11
33
F FH F
FH F FH
aa
+×
= × ⇒=−
rr
kja i
a
2
E F EF
ω
= −aa
( )
//
1
3
1
3
EF EF EF
FH EF
FH EF EF
ab
ba
aa
aa a
+×
=− + ×− +
=−+−
rr
i k ij
ij
a
ED
=aa
2
DE
ω
−
( )
/D /D
0.5297 0.5883
E DE E
DE
a
+×
=×+
rr
k ij
a
3FH EF EF DE DE
10.5883
3FH EF DE
b
aa a
+=
a
G E EF
( )
/E /E
0.5883 0.5297 2
G EF G
x y DE DE EF
aa
a aa
+ =− + ×−
i j i+ j k i
0.5883
x DE
a
a
= −
DE F DE EF
1.1831 2 40
F DE EF
F mm
aa
= −+
cos48
0.2527
E EF
Fm
a
=
( ) ( ) ( )
2.1523 2.3401 40 0.06779 0.2527 cos48 2 0.2527 sin 48 15
EF EF EF EF
m mm
a a aa
− + + °+ °=
F
F
PROBLEM 16.135
The 6–kg rod BC connects a 10–kg disk centered at A to a 5–kg rod
CD. The motion of the system is controlled by the couple M
applied to disk A. Knowing that at the instant shown disk A has an
angular velocity of 36 rad/s clockwise and no angular
acceleration, determine (a) the couple M, (b) the components of
the force exerted at C on rod BC.
PROBLEM 16.135 (Continued)
Accelerations of the mass centers.
Disk AB:
0
AB A
= =aa
/
QD
2
//Q CD Q D CD Q D
ω
=×−aαr r
[0.125
CD
a
=
2
30 ] [(28.8) (0.125)°−
60 ]°
[14.964875=
30 ] [103.60°−
60 ]°
Rod BC:
6 [1555.2 N
BC P P
m= =aa
60 ] [179.58 N°+
]
PROBLEM 16.135 (Continued)
Kinetics
Rod BC:
1
PROBLEM 16.135 (Continued)
eff
2
( ) : (524.27)(0.4 0.2sin30 ) (230.93)(0.2cos30)
(6 kg)(9.81 m/s )(0.2 0.2sin30 )
11.9719 (179.58)(0.2 0.2sin30 )
(1555.2cos30 )(0.2)
AA
MM MΣ =Σ − − + °+ °
− +°
=+ +°
−°
262.135 40.0 17.658
11.9719 53.874 269.369
M
−− + −
= +−
PROBLEM 16.136
The 6–kg rod BC connects a 10-kg disk centered at A to a 5–kg rod
CD. The motion of the system is controlled by the couple M
applied to disk A. Knowing that at the instant shown disk A has an
angular velocity of 36 rad/s clockwise and an angular acceleration
of 150 rad/s2 counterclockwise, determine (a) the couple M,
(b) the components of the force exerted at C on rod BC.
PROBLEM 16.136 (Continued)
Equate components of two expressions (1) and (2) for
.
C
a
:
30cos30 259.2cos60 0.25 cos30 207.36 cos 60°
CD
a
− °− °=− °−
2
CD
2
CD
:
30cos30 259.2sin 60 0.4 0.25 sin30 207.36sin 60
BC CD
aa
°− °+ = °− °
2
149.649 rad/s
BC
a
=
2
149.649 rad/s
BC =α
Accelerations of the mass centers.
Disk AB:
0
AB A
= =aa
Rod CD: Mass center at Point Q.
/0.125 m
QD=r
60°
2
//Q CD Q D CD Q D
ω
=×−aαr r
[0.125 CD
a
=
2
30 ] [(28.8) (0.125)°−
60 ]°
60 ]°
60 ]°