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PROBLEM 16.128 (Continued)
Kinematics:
Relative Acceleration:
(8)
Substitute into (6)→(1): (9)
BA
=vv
( )
B/A
0.2
0.8
AB
AB
ω
π
+×
= ×
= −
ωr
kj
i
( )
C/B
0.8 0.4
0.8 0.4
C B BC
C BC
C BC
v
v
πω
πω
=+×
=−×
=−+
vvωr
j i+ k j
ji
BA
/AAB B AB
/A
2
3.2
B
π
= −
j
( )
2
/B /B
22
22
3.2 1.6 0.4
0.4 - 1.6 3.2
C B BC C BC C
c BC
c BC
a
a
ω
π πa
a ππ
=− +×
=− +×
=−+
aa r r
j j - j k j
ji j
a
,4.8 m/s
c CD y
2
/BBC B BC G BC
ω
=−+aa r a
/B
22
,,
3.2 0.8
G
BC x BC y
aa
ππ
×
+=−−
r
ij j j
,
BC x
2
,
4.0
BC y
a
π
= −
2
2 9.6 75.128 N
yy
gC C
π
−− =− ⇒ =
BC
a
PROBLEM 16.129
The 4-kg uniform slender bar BD is attached to bar AB and a wheel
of negligible mass which rolls on a circular surface. Knowing that at
the instant shown bar AB has an angular velocity of 6 rad/s and no
angular acceleration, determine the reaction at Point D.
PROBLEM 16.129 (Continued)
Kinetics:
22 2
11
12 12
G
eff
( ):
BB
MMΣ=Σ
(39.24 N)(0.375 m) cos30 (0.75 m)D− +°
1
(0.1875)(56.7846) (54)(0.375) 1 N m
3
= − −⋅
0.64952 (14.715 10.6471 8.5587) N mD=+− ⋅
25.87 N=D
60°
or
25.9 N=D
60°
PROBLEM 16.130
The motion of the uniform slender rod of length L = 0.5 m and mass m = 3 kg is
guided by pins at A and B that slide freely in frictionless slots, circular and
horizontal, cut into a vertical plate as shown. Knowing that at the instant shown
the rod has an angular velocity of 3 rad/s counter-clockwise and θ = 30°,
determine the reactions at Points A and B.
SOLUTION
0.75 m/s
=
aa
=
AA
a=a
2
B
B
v
R
=a
()
By
a+
2
(0.75 m/s)
0.3 m
=
()
By
a
2
1.875 m/s=
()
By
a+
/ //
B A BA A BA t BA n
where
/
()
BA t L
a
=a
30 0.5
a
°=
30°
2
2
2
PROBLEM 16.130 (Continued)
Equating the two expressions for
B
a
gives
2
By
A
2
:
1.875 0.5 cos30 2.25
A
a
a
− = + °+
0.5 cos30 4.125
A
a
a
=− °−
//
()()
G A GA t GA n
=++aa a a
(0.5 cos30 4.125)
a
= °+
2
L
a
+
2
30 2
L
ω
°+
60°
(0.5 cos30 4.125)
a
= °+
0.5
2
a
+
2
0.5
30 (3)
2
°+
60°
(0.21651 3.00)
a
= +
(1.94856 0.125 )
a
+−
Kinetics:
(3 kg)
GG
m=aa
[0.64952 9.00]
a
= +
[5.8457 0.375 ]
a
+−
0.0625I
aa
=
eff
( ) : sin 30 ( ) cos30 ( ) sin 30
2 22
C C Gx Gy
L LL
M M mg I ma ma
a
Σ =Σ °= + °− °
0.5
(3)(9.81) sin 30 0.0625
2
0.5
(0.64952 9.00) cos30
2
0.5
(5.8457 0.375 ) sin 30
2
a
a
a
°=
++ °
−− °
2
3.67875 0.25 1.21784
9.8436 rad/s
a
a
= +
=
[(0.64952)(9.8436) 9.00]
G
m= +a
[5.8457 (0.375)(9.8436)]+−
[15.394 N]=
[2.1544 N]
PROBLEM 16.130 (Continued)
PROBLEM 16.131
At the instant shown, the 20 ft long, uniform 100-lb pole ABC has an
angular velocity of 1 rad/s counterclockwise and Point C is sliding to the
right. A 120-lb horizontal force P acts at B. Knowing the coefficient of
kinetic friction between the pole and the ground is 0.3, determine at this
instant (a) the acceleration of the center of gravity, (b) the normal force
between the pole and the ground.
G
eff
22 2
Gf
sin10 cos10 cos10
2 22
103.519 (10)(sin10 0.3cos10 ) 120(10cos10 6)
k
l ll
IN N P h
N
aµ
a
− °+ °= °−
− °− ° = °−
(2)
PROBLEM 16.131 (Continued)
2
ml ml
PROBLEM 16.132
A driver starts his car with the door on the passenger’s side wide
open
( 0).
θ
=
The 80-lb door has a centroidal radius of gyration
12.5 in.,k=
and its mass center is located at a distance
22 in.r=
from its vertical axis of rotation. Knowing that the
driver maintains a constant acceleration of 6 ft/s2, determine the
angular velocity of the door as it slams shut
( 90 ).
θ
= °
PROBLEM 16.132 (Continued)
( )
( )
( )
22
12
22
12.5 22
12 12
ft cos
ft ft
A
a
d
d
ω
ωθ
θ
=
+
f
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
2
0
0
2
(0.4124 )cos
10.41234 | sin |
2
f
f
A
A
d ad
a
ωπ
ω
π
ωω θ θ
ωθ
=
=
∫∫
A
A
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
40 lb 1.242 slugs
F E E EF
PROBLEM 16.134 (Continued)
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.
SOLUTION
BC
BC
PROBLEM 16.135 (Continued)
Accelerations of the mass centers.
CD CD
Summary of effective
forces and couples
PROBLEM 16.135 (Continued)
Kinetics
Rod BC:
PROBLEM 16.135 (Continued)
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.
SOLUTION
PROBLEM 16.136 (Continued)
12 12
CD CD CD
