978-0077687342 Chapter 15 Part 15

subject Type Homework Help
subject Pages 14
subject Words 3128
subject Authors Brian Self, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
PROBLEM 15.202
In Problem 15.201 the speed of Point B is known to be constant.
For the position shown, determine (a) the angular acceleration of the
guide arm, (b) the acceleration of Point C.
PROBLEM 15.201 Several rods are brazed together to form the
robotic guide arm shown, which is attached to a ball-and-socket joint
at O. Rod OA slides in a straight inclined slot while rod OB slides in a
slot parallel to the z-axis. Knowing that at the instant shown
vB
(9 in./s) ,=k
determine (a) the angular velocity of the guide arm,
(b) the velocity of Point A, (c) the velocity of Point C.
page-pf2
PROBLEM 15.202 (Continued)
page-pf3
PROBLEM 15.202 (Continued)
C=− +−a ij k
page-pf4
PROBLEM 15.203
Rod AB of length 25 in. is connected by ball-and-socket joints to
collars A and B, which slide along the two rods shown. Knowing
that collar B moves toward Point E at a constant speed of
20 in./s, determine the velocity of collar A as collar B passes
through Point D.
A
A
page-pf5
PROBLEM 15.204
Rod AB of length 13 in. is connected by ball-and-socket joints to collars
A and B, which slide along the two rods shown. Knowing that collar B
moves toward point D at a constant speed of 36 in./s, determine the
velocity of collar A when b = 4 in.
SOLUTION
22
22 22 2 2
page-pf6
PROBLEM 15.205
Rods BC and BD are each 840 mm long and are
connected by ball-and-socket joints to collars that
may slide on the fixed rods shown. Knowing that
collar B moves toward A at a constant speed of 390
mm/s, determine the velocity of collar C for the
position shown.
page-pf7
PROBLEM 15.206
Rod AB is connected by ball-and-socket joints to collar A and to the
16-in.-diameter disk C. Knowing that disk C rotates counterclockwise at
the constant rate
0
3
ω
=
rad/s in the zx plane, determine the velocity of
collar A for the position shown.
A
A
page-pf8
PROBLEM 15.207
Rod AB of length 29 in. is connected by ball-and-socket joints
to the rotating crank BC and to the collar A. Crank BC is of
length 8 in. and rotates in the horizontal xz plane at the constant
rate
0
10
ω
=
rad/s. At the instant shown, when crank BC is
parallel to the z axis, determine the velocity of collar A.
SOLUTION
page-pf9
PROBLEM 15.208
Rod AB of length 300 mm is connected by ball-and-socket
joints to collars A and B, which slide along the two rods
shown. Knowing that collar B moves toward Point D at a
constant speed of 50 mm/s, determine the velocity of collar A
when
80 mm.c=
SOLUTION
/
( 40 mm) (280 mm) (100 mm)
AB
=−+ −r i jk
page-pfa
PROBLEM 15.208 (Continued)
/
DC
r
A
A=vj
page-pfb
PROBLEM 15.209
Rod AB of length 300 mm is connected by ball-and-socket
joints to collars A and B, which slide along the two rods
shown. Knowing that collar B moves toward Point D at a
constant speed of 50 mm/s, determine the velocity of collar A
when
120 mm.c=
SOLUTION
page-pfc
PROBLEM 15.209 (Continued)
/
DC
r
A
A
page-pfd
PROBLEM 15.210
Two shafts AC and EG, which lie
in the vertical yz plane, are
connected by a universal joint at
D. Shaft AC rotates with a constant
angular velocity
1
ω
as shown. At
a time when the arm of the
crosspiece attached to shaft AC is
vertical, determine the angular
velocity of shaft EG.
32
12
: cos 25
ωω
= °k
(3)
1
ω
2
ω
page-pfe
PROBLEM 15.211
Solve Problem 15.210, assuming
that the arm of the crosspiece
attached to the shaft AC is
horizontal.
PROBLEM 15.210 Two shafts
AC and EG, which lie in the
vertical yz plane, are connected by
a universal joint at D. Shaft AC
rotates with a constant angular
velocity
1
ω
as shown. At a time
when the arm of the crosspiece
attached to shaft AC is vertical,
determine the angular velocity of
shaft EG.
1
EG
page-pff
PROBLEM 15.212
Rod BC has a length of 42 in.and is connected by a
ball-and-socket joint to collar B and by a clevis
connection to collar C. Knowing that collar B moves
toward A at a constant speed of 19.5 in./s, determine at
the instant shown (a) the angular velocity of the rod,
(b) the velocity of collar C.
SOLUTION
40 8 10
C Css
page-pf10
PROBLEM 15.212 (Continued)
Resolving into components,
C= −vi
page-pf11
PROBLEM 15.213
Rod AB has a length of 275 mm and is
connected by a ball-and-socket joint to
collar A and by a clevis connection to
collar B. Knowing that collar B moves
down at a constant speed of 1.35 m/s,
determine at the instant shown (a) the
angular velocity of the rod, (b) the
velocity of collar A.
SOLUTION
Geometry. Determine the position of collar A.
( ) ( )
/
, 150 mm 50 mm
150 50
AA BB B
AB A
z xy
z
= =+= +
=−−
r kr i j i j
r k ij
Length of rod AB:
2 22 2 2
275 150 50
AB A
lz= =−−
Solving for
,
A
z
225 mm
A
z=
( ) ( ) ( ) ( ) ( ) ( )
/150 mm 50 mm 225 mm 0.15 m 0.05 m 0.225 m
AB =− − + =−− +r ij k ij k
Velocity.
( )
1.35 m/s ,
B AA
v=−=v jv k
Angular velocity of collar B.
BB
ω
=jω
The axle of the clevis at B is perpendicular to both the y-axis and the rod AB.
A vector along this axle is
/AB
= ×p jr
( ) ( ) ( )
/
22
150 50 225 225 mm 150 mm
225 150 270.42 mm
AB
r
= × = ×− + =
= +=
pj j i j k i k
p
Let
λ
be a unit vector along the axle.
0.83205 0.55470
p
= = +
p
λ ik
Let
ss
ω
=ωλ
be the angular velocity of rod AB relative to collar B.
0.83205 0.55470
ss s
ωω
= +ikω
Angular velocity of rod AB.
AB B s
= +ω ωω
0.83205 0.55470
AB s B s
ωω ω
= ++ω ij k
/A B AB A B
=vv rω
1.35 0.83205 0.55470
0.15 0.05 0.225
A sB s
v
ωω ω
=−+
−−
i jk
kj
page-pf12
PROBLEM 15.213 (Continued)
Resolving into components,
A=vk
page-pf13
PROBLEM 15.214
For the mechanism of the Prob.15.204, determine the
acceleration of collar A.
PROBLEM 15.204 Rod AB of length 13 in. is connected by
ball-and-socket joints to collars A and B, which slide along the
two rods shown. Knowing that collar B moves toward point D at
a constant speed of 36 in./s, determine the velocity of collar A
when b = 4 in.
page-pf14
PROBLEM 15.214 (Continued)
/ // /
, where
A B AB AB AB AB AB AB
=+ = ×+ ×a aa a r vαω
Noting that
/AB A B
×rα
is perpendicular to
/
,
AB
r
we get
//
0
AB AB AB
⋅× =rrα
We note also that
/ / // /AB AB AB AB AB AB
⋅ × = ⋅×r v vrωω
( )
2
// /AB AB AB
v=−⋅ =vv
Then,
( ) ( )
22
// / /
0
AB AB AB AB
vv⋅=− =ra
Forming
/
,
AB A
ra
we get
( )
/ / / / //AB A AB A AB AB B AB AB
= ⋅ + = ⋅+ ra r a a ra ra
or
( )
2
// /AB A AB B AB
v= ⋅−ra ra
(2)
From (2),
( ) ( )
7.8 9.6 4 0 1521
A
a−+ − =i jk j
9.6 1521
A
a= −
( )
2
158.4 in./s
A= −aj
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