Tenth Edition
CHAPTER
16
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
Plane Motion of Rigid Bodies:
Forces and Accelerations
California Polytechnic State University
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Equations of Motion of a Rigid
Body
Angular Momentum of a Rigid
Body in Plane Motion
Plane Motion of a Rigid Body:
d’Alembert’s Principle
Axioms of the Mechanics of Rigid
Bodies
Problems Involving the Motion of a
Rigid Body
Sample Problem 16.1
Sample Problem 16.2
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Sample Problem 16.3
Sample Problem 16.4
Sample Problem 16.5
Constrained Plane Motion
Constrained Plane Motion:
Noncentroidal Rotation
Constrained Plane Motion:
Rolling Motion
Sample Problem 16.6
Sample Problem 16.8
Sample Problem 16.9
Sample Problem 16.10
16 - 2
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Vector Mechanics for Engineers: Dynamics
Rigid Body Kinetics
Early design of prosthetic legs
relied heavily on kinetics. It
was necessary to calculate the
different kinematics, loads, and
moments applied to the leg to
make a safe device.
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The forces and moments applied to
a robotic arm control the resulting
kinematics, and therefore the end
position and forces of the actuator
at the end of the robot arm.
2-3
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Vector Mechanics for Engineers: Dynamics
Introduction
• In this chapter and in Chapters 17 and 18, we will be
concerned with the kinetics of rigid bodies, i.e., relations
between the forces acting on a rigid body, the shape and mass
of the body, and the motion produced.
• Results of this chapter will be restricted to:
- plane motion of rigid bodies, and
- rigid bodies consisting of plane slabs or bodies which
are symmetrical with respect to the reference plane.
• Our approach will be to consider rigid bodies as made of
large numbers of particles and to use the results of Chapter
14 for the motion of systems of particles. Specifically,




F  ma
and
M H


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G
G
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Vector Mechanics for Engineers: Dynamics
Equations of Motion for a Rigid Body
• Consider a rigid body acted upon
by several external forces.
• Assume that the body is made of
a large number of particles.
• For the motion of the mass center
G of the body with respect to the
Newtonian frame Oxyz,


F

m
a

• For the motion of the body with
respect to the centroidal frame
Gx’y’z’,


 M G  HG
• System of external forces is
equipollent to the

 system
consisting of ma and H G .
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Vector Mechanics for Engineers: Dynamics
Angular Momentum of a Rigid Body in Plane Motion
• Angular momentum of the slab may be
computed by
n  

H G   ri viΔmi 
i 1
n
  
  ri   riΔmi 
i 1


   ri 2 Δmi

 I

• After differentiation,



H G  I  I
• Consider a rigid slab in
plane motion.
• Results are also valid for plane motion of bodies
which are symmetrical with respect to the
reference plane.
• Results are not valid for asymmetrical bodies or
three-dimensional motion.
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Vector Mechanics for Engineers: Dynamics
Plane Motion of a Rigid Body: D’Alembert’s Principle
• Motion of a rigid body in plane motion is
completely defined by the resultant and moment
resultant about G of the external forces.
 Fx  ma x  Fy  ma y  M G  I
• The external forces and the collective effective
forces of the slab particles are equipollent (reduce
to the same resultant and moment resultant) and
equivalent (have the same effect on the body).
• d’Alembert’s Principle: The external forces
acting on a rigid body are equivalent to the
effective forces of the various particles forming
the body.
• The most general motion of a rigid body that is
symmetrical with respect to the reference plane
can be replaced by the sum of a translation and a
centroidal rotation.
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Vector Mechanics for Engineers: Dynamics
Axioms of the Mechanics of Rigid Bodies


• The forces F and F  act at different points on
a rigid body but but have the same magnitude,
direction, and line of action.
• The forces produce the same moment about
any point and are therefore, equipollent
external forces.
• This proves the principle of transmissibility
whereas it was previously stated as an axiom.
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Vector Mechanics for Engineers: Dynamics
Problems Involving the Motion of a Rigid Body
• The fundamental relation between the forces
acting on a rigid body in plane motion and
the acceleration of its mass center and the
angular acceleration of the body is illustrated
in a free-body-diagram equation.
• The techniques for solving problems of
static equilibrium may be applied to solve
problems of plane motion by utilizing
- d’Alembert’s principle, or
- principle of dynamic equilibrium
• These techniques may also be applied to
problems involving plane motion of
connected rigid bodies by drawing a freebody-diagram equation for each body and
solving the corresponding equations of
motion simultaneously.
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
The free body diagram is the same as you have done in statics and
in Ch 13; we will add the kinetic diagram in our dynamic analysis.
1. Isolate the body of interest (free body)
2. Draw your axis system (Cartesian, polar, path)
3. Add in applied forces (e.g., weight)
4. Replace supports with forces (e.g., tension force)
5. Draw appropriate dimensions (angles and distances)
y
x
Include your
positive z-axis
direction too
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
Put the inertial terms for the body of interest on the kinetic diagram.
1. Isolate the body of interest (free body)
2. Draw in the mass times acceleration of the particle; if unknown,
do this in the positive direction according to your chosen axes. For
rigid bodies, also include the rotational term, IG.
F

M G 
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ma
I
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
Draw the FBD and KD for
the bar AB of mass m. A
known force P is applied at
the bottom of the bar.
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2 - 12
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
1.
2.
3.
4.
5.
6.
y
Cy
A
C
x
Cx
L/2
r
ma y
I
G
Isolate body
Axes
Applied forces
Replace supports with forces
Dimensions
Kinetic diagram
G
max
mg
L/2
B
P
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
A drum of 4 inch radius is attached
to a disk of 8 inch radius. The
combined drum and disk had a
combined mass of 10 lbs. A cord is
attached as shown, and a force of
magnitude P=5 lbs is applied. The
coefficients of static and kinetic
friction between the wheel and
ground are ms= 0.25 and mk= 0.20,
respectively. Draw the FBD and
KD for the wheel.
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2 - 14
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
1.
2.
3.
4.
5.
6.
Isolate body
Axes
Applied forces
Replace supports with forces
Dimensions
Kinetic diagram
ma y
P
4 in
I
=
8 in
max
W
F
N
y
x
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
The ladder AB slides down
the wall as shown. The wall
and floor are both rough.
Draw the FBD and KD for
the ladder.
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Vector Mechanics for Engineers: Dynamics
Free Body Diagrams and Kinetic Diagrams
1. Isolate body
3. Applied forces
5. Dimensions
2. Axes
4. Replace supports with forces
6. Kinetic diagram
NB
q
FB
ma y
I
=
max
W
y
NA
FA
x
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.1
SOLUTION:
• Calculate the acceleration during the
skidding stop by assuming uniform
acceleration.
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces.
At a forward speed of 30 ft/s, the truck • Apply the three corresponding scalar
brakes were applied, causing the wheels
equations to solve for the unknown
to stop rotating. It was observed that the
normal wheel forces at the front and rear
truck to skidded to a stop in 20 ft.
and the coefficient of friction between
the wheels and road surface.
Determine the magnitude of the normal
reaction and the friction force at each
wheel as the truck skidded to a stop.
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16 - 18
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.1
SOLUTION:
• Calculate the acceleration during the skidding stop
by assuming uniform acceleration.
v 2  v02  2a  x  x0 
ft
v0  30
s
2
x  20 ft
 ft 
0   30   2a 20 ft 
 s
a  22.5
ft
s
• Draw a free-body-diagram equation expressing the
equivalence of the external and inertial terms.
• Apply the corresponding scalar equations.
 Fy   Fy eff
 Fx   Fx eff
N A  NB  W  0
 FA  FB  ma
 mk N A  N B  
 m kW  W g a
mk 
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a 22.5

 0.699
g 32.2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.1
• Apply the corresponding scalar equations.
 M A   M A eff
 5 ft W  12 ft N B  4 ft ma
1
W  W
a
 5W  4 a    5  4 
12 
g  12 
g
N B  0.650W
NB 
N A  W  N B  0.350W
Nrear  12 N A  12 0.350W 
Nrear  0.175W
Frear  mk Nrear  0.6900.175W 
N front  12 NV  12 0.650W 
Frear  0.122W
N front  0.325W
Ffront  mk N front  0.6900.325W 
Ffront  0.0.227W
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.2
SOLUTION:
• Note that after the wire is cut, all
particles of the plate move along parallel
circular paths of radius 150 mm. The
plate is in curvilinear translation.
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces.
The thin plate of mass 8 kg is held in
place as shown.
• Resolve into scalar component equations
parallel and perpendicular to the path of
the mass center.
Neglecting the mass of the links,
determine immediately after the wire
has been cut (a) the acceleration of the
plate, and (b) the force in each link.
• Solve the component equations and the
moment equation for the unknown
acceleration and link forces.
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Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.2
SOLUTION:
• Note that after the wire is cut, all particles of the
plate move along parallel circular paths of radius
150 mm. The plate is in curvilinear translation.
• Draw the free-body-diagram equation expressing
the equivalence of the external and effective
forces.
• Resolve the diagram equation into components
parallel and perpendicular to the path of the mass
center.
 Ft   Ft eff
W cos 30  ma
mg cos 30 


a  9.81m/s 2 cos 30
a  8.50 m s 2
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60o
16 - 22
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.2
• Solve the component equations and the moment
equation for the unknown acceleration and link
forces.
 M G   M G eff
FAE sin 30250 mm   FAE cos 30100 mm 
FDF sin 30250 mm   FDF cos 30100 mm   0
38.4 FAE  211.6 FDF  0
FDF  0.1815 FAE
 Fn   Fn eff
a  8.50 m s 2
60o
FAE  FDF  W sin 30  0
FAE  0.1815 FAE  W sin 30  0

FAE  0.6198 kg  9.81m s 2
FDF  0.1815 47.9 N
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
FAE  47.9 N T
FDF  8.70 N C
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.3
SOLUTION:
• Determine the direction of rotation by
evaluating the net moment on the
pulley due to the two blocks.
• Relate the acceleration of the blocks to
the angular acceleration of the pulley.
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces on the
complete pulley plus blocks system.
A pulley weighing 12 lb and having a
radius of gyration of 8 in. is connected to
• Solve the corresponding moment
two blocks as shown.
equation for the pulley angular
Assuming no axle friction, determine the
acceleration.
angular acceleration of the pulley and the
acceleration of each block.
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16 - 24
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.3
SOLUTION:
• Determine the direction of rotation by evaluating the net
moment on the pulley due to the two blocks.
 M G  10 lb6 in   5 lb10 in   10 in  lb
rotation is counterclockwise.
note:
I  mk 2 
W 2
k
g
12 lb  8 

ft 
2
32.2 ft s  12 
2
 0.1656 lb  ft  s 2
• Relate the acceleration of the blocks to the angular
acceleration of the pulley.
a A  rA
 10
ft 
12
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aB  rB
6 
 12
ft 
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.3
• Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on the
complete pulley and blocks system.
• Solve the corresponding moment equation for the pulley
angular acceleration.
 M G   M G eff
6 
10 lb 126 ft   5 lb 10
ft   I   mB aB 12
ft  m Aa A 10
ft 
12
12
10  6  6   5 10 10 
10126   510





0
.
1656


 12  32.2 12 12
12
32.2 12
  2.374 rad s 2
I  0.1656 lb  ft  s 2
2

a A  10

ft
s
12
aB  
6

12
ft s
2
Then,
a A  rA


a A  1.978 ft s 2


aB  1.187 ft s 2
2

 10
ft
2.374
rad
s
12
aB  rB
6
 12
ft  2.374 rad s 2
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16 - 26
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.4
SOLUTION:
• Draw the free-body-diagram equation
expressing the equivalence of the external
and effective forces on the disk.
• Solve the three corresponding scalar
equilibrium equations for the horizontal,
vertical, and angular accelerations of the
disk.
A cord is wrapped around a
homogeneous disk of mass 15 kg.
The cord is pulled upwards with a
force T = 180 N.
• Determine the acceleration of the cord by
evaluating the tangential acceleration of
the point A on the disk.
Determine: (a) the acceleration of the
center of the disk, (b) the angular
acceleration of the disk, and (c) the
acceleration of the cord.
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16 - 27
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.4
SOLUTION:
• Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on the
disk.
• Solve the three scalar equilibrium equations.
 Fx   Fx eff
ax  0
0  max
 Fy   Fy eff
T  W  ma y

T  W 180 N - 15 kg  9.81m s 2
ay 

m
15 kg
 M G   M G eff
 Tr  I  
 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

a y  2.19 m s 2
12 mr 2 
2T
2180 N 

15 kg 0.5 m 
mr
  48.0 rad s 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.4
• Determine the acceleration of the cord by evaluating the
tangential acceleration of the point A on the disk.

acord  a A t  a  a A G t

 2.19 m s 2  0.5 m  48 rad s 2

acord  26.2 m s 2
ax  0
a y  2.19 m s 2
  48.0 rad s 2
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16 - 29
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.5
SOLUTION:
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces on the
sphere.
A uniform sphere of mass m and radius
r is projected along a rough horizontal
surface with a linear velocity v0. The
coefficient of kinetic friction between
the sphere and the surface is mk.
Determine: (a) the time t1 at which the
sphere will start rolling without sliding,
and (b) the linear and angular velocities
of the sphere at time t1.
• Solve the three corresponding scalar
equilibrium equations for the normal
reaction from the surface and the linear
and angular accelerations of the sphere.
• Apply the kinematic relations for
uniformly accelerated motion to
determine the time at which the
tangential velocity of the sphere at the
surface is zero, i.e., when the sphere
stops sliding.
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16 - 30
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.5
SOLUTION:
• Draw the free-body-diagram equation expressing the
equivalence of external and effective forces on the
sphere.
• Solve the three scalar equilibrium equations.
 Fy   Fy eff
N W  0
 Fx   Fx eff
 F  ma
 m k mg 
N  W  mg
a  mk g
 M G   M G eff
Fr  I
5 mk g
2 r
NOTE: As long as the sphere both rotates and slides,
its linear and angular motions are uniformly
accelerated.
mk mg r  23 mr 2 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.5
• Apply the kinematic relations for uniformly accelerated
motion to determine the time at which the tangential velocity
of the sphere at the surface is zero, i.e., when the sphere
stops sliding.
v v 0 a t v 0  mk g t
5 mk g 
t
2 r 
  0  t  0  
a  mk g

5 mk g
2 r
At the instant t1 when the sphere stops sliding,
v1  r1
 5 mk g 
v0  m k gt1  r 
 t1
2 r 
5 mk g 
 5 m k g  2 v0 
t

1 

2 r 
 2 r  7 m k g 
t1 
2 v0
7 mk g
1  
1 
5v 
v1  r1  r  0 
7 r 
v1  75 v0
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5 v0
7 r
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
Knowing that the coefficient of
static friction between the tires
and the road is 0.80 for the
automobile shown, determine the
maximum possible acceleration
on a level road, assuming rearwheel drive
• Draw the free-body-diagram and
kinetic diagram showing the
equivalence of the external forces
and inertial terms.
• Write the equations of motion for
the sum of forces and for the sum
of moments.
• Apply any necessary kinematic
relations, then solve the resulting
equations.
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2 - 33
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Given: rear wheel drive,
dimensions as shown, m= 0.80
• Find: Maximum acceleration
• Draw your FBD and KD
• Set up your equations of motion,
realizing that at maximum acceleration,
may and  will be zero
ma y
y
x
I
=
FR
mg
NF
NR
F
 max
FR  max
x
max
F
y
 may
N R  N F  mg  0
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
M
G
 IG
60
40
20
 N R ( 12
)  N F ( 12
)  FR ( 12
)0
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Vector Mechanics for Engineers: Dynamics
• Solve the resulting equations: 4 unknowns are FR, max, NF and NR
FR  max (1)
N R  N F  mg  0 (2)
60
40
20
 N R ( 12
)  N F ( 12
)  FR ( 12
)  0 (4)
(5)→(2)
N F  mg  N R  mg 
max
m
FR  m N R (3)
(1)→(3) N R 
max
m
(5)
(6)
(1) and (5) and (6) →(4)
max  60  
max   40 
 20 

    mg 
    max    0
m  12  
m   12 
 12 
Solving this equation, the masses cancel out and you get:
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
ax  12.3 ft/s
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
• Alternatively, you could have chosen to sum moments about the front wheel
ma y
y
x
=
FR
mg
I
max
NF
NR
M
F
 IG  mad
40
20
 N R ( 100
12 )  mg ( 12 )  0  max ( 12 )
• You can now use this equation with those on the previous slide to solve for the
acceleration
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2 - 36
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Vector Mechanics for Engineers: Dynamics
Concept Question
The thin pipe P and the uniform cylinder C have the same outside
radius and the same mass. If they are both released from rest,
which of the following statements is true?
a) The pipe P will have a greater acceleration
b) The cylinder C will have a greater acceleration
c) The cylinder and pipe will have the same acceleration
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 37
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Vector Mechanics for Engineers: Dynamics
Kinetics: Constrained Plane Motion
The forces at the bottom of the
pendulum depend on the
pendulum mass and mass moment
of inertia, as well as the pendulum
kinematics.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
The forces one the wind turbine
blades are also dependent on
mass, mass moment of inertia, and
kinematics.
2 - 38
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Vector Mechanics for Engineers: Dynamics
Constrained Plane Motion
• Most engineering applications involve rigid
bodies which are moving under given
constraints, e.g., cranks, connecting rods, and
non-slipping wheels.
• Constrained plane motion: motions with
definite relations between the components of
acceleration of the mass center and the angular
acceleration of the body.
• Solution of a problem involving constrained
plane motion begins with a kinematic analysis.
• e.g., given q, , and , find P, NA, and NB.
- kinematic analysis yields ax and a y .
- application of d’Alembert’s principle yields
P, NA, and NB.
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16 - 39
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Vector Mechanics for Engineers: Dynamics
Constrained Motion: Noncentroidal Rotation
• Noncentroidal rotation: motion of a body is
constrained to rotate about a fixed axis that does
not pass through its mass center.
• Kinematic relation between the motion of the mass
center G and the motion of the body about G,
at  r 
an  r  2
• The kinematic relations are used to eliminate
at and an from equations derived from
d’Alembert’s principle or from the method of
dynamic equilibrium.
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16 - 40
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Vector Mechanics for Engineers: Dynamics
Constrained Plane Motion: Rolling Motion
• For a balanced disk constrained to
roll without sliding,
x  rq  a  r
• Rolling, no sliding:
F  ms N
a  r
Rolling, sliding impending:
F  ms N
a  r
Rotating and sliding:
a, r independent
F  mk N
• For the geometric center of an
unbalanced disk,
aO  r
The acceleration of the mass center,



aG  aO  aG O



 aO  aG O  aG O

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
t 
n
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.6
SOLUTION:
• Draw the free-body-equation for AOB,
expressing the equivalence of the
external and effective forces.
mE  4 kg
k E  85 mm
mOB  3 kg
The portion AOB of the mechanism is
actuated by gear D and at the instant
shown has a clockwise angular velocity
of 8 rad/s and a counterclockwise
angular acceleration of 40 rad/s2.
Determine: a) tangential force exerted
by gear D, and b) components of the
reaction at shaft O.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
• Evaluate the external forces due to the
weights of gear E and arm OB and the
effective forces associated with the
angular velocity and acceleration.
• Solve the three scalar equations
derived from the free-body-equation
for the tangential force at A and the
horizontal and vertical components of
reaction at shaft O.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.6
SOLUTION:
• Draw the free-body-equation for AOB.
• Evaluate the external forces due to the weights of
gear E and arm OB and the effective forces.


WE  4 kg  9.81m s 2  39.2 N


WOB  3 kg  9.81m s 2  29.4 N

I E  mE k E2  4kg 0.085 m 2 40 rad s 2

 1.156 N  m

mE  4 kg
k E  85 mm
mOB  3 kg
  40 rad s 2
  8 rad/s
mOB aOB t  mOB r    3 kg 0.200 m  40 rad s 2

 24.0 N
 
mOB aOB n  mOB r  2  3 kg 0.200 m 8 rad s 2
 38.4 N
IOB 
121 mOBL2   121 3kg0.400 m2 40 rad s2 
 1.600 N  m
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16 - 43
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.6
• Solve the three scalar equations derived from the freebody-equation for the tangential force at A and the
horizontal and vertical components of reaction at O.
 M O   M O eff
F 0.120m   I E  mOB aOB t 0.200m   IOB
 1.156 N  m  24.0 N 0.200m   1.600 N  m
F  63.0 N
 Fx   Fx eff
WE  39.2 N
WOB  29.4 N
I E  1.156 N  m
Rx  mOB aOB t  24.0 N
Rx  24.0 N
 Fy   Fy eff
mOB aOB t  24.0 N
R y  F  WE  WOB  mOB aOB 
mOB aOB n  38.4 N
R y  63.0 N  39.2 N  29.4 N  38.4 N
IOB  1.600 N  m
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Ry  24.0 N
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.8
SOLUTION:
• Draw the free-body-equation for the
sphere, expressing the equivalence of the
external and effective forces.
• With the linear and angular accelerations
related, solve the three scalar equations
derived from the free-body-equation for
the angular acceleration and the normal
A sphere of weight W is released with
and tangential reactions at C.
no initial velocity and rolls without
• Calculate the friction coefficient required
slipping on the incline.
for the indicated tangential reaction at C.
Determine: a) the minimum value of
• Calculate the velocity after 10 ft of
the coefficient of friction, b) the
uniformly accelerated motion.
velocity of G after the sphere has
rolled 10 ft and c) the velocity of G if • Assuming no friction, calculate the linear
acceleration down the incline and the
the sphere were to move 10 ft down a
corresponding velocity after 10 ft.
frictionless incline.
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16 - 45
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.8
SOLUTION:
• Draw the free-body-equation for the sphere, expressing
the equivalence of the external and effective forces.
• With the linear and angular accelerations related, solve
the three scalar equations derived from the free-bodyequation for the angular acceleration and the normal
and tangential reactions at C.
 M C   M C eff
a  r
W sin q r  ma r  I
 mr r  52 mr 2 
W
  2W 2 
  r  r  
r 
g
 5 g 
a  r 

5 g sin q
7r
5 g sin 30
7

5 32.2 ft s 2 sin 30

7
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

a  11.50 ft s 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.8
• Solve the three scalar equations derived from the freebody-equation for the angular acceleration and the
normal and tangential reactions at C.
 Fx   Fx eff W sin q  F  ma
W 5 g sin q

g
7
2
F  W sin 30  0.143W
7
N  W cosq  0
 Fy   Fy 
eff

5 g sin q
7r
a  r  11.50 ft s 2
N  W cos 30  0.866W
• Calculate the friction coefficient required for the
indicated tangential reaction at C.
F  ms N
ms 
F 0.143W

N 0.866W
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
m s  0.165
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.8
• Calculate the velocity after 10 ft of uniformly
accelerated motion.
v 2  v02  2a  x  x0 


 0  2 11.50 ft s 2 10 ft 

v  15.17 ft s
• Assuming no friction, calculate the linear acceleration
and the corresponding velocity after 10 ft.

5 g sin q
7r
 M G   M G eff
0  I
 Fx   Fx eff
W 
W sin q  ma   a
g

 0

a  32.2 ft s 2 sin 30  16.1ft s 2
a  r  11.50 ft s 2
v 2  v02  2a  x  x0 


 0  2 16.1ft s 2 10 ft 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

v  17.94 ft s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.9
SOLUTION:
• Draw the free-body-equation for the
wheel, expressing the equivalence of the
external and effective forces.
A cord is wrapped around the inner
hub of a wheel and pulled
horizontally with a force of 200 N.
The wheel has a mass of 50 kg and a
radius of gyration of 70 mm.
Knowing ms = 0.20 and mk = 0.15,
determine the acceleration of G and
the angular acceleration of the wheel.
• Assuming rolling without slipping and
therefore, related linear and angular
accelerations, solve the scalar equations
for the acceleration and the normal and
tangential reactions at the ground.
• Compare the required tangential reaction
to the maximum possible friction force.
• If slipping occurs, calculate the kinetic
friction force and then solve the scalar
equations for the linear and angular
accelerations.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
16 - 49
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Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 16.9
SOLUTION:
• Draw the free-body-equation for the wheel,.
• Assuming rolling without slipping, solve the scalar
equations for the acceleration and ground reactions.
 M C   M C eff
200 N 0.040 m   ma 0.100 m   I
8.0 N  m  50 kg 0.100 m 2  0.245 kg  m 2 
I  mk 2  50 kg 0.70 m 2
 0.245 kg  m 2
Assume rolling without slipping,
a  r
 0.100 m 
  10.74 rad s 2


a  0.100 m  10.74 rad s 2  1.074 m s 2
 Fx   Fx eff

F  200 N  ma  50 kg  1.074 m s 2
F  146.3 N

 Fx   Fx eff
N W  0


N  mg  50kg  1.074 m s 2  490.5 N
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
16 - 50
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Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 16.9
• Compare the required tangential reaction to the
maximum possible friction force.
Fmax  ms N  0.20490.5 N  98.1 N
F > Fmax , rolling without slipping is impossible.
Without slipping,
F  146.3 N N  490.5 N
• Calculate the friction force with slipping and solve the
scalar equations for linear and angular accelerations.
F  Fk  mk N  0.15490.5 N  73.6 N
 Fx   Fx eff
200 N  73.6 N  50 kg a
a  2.53 m s 2
 M G   M G eff
73.6 N 0.100 m   200 N 0.0.060 m 


 0.245 kg  m 2 
  18.94 rad s 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
  18.94 rad s 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 16.10
SOLUTION:
• Based on the kinematics of the constrained
motion, express the accelerations of A, B,
and G in terms of the angular acceleration.
The extremities of a 4-ft rod
weighing 50 lb can move freely and
with no friction along two straight
tracks. The rod is released with no
velocity from the position shown.
• Draw the free-body-equation for the rod,
expressing the equivalence of the
external and effective forces.
• Solve the three corresponding scalar
equations for the angular acceleration and
the reactions at A and B.
Determine: a) the angular
acceleration of the rod, and b) the
reactions at A and B.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
16 - 52
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Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 16.10
SOLUTION:
• Based on the kinematics of the constrained motion,
express the accelerations of A, B, and G in terms of
the angular acceleration.
Express the acceleration of B as



aB  a A  aB A
With aB A  4 , the corresponding vector triangle and
the law of signs yields
a A  5.46
aB  4.90
The acceleration of G is now obtained from
 


a a G  a A  aG A where aG A  2
Resolving into x and y components,
ax  5.46  2 cos 60  4.46
a y  2 sin 60  1.732
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
16 - 53
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Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 16.10
• Draw the free-body-equation for the rod, expressing
the equivalence of the external and effective forces.
• Solve the three corresponding scalar equations for the
angular acceleration and the reactions at A and B.
 M E   M E eff
501.732  6.93 4.46  2.69 1.732  2.07
  2.30 rad s 2
1 ml 2 
I  12
1 50 lb
2


4
ft
12 32.2 ft s 2
 2.07 lb  ft  s 2
I   2.07
50
4.46   6.93
ma x 
32.2
50
1.732   2.69
ma y  
32.2
  2.30 rad s 2
 Fx   Fx eff
RB sin 45  6.932.30
RB  22.5 lb
 Fy   Fy eff

RB  22.5 lb
45o
RA  22.5cos 45  50  2.692.30
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
RA  27.9 lb
16 - 54
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
The uniform rod AB of weight W is
released from rest when Assuming that
the friction force between end A and the
surface is large enough to prevent
sliding, determine immediately after
release (a) the angular acceleration of
the rod, (b) the normal reaction at A, (c)
the friction force at A.
SOLUTION:
• Draw the free-body-diagram and
kinetic diagram showing the
equivalence of the external forces
and inertial terms.
• Write the equations of motion for
the sum of forces and for the sum
of moments.
• Apply any necessary kinematic
relations, then solve the resulting
equations.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 55
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
• Draw your FBD and KD
• Set up your equations of motion
• Kinematics and solve (next page)
SOLUTION: Given: WAB = W, b= 70o
• Find: AB, NA, Ff
y
ma y
L/2
x
=
L/2
I
max
W
70o
70o
Ff
NA
F
 max
F f  max
x
F
y
 may
N A  mg  ma y
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
M
G
 IG
 N A ( L2 cos(70 ))  FF ( L2 sin(70 ))
 121 mL2 AB
2 - 56
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
• Set up your kinematic relationships – define rG/A, aG
rG /A
L/2
1
 ( L cos(70 )i  L sin(70 ) j)
2
 (0.17101 L)i  (0.46985 L)j
2
aG  a A   AB  rG /A   AB
rG /A
L/2
 0  ( AB k )  (0.17101 L i  0.46985 L j)  0
 0.46985 L  AB i  0.17101 L  AB j
70o
• Realize that you get two equations from the kinematic relationship
ax  0.46985 L  AB
a y  0.17101 L  AB
• Substitute into the sum of forces equations
F f  max
Ff  (m)0.46985 L  AB
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
N A  mg  ma y
N A  m(0.17101 L  AB  g )
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
• Substitute the Ff and NA into the sum of moments equation
 N A ( L2 cos(70 ))  FF ( L2 sin(70 ))  121 mL2 AB
[m(0.17101 L  AB  g )]( L2 cos(70 ))  [(m)0.46985 L  AB ]( L2 sin(70 ))
 121 mL2 AB
• Masses cancel out, solve for AB
0.171012 L2  AB  0.469852 L2  AB  121 L2 AB  g ( L2 cos(70 ))
 AB
g
 0.513 k
L
• The negative sign means  is
clockwise, which makes sense.
• Subbing into NA and Ff expressions,
Ff  (m)0.46985 L  0.513 Lg 
Ff  0.241mg 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
N A  m(0.17101 L  0.513 Lg   g )
N A  0.912mg 
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Vector Mechanics for Engineers: Dynamics
Concept Question
What would be true if the floor was
smooth and friction was zero?
= 70o
NA
a) The bar would rotate about point A
b) The bar’s center of gravity would go straight downwards
c) The bar would not have any angular acceleration
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 59