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VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Eighth Edition
Ferdinand P. Beer E. Russell Johnston, Jr.
Lecture Notes:J. Walt Oler Texas Tech University
CHAPTER
2007 The McGraw-Hill Companies, Inc. All rights reserved.
16 Plane Motion of Rigid Bodies:Forces and Accelerations
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Contents
IntroductionEquations of Motion of a Rigid Body
Angular Momentum of a Rigid Bodyin Plane Motion
Plane Motion of a Rigid Body:dAlemberts Principle
Axioms of the Mechanics of RigidBodies
Problems Involving the Motion of aRigid Body
Sample Problem 16.1Sample Problem 16.2
Sample Problem 16.3Sample Problem 16.4
Sample Problem 16.5Constrained Plane MotionConstrained Plane Motion:
Noncentroidal Rotation
Constrained Plane Motion:Rolling Motion
Sample Problem 16.6Sample Problem 16.8Sample Problem 16.9Sample Problem 16.10
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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 centerG of the body with respect to the
Newtonian frame Oxyz ,am F rr =
For the motion of the body withrespect to the centroidal frame
Gxyz ,GG H M
&rr =
System of external forces isequipollent to the system
consisting of .and G H am &rr
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Angular Momentum of a Rigid Body in Plane Motion
Consider a rigid slab in plane motion.
Angular momentum of the slab may becomputed by
( )
( )[ ]
( )
r
r
rrr
rrr
mr
mr r
mvr H
ii
n
iiii
n
iiiiG
=
=
=
=
=
=
2
1
1
After differentiation,
r&r&r
I I H G ==
Results are also valid for plane motion of bodieswhich are symmetrical with respect to thereference plane.
Results are not valid for asymmetrical bodies or
three-dimensional motion.
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Plane Motion of a Rigid Body: DAlemberts Principle
I M am F am F G y y x x
===
Motion of a rigid body in plane motion iscompletely defined by the resultant and momentresultant about G of the external forces.
The external forces and the collective effectiveforces of the slab particles are equipollent (reduceto the same resultant and moment resultant) andequivalent (have the same effect on the body).
The most general motion of a rigid body that issymmetrical with respect to the reference planecan be replaced by the sum of a translation and acentroidal rotation.
dAlemberts Principle : The external forcesacting on a rigid body are equivalent to theeffective forces of the various particles formingthe body.
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Axioms of the Mechanics of Rigid Bodies
The forces act at different points ona rigid body but but have the same magnitude,direction, and line of action.
F rr
and
The forces produce the same moment aboutany point and are therefore, equipollentexternal forces.
This proves the principle of transmissibilitywhereas it was previously stated as an axiom.
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Problems Involving the Motion of a Rigid Body
The fundamental relation between the forcesacting on a rigid body in plane motion andthe acceleration of its mass center and the
angular acceleration of the body is illustratedin a free-body-diagram equation.
The techniques for solving problems ofstatic equilibrium may be applied to solve
problems of plane motion by utilizing- dAlemberts principle, or - principle of dynamic equilibrium
These techniques may also be applied to problems involving plane motion ofconnected rigid bodies by drawing a free-
body-diagram equation for each body andsolving the corresponding equations of
motion simultaneously.
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Sample Problem 16.1
At a forward speed of 30 ft/s, the truck brakes were applied, causing the wheelsto stop rotating. It was observed that thetruck to skidded to a stop in 20 ft.
Determine the magnitude of the normalreaction and the friction force at eachwheel as the truck skidded to a stop.
SOLUTION:
Calculate the acceleration during theskidding stop by assuming uniform
acceleration.
Apply the three corresponding scalarequations to solve for the unknownnormal wheel forces at the front and rearand the coefficient of friction betweenthe wheels and road surface.
Draw the free-body-diagram equationexpressing the equivalence of theexternal and effective forces.
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Sample Problem 16.1
ft20sft
300 == xv
SOLUTION: Calculate the acceleration during the skidding stop
by assuming uniform acceleration.
( )
( )ft202sft
300
22
020
2
a
x xavv
+
=
+=
sft
5.22=a
Draw a free-body-diagram equation expressing theequivalence of the external and effective forces.
Apply the corresponding scalar equations.
0=+ W N B = eff y y F F
( )( )
699.0
2.32
5.22 ===
=
=+=
g
a
a g W W
N N am
k
k
B Ak
B A
( ) = eff x x F F
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Sample Problem 16.1
W N W N B
350.0==
( )W N N Arear 350.021
21 == W rear 175.0=
( )W N N V front 650.021
21 == W N front 325.0=
( )( )W N rear k rear 175.0690.0==W rear 122.0=
( )( )W N front k front 325.0690.0==W F
front 227.0.0=
Apply the corresponding scalar equations.
( ) ( ) ( )
W N
g aW
a g W
W N
am N W
B
B
B
650.0
4512
45121
ft4ft12ft5
=
+=
+=
=+
( ) = eff A A M M
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Sample Problem 16.2
The thin plate of mass 8 kg is held in place as shown.
Neglecting the mass of the links,determine immediately after the wirehas been cut ( a) the acceleration of the
plate, and ( b) the force in each link.
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 equationexpressing the equivalence of the
external and effective forces. Resolve into scalar component equations
parallel and perpendicular to the path ofthe mass center.
Solve the component equations and themoment equation for the unknownacceleration and link forces.
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Sample Problem 16.2SOLUTION: Note that after the wire is cut, all particles of the
plate move along parallel circular paths of radius150 mm. The plate is in curvilinear translation.
Draw the free-body-diagram equation expressingthe equivalence of the external and effectiveforces.
Resolve the diagram equation into components
parallel and perpendicular to the path of the masscenter.
( ) = eff t t F F
=
=
30cos
30cos
mg
amW
= 30cosm/s81.9 2a
2sm50.8=a 60o
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Sample Problem 16.2
2sm50.8=a 60o
Solve the component equations and the momentequation for the unknown acceleration and linkforces.
( )eff GG M M =
( )( ) ( )( )( )( ) ( )( ) 0mm10030cosmm25030sin
mm10030cosmm25030sin=+
DF DF
AE AE
F F
F F
E DF
DF AE
F F
F F
1815.0
06.2114.38=
=+
( ) = eff nn F F
( )( )2sm81.9kg8619.0030sin1815.0
030sin
=
=
=+
AE
AE AE
DF AE
F W F F
W F F
T E N9.47=
( ) N9.471815.0= DF F C DF N70.8=
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Sample Problem 16.3
A pulley weighing 12 lb and having aradius of gyration of 8 in. is connected totwo blocks as shown.
Assuming no axle friction, determine theangular acceleration of the pulley and theacceleration of each block.
SOLUTION:
Determine the direction of rotation byevaluating the net moment on the
pulley due to the two blocks.
Relate the acceleration of the blocks tothe angular acceleration of the pulley.
Draw the free-body-diagram equationexpressing the equivalence of theexternal and effective forces on thecomplete pulley plus blocks system.
Solve the corresponding momentequation for the pulley angularacceleration.
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Sample Problem 16.3
Relate the acceleration of the blocks to the angularacceleration of the pulley.
( )
ft1210=
= A A r a
( )
ft126=
= B B r a
2
2
2
22
sftlb1656.0
ft128
sft32.2lb12
=
=
== k g W
k m I note:
SOLUTION: Determine the direction of rotation by evaluating the net
moment on the pulley due to the two blocks.
( )( ) ( )( ) lbin10in10lb5in6lb10 == Grotation is counterclockwise.
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Sample Problem 16.3 Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on thecomplete pulley and blocks system.
( )( ) 21262
12
10
2
sft
sft
sftlb1656.0
=
=
=
B
A
a
a
I
( ) = eff GG M M
( ) ( )
( )( ) ( )( ) ( ) ( )( )( ) ( )( )( )1210
1210
2.325
126
126
2.3210
1210
126
1210
126
1210
126
1656.0510
ftftftlb5ftlb10
+=
+=
A A B B amam I
Solve the corresponding moment equation for the pulley
angular acceleration.
2srad374.2=
( )( )2126 srad2.374ft== B B r a
2sft187.1= Ba
( )( )21210 srad2.374ft==
A A r a
2sft978.1=a
Then,
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Sample Problem 16.4
A cord is wrapped around ahomogeneous disk of mass 15 kg.The cord is pulled upwards with aforce T = 180 N.
Determine: ( a) the acceleration of thecenter of the disk, ( b) the angularacceleration of the disk, and ( c) theacceleration of the cord.
SOLUTION:
Draw the free-body-diagram equationexpressing the equivalence of the external
and effective forces on the disk.
Solve the three corresponding scalarequilibrium equations for the horizontal,vertical, and angular accelerations of the
disk.
Determine the acceleration of the cord byevaluating the tangential acceleration ofthe point A on the disk.
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Sample Problem 16.4SOLUTION: Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on thedisk.
=eff y y
F F
( )( )kg15
sm81.9kg15- N180 2=
=
=
mW T
a
amW T
y
y
2
sm19.2=
ya( ) = eff GG M M
( )
( )( )m5.0kg15
N18022
221
==
==
mr
T
mr I Tr
2srad0.48=
( ) = eff x x F F am=0 0= xa
Solve the three scalar equilibrium equations.
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Sample Problem 16.4
2sm19.2= ya
2srad0.48=
0= xa
Determine the acceleration of the cord by evaluating thetangential acceleration of the point A on the disk.
( )( )( )22 srad48m5.0sm19.2 +=
+== t G At Acord aaaar
2sm2.26=cord a
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Sample Problem 16.5
A uniform sphere of mass m and radiusr is projected along a rough horizontalsurface with a linear velocity v0. Thecoefficient of kinetic friction between
the sphere and the surface is k .Determine: ( a) the time t 1 at which thesphere will start rolling without sliding,and ( b) the linear and angular velocities
of the sphere at time t 1.
SOLUTION:
Draw the free-body-diagram equationexpressing the equivalence of the
external and effective forces on thesphere.
Solve the three corresponding scalarequilibrium equations for the normal
reaction from the surface and the linearand angular accelerations of the sphere.
Apply the kinematic relations foruniformly accelerated motion to
determine the time at which thetangential velocity of the sphere at thesurface is zero, i.e., when the spherestops sliding.
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Sample Problem 16.5
g a k =
r g k
25=
Apply the kinematic relations for uniformly acceleratedmotion to determine the time at which the tangential velocityof the sphere at the surface is zero, i.e., when the spherestops sliding.
( )t g vt avv k +=+= 00
t r g
t k
+=+= 25
00
110 25
t r g
r gt v k k
=
g v
t k 0
1 72=
=
=
g v
r g
t r g
k
k k
011 72
25
25
r v0
1 75=
==
r v
r r v 011 75
075
1 vv =
At the instant t 1
when the sphere stops sliding,
11 r v =
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Constrained Plane Motion Most engineering applications involve rigid
bodies which are moving under givenconstraints, e.g., cranks, connecting rods, andnon-slipping wheels.
Constrained plane motion : motions withdefinite relations between the components ofacceleration of the mass center and the angularacceleration of the body.
Solution of a problem involving constrained plane motion begins with a kinematic analysis.
e.g., given , , and , find P , N A, and N B.- kinematic analysis yields- application of dAlemberts principle yields
P , N A, and N B.
.and y x aa
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Constrained Motion: Noncentroidal Rotation Noncentroidal rotation : motion of a body is
constrained to rotate about a fixed axis that doesnot pass through its mass center.
Kinematic relation between the motion of the masscenter G and the motion of the body about G,
2 r ar a nt ==
The kinematic relations are used to eliminatefrom equations derived from
dAlemberts principle or from the method ofdynamic equilibrium.
nt aa and
EE
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Constrained Plane Motion: Rolling Motion For a balanced disk constrained to
roll without sliding, r ar x ==
Rolling, no sliding: N F s r a =
Rolling, sliding impending: N F s= r a =
Rotating and sliding: N F k = r a , independent
For the geometric center of anunbalanced disk,
r aO
=
The acceleration of the mass center,
( ) ( )nOGt OGO
OGOG
aaa
aaarrr
rrr
++=
+=
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EE
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Sample Problem 16.6
rad/s8=
2srad40=
kg3mm85
kg4
=
=
=
OB
E
E
mk
m
SOLUTION: Draw the free-body-equation for AOB.
Evaluate the external forces due to the weights ofgear E and arm OB and the effective forces.
( )
( )( ) N4.29sm81.9kg3 N2.39sm81.9kg4
2
2
==
==
OB
E
W
W
( )( )m N156.1
srad40m085.0kg4 222
=
== E E E k m I
( ) ( ) ( )( ) N0.24
srad40m200.0kg3 2
=
== r mam OBt OBOB
( ) ( )( )( ) N4.38
srad8m200.0kg3 22
=
== r mam OBnOBOB
( )( )
m N600.1
srad40m.4000kg3 221212
121
=
== Lm I OBOB
EE
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Sample Problem 16.6
N4.29
N2.39=
=
OB
E
W
W
m N156.1 = I ( ) N0.24=t OBOB am
( ) N4.38=nOBOB am
m N600.1 = OB I
Solve the three scalar equations derived from the free- body-equation for the tangential force at A and thehorizontal and vertical components of reaction at O.
( )eff OO =
( ) ( ) ( )( )( ) m N600.1m200.0 N0.24m N156.1
m200.0m120.0++=
++= OBt OBOB E I am I F
N0.63= F
( )eff x x F F =
( ) N0.24== t OBOB x am R N0.24= x R
eff y y F =( )
N4.38 N4.29 N2.39 N0.63 =
=
y
OBOBOB E y
R
amW W F R
N0.24= y R
E E
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Sample Problem 16.8
A sphere of weight W is released withno initial velocity and rolls withoutslipping on the incline.
Determine: a) the minimum value ofthe coefficient of friction, b) thevelocity of G after the sphere hasrolled 10 ft and c) the velocity of G ifthe sphere were to move 10 ft down a
frictionless incline.
SOLUTION:
Draw the free-body-equation for thesphere , expressing the equivalence of the
external and effective forces. With the linear and angular accelerations
related, solve the three scalar equationsderived from the free-body-equation for
the angular acceleration and the normaland tangential reactions at C .
Calculate the velocity after 10 ft ofuniformly accelerated motion.
Assuming no friction, calculate the linearacceleration down the incline and the
corresponding velocity after 10 ft.
Calculate the friction coefficient requiredfor the indicated tangential reaction at C .
h i f i iE E
d
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Sample Problem 16.8SOLUTION: Draw the free-body-equation for the sphere , expressing
the equivalence of the external and effective forces.
r a =
With the linear and angular accelerations related, solve
the three scalar equations derived from the free-body-equation for the angular acceleration and the normaland tangential reactions at C.
( ) = eff C C
( ) ( )
( ) ( )
+
=
+=
+=
2
252
5
2
sin
r
g
W r r
g
W
mr r mr
I r amr W
r
g
7
sin5 =
( )7
30sinsft2.325
730sin5
2 =
== g r a
2sft50.11=a
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V t M h i f E gi D iE i
E d
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Sample Problem 16.9
A cord is wrapped around the innerhub of a wheel and pulledhorizontally with a force of 200 N.The wheel has a mass of 50 kg and aradius of gyration of 70 mm.Knowing s = 0.20 and k = 0.15,determine the acceleration of G andthe angular acceleration of the wheel.
SOLUTION:
Draw the free-body-equation for thewheel , expressing the equivalence of theexternal and effective forces.
Assuming rolling without slipping andtherefore, related linear and angularaccelerations, solve the scalar equations
for the acceleration and the normal andtangential reactions at the ground.
Compare the required tangential reactionto the maximum possible friction force.
If slipping occurs, calculate the kineticfriction force and then solve the scalarequations for the linear and angularaccelerations.
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Sample Problem 16.9SOLUTION: Draw the free-body-equation for the wheel ,.
Assume rolling without slipping,
( )
m100.0== r a
( )( )2
22
mkg245.0m70.0kg50
=
== k m I
Assuming rolling without slipping, solve the scalarequations for the acceleration and ground reactions.
( )( ) ( )
( )( ) ( )
( )( ) 222
22
sm074.1srad74.10m100.0srad74.10
mkg245.0m100.0kg50m N0.8
m100.0m040.0 N200
==
+=
+=
+=
a
I am
( ) = eff C C
( ) = eff x x F
( )( ) N5.490sm074.1kg500
2 +===
=
mg N
W N
( ) = eff x x F ( )
N3.146
sm074.1kg50 N200 2
=
==+
F
am F
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Sample Problem 16.9
N3.146= N5.490= N Without slipping,
Compare the required tangential reaction to themaximum possible friction force.
( ) N1.98 N5.49020.0max === F
F > F max , rolling without slipping is impossible.
Calculate the friction force with slipping and solve thescalar equations for linear and angular accelerations.
( ) N6.73 N5.49015.0 ==== N F F k k
( ) =
eff GG
( )( ) ( )( )
( )2
2
srad94.18
mkg245.0
m060.0.0 N200m100.0 N6.73
=
=
2srad94.18=
( ) = eff x x F ( )akg50 N6.73 N200 = 2sm53.2=a
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Sample Problem 16.10
The extremities of a 4-ft rodweighing 50 lb can move freely andwith no friction along two straighttracks. The rod is released with novelocity from the position shown.
Determine: a) the angularacceleration of the rod, and b) thereactions at A and B.
SOLUTION:
Based on the kinematics of the constrainedmotion, express the accelerations of A, B,
and G in terms of the angular acceleration.
Draw the free-body-equation for the rod ,expressing the equivalence of theexternal and effective forces.
Solve the three corresponding scalarequations for the angular acceleration andthe reactions at A and B.
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Sample Problem 16.10SOLUTION: Based on the kinematics of the constrained motion,
express the accelerations of A, B, and G in terms ofthe angular acceleration.
Express the acceleration of B as A B A B aaa
rrr+=
With the corresponding vector triangle andthe law of signs yields
,4 = A Ba
90.446.5 == Baa
The acceleration of G is now obtained from
AG AG aaaa rrrr
+== 2where = AGa
Resolving into x and y components,
732.160sin2
46.460cos246.5==
==
y
x
a
a
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Vector Mechanics for Engineers: Dynamicsh t h i t i on
16 - 39
Sample Problem 16.10 Draw the free-body-equation for the rod , expressing
the equivalence of the external and effective forces.
( )
( )
( )
69.2732.12.32
50
93.646.42.32
5007.2
sftlb07.2
ft4sft32.2
lb50121
2
22
2121
==
==
=
=
==
y
x
am
am
I
ml I
Solve the three corresponding scalar equations for theangular acceleration and the reactions at A and B.
( )( ) ( )( ) ( )( )2srad30.2
07.2732.169.246.493.6732.150
+=
++=
( ) = eff E E
2
srad30.2= ( ) = eff x x F
( )( )lb5.22
30.293.645sin=
=
B
B
R
R
lb5.22=
B R
r
45o
=eff y y
F F
( ) ( )( )30.269.25045cos5.22 =+ Rlb9.27= R