Tenth Edition - apl100.wdfiles.comapl100.wdfiles.com/local--files/notes/chapter 16 mod.pdf · Plane Motion of Rigid Bodies: Forces and Accelerations . Contents 16 ... The free body

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. Self California Polytechnic State University

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

16 Plane Motion of Rigid Bodies:

Forces and Accelerations

Contents

16 - 2

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

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


Vector Mechanics for Engineers: Dynamics

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Rigid Body Kinetics

2 - 3

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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Rigid Body Kinetics

2 - 4

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.



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Introduction

16 - 5

• 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.



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Introduction

16 - 6

• 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,

GG HMamF and



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Equations of Motion for a Rigid Body

16 - 7

• 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,

amF



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Equations of Motion for a Rigid Body

16 - 8

• For the motion of the

body with respect to the

centroidal frame

Gx’y’z’,

GG HM

• System of external

forces is equipollent

to the system

consisting of

. and GHam



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Rigid Body Angular Momentum in Three Dimensions

18 - 9

Angular momentum of a body about its mass center,

n

iiii

n

iiiiG mrrmvrH

11

Δ

The x component of the angular momentum,

n

iiiiz

n

iiiiy

n

iiiix

n

iiixiziiyixi

n

iiyiiziix

mxzmyxmzy

mzxzxyy

mrzryH

111

22

1

1

ΔΔΔ

Δ

Δ

dmzxdmxydmzyH zyxx 22

zxzyxyxx III

zzyzyxzxz

zyzyyxyxy

IIIH

IIIH



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Rigid Body Angular Momentum in Three Dimensions

18 - 10

zzyzyxzxz

zyzyyxyxy

zxzyxyxxx

IIIH

IIIH

IIIH

With respect to the principal axes of inertia,

z

y

x

I

I

I

00

00

00

zzzyyyxxx IHIHIH



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Angular Momentum of a Rigid Body in Plane Motion

16 - 11

• Consider a rigid slab

in plane motion. Slab

like bodies have xy

as plane of symmetry

• Angular momentum of the

slab may be computed by

I

mr

mrr

mvrH

ii

n

iiii

n

iiiiG

Δ

Δ

Δ

2

1

1

• After differentiation,

IIHG



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Angular Momentum of a Rigid Body in Plane Motion

16 - 12

• 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.

• Consider a rigid

slab in plane

motion.



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Plane Motion of a Rigid Body: D’Alembert’s Principle

16 - 13

IMamFamF Gyyxx

• Motion of a rigid body in

plane motion is completely

defined by the resultant and

moment resultant about G of

the external forces.

• 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).



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Problems Involving the Motion of a Rigid Body

16 - 14

• 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.

d’Alembert’s principle



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16 - 15

principle of dynamic equilibrium



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16 - 16

• 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 free-body-diagram equation for each body and

solving the corresponding equations of motion

simultaneously.



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Free Body Diagrams and Kinetic Diagrams

12 - 17

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.

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)

1. Isolate the body of interest (free body)

5. Draw appropriate dimensions (angles and distances)

x

y

Include your

positive

z-axis

direction too



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12 - 18

Put the inertial terms for the body of interest on the kinetic diagram.

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.

1. Isolate the body of interest (free body)

m F a

G I M



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2 - 19

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 - 20

P

L/2

L/2

r

A

Cx

Cy

mg

1. Isolate body

2. Axes

3. Applied forces

4. Replace supports with forces

5. Dimensions

6. Kinetic diagram

G G

xma

yma

I

C

B

x

y



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2 - 21

A drum of 100 mm radius is attached

to a disk of 200 mm radius. The

combined drum and disk had a

combined mass of 5 kg. A cord is

attached as shown, and a force of

magnitude P=25 N 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 - 22

xma

yma

I

P

F

W

N x

y

=

1. Isolate body

2. Axes

3. Applied forces


5. Dimensions

6. Kinetic diagram

100

mm

200 mm



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2 - 23

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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2 - 24

FA

W

NA

FB

NB

xma

yma

I

1. Isolate body

2. Axes

3. Applied forces


5. Dimensions

6. Kinetic diagram

=

x

y

q



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Sample Problem 16.1

16 - 25

At a forward speed of 10 m/s, the truck brakes were

applied, causing the wheels to stop rotating. It was

observed that the truck to skidded to a stop in 7 m.

Determine the magnitude of the normal reaction and

the friction force at each wheel as the truck skidded

to a stop. The weight of the truck is WN.



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Sample Problem 16.1

16 - 26

SOLUTION:

• Calculate the acceleration

during the skidding stop by

assuming uniform

acceleration.

• Apply the three corresponding scalar equations to solve

for the unknown normal wheel forces at the front and rear

and the coefficient of friction between the wheels and

road surface.

• Draw the free-body-diagram equation expressing the

equivalence of the external and effective forces.



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16 - 27

0

m10 7m

sv x= =

• Calculate the acceleration during the skidding stop by

assuming uniform acceleration.

( )

( )

2 20 0

2

2

m0 10 2 7m

s

v v a x x

a

= + -

æ ö= +ç ÷è ø

2

m7.14

sa = -

• Draw a free-body-diagram equation

expressing the equivalence of the

external and inertial terms.



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16 - 28

• Apply the corresponding scalar equations.

0 WNN BA

effyy FF

( )

( )7.14

0.7289.81

A B

k A B

k

k

F F ma

N N

W W g a

a

g

m

m

m

- - = -

- + =

- = -

= = =

effxx FF



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16 - 29

0.341A BN W N W= - =

• Apply the corresponding scalar

equations.

( ) ( ) ( )1.5m 3.6m 1.2m

11.5 1.2 1.5 1.2

3.6 3.6

0.659

B

B

B

W N ma

W W aN W a

g g

N W

- + =

æ ö æ ö= + = +ç ÷ ç ÷è ø è ø

=

effAA MM



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16 - 30

( )1 12 2

0.341rear AN N W= =

0.1705rearN W=

( )1 12 2

0.659front BN N W= =

0.3295frontN W=

( )( )0.728 0.1705rear k rearF N Wm= =

0.124rearF W=

( )( )0.728 0.3295front k frontF N Wm= =

0.240frontF W=



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Sample Problem 16.2

16 - 31

The thin plate of mass 8 kg is held in place as

shown.

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.



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16 - 32

• 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.



• Resolve into scalar component equations parallel and

perpendicular to the path of the mass center.

• Solve the component equations and the moment equation

for the unknown acceleration and link forces.



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16 - 33



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16 - 34

• 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.



• Resolve the diagram equation into components parallel

and perpendicular to the path of the mass center.

efftt FF

30cos

30cos

mg

amW

30cosm/s81.9 2a 2sm50.8a 60o



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16 - 35

• Resolve the diagram equation into components parallel

and perpendicular to the path of the mass center.

efftt FF

30cos

30cos

mg

amW

30cosm/s81.9 2a 2sm50.8a 60o



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16 - 36

• Solve the component equations and the moment equation

for the unknown acceleration and link forces.

effGG MM

0mm10030cosmm25030sin

mm10030cosmm25030sin

DFDF

AEAE

FF

FF

AEDF

DFAE

FF

FF

1815.0

06.2114.38



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Sample Problem 16.2

16 - 37

effnn FF

2sm81.9kg8619.0

030sin1815.0

030sin

AE

AEAE

DFAE

F

WFF

WFF

TFAE N9.47

N9.471815.0DFF CFDF N70.8



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Sample Problem 16.5

16 - 38

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.



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Sample Problem 16.5

16 - 39

SOLUTION:

• Draw the free-body-diagram equation


external and effective forces on the

sphere.

• 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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Sample Problem 16.5

16 - 40

SOLUTION:

• Draw the free-body-diagram equation

expressing the equivalence of external and

effective forces on the sphere.



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Sample Problem 16.5

16 - 41

• Solve the three scalar

equilibrium equations.

effyy FF

0WN mgWN

effxx FF

ammg

amF

k

m ga km



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Sample Problem 16.5

16 - 42

m

2

32 mrrmg

IFr

k

r

gkm

2

5

effGG MM

NOTE: As long as the sphere both rotates

and slides, its linear and angular motions are

uniformly accelerated.



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Sample Problem 16.5

16 - 43

• 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.

tgvtavv km 00

tr

gt k

m

2

500

ga kmr

gkm

2

5



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Sample Problem 16.5

16 - 44

1102

5t

r

grgtv k

k

mm

g

vt

km0

17

2

g

v

r

gt

r

g

k

kk

m

mm 0

117

2

2

5

2

5

r

v01

7

5

r

vrrv 0

117

5 07

51 vv

At the instant t1 when the

sphere stops sliding,

11 rv



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Group Problem Solving

2 - 45

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 rear-wheel drive



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2 - 46

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.



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2 - 47

SOLUTION:

• Given: rear wheel drive,

dimensions as shown, m= 0.80

• Find: Maximum acceleration

• Draw your FBD and KD

NF NR

FR mg

x

y

xma

I

yma

=

x xF ma y yF ma

R xF ma 0R FN N mg

• Set up your equations of motion,

realizing that at maximum acceleration,

may and will be zero

G GM I (1.5) (1) (0.5) 0R F RN N F



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2 - 48

• Solve the resulting equations: 4 unknowns are FR, max, NF and NR

(1)

(2) (3)

(4) (1.5) (1) (0.5) 0R F RN N F

0R FN N mg

R xF ma

R RF Nm

xR

maN

m(1)→(3) (5)

xF R

maN mg N mg

m (6) (5)→(2)



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2 - 49


Solving this equation,

xR

maN

m(1)→(3) (5)

xF R

maN mg N mg

m (6)

(5)→(2)

(1) and (5) and (6) →(4)

1.5 1 0.5 0x xx

ma mamg ma

m m

23.74 m/sxa

50.5

2

x

ga

m



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2 - 50


Solving this equation,

(1) and (5) and (6) →(4)

1.5 1 0.5 0x xx

ma mamg ma

m m

23.74 m/sxa

50.5

2

x

ga

m



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2 - 51

• Alternatively, you could have chosen to sum moments about the

front wheel

NF

FR

NR

mg

x

y

xma

I

yma

=

F GM I mad (2.5) (1) 0 (0.5)R xN mg ma

• You can now use this equation with those on the previous slide to

solve for the acceleration



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Concept Question

2 - 52

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?



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Concept Question

2 - 53

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



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Kinetics: Constrained Plane Motion

2 - 54

The forces at the bottom of the pendulum

depend on the pendulum mass and mass

moment of inertia, as well as the pendulum

kinematics.



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Kinetics: Constrained Plane Motion

2 - 55

The forces one the

wind turbine blades

are also dependent

on mass, mass

moment of inertia,

and kinematics.



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16 - 56

• Most engineering

applications involve rigid

bodies which are moving

under given constraints,

e.g., cranks, connecting

rods, and non-slipping

wheels.



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16 - 57

• Constrained plane motion:

motions with definite

relations between the

components of acceleration

of the mass center and the

angular acceleration of the

body.



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16 - 58

• 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

• - application of d’Alembert’s

principle yields P, NA, and

NB.

. and yx aa



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Constrained Motion: Noncentroidal Rotation

16 - 59

• 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,

2 rara nt



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2 - 60

2 rara nt



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Constrained Motion: Noncentroidal Rotation

16 - 61

• The kinematic relations are

used to eliminate

from equations derived from

d’Alembert’s principle or

from the method of dynamic

equilibrium.

nt aa and



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Constrained Plane Motion: Rolling Motion

16 - 62

• For a balanced disk

constrained to roll

without sliding,

q rarx

• Rolling, no sliding:

NF sm ra

Rolling, sliding impending:

NF sm ra

Rotating and sliding:

ra , independent NF km



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Constrained Plane Motion: Rolling Motion

16 - 63

• For the geometric

center of an

unbalanced disk,

raO

The acceleration of

the mass center,

nOGtOGO

OGOG

aaa

aaa



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Sample Problem 16.6

16 - 64

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.

OB is welded at O.

Determine: a) tangential force

exerted by gear D, and b)

components of the reaction at

shaft O.

kg 3

mm 85

kg 4

OB

E

E

m

k

m



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Sample Problem 16.6

16 - 65

SOLUTION:

• Draw the free-body-equation for AOB,


external and effective forces.

• 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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Sample Problem 16.6

16 - 66

rad/s 8

2srad40

kg 3

mm 85

kg 4

OB

E

E

m

k

m• Evaluate the external forces due to the weights of

gear E and arm OB and the effective forces.

N4.29sm81.9kg3

N2.39sm81.9kg4

2

2

OB

E

W

W

The reaction Rx and Ry arise from pinning the gear-rod system to the wall



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Sample Problem 16.6

16 - 67

rad/s 8

2srad40

kg 3

mm 85

kg 4

OB

E

E

m

k

m

mN156.1

srad40m085.0kg4 222

EEE kmI

N0.24

srad40m200.0kg3 2

rmam OBtOBOB

N4.38

srad8m200.0kg322

rmam OBnOBOB

mN600.1

srad40m.4000kg3 22

1212

121

LmI OBOB



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Sample Problem 16.6

16 - 68

N4.29

N2.39

OB

E

W

W

mN156.1 EI

N0.24tOBOB am

N4.38nOBOB am

mN600.1 OBI

• 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 O.

effOO MM

mN600.1m200.0N0.24mN156.1

m200.0m120.0

OBtOBOBE IamIF

N0.63F



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Sample Problem 16.6

16 - 69

N4.29

N2.39

OB

E

W

W

mN156.1 EI

N0.24tOBOB am

N4.38nOBOB am

mN600.1 OBI

effxx FF

N0.24tOBOBx amR

N0.24xR



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Sample Problem 16.6

16 - 70

N4.29

N2.39

OB

E

W

W

mN156.1 EI

N0.24tOBOB am

N4.38nOBOB am

mN600.1 OBI

effyy FF

N4.38N4.29N2.39N0.63

y

OBOBOBEy

R

amWWFR

N0.24yR



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2 - 71



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2 - 72



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2 - 73

In last sample problem the of rod and gear were same and one

torque equation was enough. Here 1 and 2 are different. So we

have to separate the two parts.



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2 - 74



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2 - 75



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2 - 76



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2 - 77



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Sample Problem 16.8

16 - 78

A sphere of weight W is

released with no initial

velocity and rolls without

slipping on the incline.

Determine: a) the minimum

value of the coefficient of

friction, b) the velocity of G

after the sphere has rolled 10

ft and c) the velocity of G if

the sphere were to move 3 m

down a frictionless incline.



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Sample Problem 16.8

16 - 79

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

and tangential reactions at C.

• Calculate the velocity after 3 m of

uniformly accelerated motion.

• Assuming no friction, calculate the linear

acceleration down the incline and the

corresponding velocity after 3 m.

• Calculate the friction coefficient required

for the indicated tangential reaction at C.



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Sample Problem 16.8

16 - 80

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

and tangential reactions at C.



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Sample Problem 16.8

16 - 81

ra

q

2

2

52

5

2

sin

rg

Wrr

g

W

mrrmr

IramrW

r

g

7

sin5 q ( )2

5 sin30

7

5 9.31m s sin30

7

ga ra

°= =

°=

23.504m sa =

effCC MM



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Sample Problem 16.8

16 - 82

• Solve the three scalar equations derived

from the free-body-equation for the

angular acceleration and the normal and

tangential reactions at C.

r

g

7

sin5 q

23.504m sa ra= =

effxx FF

WWF

g

g

W

amFW

143.030sin7

2

7

sin5

sin

q

q

effyy FF

WWN

WN

866.030cos

0cos

q



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Sample Problem 16.8

16 - 83

• Calculate the friction coefficient required

for the indicated tangential reaction at C.

W

W

N

F

NF

s

s

866.0

143.0

m

m

165.0sm



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Sample Problem 16.8

16 - 84

r

g

7

sin5 q

23.504m sa ra= =

• Calculate the velocity after

3 m of uniformly

accelerated motion.

( )

( )( )

2 20 0

2

2

0 2 3.504m s 3m

v v a x x= + -

= +

4.59m sv =



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Sample Problem 16.8

16 - 85

effGG MM 00 I

• Assuming no friction, calculate

the linear acceleration and the

corresponding velocity after 3m.

effxx FF

( )2 2

sin

9.81m s sin30 4.905m s

WW ma a

g

a

qæ ö

= = ç ÷è ø

= ° =

( )

( )( )

2 20 0

2

2

0 2 9.81m s 3m

v v a x x= + -

= +5.42m sv =



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Sample Problem 16.9

16 - 86

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.



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Sample Problem 16.9

16 - 87

SOLUTION:

• Draw the free-body-equation

for the wheel, expressing the

equivalence of the external

and effective forces.

• 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.



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Sample Problem 16.9

16 - 88

• Draw the free-body-equation for the wheel,.

Assume rolling without slipping,

m100.0

ra

2

22

mkg245.0

m70.0kg50

kmI



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Sample Problem 16.9

16 - 89

• Assuming rolling without slipping,

solve the scalar equations for the

acceleration and ground reactions.

22

2

22

sm074.1srad74.10m100.0

srad74.10

mkg245.0m100.0kg50mN0.8

m100.0m040.0N200

a

Iam

effCC MM



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Sample Problem 16.9

16 - 90

effxx FF

N5.490sm074.1kg50

0

2

mgN

WN

effxx FF

N3.146

sm074.1kg50N200 2

F

amF



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Sample Problem 16.9

16 - 91

N3.146F N5.490N

Without slipping,

• Compare the required tangential reaction to the

maximum possible friction force.

N1.98

N5.49020.0

max

NF sm

F > Fmax , rolling without slipping is

impossible.



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Sample Problem 16.9

16 - 92

N3.146F

N5.490N

Without

slipping,

• Calculate the friction force

with slipping and solve the

scalar equations for linear and

angular accelerations.

N6.73N5.49015.0 NFF kk m

effxx FF

akg50N6.73N200

2sm53.2a



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Sample Problem 16.9

16 - 93

effGG MM

2

2

srad94.18

mkg245.0

m060.0.0N200m100.0N6.73

2srad94.18



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2 - 94

video

Whiplash Contre-coup in 30 seconds.mp4



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16 - 111

The extremities of a 1.2-m rod of mass 25 kg can

move freely and with no friction along two straight

tracks. The rod is released with no velocity from

the position shown.

Determine: a) the angular acceleration of the rod,

and b) the reactions at A and B.



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16 - 112

SOLUTION:

• Based on the kinematics of the constrained

motion, 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 the external and

effective forces.

• Solve the three corresponding scalar equations

for the angular acceleration and the reactions

at A and B.



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16 - 113



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2 - 114



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16 - 115

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

ABAB aaa

With the corresponding vector triangle and

the law of sines yield

1.2 ,B Aa a=

1.639 1.47A Ba aa a= =

The acceleration of G is now obtained from

AGAG aaaa

where 0.6G Aa a=

Resolving into x and y components,

1.639 0.6 cos60 1.339

0.6 sin60 0.52

x

y

a

a

a a a

a a

= - ° =

= - ° = -



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16 - 116

( )

( )

( )

22112

2

125kg 1.2m

12

3kg m

3

25 1.39 33.5

25 0.520 13.0

x

y

I ml

I

ma

ma

a a

a a

a a

= =

= ×

=

= =

= - = -

( )( )( ) ( )( ) ( )( )2

25 9.81 0.520 33.5 1.34 13.0 0.520 3

2.33rad s

a a a

a

= + +

= +

effEE MM

22.33rad sa =



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16 - 117

effxx FF

( )( )sin 45 33.5 2.33

110.4 N

B

B

R

R

° =

=

110.4NBR = 45o

effyy FF

( ) ( )( ) ( )( )110.4 cos45 25 9.81 13.0 2.33AR + ° - = -

136.6NAR =



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2 - 118

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.



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2 - 119

SOLUTION: Given: WAB = W, b= 70o

• Find: AB, NA, Ff

• Draw your FBD and KD

• Set up your equations of motion

x

y

xma

I

yma

=

L/2

L/2

Ff

NA

W

70o

x xF ma y yF maf xF ma A yN mg ma

G GM I

2 2

2112

( cos(70 )) ( sin(70 ))

L LA F

AB

N F

mL

70o

• Kinematics and solve (next page)



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2 - 120

L/2

L/2

70o

• Set up your kinematic relationships – define rG/A, aG

/

2/A /A

1( cos(70 ) sin(70 ) )

2

(0.17101 ) (0.46985 )

0 ( ) (0.17101 0.46985 ) 0

0.46985 0.17101

G A

G A AB G AB G

AB

AB AB

r L L

L L

L L

L L

i j

i j

a a r r

k i j

i j

• Realize that you get two equations from the kinematic relationship

0.46985 0.17101 x AB y ABa L a L

f xF ma A yN mg ma

• Substitute into the sum of forces equations

( )0.46985 f ABF m L (0.17101 )A ABN m L g



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2 - 121

• Substitute the Ff and NA into the sum of moments equation

• Masses cancel out, solve for AB

• Subbing into NA and Ff expressions,

( )0.46985 0.513g

f LF m L

212 2 12

( cos(70 )) ( sin(70 ))L LA F ABN F mL

2 2

2112

[ (0.17101 )]( cos(70 )) [ ( )0.46985 ]( sin(70 ))

L LAB AB

AB

m L g m L

mL

0.513AB

g

L k

2 2 2 2 2112 2

0.17101 0.46985 ( cos(70 ))LAB AB ABL L L g

• The negative sign means is

clockwise, which makes sense.

(0.17101 0.513 )g

A LN m L g

0.912AN mg 0.241fF mg



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Concept Question

2 - 122

What would be true if the floor was

smooth and friction was zero?

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

= 70o

NA



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Plane Motion of a Rigid Body: D’Alembert’s Principle

16 - 123

• 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.

• 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.



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Axioms of the Mechanics of Rigid Bodies

16 - 124

• The forces act

at different points on a

rigid body but but

have the same

magnitude, direction,

and line of action.

FF and

• 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.

Tenth Edition - apl100.wdfiles.comapl100.wdfiles.com/local--files/notes/chapter 16 mod.pdf · Plane Motion of Rigid Bodies: Forces and Accelerations . Contents 16 ... The free body

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