tenth edition - eng.sut.ac.theng.sut.ac.th/me/2014/document/engdynamics/16_lecture_ppt.pdf ·...

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. SelfCalifornia Polytechnic State University

CHAPTER

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

16Plane Motion of Rigid Bodies:

Forces and Accelerations

Plane Motion of Rigid

Bodies: Forces and

Accelerations


Vector Mechanics for Engineers: Dynamics

Ten

th

Ed

ition

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



Ten

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Rigid Body Kinetics

16 - 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.

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.



Ten

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Introduction

16 - 4

• 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.

• 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

• 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.



Ten

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Equations of Motion for a Rigid Body

16 - 5

• 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

• 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



Ten

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Angular Momentum of a Rigid Body in Plane Motion

16 - 6

• Consider a rigid slab in

plane motion.

• 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

• 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.



Ten

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

16 - 7

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).

• 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.



Ten

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Axioms of the Mechanics of Rigid Bodies

16 - 8

• 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.



Ten

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Problems Involving the Motion of a Rigid Body

16 - 9

• 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 free-

body-diagram equation for each body and

solving the corresponding equations of

motion simultaneously.



Ten

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Free Body Diagrams and Kinetic Diagrams

16 - 10

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



Ten

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16 - 11

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



Ten

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16 - 12

Draw the FBD and KD for

the bar AB of mass m. A

known force P is applied at

the bottom of the bar.



Ten

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16 - 13

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

GGxma

yma

I

C

B

x

y



Ten

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ition


16 - 14

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.



Ten

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16 - 15

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



Ten

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16 - 16

The ladder AB slides down

the wall as shown. The wall

and floor are both rough.

Draw the FBD and KD for

the ladder.



Ten

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16 - 17

FA

W

NA

FB

NB

xma

yma

I

1. Isolate body

2. Axes

3. Applied forces


5. Dimensions

6. Kinetic diagram

=

x

y

q



Ten

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Sample Problem 16.1

16 - 18

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.

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.



Ten

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Sample Problem 16.1

16 - 19

0

m10 7m

sv x= =

SOLUTION:

• Calculate the acceleration during the skidding stop

by assuming uniform acceleration.

v 2 = v02 + 2a x - x0( )

0 = 10m

s

æ

èç

ö

ø÷

2

+ 2a 7m( ) 2

m7.14

sa = -

• Draw a free-body-diagram equation expressing the

equivalence of the external and inertial terms.

• 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



Ten

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Sample Problem 16.1

16 - 20

0.341A BN W N W= - =

( )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=

• Apply the corresponding scalar equations.

- 1.5m( )W + 3.6m( )NB = 1.2m( )ma

NB =1

3.61.5W +1.2

W

ga

æ

èç

ö

ø÷ =W

3.61.5+1.2

a

g

æ

èç

ö

ø÷

NB = 0.659W

effAA MM



Ten

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ition

Sample Problem 16.2

16 - 21

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.

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.



Ten

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Sample Problem 16.2

16 - 22

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.

efftt FF

30cos

30cos

mg

amW

30cosm/s81.9 2a

2sm50.8a 60o



Ten

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Sample Problem 16.2

16 - 23

2sm50.8a 60o

• 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

effnn FF

2sm81.9kg8619.0

030sin1815.0

030sin

AE

AEAE

DFAE

F

WFF

WFF

TFAE N9.47

N9.471815.0DFF CFDF N70.8



Ten

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ition

Sample Problem 16.3

16 - 24

A pulley of mass 6 kg and having a radius

of gyration of 200 mm. is connected to two

blocks as shown.

Assuming no axle friction, determine the

angular acceleration of the pulley and the

acceleration of each block.

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.



external and effective forces on the

complete pulley plus blocks system.

• Solve the corresponding moment

equation for the pulley angular

acceleration.



Ten

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ition

Sample Problem 16.3

16 - 25

• Relate the acceleration of the blocks to the angular

acceleration of the pulley.

( )0.25m

A Aa r a

a

=

= ( )0.15m

B Ba r a

a

=

=

2

2

2

(6 kg)(0.2m)

0.24 kg-m

I mk=

=

=

note:

SOLUTION:

• Determine the direction of rotation by evaluating the net

moment on the pulley due to the two blocks.

MGå = 5kg( ) 9.81m2

s

æ

èç

ö

ø÷ 0.15m( )

- 2.5kg( ) 9.81m2

s

æ

èç

ö

ø÷(0.25m) =1.227N ×m

rotation is counterclockwise, SMG is positive.



Ten

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Sample Problem 16.3

16 - 26

• Draw the free-body-diagram equation expressing the

equivalence of the external and effective forces on the

complete pulley and blocks system.

( )

( )

2

2

2

0.24 kg m

0.25 m s

0.15 m s

A

B

I

a

a

a

a

= ×

=

=

effGG MM

( )( )( ) ( )( )( ) ( ) ( )

( ) ( )( )( ) ( )( )( )

2

2

m mss

5kg 9.81 0.15m – 2.5kg 9.81

0.25m 0.15m 0.25m

7.3575 6.1312 0.24 5 0.15 0.15 0.252 0.25 0.25

B B A AI m a m aa

a a

= + -

- = + -

• Solve the corresponding moment equation for the pulley

angular acceleration.

22.41rad sa =

( )( )20.15m 2.41rad s

B Ba r a=

=20.362m sBa =

( )( )20.25m 2.41rad s

A Aa r a=

= 20.603m sAa =

Then,



Ten

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Sample Problem 16.4

16 - 27

A cord is wrapped around a

homogeneous disk of mass 15 kg.

The cord is pulled upwards with a

force T = 180 N.

Determine: (a) the acceleration of the

center of the disk, (b) the angular

acceleration of the disk, and (c) the

acceleration of the cord.

SOLUTION:


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.

• Determine the acceleration of the cord by

evaluating the tangential acceleration of

the point A on the disk.



Ten

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Sample Problem 16.4

16 - 28

SOLUTION:


equivalence of the external and effective forces on the

disk.

effyy FF

kg15

sm81.9kg15-N180 2

m

WTa

amWT

y

y

2sm19.2ya

effGG MM

m5.0kg15

N18022

2

21

mr

T

mrITr

2srad0.48

effxx FF

xam0 0xa

• Solve the three scalar equilibrium equations.



Ten

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Sample Problem 16.4

16 - 29

2sm19.2ya

2srad0.48

0xa

• Determine the acceleration of the cord by evaluating the

tangential acceleration of the point A on the disk.

22 srad48m5.0sm19.2

tGAtAcord aaaa

2sm2.26corda



Ten

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Sample Problem 16.5

16 - 30

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.

SOLUTION:



external and effective forces on the

sphere.


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.



Ten

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Sample Problem 16.5

16 - 31

SOLUTION:


equivalence of external and effective forces on the

sphere.

• Solve the three scalar equilibrium equations.

effyy FF

0WN mgWN

effxx FF

mg

amF

km ga km

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.



Ten

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Sample Problem 16.5

16 - 32

ga km

r

gkm

2

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.

tgvtavv km 00

tr

gt k

m

2

500

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



Ten

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Group Problem Solving

16 - 33

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

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.



Ten

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16 - 34

SOLUTION:

• Given: rear wheel drive,

dimensions as shown, m= 0.80

• Find: Maximum acceleration

• Draw your FBD and KD

NFNR

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



Ten

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16 - 35

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

(1) (2) (3)

(4)

Solving this equation,

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

0R FN N mg R xF maR RF Nm

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



Ten

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ition


16 - 36

• 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



Ten

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ition

Concept Question

16 - 37

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



Ten

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Kinetics: Constrained Plane Motion

16 - 38

The forces at the bottom of the

pendulum depend on the

pendulum mass and mass moment

of inertia, as well as the pendulum

kinematics.

The forces one the wind turbine

blades are also dependent on

mass, mass moment of inertia, and

kinematics.



Ten

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Constrained Plane Motion

16 - 39

• 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

- application of d’Alembert’s principle yields

P, NA, and NB.

. and yx aa



Ten

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Constrained Motion: Noncentroidal Rotation

16 - 40

• 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

• 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



Ten

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Constrained Plane Motion: Rolling Motion

16 - 41

• 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:

NF km ra, independent

• For the geometric center of an

unbalanced disk,

raO

The acceleration of the mass center,

nOGtOGO

OGOG

aaa

aaa



Ten

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Sample Problem 16.6

16 - 42

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.

kg 3

mm 85

kg 4

OB

E

E

m

k

m

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 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.



Ten

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Sample Problem 16.6

16 - 43

rad/s 8

2srad40

kg 3

mm 85

kg 4

OB

E

E

m

k

m

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.

N4.29sm81.9kg3

N2.39sm81.9kg4

2

2

OB

E

W

W

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



Ten

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Sample Problem 16.6

16 - 44

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

effxx FF

N0.24tOBOBx amR

N0.24xR

effyy FF

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

y

OBOBOBEy

R

amWWFR

N0.24yR



Ten

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Sample Problem 16.8

16 - 45

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.

SOLUTION:

• Draw the free-body-equation for the

sphere, expressing the equivalence of the


• 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.



Ten

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Sample Problem 16.8

16 - 46

SOLUTION:

• Draw the free-body-equation for the sphere, expressing

the equivalence of the external and effective forces.

ra

• 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.

effCC MM

q

2

2

52

5

2

sin

rg

Wrr

g

W

mrrmr

IramrW

r

g

7

sin5 q

a = ra =5gsin30°

7

=5 9.31m s2( )sin30°

7

23.504m sa =



Ten

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ition

Sample Problem 16.8

16 - 47

• 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

• 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



Ten

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Sample Problem 16.8

16 - 48

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= + -

= +

v = 4.59m s

effGG MM 00 I

• Assuming no friction, calculate the linear acceleration

and the corresponding velocity after 3 m.

effxx FF W sinq =ma =

W

g

æ

èç

ö

ø÷a

a = 9.81m s2( )sin30° = 4.905m s2

( )

( )( )

2 20 0

2

2

0 2 9.81m s 3m

v v a x x= + -

= +v = 5.42m s



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Sample Problem 16.9

16 - 49

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.

SOLUTION:

• Draw the free-body-equation for the

wheel, expressing the equivalence of the


• 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 - 50

SOLUTION:

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

Assume rolling without slipping,

m100.0

ra

2

22

mkg245.0

m70.0kg50

kmI

• 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

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 - 51

N3.146F N5.490N

Without slipping,

• Compare the required tangential reaction to the

maximum possible friction force.

N1.98N5.49020.0max NF sm

F > Fmax , rolling without slipping is impossible.

• Calculate the friction force with slipping and solve the

scalar equations for linear and angular accelerations.

N6.73N5.49015.0 NFF kk m

effGG MM

2

2

srad94.18

mkg245.0

m060.0.0N200m100.0N6.73

2srad94.18

effxx FF

akg50N6.73N200 2sm53.2a



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16 - 52

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.

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,




equations for the angular acceleration and

the reactions at A and B.



Ten

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16 - 53

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 - 54

• Draw the free-body-equation for the rod, expressing

the equivalence of the external and effective forces.

( )

( )

( )

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

= =

= ×

=

= =

= - = -

• Solve the three corresponding scalar equations for the

angular acceleration and the reactions at A and B.

( )( )( ) ( )( ) ( )( )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 =

effxx FF

RB sin45° = 33.5( ) 2.33( )

RB =110.4N RB =110.4N 45o

effyy FF

RA + 110.4( )cos45°- 25( ) 9.81( ) = - 13.0( ) 2.33( )136.6 NAR =



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16 - 55

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.



Ten

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16 - 56

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 ma

f 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)



Ten

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16 - 57

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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16 - 58

• 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

16 - 59

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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