plane motion of rigid bodies: forces and accelerations

VECTOR MECHANICS FOR ENGINEERS: DYNAMICSDYNAMICS

Tenth Tenth EditionEdition

Ferdinand P. BeerFerdinand P. Beer

E. Russell Johnston, Jr.E. Russell Johnston, Jr.

Phillip J. CornwellPhillip J. Cornwell

Lecture Notes:Lecture Notes:

Brian P. SelfBrian P. SelfCalifornia Polytechnic State UniversityCalifornia Polytechnic State University

CHAPTER

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

16Plane Motion of Rigid Bodies:

Forces and Accelerations


Vector Mechanics for Engineers: DynamicsVector Mechanics for Engineers: Dynamics

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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



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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.

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 - 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.



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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



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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.



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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.



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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.



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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.



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Free Body Diagrams and Kinetic Diagrams

12 - 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



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12 - 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



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2 - 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.



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2 - 13

P

L/2

L/2r

A

Cx

Cy

mg

1. Isolate body2. Axes

3. Applied forces

4. Replace supports with forces

5. Dimensions6. Kinetic diagram

GGxma

ymaI

C

B

x

y



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2 - 14

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 s= 0.25 and k= 0.20, respectively. Draw the FBD and KD for the wheel.



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2 - 15

xma

yma

I

P

F

W

N x

y

=

1. Isolate body2. Axes

3. Applied forces


5. Dimensions6. Kinetic diagram

4 in

8 in



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2 - 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.



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2 - 17

FA

W

NA

FB

NB

xma

yma

I

1. Isolate body

2. Axes

3. Applied forces


5. Dimensions

6. Kinetic diagram

=

x

y

0.22

5 m

0.22

5 m



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Sample Problem 16.1

16 - 18

At a forward speed of 30 ft/s, the truck brakes were applied, causing the wheels to stop rotating. It was observed that the truck to skidded to a stop in 20 ft.

Determine the magnitude of the normal reaction and the friction force at each wheel as the truck skidded to a stop.

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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Sample Problem 16.1

16 - 19

ft20s

ft300 xv

SOLUTION:

• Calculate the acceleration during the skidding stop by assuming uniform acceleration.

ft202s

ft300

22

020

2

a

xxavv

s

ft5.22a

• 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

699.02.32

5.22

g

a

agWW

NN

amFF

k

k

BAk

BA

effxx FF



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Sample Problem 16.1

16 - 20

WNWN BA 350.0

WNN Arear 350.021

21 WNrear 175.0

WNN Vfront 650.021

21 WN front 325.0

WNF rearkrear 175.0690.0WFrear 122.0

WNF frontkfront 325.0690.0WFfront 227.0.0

• Apply the corresponding scalar equations.

WN

g

aWa

g

WWN

amNW

B

B

B

650.0

4512

4512

1

ft4ft12ft5

effAA MM



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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.



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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.


• 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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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



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Sample Problem 16.3

16 - 24

A pulley weighing 12 lb and having a radius of gyration of 8 in. 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.

• Draw the free-body-diagram equation expressing the equivalence of the external and effective forces on the complete pulley plus blocks system.

• Solve the corresponding moment equation for the pulley angular acceleration.



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Sample Problem 16.3

16 - 25

• Relate the acceleration of the blocks to the angular acceleration of the pulley.

ft1210

AA ra

ft126

BB ra

2

2

2

22

sftlb1656.0

ft12

8

sft32.2

lb12

kg

WkmInote:

SOLUTION:

• Determine the direction of rotation by evaluating the net moment on the pulley due to the two blocks.

lbin10in10lb5in6lb10 GM

rotation is counterclockwise.



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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.

2126

21210

2

sft

sft

sftlb1656.0

B

A

a

a

I

effGG MM

1210

1210

2.325

126

126

2.3210

1210

126

1210

126

1210

126

1656.0510

ftftftlb5ftlb10

AABB amamI

• Solve the corresponding moment equation for the pulley angular acceleration.

2srad374.2

2126 srad2.374ft

BB ra2sft187.1Ba

21210 srad2.374ft

AA ra2sft978.1Aa

Then,



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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:

• 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.

• Determine the acceleration of the cord by evaluating the tangential acceleration of the point A on the disk.



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Sample Problem 16.4

16 - 28

SOLUTION:• Draw the free-body-diagram equation expressing the

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

221

mr

T

mrITr

2srad0.48

effxx FF

xam0 0xa

• Solve the three scalar equilibrium equations.



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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



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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 k.

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:

• Draw the free-body-diagram equation expressing the equivalence of the 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 - 31

SOLUTION:

• Draw the free-body-diagram equation expressing the equivalence of external and effective forces on the sphere.

• Solve the three scalar equilibrium equations.

effyy FF

0 WN mgWN effxx FF

mg

amF

k ga k

2

32 mrrmg

IFr

k

r

gk2

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 - 32

ga k

r

gk2

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 k 00

tr

gt k

2

500

110 2

5t

r

grgtv k

k

g

vt

k0

1 7

2

g

v

r

gt

r

g

k

kk

0

11 7

2

2

5

2

5

r

v01 7

5

r

vrrv 0

11 7

5 075

1 vv

At the instant t1 when the sphere stops sliding,

11 rv



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Group Problem Solving

2 - 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.



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2 - 34

SOLUTION:

• Given: rear wheel drive, dimensions as shown, = 0.80

• Find: Maximum acceleration

• Draw your FBD and KD

NF

FR

NR

mg

x

y

xma

I yma

=

x xF ma y yF maR 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 60 40 2012 12 12( ) ( ) ( ) 0R F RN N F



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2 - 35

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

(1) (2) (3)

(4)

Solving this equation, the masses cancel out and you get:

60 40 2012 12 12( ) ( ) ( ) 0R F RN N F

0R FN N mg R xF ma R RF N

xR

maN

(1)→(3) (5)

xF R

maN mg N mg

(6)

(5)→(2)

(1) and (5) and (6) →(4)

60 40 200

12 12 12x x

x

ma mamg ma

12.3 ft/sxa



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2 - 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 100 40 2012 12 12( ) ( ) 0 ( )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 - 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 accelerationb) The cylinder C will have a greater accelerationc) The cylinder and pipe will have the same acceleration



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Kinetics: Constrained Plane Motion

2 - 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.



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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 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 - 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 eliminatefrom 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 - 41

• For a balanced disk constrained to roll without sliding,

rarx

• Rolling, no sliding:NF s ra

Rolling, sliding impending:NF s ra

Rotating and sliding:NF k ra , independent

• 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 - 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, expressing the equivalence of the 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 - 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

N2.39sm81.9kg42

2

OB

E

W

W

mN156.1

srad40m085.0kg4 222

EEE kmI

N0.24

srad40m200.0kg3 2

rmam OBtOBOB

N4.38

srad8m200.0kg3 22

rmam OBnOBOB

mN600.1

srad40m.4000kg3 221212

121

LmI OBOB



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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

N0.24 tOBOBx amRN0.24xR

effyy FF

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

y

OBOBOBEy

R

amWWFR

N0.24yR



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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 10 ft down a frictionless incline.

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 10 ft of uniformly accelerated motion.

• Assuming no friction, calculate the linear acceleration down the incline and the corresponding velocity after 10 ft.

• Calculate the friction coefficient required for the indicated tangential reaction at C.



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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

2

252

5

2

sin

rg

Wrr

g

W

mrrmr

IramrW

r

g

7

sin5

7

30sinsft2.325

7

30sin5

2

gra

2sft50.11a



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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

2sft50.11 ra

effxx FF

WWF

g

g

W

amFW

143.030sin7

2

7

sin5

sin

effyy FF

WWN

WN

866.030cos

0cos

• Calculate the friction coefficient required for the indicated tangential reaction at C.

W

W

N

F

NF

s

s

866.0

143.0

165.0s



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Sample Problem 16.8

16 - 48

r

g

7

sin5

2sft50.11 ra

• Calculate the velocity after 10 ft of uniformly accelerated motion.

ft10sft50.1120

22

020

2

xxavv

sft17.15v

effGG MM 00 I

• Assuming no friction, calculate the linear acceleration and the corresponding velocity after 10 ft.

effxx FF

22 sft1.1630sinsft2.32

sin

a

ag

WamW

ft10sft1.1620

22

020

2

xxavv

sft94.17v



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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 s = 0.20 and k = 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 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 - 50

SOLUTION:

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

Assume rolling without slipping,

m100.0ra

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

02

mgN

WN

effxx FF

N3.146

sm074.1kg50N200 2

F

amF



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Sample Problem 16.9

16 - 51

N3.146F N5.490NWithout slipping,

• Compare the required tangential reaction to the maximum possible friction force.

N1.98N5.49020.0max NF s

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

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 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.

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, 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 - 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 signs yields

,4ABa

90.446.5 BA aa

The acceleration of G is now obtained from

AGAG aaaa

2 where AGa

Resolving into x and y components,

732.160sin2

46.460cos246.5

y

x

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.

69.2732.12.32

50

93.646.42.32

50

07.2

sftlb07.2

ft4sft32.2

lb50

12

1

2

22

2121

y

x

am

am

I

mlI

• Solve the three corresponding scalar equations for the angular acceleration and the reactions at A and B.

2srad30.2

07.2732.169.246.493.6732.150

effEE MM

2srad30.2

effxx FF

lb5.22

30.293.645sin

B

B

R

R

lb5.22BR

45o

effyy FF

30.269.25045cos5.22 AR

lb9.27AR



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2 - 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.



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2 - 56

SOLUTION: Given: WAB = W, = 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 - 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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2 - 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 0.513 gf LF m L

212 2 12( cos(70 )) ( sin(70 ))L L

A 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 20.17101 0.46985 ( cos(70 ))L

AB AB ABL L L g

• The negative sign means is clockwise, which makes sense.

(0.17101 0.513 )gA LN m L g

0.912AN mg 0.241fF mg



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Concept Question

2 - 59

What would be true if the floor was smooth and friction was zero?

a) The bar would rotate about point Ab) The bar’s center of gravity would go straight downwardsc) The bar would not have any angular acceleration

= 70o

NA

plane motion of rigid bodies: forces and accelerations

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