Solution manual vector mechanics engineers dynamics 8th beer chapter 17

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P.. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P.. Vector Mechanics for Engineers: Statics

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Kinetic energy Position 1 ω1 =3600 rpm =120 rad/sπ

8690 rev

Rotation angle θ θ2− =1 560 rev 1120 radians= π

Let M f be the couple due to friction

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

U M

13.27 rev

2500

15.528 lb s /ft32.2

W m g

15.528

I k m

2 17.905 100

70686 radians

2 12.5

I M

πω

θ

−

11250 revθ

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Kinetic Energy. Cylinder A: ( ) 2 ( ) 2

A

k

r g

ωθ

πµ

continued

Principle of work and energy for cylinder B

B

k

r g

ωθ

ωθ

πµ

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Moments of inertia

12

6 2 32

10 2

1 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Let v = speed of block A,A v = speed of block B,B ω= angular speed of pulley

Kinematics vA =rAω = 0.250 ω vB = rBω =0.150ω

0.250 0.150

s = rθ = θ s =rθ = θ(a) Cylinder A falls to ground

0.1500.900 m 0.900 0.540 m

0.250

Work of weight A: U1 2→ = m gsA A = (11.5 9.81 0.900)( )( ) =101.534 J

Normal contact force acting on block B: N =m gB =( )(9 9.81) =88.29 N

Friction force on block B: Ff = µkN =(0.25 88.29)( ) = 22.0725 N

Work of friction force: U1 2→ = −F sf B = −(22.0725 0.540)( ) = −11.919 J

2.93 m/s

A =

continued

(b) Block B comes to rest

Principle of work and energy T3+U3 4→ =T4:

40.159 22.0725− sB′ = 0 s′B =1.819 mTotal distance for block B d =sB +s′B:

0.540 1.819

2.36 m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

107.445 306.99 N0.35

f k

FNµ

107.445 306.99 N0.35

f k

FNµ

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Moments of inertia

Gears Aand :B

2

2 5 3 9.7050 10 lb s ft3 232.2 12

Position 2 ωC = 450 rpm =15 rad/s; π ωA =ωB =37.5 rad/sπ

2 1 9.7050 10 37.5 67.348 ft lb2

A

2 1 9.7050 10 37.5 67.348 ft lb2

B

2 1 303.28 10 15 336.741 ft lb2

Rotation of gear A θA =( )(2.5 66.393) =165.98 radians

Principle of work and energy for gear A.( )T1 A +MA Aθ =( )T2 A:

MFr

1.157 lbt

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Position 2 ωC = 450 rpm =15 rad/s; π ωA =ωB =37.5 rad/sπ

2

19.7050 10 37.5 67.348 ft lb2

Principle of work and energy for gear A ( )1 A A ( )2 :

M F r

2.89 lb

t

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Position 1 v = 0, ω =0 T1 = 0Elevation: h =0 V1 = mgh = 0

2112

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Mass center G lies at the midpoint of AC m =3 kg

Velocity of mass center: v G =l BGω =0.225ω

Datum for potential energy:

2

V =mgh+ ke Motion is a rotation about a fixed axis through B

Position 2 θ = ° , 90 Point G is directly above B

10.935+2.34=7.83675+0.14625ω

(a) Angular velocity at θ = ° 90 ω2 =6.10 rad/s !

Position 3 θ =180° Bar is horizontal

10.935+2.34=10.935+0.14625ω

(b) Angular velocity at θ =180 ° ω3 =4.00 rad/s !

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Let m be the mass and r be the radius Let point G be the mass center

Moment of inertia about the mass center 2 2

=5

Datum for potential energy: level line through A

Potential energy: V =mgh=mgrcosβ

2

10(1 cos )7

v

g

Equivalence of external and effective forces

When contact is lost, N = 0 and F t =0

17

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

1 sin 30 cos30

1 sin 30 cos302

g l

2 6.257016 rad/s

0.91416

O

V I

ω =ω − − π − ω =

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

2 6.309246 rad/s

0.91416

O

V I

Chapter 17, Solution 19

Position 1 (Directly above the bar)

Elevation: h1 =1.1 mPotential energy: V1 = mgh1 =( )(73 9.81 1.1)( )= 787.74 J Speeds: ω1= 0, v1 =0 Kinetic energy: T1 = 0

(a) Position 2 (Body at level of bar after rotating 90 ).°

Elevation: h2 = 0 Potential energy: V2 = 0

Speeds: 2 1.1 2 Kinetic energy: 2 1 22 1 2 22

7.6055 rad/s a t 8.3660 m/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

0+787.74 =51.788ω −787.742

3 30.422

Kinematics: a n =( )(1.1 30.422) =33.464 m/s2 From ΣM0 = Σ( )M0 eff and ΣF x =0,

=

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Since the cylinder rolls without slipping, the point of contact with the ground is the instantaneous center

Work: U1 2→ = Ps =( )( )100 1.8 =180 J F f does no work

(a) Principle of work and energy T1+U1 2→ =T2:

2.7778 m/s

G G

v a s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Chapter 17, Solution 23

The point of contact with ground is the instantaneous center

Position 1 Point B is at the top

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Instantaneous centers of rolling disks

Rod AB is in curvilinear translation v A=v B

Let v be the velocities of centers of the disks 2

0+0.62175=0.37125ω −13.2435 2 ( )2

2 53.509 rad/s

ω =

Since the potential energy attains its minimum value, in position 2, the kinetic energy attains its maximum value in that position dT 0

AB = A =

Equivalence of external and effective forces

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Let v be the velocity of the bar (v = v↓), v′ be the velocity of the mass center G of the upper left disk,

(v′=v′ ) and ωω be its angular velocity

For all three arrangements, the magnitudes of mass center velocities are the same for all disks Likewise, the angular speeds are the same for all disks

Moment of inertia of one disk = 1 2

Position 1 Initial at rest position T1 = 0

Position 2 Bar has moved down a distance h All the disks move down a distance h′

Work U1→2 =mgh+4m gh′ ′=5gh+8gh′

Kinematics and kinetic energy for case (a)

The mass center of each disk is not moving

0,

v′ = h′ = 0

v r

Kinematics and kinetic energy for case (b)

The instantaneous center C of a typical disk lies at its point of contact

with the fixed wall

2

v r

ω =

1,2

Kinematics and kinetic energy for case (c)

The mass center of each disk moves with the bar

,

v′ = v h′′ = h The instantaneous center C of a typical disk lies at its point of contact

with the fixed wall

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

21.2

Position 2 Point B is directly below A

Position 2 Point B is directly below A

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Data: 500 mm = 0.500 m,l AB= r A= = =r B r 120 mm = 0.120 m

212

body) lies at point C

°sin 75

1.93185sin 30

4mr ωA ωB

continued

Position 1 β = ° 5 , ωA=ωB= 0

1 1

sin 5 0.08715570

0.120

AB gl r

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Chapter 17, Solution 32

Kinematics Locate the instantaneous center at point I Let

9 in 0.75 ft and 24 in 2 ft

Draw velocity diagram

1 5 2 sin 45 3.5355 ft lb2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

=( )( )2

1

5 2 sin 30 2.5 ft lb2

Conservation of energy T1+V1=T2+V2:

2 2

0+3.5355 = 0.56936ω +2.5 ω2 =1.349 rad/s !

Chapter 17, Solution 33

Kinematics Locate the instantaneous center at point I Let

9 in 0.75 ft and 24 in 2 ft

Draw velocity diagram

Law of Cosines: v G2 = v2A +v G A2/ −2v v A G A/ cosβ

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

=

2

2 0.751

0+5.0981 0.113510= ω +5.03553

2=0.742 rad/s

Position 2 Point B is directly below C

( )2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

+

2 20.320913ω

=Conservation of energy T1+V1 =T2 +V2:

0.403729ω +11.3137 = 0.320913ω +4.00 (1) Angular speed data: ω1 =60 rpm = 2 rad/sπ

Solving Equation (1) for ω2, ω2 =8.5121 rad/s

2 =81.3 rpm

Position 2 Point B is directly above C

( )2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

+

2 20.320913ω

=Conservation of energy T1+V1 =T2 +V2:

0.403729ω +11.3137 = 0.320913ω +20.0 (1) Angular speed data: ω2 =ω1

Then, 0.082816ω12 =8.6863 ω1 =10.241 rad/s

1 =97.8 rpm

angular velocity of rod BE

0.1002

Potential energy Datum is a level line through points D and E

( )(1.5 9.81 0.100sin)( ) ( )(1.5 9.81 0.100 sin)( ) ( )(5 9.81 0.200sin)( 0.120)

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

B AB AB

v l

(0.075 6)( ) 0.45 m/s2

Position 2 Rod AB has rotated 90 °

Kinematics Rod AB rotates about point A

0.150

AB AB

l

0.52

B B

v

v v

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

A

=      = × ⋅ ⋅

Potential energy Datum = level line through C

Point H is the instantaneous center of wheel A

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

 Choose the larger value for minωAB

minωAB =121.38 rad/s

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Data: Moment of inertia:

2

2

400 14

16.9082 lb s ft32.2 12

  Angular velocity: ω =2400 rpm=80 rad/sπ

Chapter 17, Solution 44

0≤ ≤t 28s, 300 12.5M = − =287.5 N m⋅

At 28s,t = 3000 rpm = 100 rad/sω = π

Principle of impulse and momentum

Moments about axle: I( )0 +Mt =Iω

(287.5 28)( ) 2

25.624 kg m100

Mt I

Principle of impulse and momentum

Moments about axle: Iω1+M t( 2 −28)= 0

M

πω

−

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

Moments about axle: 0+Mt =Iω

0+ 37.5 lb ft 0.10 s⋅ = 1.0565 lb s ft⋅ ⋅ ω

3.55 rad/s

πω

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

v r

g

µ

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P Beer, E Russell Johnston, Jr.,

Elliot R Eisenberg, William E Clausen, David Mazurek, Phillip J Cornwell

© 2007 The McGraw-Hill Companies

2

v r