Solution manual vector mechanics engineers dynamics 8th beer chapter 18

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

Chapter 18, Solution 1

Total length of rod l =2a+2b=1.8 m

1.8

m l

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 18, Solution 3

23.6

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

Pr incipal moments of inertia:

9.7

θ = ° !

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

θ = ° !

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 18, Solution 9

2

601.86335 lb s /ft, 2.4 in 0.2 ft, 10 in 0.83333 ft

G x x

x

H I

0.013910 rad/s1.29400

G y y

y

H I

ωω

(a) Rate of spin

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

Note that u must be either perpendicular to λ or equal to zero But, if u is perpendicular to

λ, × uλ cannot be equal to zero

/

rG A is parallel to λ and point G lies on the fixed axis

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

( ) 12.5664cos

13.9085

θ = A z =

A H

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

( ) 12.5664cos

13.9085

θ = − A z = −

A H

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 18, Solution 17

210

0.31056 lb s /ft32.2

W m g

Chapter 18, Solution 18

210

0.31056 lb s /ft32.2

W m g

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 18, Solution 19

Since the body lies in the xy plane, I yz= 0

( )I mass=ρ( )I area.For the pairs of area elements shown,

Angular velocity ωx=20 rad/s, ωy=ωz= 0

Angular momentum about G

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

Impulse-momentum principle: Before impact, v =0, HG = 0

Solving, ωx = −3.6087 rad/s, ωy = −3.1952 rad/s, ωz = 9.5855 rad/s

(3.61 rad/s) (3.20 rad/s) (9.59 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

Impulse-momentum principle: Before impact, v =0, HG = 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

Linear momentum components

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

Linear momentum components

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

Constraint of the supporting cable: v y = 0

Impulse-momentum principle: Before impact, v =0, HG =0

(a) Linear momentum: F( )∆ + ∆ =t T tj mv Resolve into components

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

( )H G x = I x xω −I xy yω −I xz zω =0.35ma2ωx +0.3ma2ωy −0.2ma2ωz

( )H G y = −I xy xω +I y yω −I yz zω = 0.3ma2ωx +0.66667ma2ωy +0.2ma2ωz

( )H G z = −I xz xω −I yz zω +I z zω = −0.2ma2ωx +0.1ma2ωy +0.75ma2ωz

Constraint of the supporting cable: v y = 0

Impulse-momentum principle: Before impact, v =0, HG =0

(a) Linear momentum: F( )∆ + ∆ =t T tj mv Resolve in components

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

about the origin for a system of particles consisting of the probe plus the meteorite:

Initial linear momentum of the space probe, (kg m/s :⋅ ) m′ ′ =v0 0

Final linear momentum of the space probe, (kg 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

Conservation of linear momentum of the probe plus the meteorite, (kg 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

Substitute values for , , and I I x y I and resolve into components z

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

Substitute values for , , and I I x y I and resolve into components z

2ma ωz

=

k ωz = 0Solving (1), (2) and (3) simultaneously for ω ωx, y and B t∆ ,

∆ = −

Positions of jets A and B relative to mass center

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

Solving (1) and (2) simultaneously,

F t t

F t t

Chapter 18, Solution 34

Data from Problem 18.33

Mass of satellite:m=1200 kg

Initial angular velocity: ω0=(0.050 rad/s) (i + 0.075 rad/s)j

Principal radii of gyration:

Jet thrust =50 N parallel to the y axis

Principal moments of inertia

Positions of jet B relative to mass center rB G/ =(1.2 m) (i − 1.6 m)j

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 (F t∆)B is negative, the 50-N thrust of jet B acts in the negative y direction

(a) Operating time of jet B

F t t

1:

3m c a ωy

2 0

1:

amv = ma ω −mav

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

After using (2) to eliminate ,v (3) and (4) become y 4 0

3cωx +aωz = v

0

43

1:

2 0

1:

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

After using (2) to eliminate v y, (3) and (4) become 4 0

3cωx +aωz = v

04

But T >0, 0, 0H0 > ω >

cosθ >0 θ <90°

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) Each particle of mass ( )∆m idescribes a circle of radius ρi

The speed of the particle is v i = ρ ωi

Its kinetic energy is ( ) 1( ) 2 1( ) 2 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

T = ma ω "

Chapter 18, Solution 41

23.6

= 0.1118 lb s /ft 24 in 2 ft32.2

(1200 2)( )

125.664 rad/s60

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

T = mr ω "

T = ma 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 20.203

T = maω "

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 18, Solution 48

2

100.31056 lb s /ft32.2

W m g

6.47 ft lb

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

F t T

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

Let HA be the angular momentum of the probe and m′ be its mass Conservation of angular momentum

about the origin for a system of particles consisting of the probe plus the meteorite:

Its moment about the origin, ( 2 )

kg m /s :⋅

A× m = − v y − v z − v x + v y

Initial linear momentum of the space probe, (kg m/s :⋅ ) m′ ′ =v0 0

Final linear momentum of the space probe, (kg 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

2ma ωz

=

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

Solving (1), (2) and (3) simultaneously for ω ωx, y and B t∆

0

1,6

1:

2 0

1:

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

After using (2) to eliminate v y,(3) and (4) become

04

3cωx +aωz =v

04

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 = maω

H& i!

Chapter 18, Solution 57

23.6

0.1118 lb s /ft 24 in 2 ft32.2

(1200)( )2

125.664 rad/s60

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

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sin cos

2 21

sin cos4

G = mr ω β β

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 18, Solution 63

Area of sheet metal: 1 2 2 2 1 2 3 2 0.0432 m2

For calculation ofI xz,use pairs of elements dA and 1 dA2: dA2 = dA 1.

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

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2 3

Chapter 18, Solution 64

34.722 kg/m mass per unit area

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

For calculation ofI xz,use pairs of elements dA1 and dA2: dA2 =dA1

( ) ( )

2 3

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

y

mr A

2 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

12

0.0020940 lb s ft2.76106

Chapter 18, Solution 67

The mass center G lies on fixed axis of rotation so that a=0