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COSMOS: Complete Online Solutions Manual Organization System 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 19, Solution 1. 18 in. 1.5 ft = sin 1.5sin m n n x x t t ω ω = = 1.5 cos , 1.5 6 ft/s n n n x t ω ω ω = = & 6 4 4 rad/s Hz 1.5 2 n ω π = = = 2 1.5 sin 24sin n n n x t t ω ω ω =− =− && 2 Max Acc. 24 ft/s = ! 1 Period 1.571 s 4 2 2 π π = = = !
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COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 1.

18 in. = 1.5 ftx = xm sin nt = 1.5sin nt& x = 1.5 n cos nt ,

1.5 n = 6 ft/s

n =

6 4 = 4 rad/s = Hz 1.5 2

2 && = 1.5 n sin nt = 24sin nt x

Max Acc. = 24 ft/s 2 !

Period =

1 = = 1.571 s ! 4 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 2.

Eq. 19.15: Given data

vm = xm n vm = 0.2 m/s

2 am = xm n

am = 4 m/s 2

vm = xm n :2 am = xm n :

0.2 m/s = xm m2 4 m/s 2 = xm m

(1) (2)

Divide Equ. (2) by Equ. (1): Eq. (1):

n =

4 m/s 2 = 20 rad/s 0.2 m/s

0.2 m/s = xm ( 20 rad/s )xm = 0.01 m xm = 10 mm !f n = 3.18 Hz !

Frequency

fn =

n 20 rad/s = 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 3.

x = xm sin nt , x = xm sin 12 t

n = 6

cycle 2 rad s cycle

& x = 12 xm cos12 t && = 144 2 xm sin12 t x

12 xm = 4 ft/sxm =

4 = 0.1061 ft 12xm = 1.273 in. !Max Acc. = 144 2 ( 0.1061) = 150.8 ft/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 4.

Simple Harmonic Motion

s =

W 20 lb = = 0.2222 in. k 90 lb/in.

n =

k = m

( 90 )(12 ) 20 32.2

= 41.699 rad/s = 2 f

(a)

Amplitude = s = xm = 0.222 in.xm = 0.222 in. ! f = 41.699 rad/s = 6.6366 2 f = 6.64 Hz !

(b)

vm = n xm = ( 41.699 rad/s )( 0.2222 in.) = 9.2655 in./s vm = 9.27 in./s !2 am = n xm = ( 41.699 rad/s ) ( 0.2222 in.) = 386.36 in./s 2 2

= 32.197 ft/s 2 am = 32.2 ft/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 5.

Simple Harmonic Motion (a)x = xm sin ( nt + )

n =

k = m

( 70 lb ) ( 32.2 lb/s2 )

9000 lb/ft

n =

2 = 0.90765 s n

n = 0.0977 s !fn = 1

n

= 10.240 Hz f n = 10.24 Hz !

(b) At

& t = 0: x0 = 0, x0 = v0 = 10 ft/sx0 = 0 = xm sin ( n ( 0 ) + ) = 0

& x0 = v0 = xm n cos ( n ( 0 ) + ) = xm n

Substituting or

10 ft/s = xm 64.343 rad/s xm = 0.1554 ft = 1.865 in. xm = 1.865 in. !

2 am = xm n = ( 0.15542 ft )( 64.343 rad/s )

2

= 643.4 ft/s 2 am = 643 ft/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 6.

In Simple Harmonic Motion (a) Substituting or2 am = xm n2 50 m/s 2 = ( 0.058 m ) n

2 n = 862.07 ( rad/s )

2

n = 29.361 rad/sNow

fn =

n 29.361 rad/s = = 4.6729 Hz 2 21 cycle 1 = Hz (1 min )( 60 s/min ) 60

Then

f in Hz =

So

(

1 60

f Hz 4.6729 Hz = = 280.37 r/min 1 Hz 60

)

and (b)

280 rpm vm = xm n = ( 0.058 m )( 29.361 rad/s ) = 1.7029 m/svm = 1.703 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 7.

Simple Harmonic Motion (a)

= m sin ( nt + ) n =2

n

=

2 (1.35 s )

& = m n cos ( nt + )

&m = m n& vm = l m = l m n

Thus, For a simple pendulum

m =

vm l n

(1)

n =Thus,l = g2 n

g l

=

9.81 m/s 2

( 4.833 rad/s )

2

= 0.420 m

From (1)

m =

vm 0.4 m/s = l n ( 0.42 m )( 4.833 rad/s ) = 0.197 rad11.287

or

m = 11.29 !

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.

COSMOS: Complete Online Solutions Manual Organization System

(b) Now&& at = l && Hence, the maximum tangential acceleration occurs when is

maximum.2 && = m n sin ( nt + ) 2 && m = m n

( at )mor

2 = l m n 2

( at )m

= ( 0.42 m )( 0.197 rad )( 4.833 rad/s ) = 1.9326 m/s 2

( at )m

= 1.933 m/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 8.

Simple Harmonic Motion:

n =

k = m

60 lb/ft = 6.2161 rad/s 50 lb 32.2 ft/s 2

fn =

n 6.2161 = Hz 2 2xm = 0.2 ft ! f n = 0.989 Hz !

(a)

xm = 2.4 in. = 0.2 ft

2 am = xm n = ( 0.2 ft )( 6.2161 rad/s )

2

= 7.728 ft/s 2

(b) or

F f = mam = s mg

s =

am 7.728 ft/s 2 = = 0.240 ! g 32.2 ft/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 9.

k = 6 lb/in. = 72 lb/ft, F = ma :&& kx = mx

vm = 55 in./s,&& + x

W = 4 lb.

k x=0 m

Thus:

2 =

k 72 = = 579.6 m 4 32.2 vm = xm

Eq. (19.15):

55 in./s = xm ( 24.025 rad/s )

xm = 2.2845 in.am = xm 2 = ( 2.2845 in.) 579.6 rad 2 /s 2 am = 1324.1 in./s 2

xm = 2.28 in. !

(

)am = 110.3 ft/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 10.

x = 60 cos (10 t ) + 45sin 10 t 3 = 60 cos (10 t ) + 45 sin10 t cos cos10 t sin 3 3 = 22.5sin10 t + 21.02886 cos10 t

(1)

Nowxm sin(10 t + ) = xm sin10 t cos + xm cos10 t sin

(2)

Comparing (1) and (2) gives22.5 = xm cos , 21.02866 = xm sin

(a)2 xm = (22.5)2 + (21.02866) 2

n =

2 2 = = 0.2 s ! n 10

(b)

xm = 30.8 mm !

(c)

tan =

21.02866 22.5

= 0.7516 rad = 43.1 !

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 11.

At both 600 rpm and 1200 rpm, the maximum acceleration is just equal to g. (a)

= 600 rpm = 62.832 rad/sEq. (19.15):

am = xm 2 9.81 = 2.4849 103 m

xm =

g

( 62.832 )2xm = 2.48 mm

SI:

xm = xm =

( 62.832 )32.2

2

US: (b)

( 62.832 )2

= 0.008156 ft

xm = 0.0979 in.

= 1200 rpm = 125.664 rad/sEq. (19.15):

am = xm 2

xm =

g

(125.664 )2xm = 0.621 mm

SI:

xm =xm =

9.81

(125.664 )32.2

2

= 621.2 106 m = 0.002039 ft

US:

(125.664 )2

xm = 0.0245 in.

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 12.

Simple Harmonic Motion, thusx = xm sin ( nt + )

n =Now Then

k = m

400 N/m = 16.903 rad/s 1.4 kg

x(0) = 0 = xm sin(0 + ) = 0& x(0) = xm n cos(0 + 0)

or Then (a) At

2.5 m/s = ( xm )m(16.903 rad/s) xm = 0.14790 m x = ( 0.14790 m ) sin (16.903 rad/s ) t x = 0.06 m: 0.06 m = (0.14790 m)sin (16.903 rad/s ) t

or

0.06 m sin 1 0.14790 m = 0.02471 s t = 16.903 rad/s t = 0.0247 s !

(b)

Now& x = xm n cos ( nt )2 && = xm n sin ( nt ) x

Then, for t = 0.024713 s& x = ( 0.1479 m )(16.903 rad/s ) cos (16.903 rad/s )( 0.024713 s )

= 2.285 m/s

& x = 2.29 m/s !2

And&& = ( 0.1479 m )(16.903 rad/s ) sin (16.903 rad/s )( 0.024713 s ) x

= 17.143 m/s 2

&& = 17.14 m/s 2 ! x

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 13.

Referring to the figure of Problem 19.12x = xm sin ( nt + )& x = xm n cos ( nt + )2 && = xm n sin ( nt + ) x

Using the data from Problem 19.13:

= 0, xm = 0.1479 m, n = 16.903 rad/s& x x, x, && are

And So, at t = 0.9 s,

x = ( 0.14790 m ) sin (16.903 rad/s ) t

x = ( 0.1479 m ) sin (16.903 rad/s )( 0.9 s ) = 0.0703 m x = 70.3 mm !

& x = ( 0.1479 m )(16.903 rad/s ) cos (16.903 rad/s )( 0.9 s ) = 2.19957 m/s

& x = 2.20 m/s !2

&& = ( 0.1479 m )(16.903 rad/s ) sin (16.903 rad/s )( 0.9 s ) x = 20.083 m/s 2

&& = 20.1 m/s 2 ! x

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 14.

(a)x = xm sin( nt + )

n =

k = m

(32.2 lb/s )2

9000 lb/ft ( 70 lb )

n =With the initial conditions:

2

n

= 0.9765 s& x(0) = 15 in. = 1.25 ft, x(0) = 0

1.25 ft = xm sin ( 0 + )& x(0) = 0 = xm n cos(0 + ) =

2

xm = 1.25 ft

Then

x(t ) = (1.25 ft) sin 64.343t + 2 x(1.5) = (1.25 ft) sin 64.343(1.5 s) + = 0.80137 ft 2 & x(1.5) = (1.25 ft)(64.343) cos 64.343 (1.5 s ) + = 61.726 ft/s 2 In 1.5 s, the block completes1.5 s = 15.361 cycles 0.09765 s/cycle

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.

COSMOS: Complete Online Solutions Manual Organization System

So, in one cycle, the block travels4(1.25 ft) = 5 ft

Fifteen cycles take15(0.09765 s/cycle) = 1.46477 s

Thus, the total distance traveled is15(5 ft) + 1.25 ft + (1.25 0.80137)ft = 77.1 ft Total = 77.1 ft !

(b)

&&(1.5) = (1.25 ft)(64.343 rad/s) 2 sin (64.343 rad/s)(1.5 s) + x 2 = 3317.68 ft/s 2&& = 3320 ft/s 2 ! x

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 15.

m=

10 lb = 0.31056 lb s 2 /ft 2 32.2 ft/s

With the given properties:

n =

k = m

50 lb/ft = 12.6886 rad/s 0.31056 lb s 2 /ft

From free fall of the collar v0 = 2 gh = 2 g(1.5 ft ) = 3g = 9.82853 ft/s

The free-fall time is thus:

t1 =

2y = g

2 (1.5 ft ) g

=

3 = 0.30523 s g

Now to simplify the analysis we measure the displacement from the position of static displacement of the spring, under the weight of the collar: Note that the static deflection is: W 10 lb st = = = 0.2 ft k 50 lb/ft Then mx + kx = 0, where x is measured positively up from the position of static deflection. The solution is: x = xm sin ( nt + ) , with velocity x = xm n cos ( nt + ) Now to determine xm and , impose the conditions at impact and count the time from there. Thus: At impact: t = 0, x = st = 0.2 ft and v = 9.82853 ft/s (down) or0.2 ft = xm sin

9.82853 ft/s = (12.6886 rad/s ) xm cos

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.

COSMOS: Complete Online Solutions Manual Organization System

Solving for xm and xm = 0.800 ft

= 0.25268 radSo, from time of impact, the time of flight is the time necessary for the collar to come to rest on its downward motion. Thus, t2 is the time such thatx ( t2 ) = 0 12.6886t2 + =

2

or

12.6886t2 0.25268 =

2

(a)

Hence, t2 = 0.14371 s Thus, the period of the motion is

= 2 ( t1 + t2 ) = 2 ( 0.30523 s + 0.14371 s )= 0.89788 s

= 0.898 s

(b)

After 0.4 seconds, the velocity is

x ( 0.4 ) = xm n cos n ( 0.4 t1 ) + = (12.6886 rad/s )( 0.8 ft ) cos (12.6886 rad/s )( 0.4 0.30523 s ) 0.25268 rad ] = 5.91 ft/s v ( 0.4 ) = 5.91 ft/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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 16.

= m sin nt , n =

& = m n cos nt, n =& =

g l

9.81 = 2.8592 rad/s, t = 0: 1.2

0.18 = m n 1.2

m = 0.052462 radians

At (a) (b)

t = 1.5 s, = 0.052462 sin ( 2.8592 )(1.5 )

= 0.047826 radians = 2.74 !v = 1.2 ( 0.052462 )( 2.8592 ) cos ( 2.8592 )(1.5 ) v = 74.0 mm/s !

a = 1.2 ( 0.052462 )( 2.8592 ) sin ( 2.8592 )(1.5 )2

a = 469 mm/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 17.

(a)x = xm sin ( nt + ) x0 = xm sin ( 0 + ) = 0.75 ft

& x0 = 0 = xm n cos ( 0 + ) , = xm = 0.75 ft

2

When the collar just leaves the spring, its acceleration is g (downward) and v = 0. Now

& x = ( 0.75 ft ) n cos nt + 2 & x ( 0 ) = v = 0 = ( 0.75 ft ) n cos nt + , nt + = 2 2 2 And 2 a = g = ( 0.75 ft ) n sin nt + 2 or2 g = ( 0.75 ft ) n

n =Then

32.2 ft/s 2 = 6.5524 rad/s 0.75 ft

n =

k 10 lb 2 2 , k = m n = 6.5524 rad/s ) 2( m 32.2 ft/s

= 13.333 lb/ftk = 13.33 lb/ft !

(b)

n = 6.5524 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.

COSMOS: Complete Online Solutions Manual Organization System

At t = 1.6 s:

x = ( 0.75 ft ) sin ( 6.5524 rad/s )(1.6 s ) + = 0.36727 ft 2 x = 0.367 ft above equilibrium ! & v = x = ( 0.75 ft )( 6.5524 rad/s) cos ( 6.5524 rad/s)(1.6 s) + = 4.2848 ft/s 2

v = 4.28 ft/s !

2 a = && = ( 0.75 ft )( 6.5524 rad/s) sin ( 6.5524 rad/s)(1.6 s) + = 15.768 ft/s2 x 2 a = 15.77 ft/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 18.

Determine the constant k of a single spring equivalent to the three springs shown. Springs 1 and 2:

= 1 + 2 ,Hence

and

P P P 1 = 1 + 1 k k1 k2

k =

k1k2 k1 + k2

Where k is the spring constant of a single spring equivalent of springs 1 and 2. Springs k and 3 Deflection in each spring is the same So Now

P = P + P2 , 1

and

P = k , P = k , P2 = k3 1

k = k + k3

k = k + k3 =or

k1k2 + k3 k1 + k2

k =

( 3.5 kN/m )( 2.1 kN/m ) + 2.8 kN/m ( 3.5 kN/m ) + ( 2.1 kN/m )

= 4.11 kN/m = 4.11 103 N/m

(a)

n =

2 = k mfn =

2 4.11 103 N/m 13.6 kg1

= 0.361 s

n

=

1 = 2.77 Hz 0.3614 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.

COSMOS: Complete Online Solutions Manual Organization System

(b) And

x = xm sin ( nt + )xm = 44 mm = 0.044 m

n = 2 f n =And

( ( 2 ) 2.77 Hz ) = 17.384 rad/s

x = ( 0.044 m ) sin (17.4 rad/s ) t + & x = ( 0.044 m )(17.4 rad/s ) cos (17.4 rad/s ) t + Then

vmax = ( 0.044 m )(17.4 rad/s ) = 0.766 m/s&& = ( 0.044 m )(17.4 rad/s ) sin (17.4 rad/s ) t + x 2

Then

amax = ( 0.044 m )(17.4 rad/s ) = 13.3 m/s 2vmax = 0.765 m/s

2

amax = 13.31 m/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 19.

(a) First, calculate the spring constantP = ( 24 kN/m ) + (12 kN/m ) + (12 kN/m ) = ( 48 kN/m )

k = 48 kN/m

Then

n =

k = m

40 103 N/m = 30.984 rad/s 50 kg 2

n =

n

= 0.20279 s 1 = 4.9312 Hz

fn =

n

n = 0.203 s !f n = 4.93 Hz !

(b) Nowx = xm sin ( nt + )

And, since

x0 = 0.060 m x = ( 0.060 m ) sin ( 30.984 rad/s ) t +

& x = ( 0.060 m )( 30.984 rad/s ) cos ( 30.984 rad/s ) t + && = ( 0.060 m )( 30.984 rad/s ) sin ( 30.984 rad/s ) t + x 2

Hence

vmax = 1.859 m/s ! amax = 57.6 m/s 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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 20.

Equivalent spring constantk = 2k + 2k = 4k

(Deflection of each spring is the same.)

( n )1 = 6.8 s =

2 k 10 lb 32.2 ft/s 22

k 2 = 10 6.8 32.22 4k 6 lb 32.2 ft/s 2

k = 0.2651 lb/ft

( n )2

=

=

2 6 lb

4 ( 0.2651 lb/ft ) 32.2 ft/s 2

(

) n = 2.63 s

= 2.633 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.

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Chapter 19, Solution 21.Equivalent Springs Series: Parallel:

ks =

k1k2 k1 + k2

k p = k1 + k2

s =

2

s

=

2 ; ks m

p =

2

p

=

2 kp m

2 kp ( k + k2 )2 k + k2 5 = 1 = 1 s = = p ks k1k2 ( k1k2 ) 2 ( k1 + k2 )

2

( 6.25)( k1k2 ) = k12 + 2k1k2 + k22k1 =

( 4.25) k2 m ( 4.25)2 k22 4k222k1 = 2.125 m 3.516 k2 k1 1 = 4 or k2 4

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.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 19, Solution 22.

For a static load P the total elongation of the spring is:

=

P P P + + k1 k2 k3

1 1 1 = P + + 8 kN/m 12 kN/m 16 kN/m

= Pk= P

13 6 + 4 + 3 = P 48 48 = 48 = 3.6923 kN/m 13 3.6923 103 N/m = 12.153 rad/s 25 kg 2 = 0.517 s 12.153

(a )

= =f =

k = m 2

=

= 0.517 s !f = 1.934 Hz !

12.153 = = 1.9342 Hz 2 2

(b)

For xm = 30 mm = 0.03 mvm = xm = ( 0.03 m )(12.153 rad/s ) am = xm 2 = ( 0.03 m )(12.153 rad/s )2

vm = 0.365 m/s ! am = 4.43 m/s 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.

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Chapter 19, Solution 23.

For equilibrium Equal stretch

A =

12 N = C , k

B =

16 N k

TA = TC = 12 N + kx,

TB = 16 N + kx

&& mx = 4.08 ( 9.81) 2 (12 + kx ) (16 + kx ) && mx + 3 kx = 0

x = ( 0.0125 ) cos

3k t m

(a) Then

TB = 16 + k ( 0.0125 )( 1) = 0 k = 1.280 kN/m !

(b)

2 n =

3 (1280 ) 3k = = 941.76 rad 2 /s 2 m ( 40 9.81)

n = 30.688 rad/s f = 4.88 Hz !(c)TA = 12 + 1280 ( 0.0125 ) cos = Minimum when cos 3k t m

3k t = 1 m

(TA )min

= 12 1280 ( 0.0125 ) = 4 Max Comp: 4 N !

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Chapter 19, Solution 24.

Springs in parallel,

keq = 2k1 + k2

n =After removal

2k1 + 16, 000 2 2 = = = 10 m 0.2 2k1 2 = = 8 m 0.25 2k1 + 16, 000 m 2k1 m 16, 000 , m m = 45.0 kg !

n =Elimination:100 2 = 64 2 =

(a) (b)

100 2 = 64 2 2k1 , 45.0316

64 2 =

k1 = 14, 222 N m k1 = 14.22 kN/m !

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Chapter 19, Solution 25.

The equivalent spring constant for springs in series is:

ke =For k1 and k2

k1k2 k1 + k2

=

2 = ke mA

2 k1k2 mA ( k1 + k2 )2 k1 mA

For k1 alone

=

(a)

k2 = k1 + k2 0.2 s = = 0.16667 0.12 sAnd with k2 = 3.5 kN/m

2

( 3.5 kN/m )(1.6667 )2or (b)

= k1 + 3.5 kN/mk1 = 6.22 kN/m

=

2 k1 mA

mA =

WA g

mA =And with k1 = 6.22 kN/m = 6.22 103 N/m

(1 )2 k1 ( 2 )2= 2.2688 Ns 2 /m = 2.2688 kgmA = 2.27 kg

mA =

( 0.12 s )2 ( 6.22 103 N/m ) ( 2 )2

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Chapter 19, Solution 26.

The equivalent spring constant, accounting for the springs in series, iskeq = k + k 2 = 2 3 k /2

= 0.4 s =

(100 lb ) ( 32.2 ft/s 2 )It follows that k = 510.849 lb/ft

2 keq

(100 lb ) ( 32.2 ft/s2 )k = 511 lb/ft !

After the changes

= 0.4 s =

2 keq 120/32.2

Thus Since

keq = 919.528 lb/ft

keq = k +

kk A k + kA

orkA = k keq k 2k keq

(

) = 510.849 ( 919.528 510.849 )2 ( 510.849 ) 919.528 = 2043.4 lb/ft k A = 2040 lb/ft !

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Chapter 19, Solution 27.

Initially

=

2 = 1.6 s k mA

After the 14 lb weight is added to A,

=

2 = 2.1 s k m A + mB m A + mB mA2

(a)

=

mA + mB 2.1 1.6 = mA

(1.7227 )( mA ) = mA + mB

Note mB =

14 lb 32.2 ft/s 2

14 lb m A = 1.3838mB = 1.3838 2 32.2 ft/s 14 lb WA = mA 32.2 ft/s 2 = 1.3838 = 19.373 lb 2 32.2 ft/s

(

)

WA = 19.37 lb !

(b)

=

2 k mA

k = ( 2 )

2

( mA ) 2 ( )

19.373 lb 2 2 32.2 ft/s k = ( 2 ) (1.6 s )2 k = 9.278 lb/ft k = 9.28 lb/ft !

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Chapter 19, Solution 28.

xs = x

d , h

2F = 2kxs = 2k

d x h

&& M A = ( M A )eff : 2Fd mgx = ( mx ) h

d && 2k x d mg x = mxh h 2kd 2 g && + x = 0 x 2 h mh

2 = Data: (a) For

2kd 22

mh

g , h

2 =

2k d g m h h

2

(1)

d = 0.3 m; h = 0.7 m; m = 20 kg

= 1 s:4 2 =2

=2k 0.3 9.81 20 0.7 0.7

2

;

1s =

2

;

2 = 4 2

Eq. (1):

k = 2912.4 N/m(b) For

k = 2.91 kN/m

= Infinite: 2 = 0 =2

=

2

;

=0

Eq. (1):

2k 0.3 9.81 20 0.7 0.7k = 763.0 N/m k = 763 N/m

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Chapter 19, Solution 29.

(a) The equation of motion is: && M A : ( 0.75 )( 0.75k )( 2 ) = (1.2 m ) 1.2

(

)

&& 1.125k = 1.44m && 1.44m + 1.125k = 0&& + 1.125 1.35 103 1.44m

(

) = 0

&& +Then2 n =

1054.69 =0 m1054.69 2 , and n = = 4s n m2

2 = 4

1054.69 1054.69 2 = or m m 4 m = 427.45 kg

m = 427 kg(b)

= m sin ( nt + )At t = 0

=

0.06m = 0.03636 rad 1.65m

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And Then

&& =00.03636 = m sin

0 = m n cos ( nt + )Thus Now

=

2

, m = 0.0363 rad

=

0.06m = 0.03636 rad 1.65m

& = m n cos ( nt + ) n =Then

1054.69 = 1.5708 rad/s 427.450.06m = 0.03636 rad 1.65m

=

& = ( 0.03636 rad )(1.5708 rad/s ) cos (1.5708 rad/s ) t + &max = ( 0.03636 rad )(1.5708 rad/s ) = 0.05712 rad/s

2

( vc )max

= (1.2m )( 0.05712 rad/s ) = 0.06854 m/s

( vc )max

= 68.5 mm/s

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Chapter 19, Solution 30.

(a)

6 2 2 2 3 4 48EI 48 30 10 lb/in. 144 in. /ft 2 10 ft k= = 3 = L (15 ft )3

P

(

)(

)(

)k = 1.229 105 lb/ft

= 122880 lb/ft(b)

n =

k = mor

(1500 lb 32.2 ft/s )2

122880 lb/ft

fn =

1 2

k 51.36 = n = = 8.17 Hz m 2 2

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Chapter 19, Solution 31.

(a)

P = ke , =

PL AE ,P= AE L

( 0.32 in.)2 29 106 psi 4 AE ke = = = 129.57 103 lb/in. L 18 in.

ke = 129.6 103 lb/in. !

(129.57 10(b)ke fn = m = 2 2

3

lb/in.2

(32.2 ft/s )

(16 lb )

)= 81.272 Hzf n = 81.3 Hz !

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Chapter 19, Solution 32.

k =

W

stW g stW g W

m=

2 n =

k = m

=

g

stFn =

n 1 = 2 2

g

st

!

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Chapter 19, Solution 33.

(a)

mg ( 0.10 kg ) 9.81 m/s 2 = 0.981 N1 0.981 N 2 F = mg = 4 x0 x0 = 4

(

)

2

= 0.06015 m

x0 = 60.1 mm !

(b) At x0 ,

4 1 1 dF 2 2 = x0 = 2 ( 0.06015 ) dx x0 2 = 8.1549 N/m

dF ke = = 8.1549 N/m dx x0ke Fn = m = 28.1549 N/m 0.10 kg = 1.4372 Hz 2

Fn = 1.437 Hz !

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Chapter 19, Solution 34.

Using the Binomial Theorem we write1 1 sin 2 m sin 2 2 = 1 sin 2 m sin 2 1 2

=1+

1 2 m sin sin 2 + 2 2 1 (1 cos 2 ) , we have 2

Neglecting terms of order higher than 2 and setting sin 2 =

n = 4

l 2 1 2 m 1 0 1 + 2 sin 2 2 (1 cos 2 ) d g l 2 1 2 m 1 2 m 0 1 + 4 sin 2 4 sin 2 cos 2 d g

=4

2 l 1 2 m 1 2 m =4 sin 2 + sin sin g 4 2 8 2 0=4

l 1 2 m + sin + 0 g 2 4 2 2

n = 2

l g

1 2 m 1 + sin ! 4 2

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Chapter 19, Solution 35.

For small oscillations:

0 = 2 = 1.01 0 = (1.01) 2l g

l g

We want

Using the formula of Prob. 19.34, we write

= 0 1 +sin 2

1 2 m sin = 1.01 0 4 2 sin

m2

= 4 (1.01 1) = 0.04;

m2

= 0.2;

m2

= 11.54

m = 23.1 !

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Chapter 19, Solution 36.

Eq. (19.20) Forl = 0.8 m: 2

=

2k l 2 g

l 0.8 m = 2 = 1.7943 s g 9.81 m/s 2

Thus: Using Table 19.1: (a)

=

2k

(1.7943 s )2k

For small oscillations:

m = 0;

=1

= 1.7943 s !(b) For m = 30:2k

= 1.017

= (1.017 )(1.7943 s ) = 1.825 s !(c) For m = 90:2k

= 1.180

= (1.180 )(1.7943 s ) = 2.12 s !

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Chapter 19, Solution 37.

From Equation 19.20:

n = 2 g For

2k

l

m = 60, k = 1.686

( 3 s ) = ( 2 )(1.686 )( 2 )l 3 = = 0.44484 32.2 6.744

l 32.2

l = 0.19788 32.2

Thus

l = 6.3718 ft = 76.462 in. l = 76.5 in. !

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Chapter 19, Solution 38.

FA = k 0.6l + ( st ) A , FB = k 0.4l + ( st ) B (a)

M C = ( M C )eff : 0.6lk 0.6l + ( st ) A + 0.1lmg 0.4lk 0.4l + ( st ) B = I + 0.1lmat Equilibrium position: (1)

=00.6lk ( st ) A + 0.1lmg 0.4lk ( st ) B = 0(2)

Substitute (2) into (1)

I + 0.1lmat + ( 0.6l ) k + ( 0.4l ) k = 0or But So or

2

2

I + 0.1lmat + 0.52l 2k = 0I = ml 2 , 12 at = 0.1l , && = ,

ml 2 2 && + ( 0.1l ) m + 0.52l 2k = 0 12 && 0.09333 + 0.52l 2k = 0 && + 5.5714l 2k = 0

&& + ( 5.5714 )

850 N/m = 0 9 kg

&& + 526.19 s 2

=0

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Thus

n = 526.19 s 2 = 22.939 rad/s,

fn =

n = 3.6508 Hz 2f n = 3.65 Hz

(b)

= m sin ( nt + ) ,

& = m n cos ( nt + )

&m = m n& ( x A )m & = 0.6l m

( )

0.0011 m/s = 0.6 ( 0.6 m ) ( m ) ( 22.939 rad/s )

m = 0.0001332 rad = 0.0076 m = 0.0076

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Chapter 19, Solution 39.

M B = ( M B )eff :mg L L L cos kxr = I AB + m + I Disk 2 2 2 where from statics mg L = k st r 2

(1)

where x = r + st

into (1) and assuming small angles ( cos 1)2 L L mg kr 2 kr st = I AB + m + I Disk 2 2

&& mL2 + I Disk + kr 2 = 0 so I AB + 4 2 so n =

kr 2 kr 2 = 1 mL2 mL2 1 I AB + mL2 + + I Disk + M Disk r 2 4 12 4 2 m = 7.5 kg r = 0.25 m M Disk = 6 kg k = 5 kN/m

(2)

Data :

L = 0.9 m

into (2)

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2 n =

12

( 5000 N/m )( 0.25 m ) 1 1 1 ( 7.5 kg ) ( 0.9 m 2 ) + ( 7.5 kg ) ( 0.9 m 2 ) + ( 6 kg ) ( 0.25 m 2 )2

4

2

2 n = 141.243 rad/s 2

n = 11.8846 rad/s(a) (b) so =2

n

=

2 11.8846

= 0.529 s !

xm = 0.020 m vm = xm n = 0.02 (11.8846 ) = 0.2377 m/s vmax = 238 mm/s !

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Chapter 19, Solution 40.

At equilibrium, spring tension = ( 20 lb ) sin15stretch =

( 20 lb ) sin15 = x50 lb/ft

0

M P = 20 sin15

(

4 4 ) 12 50 x0 + x 12 2

=

1 20 4 2 32.2 12

!! 20 4 x !! x 16.667 x = 0.310559!! x 4 + 32.2 12 12

0.310559!! + 16.667 x = 0, n = 7.326 rad/s x (a)(b)x = A sin nt ,

= 0.858 s !! x = A n cos ntA n = 0.5 ( 7.326 ) = 3.66 in./s !

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Chapter 19, Solution 41.

Kinematics& && a A = l 2 cos + l sin + Non-linear terms ml && sin ma A Fx : N = 2 M A = && kl 2 ml 2 l sin cos + Nl sin = ma A sin 2 3 2sin cos = , sin = 0, since

mr 2 2

n = 5.9461 rad/s

=

2

n

= 1.057 s !

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Chapter 19, Solution 77.

Since vibration takes place about the position of equilibrium, we shall neglect the effect of weight and the static deflection.

Position of Max Displacement Spring ElongationsxB = l m 2xC = l m

Position 1:

T1 = 02 V1 = k B xB +

1 1 l 1 2 2 kC xC = k B m + kC ( l m ) 2 2 2 2

2

V1 =

11 2 2 4 k B + kC l m 2 2

Position 2:

V2 = 0

1 2 1 1 1 2 1 l 2 T2 = I m + mvm = ml 2 m + m m 2 2 2 12 2 2 T2 =

1 2 2 ml m 6

Conservation of Energy11 2 2 1 2 2 4 k B + kC l m = 6 ml m 2 11 2 2 1 For simple harmonic motion, m = n m ; k B + kC l 2 m = ml 2 ( n m ) 24 6 T1 + V1 = T2 + V2 : 0 +2 n =

3 1 k B + kC m4

(1)

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Note: Result is independent of length of the rod. DataW = 6 lb,2 n =

k = 3 lb/in. = 36 lb/ft, kC = 5 lb/in. = 60 lb/ft.

3 36 lb/ft + 60 lb/ft = 1110.9 6 lb 4 32.2 ft/s 2 f =

n = 33.33 rad/s

n 33.33 = 2 2

f = 5.30 Hz !

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Chapter 19, Solution 78.

1 l l 2 V = k + mg 2 2 4

2

T =

& & 1 L2 2 1 mr 2 L2 2 m + 2 2 4 2 2 4r & 3mL2 2 16 kL2 mgL + 2 4 n = 8 3mL2 16 = 2 k 2g m + L 3 fn = 1 2 2k 4g + Hz ! 3m 3L

=

Then

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Chapter 19, Solution 79.

(a)

Position 1

T1 = 0 Position 2

V1 =

1 2 k ( st + r m ) 2

T2 =

1 !2 1 2 J m + mvm 2 2

V2 = mgh +

1 2 k ( st ) 2

T1 + V1 = T2 + V2

0+

1 1 !2 1 1 2 2 2 k ( st + r m ) = J m + mvm + mgh + k ( st ) 2 2 2 2 (1)

2 2 2 2 !2 k st + 2k st r m + kv 2 m = J m + mvm + 2mgh + k st

When the disk is in equilibriumM C = 0 = mg sin r k st r

Also Thus

h = r sin m

mgh k st r = 02 2 !2 kr 2 m = J m + mvm

(2)

Substitute (2) into (1)

!m = n m

! vm = r m = r n m

2 2 2 kr 2 m = J + mr 2 m n

(

)

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2 n =

kr 2 J + mr 2

J =2 n =

1 2 mr 2=

kr 2

1 2 mr + mr 2 2 2 ( 800 N/m ) 3 7 kg 2

2 k 3m

n =

2

n

=

= 0.71983 s

n = 0.720 s !(b)! vm = r m

!m = m nvm = r m n r m = 0.01 m 2 vm = ( 0.01 m ) = 0.0873 m/s ! 0.720 s

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Chapter 19, Solution 80.

sin 1 cos m = 2sin 2

m2

2 m

22

Position11 2 1 l T1 = 2 J &m + m &m 2 2 2 V1 = 0

Position 2T2 = 0 V2 = W V2 = l 2 l (1 cos m ) + k m 2 2 2 2

2 Wl m kl 2 2 m + 2 2 4

Conservation of EnergyT1 + V1 = T2 + V22 1 1 l 2 &2 wl m kl 2 2 2 2 J &m + m m + 0 = 0 m + 2 2 4 2 2 4

( )

&m = n m

J =

1 W 2 l 12 g

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W W 2 2 2 Wl kl 2 2 + l n m = 2 + 2 m 6g 4g W + 2 n = 2 5 W 12 g = kl k 6 g 2 = + W 5 l l g

6 9.81 m/s 2 120 N/m + 5 0.160 m 0.6 kg

= 166.43 s 2

n = 12.901 rad/sfn =

n = 2.0532 s 1 2f n = 2.05 Hz !

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Chapter 19, Solution 81.

V =

l 2 1 2 k ( l ) + mg 4 2 2

1 1 l T = ml 2 + m 2 12 2 kl 2 mgl + 2 n = 2 2 4 ml 62 n =

& 2

3k 3g + m 2l l = 0.6 m, m = 10 kg

With

k = 1500 N/m,

2 n = 474.525 rad 2 /s 2

=

2

n

= 0.288 s !

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Chapter 19, Solution 82.

Let m be the mass of rod and mC be mass of each collar Then kl 2 2 mgl 2 l 2 V = + + mC g 2 4 2 T = 1 l 2 &2 1 & m + mC l 2 3 2

( )

2

2 n

kl 2 + mgl + mC gl 4 2 = 2 2 ml 2 + mC l 6 2 k + mg + mC g 4l 2l = 2 m + mC 6 2

( 5 kg )( 9.81 m/s2 ) ( 2.5 kg )( 9.81 m/s2 ) + 1500 N/m + 2 4( 0.6 m ) 2( 0.6 m ) 2 n = 5 kg 2.5 k 6 + 2 = 397.62 s 2

n = 19.4838 rad/s =2 = 0.322 s ! n

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Chapter 19, Solution 83.

V1 = 0, V2 = mgl

2

l 2 + 2mg 2 2 2

= mgl 2 T2 = 0, T1 = =

1 ml 2 & 2 1 2 &2 ml + 2 2 2 3 & 5ml 2 2 6

2 & 2 = n 2 : mgl =

5ml 2 2 6g 2 n , n = 6 5l

=

2

n

=

2 = 1.586 s ! 6 ( 9.81 m/s ) 5 ( 0.75 m )

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Chapter 19, Solution 84.

WA = mA g = 6.87 N WC = mC g = 4.91 N WAC = m AC g = 9.81 N

Position 1

T1 = I AC =

1 & m A 0.1 m 2

(

)

2

+

1 & mC 0.16 m 2

(

)

2

+

1 & mAC 0.03 m 2

(

)

2

+

1 &2 I AC m 2

1 2 m AC ( 0.26 ) 12 1 1 2 2 2 2 &2 0.7 ( 0.1) + 0.5 ( 0.16 ) + 1( 0.03) + 12 (1)( 0.26 ) m 2 1 &2 0.02633 kg m 2 m 2V1 = 0

So

T1 =

=

(

)

Position 2T2 = 0, V2 = WA ( 0.1) (1 cos m ) + WC ( 0.16 ) (1 cos m ) + WAC ( 0.03) (1 cos m )

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With

1 cos m = 2sin 2

m2

"

2 m

2

V2 = ( 6.87 )( 0.1) + ( 4.91)( 0.16 ) + ( 9.81)( 0.03) m = 0.3929 m 2 2 T1 + V1 = T2 + V2 :

2

2

1 1 2 2 ( 0.02633)&m + 0 = 0 + ( 0.3929 ) m 2 2 2

&m = n mSo2 n =

0.3929 = 14.922 s 2 0.02633 2

n =

n

=

2 = 1.6266 s 14.922

n = 1.627 s !

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Chapter 19, Solution 85.

Position 1T1 = IG =

1 1 1 1 2m l 2 & 2 2 2 &2 m ( v A )m + m ( vC )m + ( 2 I 0 ) m + m 2 2 2 2 2 4

( )

1 m 2 & l ( v A )m = ( vC )m = l m 12 2

2 2 & 2 ml + ml = 7 ml 2 & T1 = ml 2 m + &m m 24 8 6

l 5 V1 = 2mgl cos mg cos = mgl cos 2 2

Position 2T2 = 0 V2 = mgl cos ( m ) m l g cos ( m ) 2 2 ml cos ( + m ) 2 2

mgl cos ( + m )

5 V2 = mgl [ cos cos m + sin sin m + cos cos m sin sin m ] 4

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5 V2 = mgl cos cos m 2 cos m 1 2 m

2

( small angles )

2 5 V2 = mgl cos 1 m 2 2 T1 + V1 = T2 + V2

7 2 &2 5 5 2 ml m mgl cos = 0 mgl cos 1 m 6 2 2 2

&m = n m7 2 2 5 2 l n m = g cos m 6 42 n =

15 g cos 14 l

2 n

15 32.2 ft/s 2 cos 40 = 14 25 in. 12 in./ft = 12.686 s 2

n = 3.5617 s 1fn =

n = 0.56686 s 1 2f n = 0.567 Hz !

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Chapter 19, Solution 86.

D = Disk R = Rodl AB = 2r

For small oscillations: Position 1:

h = r (1 cos m ) = T1 = 0 V1 = WR h =

1 2 r m 2

1 2 mR gr m 2

Position 2:

V2 = 0 T2 =

1 1 1 2 2 2 I D m + mD ( vD )m + I R m 2 2 2 11 1 1 1 2 2 2 2 2 2 2 mD r m + 2 mD r m + 2 12 mR ( 2r ) m 2 13 1 2 2 2 mD + 3 mR r m 2

= =

Conservation of Energy.T1 + V1 = T2 + V2 :

0+

1 13 1 2 2 mR g m = mD + mR r 2 m 2 22 3

But for simple harmonic motion:

m = n m1 13 1 2 2 mR gr m = mD + mR r 2 ( n m ) 2 2 2 3

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2 n =

mR g 3 1 mD + mR r 2 3 WR g 3 1 WD + WR r 2 3 r =

or

2 n =

(1)

Data:

WR = 3 lb, WD = 5 lb,

4 ft 12

2 n

32.2 = 4 = 34.094 n = 5.839 rad/s 3 1 5 ) + ( 3) ( 2 3 12 3 2

=

n

=

2 5.839

= 1.076 s !

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Chapter 19, Solution 87.

Position 1: Position 2:

T1 = 0, V2 = 0, T2 = T2 =

V1 = T2 =

1 2 kxm 2 1 1 1 2 2 2 m ABvm + 2 I m + mDisk vm 2 2 2 2

1 1 v 2 2 m ABvm + mDisk r 2 m + mDisk vm r 2 2 1 2 ( mAB + 3mDisk ) vm 2 1 2 1 2 kxm = ( m AB + 3mDisk ) vm 2 2

Conservation of EnergyT1 + V1 = T2 + V2:

0+

But for simple harmonic motion,

vm = n xm :

1 2 1 2 kxm = ( m AB + 3mDisk )( n xm ) 2 22 n =

k mAB + 3mDisk m AB = 9 kg,

Note: Result is independent of rmDisk = 6 kg

Data:

k = 5 kN/m,2 n =

5000 N/m = 185.185 9 kg + 3 ( 6 kg )

n = 13.608 rad/sf = 2.17 Hz !

f =

n 13.608 = 2 2

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Chapter 19, Solution 88.

4r V1 = mgh = mg (1 cos ) 3 1 cos V1 = 2mgr2 m

2

2 m 3

T2 =2

1 1 11 2 2 I A A + I B B + mr 2 2 2 2 22

Where and

1 m r mr 2 I A = I B = = 2 8 4 256

A = B = 4 mr 2 mr 2 2 5mr 2 2 T2 = 16 + 4 = 16

V1 = T2 ,2 n =

2 2 m gr m

3

=

2 2 5 m r 2 n m

16 1 32 g fn = 2 15 r

32 g , 15 r

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Chapter 19, Solution 89.

Kinematics: Data

vm = r m = 0.15 m ,Conservation of EnergyT1 + V1 = T2 + V2

v AB = b m = 0.05 m

mAB = 10 kg mD = 4 kg(1)

Where

Position 1, Max. Displacement V1 = mAB g ( r b ) cos m = mAB g ( 0.10 ) cos m = 9.81cos m1 1 1 2 2 2 T2 = 2 mDvm + I D m + mABv AB 2 2 2 1 1 1 1 2 2 2 2 = 2 ( 4 ) ( 0.15 m ) + ( 4 )( 0.15 ) m + (10 ) ( 0.05 m ) 2 2 2 22 = 0.1475 m

T1 = 0

V2 = mAB g ( r b ) = mAB g ( 0.10 ) = 9.81Into (1)2 0 9.81cos m = 0.1475 m 9.812 0.1475 m = 9.81(1 cos m )

But

1 cos m

1 2 m 2(2)

2 2 0.1475 m = 4.905 m

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m = n mInto (2)2 2 2 0.1475 n m = 4.905 m

(

)

So2 n =

4.905 = 33.254 0.1475

n = 5.7666 rad/sf =

n = 0.9178 Hz 2

f n = 0.918 Hz

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Chapter 19, Solution 90.

& x2 1 1 & & T = 2 mx 2 + my 2 = 7m 6 2 2

Equilibrium of BC:M D = 0 = 3l mg l ku 2 2 2

WhereVk = Vg = mgy,

u =

mg , l0 = natural length = l + u 2 3k

2 1 k 2 k ( l 2 x ) l0 = ( 2 x + u ) 2 22 l 3 l + y l = x + 2 2 2 2

where

And

y + 3ly + x lx = 0, y =2 2

(

)

3l + 3l 2 4 x 2 lx 2

(

)

So

1 8 2 1 y = x + x + H.O.T. 2 3 3l 3 V = 2kx 2 + 2kux + constant mg 2k + 4mg 3 3l 7m 6 fn = 1 2 12k 8g + Hz ! 7m 7 3l x 4mgx 2 + 3 3 3l

Then

2 n

=

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Chapter 19, Solution 91.

Position 11 2 &2 1 &2 1 &2 T1 = 2 I D m + 2 mD ( r c ) m + mr r 2 m 2 2 2

For one disk

I D = ( I D ) A mD c 2 =mD ( r c )2

2 r 2 16r 2 1 4r 2 mD r 2 mD = mD = 0.31987mD r 2 3 2 9 2 2 2

4 = mD r 1 = 0.3313mD r 2 3

&2 T1 = ( 0.3199 + 0.3313) mD r 2 + 0.5mr r 2 m &2 = ( 0.6512mD + 0.5mr ) r 2 m

But

mD =

WD W , mr = r g g 1 6 &2 &2 0.65117 ( 6 ) + 0.5 ( 4 ) m = 0.045866 m 12 32.2 2

So

T1 =

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Position 2

2r 2 T2 = 0, V2 = 2mD gc 1 cos m , c = , 1 cos m m ( m " 1) 2 3 2 & 4 4 6 2 4r 2 2 2 V2 = 2mD g m = WD r m = ( 6 ) m = 1.27324 m 3 3 3 2 12 2 &2 T1 + V1 = T2 + V2 : 0.045866 m + 0 = 0 + 1.27324 m

2 &m = n m : n =

1.27324 = 27.747 0.045866 2

n = 5.2677, n =

n

= 1.192 s

n = 1.192 s !

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Chapter 19, Solution 92.

With m =

4 lb , 32.2 ft/s 2

M =

6 lb , 32.2 ft/s 2

k = 20 lb/ft,

40 l = ft 12

2 2 1 l l 2 V = k + mg + Mgl 2 2 2 2 2

And for small :11 1 & & T = ml 2 2 + M l 23 2

( )

2

& 2 11 2 l + Mr 2 2 r

So

kl 2 l l + mg + Mg 2 4 2 = 23.05935 n = 8 l2 l2 m + 3M 6 4

Then

=

2

n

= 1.308 s !

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Chapter 19, Solution 93.

T2 =Ignoring static terms:

1 1 & ml 2 2 2 12 2 2

V1 =

1 a 1 a k + k 2 2 2 2

= ka 2T2 = V1 : a2 2 1 2 2 ml 2 n m = k m 4 24 2 n =

24

6ka 2 ml 2

fn =

a 2 l

6k m

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Chapter 19, Solution 94.

AA = BB =

a m = l m 2

m =

a m 2l

Position 1T1 = 0 V1 = mgyc = mgl (1 cos )

For small angles

1 cos m = 2sin

m2

2 m

2

=

a2 2 m 8l 2

a2 2 V1 = mgl 2 m 8l

Position 2

T2 =

1 &2 1 1 &2 I m = ma 2 m 2 2 12

V2 = 0

&m = n mT1 + V1 = T2 + V2

a2 1 2 2 mgl 2 + 0 + ma 2 n m 8l 24 2 n =

3g lfn =

1 2

3g ! l

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Chapter 19, Solution 95.

n = r (1 cos m ) =

1 2 rm 21 2 mgr m 2

vm = ( r r ) m

Position 1: Position 2:

T1 = 0, V1 = Wh = V2 = 0 T2 =

1 1 1 1 2 2 2 2 2 I m + mvm = mk 2 m + m ( r r ) m 2 2 2 2

Conservation of energyT1 + V1 = T2 + V2

0+

1 1 2 2 2 mgr m = m k 2 + ( r r ) m 2 2

But for simple harmonic motion,

m = n m

1 1 2 2 2 mgr m = m k 2 + ( r r ) ( n m ) 2 2 2 n =

r k + (r r )2 2

g

(1)

For half section of piper =

2r

2 r r = r 1 I 0 = I + mr 2 2r mr 2 = I + m 2

Parallel-Axis Theorem:I 0 = mr 2 ;

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4 I = mr 2 1 2 2r2 n =

4 k 2 = r 2 1 2

Eq. (1):

4 2 r 2 1 2 + r 2 1

2

g

2r2 n =

2 g = g 4 r 2

4 4 4 r 2 1 2 + 1 + 2 2

2 n =

( 2

1 g g g g = = 4 ) r 2 4 r 2 r 2

=

n

= 2

(

2) r ! g

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Chapter 19, Solution 96.

( r sin m ) sin m

2 r m 2 m

r (1 cos m ) r

2

Position 1 maximum deflection

T1 = 0 V1 = Wym = mgr2 m

2

Position 2 ( = 0 )T2 = 1 &2 1 1 &2 I m = ml 2 m 2 2 12

&m = n mT2 = 1 1 2 2 2 ml n m 2 12

T1 + V1 = T2 + V2 0+ 1 1 1 2 2 2 mgr m = ml 2 n m 2 2 12 12 gr l2

2 n =

n =

2

n

= 2

l2 12 gr

n =

l3gr

!

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Chapter 19, Solution 97.

This is not a damped vibration. However, the kinetic energy of the fluid must be included (a) Position 2V2 = T2 = 0 1 2 kxm 2 1 1 r 2 2 msvm + Vvm 2 4 g

Position 1T1 = Tspere + Tfluid =

V1 = 0 T1 + V1 = T2 + V2 1 1 r 2 1 2 2 msvm + Vvm + 0 = 0 + kxm 2 4 g 2 & vm = xm = xm n

1 1 r 2 2 1 2 2 k ms + V xm n = kxm , n = r 2 2 g 2 ms + V g 1 r 1 62.4 lb/ft 3 4 4 in. 2 V = = 0.15032 lb s /ft 2 2g 2 32.2 ft/s 3 12 in./ft 2 n =

40 lb/ft 1 lb + 0.15032 lb s 2 /ft 2 32.2 ft/s

= 220.54 s 2

n = 14.850 s 1 n =2 = 0.4231 s n

n = 0.423 s !(b) The acceleration does not change mass n = 0.423 s !

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Chapter 19, Solution 98.

Eq. (19.33)Pm k xm = 2 f 1 n

n =

k = m

450 N/m = 10.607 s 1 4 kg

Pm 13 N = = 0.28889 m k 450 N/m xm = 0.28889 m f 1 10.607 2 2

(a)

f = 5: xm =

0.28889 m 5 1 10.607

= 0.03714 m

( In Phase )(b)

xm = 37.1 mm !

f = 10: xm =

0.28889 m 10 1 10.607 2

= 0.25984 m

( In Phase )

xm = 260 mm !

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Chapter 19, Solution 99.

Eq. (19.33)Pm k 2 k , n = xm = 2 m 1 f n xm =

(k

Pm m 2 f

)

,

or

k =

Pm xm + m 2 f

(a) In phasek = 9N 0.15 m + ( 4 kg ) 5 s 1

(

)

= 160.0 N/m !

(b) Out of phase

xm = 0.15 m k =9N

0.15 m + ( 4 kg ) 5 s 1

(

)

= 40.0 N/m !

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Chapter 19, Solution 100.

Eq. (19.33)

xm =

Pm k 1

2 f 2 n

Pm = st k

2 n =

k m

st 3 st 2 f 1 2 n1

2 f2 n

1 3

2 f2 n

>

2 3

Also

st < 3 st 2 f 1 2 n1

2 f2 n

1

2 f245.42

In phase

2 f > 10 + 20 1 245.25

2 < f

245.25 2

f < 11.0736 rad/s f < 11.07 rad/s !

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Out of phase

2 f > 10 20 1 245.25

2 > ( 245.25 ) f 2 f > 19.1801 rad/s f > 19.18 rad/s !

3

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Chapter 19, Solution 105.

From Eq. (19.33 )

xm =

2 1 f n

m

(a)

2 n =

k 117 N/m = 2 kg m

2 n = 58.5 s 2

xm =

0.1 m = 0.17463 m 25 1 58.5 xm = 174.6 mm !

2 (b) Since 2 < n , x and are in phase and net spring deflection is f

x and F = k ( x ) Fm = (117 N/m )( 0.17463 m 0.10 m ) Fm = 8.73 N !

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Chapter 19, Solution 106.

xm =

m 2 f 1 2 n

2 n =

k 130 lb/ft = = 261.63 s 2 16 lb m 32.2 ft/s 2

In phase 1 = k m 1 2 f 1 2 n

Fm = k ( xm m )

Fm < 30 lb

So1

1

2 f 2 n

1 0.68241

2 f2 n

< 1 0.68241 = 0.31579

2 2 < 0.31579 n = ( 0.31579 )( 216.63) f

2 < 68.409 s 2 fSo

f < 9.0896 s 1 f < 9.09 rad/s !

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Out of phase6 in. Fm = k ( xm m ) = 130 lb/ft xm = 130 xm + 65 > 30 12 in./ft

There is no value of xm for whichFm < 30 lb when x and are out of phase !

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Chapter 19, Solution 107.

5 && l = kl + 2k m sin f t 2l g

(

)

7.5 && 0.2 32.2 + 30 = 8 12 sin10t = 0.1333sin10t

m =(a) (b)

( 23.29 + 30 )

0.1333

= 0.0198659 radians&& max l = 1.5 ( m ) 2 = 2.98 ft/s 2 ! f

( )

0.2 max R = 3 ( 4 ) 2 (1.5 )( 0.01988 ) = 0.516 lb ! 12

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Chapter 19, Solution 108.

&& M A ; ml 2 =

l kl k m sin f t 2 2

&& ml 2 +

kl 2 kl m = sin f t , 4 22 n =

= m sin f t

So

k ( 3600 N/m ) = 450 s 2 = 4m 4 ( 2 kg )

m

( 3600 N/m )( 0.003 m ) k m 2 ( 2 kg )( 0.4 m ) = 22ml 2 = n f ( 450 225) s 2= 0.03 rad = 1.719

(a) (b)

f m = 0.450 rad/s !l 2 m = 2.70 m/s 2 ! f

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Chapter 19, Solution 109.

F = ma

&& mxc = k ( xc ) &&c + x

=

2

k xc = k = k m sin f t 2 2 m

From Equation (19.31 and 19.33)

( xc )m

=

m2

2 f 1 2 n

Thus

( xc )m

0.010 m g 9.81 m/s 2 2 2 = n = = ( ST )c 0.015 m (18)2 1 654

( xc )m

= 0.009909 m

2 n = 654 s 2

X c = ( xc )m sin f t

&& X c = ( ac )m = ( xc )m 2 f

( ac )m

= ( 0.009909 m ) 18 s 1

(

)

2

( ac )m

= 3.21 m/s 2

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Chapter 19, Solution 110.

2 n =

g l

From Eq. (19.33'):

mxm

xm

=

1 1 n 2

We want

m

< 1 or

mxm

> 1 or 1 >2

mxm

>1

Thus we have

1 > 1 >1 n 2 > >0 n 0> >2 n 2 2

2 Since n =

g : l2

> 2 < 0 (Impossible) n

g l

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Chapter 19, Solution 111.

From Eq. (19.33'):

m

xm

=

1 1 n 2 or

mxm

1 1 1 or < m < 2 2 xm 22

Thus we must have

1 1 1, with ( xm )2 being < 1 and ( xm )1 > 0. However, ( n )1 ( n ) 2

> , we must have ( n )2 ( n )1

For one collar:

( xm )1 =

m

1 n 1

2

j

+ 15 mm =

m

1 n 1

2

(1)

For two collars:

( xm )2 =

m

1 n 2

2

j

+ 18 mm =

m

1 2 n 1

2

(2)

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Dividing (2) by (1), member by member:

1.2 =

1 ; we find = 2 n 1 7 1 2 n 1

1 n 1

2 2

Substituting into (1): For three collars:

m = (15 mm ) 1

1 90 mm = 7 7

90 mm 90 7 = = mm, ( xm )3 = 2 1 4 1 3 1 3 7 n 1

m

( xm )3 = 22.5 mm

2. Assuming ( xm )2 < 0: For two collars, we have:

18 mm =

m

1 2 n 1

2

(3)

Dividing (3) by (1), member by member:

1.2 =

1 n 1

2

1 2 n 12

2

1.2 + 2.4 =1 n 1 n 1

2

2.2 1.1 = = n 1 3.4 1.7

2

1.1 9 Substitute into (1): m = (15 mm ) 1 mm = 1.7 1.7 For three collars:

9 9 mm 1.7 = = , ( xm )3 = 2 1.1 1.6 1 3 1 3 1.7 n 1

m

( xm )3 = 5.63 mm

( out of

phase )

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Points corresponding to the two solutions are indicated below:

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.

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Chapter 19, Solution 117.

(a) One collar: (b) Two collars:

( xm )1 = 9 mm( xm )2 = 3 mm

2 ( n )1 =

k m = 2 n 2 n 1

( n ) 2 = 22 ( n )3 =

k 1 2 = ( n )1 , 2m 22

(c) Three collars:

( xm )3 = unknown

k 1 = , 3m 3 n 1

= 3 n 3 n 1

We also note that the amplitude m of the displacement of the base remains constant.

Referring to Sec. 19.7, Fig. 19.9, we note that, since ( xm )2 < ( xm )1 and > , we must n 2 n 1 have > 1 and ( xm )2 < 0. However, n 2 correspondingly either > 0 or < 0.1. Assuming ( xm )1 > 0: For one collar:

may be either < 1 or > 1, with ( xm )1 being n 1

( xm )1 =

m

1 n 1

2

j

+ 9 mm =

m

1 n 1

2

(1)

For two collars:

( xm )2 =

m

1 n 2

2

j

3 mm =

m

1 2 n 1

2

(2)

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Dividing (1) by (2), member by member:

4 3 = ; we find, = 2 n 1 5 1 n 1From Eq. (1) For three collars:

1 2 n 1

2 2

m = 1.8 mm1.8 mm = 1.286 mm 4 1 3 5

( xm )3 =

m

1 3 n 1

2

=

( xm )3 = 1.286 mm(out of phase) 2. Assuming ( xm )1 < 0: For one collar, we have now:

9 mm =

m

1 n 12

2

(3)

Dividing (3) by (2), member by member:

3=

1 2 n 1 1 n 12

3 3 =1 n 1

2

2 , n 1

2

=2 n 1

2

Substituting into (3): m = ( 9 mm )(1 2 ) = 9 mm For three collars:

( xm )3 =

m

1 3 n 1

2

=

9 mm = 1.800 mm 1 3( 2)

( xm )3 = 1.800 mm

(out of phase)

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.

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Points corresponding to the two solutions are indicated below:

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.

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Chapter 19, Solution 118.

Pm k xm = 2 f 1 n m=2 n =

Pm = mr 2 f

0.8 lb s 2 /ft, 16 ( 32.2 )20, 000 400 32.2

r = 0.5 ft,2 n = 1610 rad/s 2

k = 20, 000 lb/ft

x = xm sin f t ,

&& = 2 xm sin f t x fmr 2 f

acc. amplitude = 2 xm = f

k 2 f 1 2 n

( )

2

mr 2 f 0.4 ft/s 2 =

k = 2 2 f f 1 1 ( 20, 000 ) 1610 1610

( )

2

0.0007764 2 f

( )

2

+) 0.0007764 2 f

( )

2

+ 4.9689 2 8000 = 0 f

2 = 1332.55 f) 0.0007764 2 f

f = 36.5 rad/s

( )

2

4.9689 2 + 8000 = 0 fThere are no real roots, acc. > 0.4 ft/s 2 Answer: f < 36.5 rad/s

f < 349 rpm

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.

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Chapter 19, Solution 119.

2mr 2 fPm = 2mr 2 , f xm = k

1 f n

2

,

2 n =

2k M

With

2kxm = 36 lb =

2mr 2 f 1M 2 f

,

f = 40 rad/s

2k

Solving for r M 2 f 2kxm 1 2k r = 2 2m f

Data :

2kxm = 36 lbk = 3400 lb/in. = 40.8 103 lb/ft M =

600 lb 32.2 ft/s

3.5 oz 16 oz/lb m= = 6.79348 103 lb s 2 /ft 32.2 ft/s 2

f = 40 rad/s 600 lb 1 40 s 32.2 ft/s 2 ( 36 lb ) 1 2 40.8 103 lb/ft

(

)

2

(

)

r=

2 6.79348 103 lb s 2 /ft 40 s 1

(

)(

)

2

= 0.43725 ft r = 5.25 in. !

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Chapter 19, Solution 120.

(1) From Equation (19.33)

Pm k xm = 2 f 1 n Pm k = kxm = k 2 f 1 n

Force transmitted,

( PT )m

Thus Transmissibility =

( PT )mPm

=

1 1 f n 2

!

(2) From Equation (19.33 ) Displacement transmittedxm =

m 1 f n 2

Transmissibility =

m

xm

=

1 1 f n 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.

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Chapter 19, Solution 121.

G describes a circle about the axis AB of radius r + e. Thus Deflection of the shaft is. ThusF = kr

an = ( r + e ) 2 f

F = man kr = m ( r + e ) 2 f2 n =

k m k2 n

m=

k2 n

kr =

( r + e ) 2 f

2 f 2 n r = 2 f 1 2 ne

(a) Resonance occurs when

f = n , i.e., r k = m 580000 N/m = 146.57 rad/s = 1399.6 rpm 27 kg

n =

n = f = 1400 rpm !

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.

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(b)

r =

(150 10

1200 m 1399.6 = 416.28 106 m 2 1200 1 1399.6 6

)

2

r = 0.416 mm !

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.

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Chapter 19, Solution 122.

Total spring constant k = 2 ( 350 lb/ft ) = 700 lb/ft (a)2 n =

k 700 lb/ft = = 45.08 s 2 500 lb m 32.2 ft/s 2

= 15 ft, m = 2 in. = 0.16667 fty = m sin x

, x = vt

v y = m sin t , = v so

and

f = f =2 v

2

=

,2 v 15

y = 0.16667sin f t

where

From Equation (19.33 ) ,

xm =

m 2 f 1 2 n 45.08 s 1,

Resonance:

f =

2 v = n = 15

v = 16.029 ft/s = 10.93 mph !

(b)

5280 2 ( 30 ) 3600 = 18.431 s 1 f = 15xm =0.16667 = 0.0255 ft ! (18.431) 1 45.08

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.

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Chapter 19, Solution 123.

In steady state vibration, block A does not move and therefore remains in its original equilibrium position (a) Block AF = 0 kx = Pm sin f t

(1)

Block B

F = mB &&, x mB && + kx = 0 x2 X = xm sin nt , n = k /mB

From (1)

kxm sin nt = Pm sin f t

n = f = 2 rad/s, kxm = Pmk = 2 s 1 MB

44 lb k = 4 s 2 = 5.4658 lb/ft 2 32.2 ft/s k = 5.47 lb/ft !

(b)

kxm = Pm, xm =

5 lb = 0.91477 ft 5.4658 lb/ft xm = 0.915 ft !

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.

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Chapter 19, Solution 124.

m x= 2 1 f 2 n

sin t f

y = m sin f t z = relative motion m sin t z = x y = m f 2 f 1 2 n 2 f m 2 1 n 1 = zm = m 2 2 f f 1 2 1 2 n n zm2 2 f 750 2 25 n 150 = = = = 1.0417 ! 2 2 24 f 750 1 2 1 n 150

(a)

m

Error = 4.17%

(b)

m

zm

2 f 2 n =1= 2 f 1 2 n2 2 fn = (150 ) = 106.07 Hz 2 2 f n = 106.1 Hz !

1= 2

2 f2 n

ff =

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.

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Chapter 19, Solution 125.

From Problem 19.124

2 f 2 n zm = m 2 f 1 2 nNowam = m 2 , f

or

2 zm n =

m 2 f1

2 f 2 n2 f 2 n

so

2 zm n =

am 1

2 f 480 = = 0.07438 1760 n 2 zm n 1 = = 1.0803 am 1 0.07438

2

Then

Error = 8.03% !

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.

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Chapter 19, Solution 126.

Since c > cc we use equation (19.42), where

1 < 0, 2 < 0x = c1e1t + c2e2t v= dx = c11e1t + c22e2t dt

(1) (2)

(a)

t = 0, x = x0

v= 0 x0 = c1 + c2

From (1) and (2)

0 = c11 + c22

Solving for c1 and c2c1 =

2 x 2 1 0

c2 =

1 x 2 x1 0

Substituting for c1 and c2 in (1)x=

2 1

x2

2e1t 1e2t

For x = 0 When t , we must have

1e2t 2e1t = 0Recall that

2 t = e( 2 1 ) 1

(3)

1 < 0, 2 < 0. Choosing 1 and 2 so that 1 < 2 < 0, we have0 0 for Equation (3) cannot exist since it would require that e raised to a positive power be less than 1, which is impossible. Thus x is never 0. The x t curve for this case is shown !

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(b) t = 0, x = 0, v = v0 Equations (1) and (2), yield0 = c1 + c2v0 = c11 + c22 c1 = v0 v2 , c2 = 2 1 2 1

Solving for c1 and c2 , Substituting into (1)x=

2 1

v0

e2t e1t t

For x = 0,

e2t = e1t

For c > cc , 1 2 ; Thus no solution can exist for t, and x is The x t curve for this motion is as shown

never

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.

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Chapter 19, Solution 127.

Substitute the initial conditions, t = 0, x = x0 , v = v0 in Equations (1) and (2) of Problem 19.126x0 = c1 + c2 v0 = c11 + c22 c2 =

Solving for c1 and c2 ,

c1 = x=

( v0 2 x0 ) ,2 1

( v0 1x0 )2 1

And substituting in (1) For x = 0, t

1 ( v 1x0 ) e2t ( v0 2 x0 ) e1t 2 1 0

( v0 1x0 ) e2te(2 1 )t

= ( v0 2 x0 ) e1t

=

( v0 2 x0 ) ( v0 1x0 )ln

t =

( 2 1 )

1

v0 2 x0 v0 1x0

This defines one value of t only for x = 0, which will exist if the natural log is positive, v 2 x0 i.e., if 0 > 1. Assuming 1 < 2 < 0 v0 1x0 This occurs if v0 < 1x0

!

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Chapter 19, Solution 128.

For light damping, Equation (19.46) At given maximum displacement,x = x0e

c < cc c t 2m

sin ( 0t + )

t = tn , x = xn sin ( 0tn + ) = 1, xn = x0 c tn e 2m

At next maximum displacement,

t = tn + 1, x = xn + 1 sin ( 0tn +1 + ) = 1, xn + 1 = x0e c t n + 1 2m

But

D t n + 1 D t n = 2t n + 1 tn = 2

D

Ratio of successive displacements: ( tn tn + 1 ) + 2 m D xn x e 2m = 0 c = e 2m =e xn +1 tn + 1 x0e 2m c c tn c 2

Thus

ln

xn c = xn +1 m D

(1)2

From Equations (19.45) and (19.41)

D = n

c 1 cc

D

c c = c 1 2m c

2

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Chapter 19, Solution 129.

As in Problem 19.128, for maximum displacements xn and xn + k at tn and tn + k , sin ( 0tn + ) = 1 and sin ( ntn + k + ) = 1.xn = x0e 2c tn m

( )

xn + k = x0ec ( 2m )tn ( c )tn + k x e 2m

2c ( tn + k ) m

( )

Ratio of maximum displacementsxn xn + k = x0e0

= e 2m

c (t t n n+k )

But

Dtn + k Dtn = k ( 2 )xn xn + k =+

tn tn + k = k

2

D(2)

Thus But from Problem 19.128 Equation (1)

c 2k xn c =k ; ln xn + k m D 2m D xn c = xn +1 m D

log decrement = ln

Comparing with Equation (2)log decrement = 1 x ln n Q.E.D. ! k xn + k

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Chapter 19, Solution 130.

Fig 19.11 Eq. (19.46)

x = x0e& (a) Maxima (positive or negative) when x = 0

2c t m

( )

sin ( Dt + )

( c )t c ( 2cm )t & sin ( Dt + ) + x0 De 2m cos ( Dt + ) x = x0 e 2m

Thus zero velocities occur at times when

& x = 0,The time to the first zero velocity, t1, is

or

tan ( Dt + ) =

2m D c

(1)

1 2m D tan c t1 =

D

(2)

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The time to the next zero velocity where the displacement is negative is

1 2m D tan c + t1 =

D

(3)

Subtracting (2) from (3)

t1 t1 =(b) Zero displacements occur when

D D = = Q.E.D D 2 2or at intervals of

sin ( t + ) = 0

Dt + = , 2 , nThus,

( t1 )0 = (D2

)

and

( t1 )0

=

( 2

D

)

Time between 0s = ( t1 )0 ( t1 )0 =

D

=

D D Q.E.D = 2 2

Plot of Equation (1) (c) The first maxima occurs at 1, The first zero occurs at From the above plot or

( Dt1 + )

( D (t1 )0 + ) =

( D (t1 )0 + ) ( Dt1 + ) > 2( t1 )0 t1 >2 D

( t1 )0 t1 >

D4

Q.E.D

Similar proofs can be made for subsequent max and min

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Chapter 19, Solution 131.

(a) From Problem 19.128,

ln

xn = xn +1

c 2 cc c 1 cc 2

For

xn = 1.25 in.

and

xn +1 = 0.75 in. c 2 cc c 1 cc 2

ln

1.25 = 0.5108 = 0.75

c cc 2

2

2 2 + 1 = 1 0.5108

c 1 = 0.006566 = 151.3 + 1 cc

c = 0.0810 cc(b)

cc = 2m

k = 2 km m

36 lb = 2 (175 lb/ft ) = 25.022 lb s/ft 2 32.2 ft/s

c = ( 0.0810 )( 25.022 ) = 2.037 c = 2.04 lb s/ft

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Chapter 19, Solution 132.

n =

k = 8 rad/s, m2

cc = 2m n = 32 N s/m,

c = 0.01875 cc

d = n 1 At t = 0, First peak at

c = 7.99859363 rad/s, cc & x = c1 d = 0.4 m/s,

x = c1e 0.15t sin d t ( c1 = 0.050008791 m

)

t=

1 2 = 0.196384065 4 d 0.15( 0.19638)

x1 = c1e

= 0.04855714 m,

(a)

Log decrement =

2 ( 0.01875 ) 1 ( 0.01875 )2

Log decrement = 0.1178 !

(b)

x1 = 1.2657 x3 x3 = 0.0384 m = 38.4 mm !

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Chapter 19, Solution 133.

(a) A critically damped system regains its equilibrium position in the shortest time.c = cc = 1320 = 2m2

k m2

Then

cc 1320 2 2 k = = = 7790 lb/ft ! 1800 lb m 32.2 ft/s 2 x = ( c1 + c2t ) e nt

(b) For a critically damped system: Equation (19.43):

We take t = 0 at maximum deflection x0 Thus& x ( 0 ) = 0,x ( 0 ) = x0 = ( c1 + 0 ) e0 , x ( 0 ) = x0 so c1 = x0

Using the initial conditions

x = ( x0 + c2t ) e nt

And

& x = n ( x0 + c2t ) e nt + c2e nt & x ( 0 ) = 0 = n x0 + c2 ,

so c2 = n x0

Thus For We obtainx= x0 , 3

x = x0 (1 + nt ) e nt 1 = (1 + nt ) e nt , 3 with n = k m

n =

7790 lb/ft 1800 lb 2 32.2 ft/s

= 11.807 s 1

And, sincet =

nt = 2.2892.289 = 0.19387 11.807

t = 0.1939 s !

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Chapter 19, Solution 134.

From the given data

n =

2k = 150 rad/s, m

cc = 2m n = 58.7878 N s/m

Stretch at equilibrium =

( 2.4 kg ) ( 9.81 m/s2 ) 2 (180 N/m )

= 0.0654 m

Initial stretch =

( 3.3 kg ) ( 9.81 m/s 2 ) 2 (180 N/m ) 7.5

= 0.089925 m2

With

c < cc ,

d = 150 1 = 12.1474 rad/s 58.7878 1 1 2 = 0.25862 s D = 2 2 12.1474 Minimum x = 0.024525e1.5625( 0.25862 )

From Figure 19.11

= 0.001637 mMinimum force in one spring = (180 N/m )( 0.0654 0.01637 )

= 8.825 N 8.82 N !

Vec