Solution Manual - Vector Mechanics Engineers Dynamics 8th Beer Chapter 19 PDF

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COSMOS: Complete Online Solutions Manual Organization System Chapter 19, Solution 2. Eq. 19: vm n am n2 Given data vm 0 am 4 2 vm n : 0 m (1) am n2 : 4 2 m2 (2) Divide Equ. (2) Equ. (1): Eq. (1): 4 2 20 0 0 xm ( 20 ) xm 0 m Frequency fn n 20 Vector Mechanics for Engineers: Statics and Dynamics, Ferd...

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Chapter 19, Solution 2.

Eq. 19.15:

v m = xmωn

am = xmωn2

Given data

v m = 0.2 m/s

a m = 4 m/s2

v m = xmω n :

0.2 m/s = xmωm

(1)

am = xmωn2 :

4 m/s 2 = xmωm2

(2)

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

ωn =

Eq. (1):

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

0.2 m/s = x m ( 20 rad/s) x m = 0 .01 m

Frequency

fn =

ωn 20 rad/s = 2π 2π

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.

xm = 10 mm ! fn = 3.18 Hz !

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Chapter 19, Solution 3.

x = xm sin ωnt ,

ωn = 6

cycle 2π rad × s cycle

x = xm sin 12π t

&x = 12π x m cos12π t x&& = −1 44π 2 xm sin12πt

12π x m = 4 ft/s xm =

4 = 0.1061 ft 12π xm = 1.273 in. ! Max Acc. = 144π 2 (0.1061) = 150.8 ft/s2 !

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 4.

Simple Harmonic Motion

δs=

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

ωn =

(a)

k = m

( 90 )(12 )  20   32.2   

= 41.699 rad/s = 2π f

Amplitude = δ s = xm = 0.222 in. x m = 0.222 in. ! f =

41.699 rad/s = 6.6366 2π f = 6.64 Hz !

(b)

v m = ω nxm = ( 41.699 rad/s)( 0.2222 in.) = 9.2655 in./s vm = 9.27 in./s ! 2

am = ω n2 xm = (4 1.699 rad/s)

( 0.2222 in.) =

386.36 in./s2

= 32.197 ft/s 2 2

am = 32.2 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.

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Chapter 19, Solution 5.

Simple Harmonic Motion x = xm sin (ω nt + φ )

(a)

k = m

ωn =

9000 lb/ft

(

( 70 lb) 32.2 lb/s2

)

= 64.343 rad/s

2π = 0.90765 s ωn

τn =

τn = 0.0977 s ! fn =

1

τn

= 10.240 Hz fn = 10.24 Hz !

t = 0: x 0 = 0, x& 0 = v 0 = 10 ft/s

(b) At

x 0 = 0 = x m sin (ω n ( 0) + φ ) ⇒ φ = 0 x& 0 = v0 = xmω n cos ( ω n (0 ) + φ ) = xmωn

Substituting or

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

am = xm ωn2 = ( 0.15542 ft)( 64.343 rad/s)

2

= 643.4 ft/s2 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.

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Chapter 19, Solution 6.

In Simple Harmonic Motion (a)

am = xmω n2

Substituting

50 m/s = ( 0.058 m ) ωn

or

2 ωn2 = 862.07( rad/s)

2

2

ω n = 29.361 rad/s Now

fn =

Then

So

(

ω n 29.361 rad/s = = 4.6729 Hz 2π 2π

f in Hz =

1 cycle 1 = Hz (1 min )(60 s/min ) 60

f Hz

4.6729 Hz

1 60

)

Hz

=

1 60

= 280.37 r/min

280 rpm ⊳

and (b)

vm = xmω n = ( 0.058 m )( 29.361 rad/s) = 1.7029 m/s v m = 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.

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Chapter 19, Solution 7.

Simple Harmonic Motion

θ = θ m sin (ωn t + φ )

(a)

ωn =

2π

τn

=

2π ( 1.35 s )

= 4.833 rad/s

θ& = θm ωn cos ( ωn t + φ )

θ&m = θm ωn v m = lθ&m = lθmωn

Thus,

θm =

vm lωn

ωn =

g l

(1)

For a simple pendulum

Thus, l=

g

ω n2

=

9.81 m/s2

( 4.833 rad/s)2

= 0.420 m

From (1)

θm =

vm 0.4 m/s = l ωn ( 0.42 m)( 4.833 rad/s) = 0 .197 rad

or

11.287°

θ 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.

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(b) Now at = lθ&&

Hence, the maximum tangential acceleration occurs whenθ&& is maximum.

θ&& = −θ mω2n sin ( ωnt + φ ) θ&&m = θ mω n2

( at )m or

( at )m

= lθ mωn2

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

2

= 1.9326 m/s2

( a t ) 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:

ωn =

fn =

k = m

60 lb/ft = 6.2161 rad/s 50 lb 32.2 ft/s2

6.2161 ωn = Hz 2π 2π x m = 0.2 ft !

xm = 2.4 in. = 0.2 ft

(a)

f n = 0.989 Hz ! 2

2 am = xmωn = (0.2 ft )( 6.2161 rad/s )

= 7.728 ft/s 2 F f = mam = µ s mg

(b)

or

µs =

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.

am 7.728 ft/s2 = = 0.240 ! g 32.2 ft/s 2

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Chapter 19, Solution 9.

k = 6 lb/in. = 72 lb/ft,

v m = 55 in./s,

&&x + k x = 0 m

&& − kx = mx

F = ma:

ω 2=

Thus:

Eq. (19.15):

72 k = m  4  32.2 

W = 4 lb.

  

ω = 24.025 rad/s

= 579.6

vm = x mω

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

xm = 2.28 in. !

x m = 2.2845 in.

(

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

)

am = 1324.1 in./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.

am = 110.3 ft/s2 !

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Chapter 19, Solution 10.

π  x = 60 cos (10π t) + 45sin 10π t −  3  π π  = 60 cos (10πt ) + 45 sin10 πt cos − cos10 πt sin  3 3  = 22.5sin10πt + 21.02886 cos10πt

(1)

Now xm sin(10π t + φ ) = xm sin10π tcos φ + xm cos10π tsin φ

(2)

Comparing (1) and (2) gives 22.5 = xm cosφ ,

21.02866 = xm sin φ

(a)

(b)

τn =

2π

ωn

=

2π = 0.2 s ! 10π

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

x m = 30.8 mm !

(c)

tan φ =

21.02866 22.5

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.

φ = 0.7516 rad = 43.1° !

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Chapter 19, Solution 11.

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

ω = 600 rpm = 62.832 rad/s a m = x mω

Eq. (19.15):

(b)

SI:

xm =

US:

xm =

9.81 2

( 62.832 ) 32.2

( 62.832 )2

2

xm =

g

(62.832 )2

= 2.4849 × 10 −3 m

x m = 2.48 mm ⊳

= 0.008156 ft

xm = 0.0979 in. ⊳

ω = 1200 rpm = 125.664 rad/s a m = x mω

Eq. (19.15):

SI:

xm =

US:

xm =

9.81 2

( 125.664) 32.2

( 125.664)2

2

xm =

g

(125.664 )2

= 621.2 × 10 −6 m

xm = 0.621 mm ⊳

= 0.002039 ft

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.

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Chapter 19, Solution 12.

Simple Harmonic Motion, thus x = xm sin (ω nt + φ )

ωn =

k = m

400 N/m = 16.903 rad/s 1.4 kg

x (0) = 0 = xm sin(0 + φ) ⇒ φ = 0

Now Then

x& (0) = xmω n cos(0 + 0) 2.5 m/s = ( xm )m(16.903 rad/s) ⇒ xm = 0.14790 m

or

x = (0 .14790 m ) sin  (16.903 rad/s ) t

Then (a) At

x = 0 .06 m: 0.06 m = (0.14790 m) sin  (16.903 rad/s ) t

 0.06 m  sin −1    0.14790 m  = 0.02471 s t = 16.903 rad/s

or

t = 0.0247 s !

(b)

Now x& = xmωn cos ( ωn t) x&& = −x mω n2 sin (ω nt )

Then, for t = 0.024713 s &x = ( 0.1479 m)( 16.903 rad/s) cos ( 16.903 rad/s)( 0.024713 s)  &x = 2.29 m/s !

= 2.285 m/s

And 2

&&x = − ( 0.1479 m)( 16.903 rad/s) sin ( 16.903 rad/s)( 0.024713 s )

= −17.143 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.

x&& = 17.14 m/s2

!

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Chapter 19, Solution 13.

Referring to the figure of Problem 19.12 x = xm sin (ω nt + φ ) x& = xmωn cos( ωnt + φ ) x&& = − xmωn2 sin (ωn t + φ )

Using the data from Problem 19.13:

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

And

x = ( 0.14790 m) sin ( 16.903 rad/s) t

So, at t = 0.9 s, 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 !

&&x = ( 0.1479 m )(16.903 rad/s)2 sin (16.903 rad/s)( 0.9 s)  = −20.083 m/s2

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.

x&& = 2 0.1 m/s2

!

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Chapter 19, Solution 14.

(a) x = xm sin(ω nt + φ )

k = m

ωn =

9000 lb/ft ( 70 lb )

(32.2 lb/s ) 2

= 64.343 rad/s

τn = With the initial conditions:

2π

ωn

= 0 .9765 s x(0) = 15 in. = 1.25 ft, x& (0) = 0

1.25 ft = xm sin( 0 + φ ) x& (0) = 0 = x mωn cos(0 + φ ) ⇒ φ =

π 2

x m = 1.25 ft

Then

π  x (t ) = (1.25 ft) sin  64.343t +  2  π  x (1.5) = (1.25 ft) sin 64.343(1.5 s) +  = − 0.80137 ft 2  π  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 completes 1.5 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 travels 4(1.25 ft) = 5 ft

Fifteen cycles take 15(0.09765 s/cycle) = 1.46477 s

Thus, the total distance traveled is 15(5 ft) + 1.25 ft + (1.25 − 0.80137)ft = 77.1 ft Total = 77.1 ft !

(b)

π  && x(1.5) = −(1.25 ft)(64.343 rad/s)2 sin (64.343 rad/s)(1.5 s) +  2  = 3317.68 ft/s2

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.

&& x = 3 320 ft/s2 !

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Chapter 19, Solution 15.

m=

10 lb 32.2 ft/s2

= 0.31056 lb⋅ s2 /ft

With the given properties:

ωn =

k = m

50 lb/ft = 12.6886 rad/s 0.31056 lb⋅ s2 /ft

From free fall of the collar v0 = 2 gh =

2 g(1.5 ft ) = 3 g = 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 δ st = = = 0.2 ft 50 lb/ft k  + kx = 0, where x is measured positively up from the Then mx position of static deflection. The solution is: x = xm sin ( ω n t + φ ) , with velocity x = xmωn cos (ωn t + φ ) Now to determine x m and φ , impose the conditions at impact and count the time from there. Thus: At impact: t = 0, x = δ st = 0.2 ft and v = −9.82853 ft/s (down) or

0.2 ft = xm sinφ

−9.82853 ft/s = (12.6886 rad/s) x m cos φ

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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Solving for x m and φ x m = −0.800 ft

φ = − 0.25268 rad So, 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 that x (t2 ) = 0 ⇒ 12.6886t2 + φ = 12.6886t 2 − 0.25268 =

or

(a)

π 2

π 2

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

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

(b)

τ = 0.898 s W

After 0.4 seconds, the velocity is

x( 0.4) = xmωn cos⎡⎣ω n ( 0.4 − t1) + φ ⎤⎦ = ( 12.6886 rad/s)( − 0.8 ft) cos⎡⎣ ( 12.6886 rad/s) ( 0.4− 0.30523 )s − 0.25268 rad ] = − 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.

v( 0.4) = 5.91 ft/s

W

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Chapter 19, Solution 16.

θ = θ m sin ω nt , ωn =

θ& = θ mω n cos ω nt , ω n =

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

θ& =

g l

0.18 = θ mω n 1.2

∴ θ m = 0.052462 radians

At

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

θ = − 0.047826 radians = − 2.74 ° !

(a) (b)

v = 1.2( 0.052462)( 2.8592) cos( 2.8592)( 1.5) v = 74.0 mm/s ! 2

a = 1.2 ( 0.052462)( 2.8592) sin ( 2.8592 )(1.5 ) a = 469 mm/s2 !

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 17.

(a) x = xm sin ( ω nt + φ ) x 0 = xm sin (0 + φ ) = 0.75 ft x& 0 = 0 = x mωn cos ( 0 + φ ) , ⇒ φ =

π 2

∴ xm = 0.75 ft

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

π  x& = ( 0.75 ft) ω n cos ω n t +  2  π π π  = x& ( 0 ) = v = 0 = ( 0.75 ft )ω n cos  ω nt +  , ⇒ ω n t + 2 2 2  And π  a = −g = − ( 0.75 ft ) ω 2n sin  ω nt +  2  −g = −( 0.75 ft) ω2n

or

ωn =

32.2 ft/s2 = 6.5524 rad/s 0.75 ft

Then

ωn =

k 10 lb , ⇒ k = m ωn2 = ( 6.5524 rad/s) 2 m 32.2 ft/s 2

= 13.333 lb/ft k = 13.33 lb/ft !

(b)

ωn = 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.

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At t = 1.6 s:

π  x = (0.75 ft ) sin  ( 6.5524 rad/s )(1.6 s ) +  = − 0.36727 ft 2  x = −0.367 ft above equilibrium !

π  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 !

π 2  a = &&x = − (0.75 ft )(6.5524 rad/s) sin ( 6.5524 rad/s)( 1.6 s) +  = 15.768 ft/s2 2  a = 1 5.77 ft/s2

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 18.

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

δ = δ1 + δ 2 ,

P1 P P = 1+ 1 k ′ k1 k2

and

Hence

k′ =

k1k2 k1 + k 2

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

P = P1 + P2,