Solution manual vector mechanics engineers dynamics 8th beer chapter 19

2 2 10 lb 0.31056 lb s /ft32.2 ft/s n k m ⋅From free fall of the collar Now to simplify the analysis we measure the displacement from the position of static displacement of the spring, u

Simple Harmonic Motion

( )( )90 12

41.699 rad/s 220

32.2

n

k

f m

64.343 rad/s

=

20.90765 s

n n

πτω

n n

In Simple Harmonic Motion

Hz 4.6729 Hz 280.37 r/min Hz

v l

θω

For a simple pendulum

n

g l

n

g l

v l

θω

k m

6.2161Hz

n n

7.728 ft/s

0.24032.2 ft/s

m s

a g

k m

At both 600 rpm and 1200 rpm, the maximum acceleration is just equal to g

m

k m

Now x(0) = =0 x msin(0+φ)⇒φ = 0Then

(0) m ncos(0 0)

x& = x ω +

or 2.5 m/s =(x m)m(16.903 rad/s) ⇒ x m = 0.14790 mThen x =(0.14790 m sin 16.903 rad/s) ( )t

(a)

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

or

1 0.06 msin

0.14790 m

0.02471 s16.903 rad/s

Referring to the figure of Problem 19.12

n n

πτω

4(1.25 ft) =5 ft Fifteen cycles take

15(0.09765 s/cycle) =1.46477 sThus, the total distance traveled is

Chapter 19, Solution 15

2 2

10 lb

0.31056 lb s /ft32.2 ft/s

n

k m

⋅From free fall of the collar

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:

Then mx+kx =0, where x is measured positively up from the

position of static deflection The solution is:

−

Determine the constant k of a single spring equivalent to the three springs shown

1 2

k kk

′ =+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

3

0.361 s4.11 10 N/m

f

τ

(a) First, calculate the spring constant

20.20279 s

n n

πτω

14.9312 Hz

n n

( )1

2

26.8 s

10 lb32.2 ft/s

6 lb32.2 ft/s

ττ

P k

m

0.517 s12.153

τω

( )T A min =12−1280 0.0125( )= − 4

Max Comp: 4 N !

k m

n

k m

The equivalent spring constant for springs in series is:

A A

Wmg

=

( ) ( )

0.12 s 6.22 10 N/m

2.2688 Ns /m 2.2688 kg2

Initially 2 1.6 s

A

k m

2

2.11.6

A B A

π

τ =

( ) ( ) ( )

2 2

2

19.373 lb32.2 ft/s2

2 2

hmh

(a) The equation of motion is:

mm

mm

129.57 10 lb/in.

18 in

e

AE k L

81.272 Hz

e n

k m f

W k

δ

=

W m g

0.981 N4

x

dF

x dx

1.4372 Hz

e n

k m F

1.437 Hz

n

Using the Binomial Theorem we write

l

d g

l g

θ

Chapter 19, Solution 37

From Equation 19.20:

22

6 kg

5 kN/m

m r M k

( )( )

2 2

At equilibrium, spring tension=(20 lb sin15) °

0

20 lb sin15stretch

θ&& = −ω θ = −ω

( ) ( )

2 max

0.8

3 276

98.747 rad/s6

Now ( ) ( )

( )

2 2

k M m

++

++

2 2

42

0.021440 m16

2

r b

r r

π

ππ

n

πτω

(b)

2 2

πτω

12

n

g r

3

n n

r g g

r

τω

θ&&=

2

23

n

gb r

ω =

2

7.3816

2 1

93

n n

r g g

r

τω

π

+

!

( 2 2)

112

12

2 2

c b

Consider general pendulum of centroidal radius of gyration k

Chapter 19, Solution 51

2 2

2

26 12

32.2 ft/s 14.861 sft

πτω

1.630 s

n

1.689 s

n

See Solution to Problem 19.52 for derivation of

+ a minimum

2

2 2

Thus the length of the equivalent simple pendulum is the same as in Problem 19.52 It follows that the period is the same and that the new

center of oscillation is at O (Q.E.D)

mg k l m

2 2

(b) For τn → ∞ , ωn → oscillations will not occur 0

From Equation (1)

2 0.47266 2.6667

00.05873

π

Determine location of the centroid G

Then total mass m = ρ(2r+πr) = ρr(2+π)

π

=+

2 9.81 m/s2

( )2 2

( )vA max =lθ&max =lθ ωm n

=

( )vA max =3.50 in./s
(b) For disk riveted at A (Iα included)

gll

ω =

+sin

( )1

0.332.2

θ&&+ θ =

n

K I

ττ

R S

I I

I k m

θ&&+ θ =

2

n

K I

T = Kθ = Kθ

2 1

=+

6.4ft,12

n

τω

0.400 s

(b) θ θ= 0sinωn t; θ θ ω& = 0 ncosωn t; θ&max =θ ω0 n

So vmax=bθ&max=bθ ω0 n

27 N m/rad 3.8 s

3.31 N m s4

Geometry

3

b l

θ

φ =

mg b F

(b)

3

mx&&= − F

mgx l

= −

2

n

g l

ω =

2 l

g

( 2 2) 2

12

T = m b +c θ&

2 2

12

2 2

0

maxv D =cω θn = 0.351 m/s

maxv D =0.351 m/s!

858

n

kl ml

ω =

5

k m

=

1 Hz

n

k f

m

π

Chapter 19, Solution 72

Equilibrium:

0 0 0

22

20 lb1ft9

k y T

9

n m

Datum at 1 Position 1

2

1 0

0

2 0

44

n

J mgr

ππ

Chapter 19, Solution 74

Find ωn as a function of c

Datum at 2 Position 1 T1 =0 V1= mgh

2 2

12

n

gc l c

2 2 2

2 2

212

0 12

l c

12

l

a a

b b

2π

n

πτω

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

Note: Result is independent of length of the rod

6 lb32.2 ft/s

= &

Then

2 2

2

316

2

12

2 2

3 7 kg

n n

τω

Chapter 19, Solution 80

sinθ θ≈

2 2

6g 4g  2 2 

55

l

g g

n n

πτω

Let m be the mass of rod and m C be mass of each collar

2 2

C n

C

kl

m l ml

k

m m

n

πτω

60.75 m5

n

τω

With

2 2

n n

τω

1.627 s

n

5cos cos

5cos 1

n

g l

n n

12

R n

R n

3 376

n

mg k

l m

Chapter 19, Solution 92

With 4 lb 2, 6 lb 2, 20 lb/ft, 40 ft

1232.2 ft/s 32.2 ft/s

2 2

1024

ω =

1 32

n

g f

l

π

2

41

ππ

r r

ω

ππ

(rsinθm)sinθm ≈rθm2(1 cos ) 2

2

m m

Position 1 maximum deflection T1= 0

2 1

2

m m

π

1 1 62.4 lb/ft 4 4 in.

0.15032 lb s /ft

2 2 32.2 ft/s 3 12 in./ft

r V

40 lb/ft

220.54 s

1 lb

0.15032 lb s /ft32.2 ft/s

n n

πτω

Eq (19.33)

2

1

m m

f n

P k x

ωω

Chapter 19, Solution 99

Eq (19.33)

2

2, 1

m

f n

P

k k

x

m

ωωω

k

−

Eq (19.33)

2 2

1

m m

f n

P k x

ωω

31

st

st f n

ωω

≥

−

2 2

11

3

f n

ωω

2 2

23

f n

ω

ω >

Also

2 2

31

st

st f

n

ωω

ω

ωω

P l mb

Referring to the figure and solution for Problem 19.101

2 2

32140

4032.2

m m

n f

P l mb b

b =

Upper frequency 3.00533b = −9.3499+2.625b2

2

2.625b −3.00533b−9.3499 = 02.5446 ft

b =

16.80 in.< <b 30.5 in.!

2

245.25 s0.040 m

e n

1245.42

From Eq (19.33′ ) 2

1

m m

f n

ωω

158.5

Chapter 19, Solution 106

2 2

1

m m

f n

ωω

n

k m

f n

ωω

12 in.ft

f n

ωω

1 f 0.68241

n

ωω

− >

2

2 1 0.68241 0.31579

f n

There is no value of x m for which

30 lb when and are out of phase

m

450 225 s

m m

n f

k ml

δθ

c m

f n

x

δωω

m m

<

  (Impossible)

Chapter 19, Solution 111

From Eq (19.33'):

2

11

m m

f n

mrMx

ωωωω

2

30 10 m1.5 mm: 0.0015

1

f m

n f

n

ωω

f n

ωω

ω

ω ≤

30 10 m1.5 mm: 0.0015

1

f m

n f

n

ωω

18 kg

Four springs each of constant 40 kN/m

We note that the motor is constrained to move vertically

n

P k x

ωω

n

P x k

ωω

We have found: P m=186.4 N

For an unbalanced rotor of mass m R=4 kg, rotating at ω =1200 rpm=125.664 rad/s, with the mass

center at a distance r from the axis of rotation, we have,

P r

f n

m r M x

ωωωω

1

f m

f n

mr k x

ωωω

ft12

n s

n

2 2

0.9lb

16 0.0017469 lb s /ft32.2 ft/s

ω′ =

′

Since amplitude of vibration is to be less than 2.16×10−3in for motor speeds above 300 rpm

( 9 2) 2

1 3

2

2151.64 10 ft s 300

60 s2.16 10

ft

300601

ππ

2

149.663 10: 0.18 10

986.961

′

7200

13.361538.89

f n

p k x

ωω

n

mr p

f n m

f n

m M x

ω ω ω

=

−

!

We also note that the amplitude δm of the displacement of the base remains constant

Referring to Sec 19.7, Fig 19.9, we note that, since ( ) ( )2 1 ( ) ( )

2 1

2

1 1

ω

ωω

n

ωω

δωω

11.2

1 2

n n

ωωωω

2.2 1.13.4 1.7

n

ωω

Points corresponding to the two solutions are indicated below:

We also note that the amplitude δm of the displacement of the base remains constant

Referring to Sec 19.7, Fig 19.9, we note that, since ( ) ( )2 1

ωω

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

2

2 1

2

1 1

4

51

n

n n

ω

ωω

1

1.8 mm 1.286 mm4

1 3

m m

n

ωω

2 Assuming ( )xm 1< 0:

For one collar, we have now:

2 1

9 mm

1

δωω

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

2

1 2 1

1 231

n n

ωωωω

2

n

ωω

=

ωω

2 2

mrkx

ωω

ωω

2 2 2

0.00077640.4 ft/s

k k

2

12

f

f

mr kx

M k

f m

f

M kx

k r

m

ωω

f n

P k x

ωω

P k

ωω

f n

ωω

G describes a circle about the axis AB of radius r + e.

1

f n f n

e r

ωωωω

1400 rpm

n f

(b) ( ) 2

6

6 2

1200

150 10 m

1399.6

416.28 10 m1200

11399.6

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

n

k m

f n

ωω

145.08

m

−

!

Chapter 19, Solution 123

In steady state vibration, block A does not move and therefore remains in

its original equilibrium position

2 2

sin1

m

f f

n

ωω

sin1

m

f n

ωω

11

f m n

1.041724

11

106.1 Hz

n

Chapter 19, Solution 125

From Problem 19.124

2

2 2

a

ωω

f n

ωω

z a

−

Error =8.03%!

Sincec > c c we use equation (19.42), where

Thus a positive solution for t > for Equation (3) cannot exist since it would require that e raised to a 0

positive power be less than 1, which is impossible Thus x is never 0

The x− curve for t

this case is shown

!

(b) t =0, x =0, v =v0 Equations (1) and (2), yield

The x− curve t

for this motion is

as shown

!

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

+ +

− +

ω =ω −   

 

2

12

c D

ω = −    

As in Problem 19.128, for maximum displacements x n and x n k+ at t n and , t n k+ sin(ω0t n+φ) = 1

0 0

c n

n n k m c

n k m

t

t t n

t

n k

e x

x e

−

− +

+

Comparing with Equation (2)

1log decrement ln n Q.E.D

n k

x

D

mct

1 1

2

D

mct

(c) The first maxima occurs at 1, (ωDt1+φ)

The first zero occurs at (ωD( )t1 0 +φ) = π

From the above plot ( D( )t1 0 ) ( Dt1 ) 2

Chapter 19, Solution 131

(a) From Problem 19.128,

2 1

2ln

1

c n

n

c

ccx

3 0.0384 m 38.4 mm

c k m

From the given data

For small angles:

sinθ θ! , cosθ !10.15 , 0.45 , 0.15

Since the roots are complex conjugates (light damping) the solution of the differential equation is:

40.6

c c

1

m m

x

c c

P k x

c c

For part (a) with P m =125 lb and ω ω= n

c c

P k x

c c

For part (a) with P m =125 lband ω ω= n

c

cc

ωω

and the magnification factor will decrease as

c

c

Chapter 19, Solution 141

From Eq (19.53′)

( )2 2 ( ) ( ) 2

1Magnification factor

ω

ω for which m

m

x P k

ωω

From Eq (19.52)

m m

P x

st

W k

c c

n st

f n

ωω

2 2

0.11067 10 m0.25 10 m

1 0.60365 4 0.60365

c

c c

From solution of Prob 19.113, we have

P k x

c c

4 kg 125.664 rad/s

m m

4 2 2

2 2 2

2

10.009

4

m

n f n c

P k c

c

ωωω

From Eq (19.52)

m m

P x

Chapter 19, Solution 147

From Equation (19.48), the motion of the machine is,x = x msin(ωf t −φ)

The force transmitted to the foundation is,

c c

( )

2

2 2

c P

k F

c c

ωω

ωω

3.28462.7735

f n

Energy is dissipated by the dashpot

From Equation (19.48) the deflection of the system isx = x msin(ωf t −φ)

The force on the dashpot, F D = & cx

π ω

ωτ

f m

Chapter 19, Solution 151

Since the origins of coordinates are chosen from the equilibrium position, we may omit the initial spring

compressions and the effect of gravity

For a mechanical system oscillations take place if c c< c(lightly damped)

But from Equation (19.41),

<

Chapter 19, Solution 153

The mechanical analogue of closing a switch S is the sudden application of a constant force of magnitude P to

the mass

(a) Final value of the current corresponds to the final velocity of the mass, and since the capacitance is zero,

the spring constant is also zero

FΣ = ma

2 2

C

=

2 2

Chapter 19, Solution 154

We note that both the spring and the dashpot affect the motion of point A Thus one loop in the electrical circuit should consist of a capacitor 1

kc

 and a resistance (c ⇒ R)

The other loop consists of (Pmsinωft sin→ Em ωft), an inductor

(m → L)and the resistor (c→ R)

Since the resistor is common to both loops, the circuit is

x→ q

P → E

For simple harmonic motion and 1= 40 in.=3.333 ft:

2

32.2 ft/s 3.1082 rad/s3.333 ft

n

g l

(a) First, calculate the spring constant

(b) θ θ= msin(ωnt +φ), θ θ ω& = m ncos(ωnt +φ)

α θ= &&

2

12

0

44

2 2

AB n

2

0 0

11

22

n

AB AB

132.2 ft/s 25 lb/ft 0.33333 ft 6 lb 1.25 ft

πτω

0.831 s

n

mg

Pk

ωω

1000 lb/ft 22.43 s

64 lb32.2 ft/s

 

&&

From Equation (19.31 and 19.33 )′

2 2

1

m m

f n

ωω

Deflection of the cord is xm −δm =0.21070 0.20 0.01070 mm− =Which is less than the static deflection of 50.5 mm

211 mm

m

Chapter 19, Solution 169

In equilibrium the force in the spring is mg

For small angles

sinθ θ≈ cosθ ≈ 1

2

B

l y

(b) Substituting θ =eλt into the differential equation obtained in (a),

we obtain the characteristic equation,

04