dynamics for engineers

SOLUTIONS MANUAL

. DYNAMICS FOR ENGINEERS

B.B. Muvdi A. W. AI-Khafaji J.W. McNabb

Springer .

Springer © 1997 Springer-Verlag, Inc. Springer-Verlag New York, Inc. 175 Fifth Avenue New York, NY 10010 All rights reserved. 10 9 8 7 6 5 4 3 2 1

Printed in the United States of America.

CONTENTS CHAPTER 13. Kinematics of Particles

1

CHAPTER 14. Particle Kinetics: Force and Acceleration

102

CHAPTER 15. Particle Kinetics: Energy

172

CHAPTER 16. Particle kinetics: Impulse-Momentum

200

CHAPTER 17 Two-Dimensional Kinematics of Rigid Bodies

303

CHAPTER 18. Two-Dimensional Kinetics of Rigid Bodies: Force and Acceleration

393

CHAPTER 19. Two-Dimensional Kinetics of Rigid Bodies: Energy

413

CHAPTER 20. Two-Dimensional Kinetics of Rigid Bodies: Impulse-Momentum

460

CHAPTER 21. Three-Dimensional Kinematics of Rigid Bodies

525

CHAPTER 22. Three-dimensional Kinetics of Rigid Bodies

583

CHAPTER 23. Vibrations

669

To the Instructor: We have prepared this solutions manual to serve as an aid to instructors using the text DYNAMICS FOR ENGINEERS by B.B. Muvdi, A.W. AI-Khafaji and J.W. McNabb. Complete solutions are provided for the more than 1250 problems in the text. Intermediate algebraic and arithmetic steps are often omitted but all significant equations as well as intermediate results for lengthy solutions are included. Note that the answers to the odd numbered problems are given at the end of the text. All of the solutions have been typewritten to make it easier for the instructors to read. The lettering on the figures, however, was done by hand but great care was taken to insure that it is clear and understandable. Thus, while vector quantities in the typewritten solutions are shown in bold type, they are indicated by a letter with an arrow on top in the figures. In the belief that the transition in this country from the U.S. Customary system of units to SI will be a slow process, we have decided to use both of these systems of units. Thus, approximately one-half of the problems are stated in U.S. Customary units and the other half in SI units. The authors would appreciate hearing from instructors using the text DYNAMICS FOR ENGINEERS and this companion solutions manual. If you desire a more complete solution for a given problem or find errors in our solutions, we will gladly review our work and provide you with expanded or revised solutions as the case may be. Furthermore, general comments about the text and suggestions for improvements in future editions will be much appreciated. A suggested outline for the teaching of a course in dynamics from the text DYNAMICS FOR ENGINEERS is provided in page v. The outline indicates a suggested number of periods for the various topics normally covered in the course. The instructor may wish to modify the topics and periods depending upon specific needs.

B.B. Muvdi A.W. Al-Khafaji J.W. McNabb Bradley University Peoria, Illinois

Suggested Course Outline for Dynamics for Engineers (Three fifty-minute periods/week for 14 weeks)

Sections

Suggested No. of Periods

13.1 through 13.5 13.6 through 13.9 14.1 through 14.5 15.1 through 15.7 16.1 through 16.10 17.1, 17.2; 17.4 through 17.6 18.1 through 18.6 19.1 through 19.5 20.1 through 20.4 Three tests Total

4.0 3.5 4.5 4.5 6.0 4.0 5.0 4.0 3.5 3.0 42.0

Note that the authors assign two to three problems for each period depending upon the topic and the complexity of the problems. Note also that a sufficient number of problems is available to allow the instructor to use different problems in each of at least four semesters.

PROBLEM 13.1

2 v = ds dt = 6t m/s

ANS .

a = dv = 12t m/S 2 dt

ANS

.

************************************************************* PROBLEM 13.2 s=(4t2 + 12t+30)ft v = ~~ = 8t+ 12 ft/s a = ~~ = 8 ft/s 2

ANS.

ANS.

************************************************************* PROBLEM 13.3

v =~ = 2t- 4 m/s

ANS.

v=2t-4=0 t = 2 sANS.

************************************************************* PROBLEM 13.4 s = (2t2 - 8t + 12) ft t = 0 ~ s = 12.00 ft ANS. (b) t = 4 s ~ s = 12.00 ft ANS. (c) 6s = S4 - SO = 0 ANS. (d) v =~= 4t- 8 =0 dt t = 2.00 sANS.

(a)

(e) t = 2 s ~ s = 4 ft (DT)04 = (DT)o-2 + (DTh-4 = (12 - 4) + (12 - 4) = 16.00 ft

1

ANS.

PROBLEM 14.1 P = 100 lb W= 300lb 8 = 30° Il = 0.20

2:Fy = m ay = 0 N - 300 - 100 sin 30° = 0 N = 350 Ib

0 300 0.2 (350) - 100 cos 30 = (32.2) ax

ax = - 1.782 ft/s2 V

2

2

= Vo + 2 ax L\s

= 0 + 2 (1.782) (20) v = 8.44 fils

ANS.

**************************************************************************** PROBLEM 14.2 Refer to the figure in problem 14.1. LF'y=may=O N-W-Psin8=0 N=W+Psin8 LF'x =max Il (W + P sin 8) - P cos 8 = W ax g ax =

~ [W + P(sin 8 _1 cos 8)] Il

v = Vo + ax t

= 0 + ax t v

t=-

ax

v =--------------------

~[W .+.

P(sin 8 -

L cos8) Il

=

1 [W + P(sin 8 - - cos e)] Il

ANS.

102

PROBLEM 15.1 '\T-

n- 2DO

:EFy =may =0 N -200 =0

~

lh

~

MDnO~ ~Olb ~ t

N = 200 lb

--c

o.2N

U 1 -+ 2 =- 50 (10) - 0.2 (200) (10)

=- 900 lb·ft

N

ANS.

**************************************************************************** PROBLEM 15.2 :EFy =ma y =0 N-600=0

~

•

U 1 -+ 2 = 100 (4)-0.1 (600)(4) = 160.0 N·m

I P=/ooll

MOT/oil

N=600N

t;

I 0.01

ANS.

**************************************************************************** PROBLEM 15.3

W=5DO/6

LFy =may =0

1'\n~ ~

N - 500 cos 30° = 0 ~ N = 433 lb U 1 -+ 2 = (500 sin 30°)(30)-0.1 (433)(30) = 6200 lb·ft

~Y~ ~ ~.It.J N

~

ANS,

**************************************************************************** PROBLEM 15.4

W

12D(r.fJf)

r\ . : ~ :1177N

~ \ 0./ rJ

N - 1177 cos 20° = 0 ~ N = 1106 N

U 1-+2 = 600 (6) - (1177 sin 20°) (6) - 0.1 (1106) (6) =521 N·m

ANS.

172

N

-

PROBLEM 16.1 s = 1t(t2 + 8t) S= v = 1t(2t + 8) PosmON 1

s = 0 = 1t(e + 8t) => t = 0 s= v= 81t = 25.13 v = 25.13j L = mv = 2(25.13)j =(50.3 Ib·s)l ANS. PosmON2 1 1

s = -(21tr) = -(21t)(2) = 1t

4

4

:. 1t = 1t(e + 8t)

e +8t - 1·= 0 => t = 0.123s . s = v = 1t[2(0.123) + 8] = 25.91 v=-25.91j L = mv = -2(25.91) i = -(5l.8Ib· s) i

ANS.

************************************************************** PROBLEM 16.2 x=e+e+lO X=v = 3t 2 + 2t t = 1 s: X = v = 5 v = 5i L= mv=2(5)i = (lO.oo N . s)i

ANS.

t=2s:x=v=16 v = 16i L = mv = 2(16)i = (32.0 N· s)i

ANS.

200

PROBLEM 17.1 (a)

v 2 =v o2 +2a& (100 x 103 )2 = 0 +2a(175) 3600

a =2.20m/s 2

ANS. .

(b)

v=vo+at 0= (l00x 103) -3t 3600

t = 9.26 s A N S .

************************************************************* PROBLEM 17.2

t'\f:.

0 AT MAX, ELEYAT/tJN

f

(a)

v=vo+at

L Yl,

. . .

PACKA~

O=IO-32.2t t=031ls s= So +v ot+!at 2

BEfStINS · IT..5

2

.

MOT~ON

hI = 0+ 10(0.311) - .!:..(322)(0.31l)2 = 1.553 ft

2 hz = 0 + 0 + !(32. 2)( 4 - 0.311)2 = 220.17

+ I I I

0

. L ~3

~

, ~L

I

I

I / / / / I

2 = 220 ft

I I

ANS

(b)

h3 =h z -h 1 =219ft

ANS.

*************************~~**********************************

PROBLEM 17.3 (a)

vR=vA+at 21.375 = 2.625 + a(25) a = 0.75 m / s

2

ANS.

303

PROBLEM 18.1 .2

52

2

=IG + m =(5) m R

Iy

(J

EaR

52

2 2 2 IG = (5) m R - m [(3 R) + R ]

.j-

'3R

2

I z = IG + m d

~4-K/

2 2 2 2 = (3) m R + m [(4R) + R ] ANS.

=(~)R

o R

6/

ANS.

**************************************************************************** PROBLEM 18.2 Refer to the figure in problem 18.1. (a) 2

Ix = IG + m d 0.08 + 5 [(4 X 0.20)2 + (3 2 5.08 kg.m

= =

=~

kx

X

0.20)2]

= 1.008 m

ANS.

(b)

Iy

=IG + m d2

=0.08 + 5 [(3 X 0.20)2 + (0.20)2] =2.08 kg.m2

ky

=~

= 0.645 m

ANS.

393

PROBLEM 19.1 LMA=O:

N(15)-~N(4)-60(40)=0

N= 2400 15 - 4~

UI _

2

(1)

=- ~N [40 (23t) (-H)] = - 251.327

(1) into (2) Ul -

2=-

~N

~

603,184.8 (

15~ 4~)

T2=O

=15,325.473 Ib.ft

. 603,184.8 ( 15 ~4

~) = 0

- 15,325.473

ANS.

11= 0.346

************************************************************************ PROBLEM 19.2

TI = Ia oi2

1

=2 IG (

1800 (2:n:) 60 )

= 17,765.29 IG Ib.ft T2 =0 Ul _ 2 = - 40 (4000 (2:n:)) lb.ft

Now,

·40(4000 (2:n:))

=0 - 17,765.29 IG

IG =56.6 Ib.s2 .ft

ANS.

513

PROBLEM 20.1 (a)

Jl=O L

=>JJlNdt=O

J

Fx dt=-

SWJ.t

J~ (5t+ 20) cos 10" dt

=- 2.462 t2

-

19.696 t

SQJt

~-:-=c~bJo_ = /1I1\IG.'t°1 _ '

t

Lu = (j~) (5) = 15.5281b.s ~x=O

Since L

SI'NJt

.SNt1t

J

- 2.462 t2

/000

Fx dt= (L2 - L1)xl we obtain, -

19.696 t = 0 - 15.528

from which t2 + 8 t - 6.307 = 0 Thus, t = 0.723 s

ANS.

(b)

LJ F dt=(L -L )y=0 2

y

1

J J Ndt-

W dt=O

Thus, N= W= 100 Ib Also,

LJFx dt=- J~ (5t+ 20) cos 10" dt - J (0.3) (100) dt =- 2.462 t2 - 49.696 t

As in part (a)

Ll x = 15.528 Ib.s

L2x=0 Thus, since L

JFx dt= (L2 - L1)Xl we have - 2.462 t2 - 49.6% t = 0 - 15.528, from

which t2 + 20.185 - 6.307 = 0 Thus t = 0.308 s

ANS.

560

PROBLEM 21.1 (A)

8z +8 y +8 x +8 z

ANS.

(B)

-8 z -8 z -8 x +8 y

ANS.

************************************************************ PROBLEM 21.2 (A) _

-8z -8 z +8 y -8 x

ANS.

(B)

8z -8 x +8 y -8 z

ANS.

************************************************************ PROBLEM 21.3 (a)

= (60.Oi + 20.0k) rad I sANS. (b) rp/o

= (0. 15cos600)j + (0.15sin600)k = (0.075j+ 0.130k) m

vp=vo+(Oxrp/O i

=0+ 60.0

o

j

k

o

20.0 0.075 0.130

= (-1.500i -7.80j+4.50k) m I sANS.

625

PROBLEM 22.1

Ix =(lxh + (lxh + (I.xh - [1 m (022) _ m (2 x 0.2)2] + m [06+ 2 (0.2)]2 .

- g +

1·

1

31t

31t

1·

-h m2 (0.1 2 + 0.22) + m2 (0.552)

+ {2 m3 (0.5 2 + 0.062) + m3 (0.2s2)

=0.41593 mi + 0.306667 m2 + 0.083633m3 ml =P VI = 0.015708 P m2 =P V2 = 0.020 P m3 =P V3 = 0.00120 P

Ix =0.01277 P

kg.m2

ANS.

************************************************************************ PROBLEM 22.2

Ix =1.75926 mi + 0.79630 m2 + 0.38778 m3 + 0.18287 m4 + 0.03877 ms + 2.11052 m6 - 2.01100m 7

683

PROBLEM 23.1

p=~=~12~ = 20.0 rad! sANS. 21t 21t 't=- = - =0.314s P 20 1 f = - = 3.18 cps

ANS.

ANS.

't XM

= Xo = 1 in. = 0.0833 ft

ANS.

VMAX

= P xM = 1.667 ft! sANS.

aMAX

= p2

xM

= 33.3 fi! s2

ANS.

*************************************************************

PROBLEM 23.2

p=~=~6~ = 27.4 rad ! sANS. 21t 21t 't=-=--=0.229s p 27.4 1 f = - = 4.37 cps

ANS.

ANS.

't XM

= Xo = 0.0833 ft = 0.0254 mANS.

VMAX

= P xM = 0.696 m! sANS.

769