TWELFTH EDITION
R. C. HIBBELER
1M! design of thiS fOC~et and gant:)' structure requires 11 basIc ~nowledge of
both statics and dynamiCS. which form the subject matter of engineering
mechanil::s.
General Principles
CHAPTER OBJECTIVES
• To provide an introduction to the basic quantities and ideali zations
of me<:hanic5.
• To give a statement of Newton's Laws of Motion and Gravitation.
• To review the principles for applying the 51 system of units.
• To examine the standard procedures for performing numerical
calculations.
• To present a general guide for solving problems.
1.1
Mechanics
M~cJllmics is a branch of Ihc ph ysical sciences lh:I' is concerned w;lh thc
Slal e of rest o r mOl ion of bodies Ihal an! subjected 10 [he :lelioll of force$.
In general. this subject c:In be subdi\'idcd into thrcc branches: rigid-body
/IIt'dllmics. df'jumlllblt'.hoffy //ICc/lOll irs. andf/ilitilllt'rlwllics. In Ihis book
we will study rigid-body mechanics since it is a basic requirement for th c
study oflhe mechanics of dcfonnablt: bodies and Ihe mechanics of Ouids.
Funhermore. rigid.body mechanics isesscmial (or the design and analysis
of many types of structuT<llmcmbcrs. mechanical components.or eli:'ctrkal
devices encountcred in I.'ngineering.
Rigid-body mechanics is divided inlo two areas: statics and dynamics.
Sfllliey deals with th e equilibrium of bodies. that is. those that arc eit her
al rCSI or move with <I constant velocity: whereas (iYIWlllieS is cona.'rncd
with thc act'Clcraled motion of bodics. We can consider statics as a
special case of dynamics. in which the acceleration is zero; however,
statics dese rves se parale trealment in engineering educa tion si nce many
objecls Me designed wilh Ihe inll.'nliol1 that they remain in equilibrium.
4
CH"'PfE~
1
GEN~Ir"" l
PR INCI PLES
Historical Development. The subject of statics developed
I' Cry
carly in history because ils principles can be (onnulalcd sim pl)' (rom
measurements of geome try and fon;:o;. r-or exa mple. Ill..: writings of
Archimedes (287- 2]2 It.C. ) deal wilh the principle of the le\'er. Studies of
the pulley, inclined plane, and wrench arc also recorded in ancient
I'.Ti tings - al limcs when the requirements for cnginc~'ring were limited
prim ari ly \0 building conmuclion.
Since the principles of dynamics depen d on an accurate measu re men t
of lime. Ihis s ubject developed much late r. Ga lileo Qalilci (156+-1642)
was OIH: a rlh.:: first major COnl ri bmors [0 this field. "lis work consiSted of
experiments using pendul ums and falling bodies. The most significant
cont ributions in dynamics. howc\'cr, wcrc made by Isaac Ne wton
(1~2- 1727), who is nOlcd for his formulation of Ihe Ih ree fundamenlal
1:lws of motion and Ihe law of uni\'ers.11 gr:wil:llional all raction. Shonl)'
3fter these laws were postulated. important tech niq ues for their
3pplkation were developed by such nOl:lblcs as Eukr, D'Alem!len,
ugrangc. and others.
1.2
Fundamental Concepts
Before we begin ou r study of engineering mechanics. it is important to
understand the meaning of certain fUnd:lment:l1 concepts and prinCiples.
Basic Quantities. The foll0\\1ng four quantities arc us<.>d throughout
mechanics..
Length. Lmgtlr is US('d lQ loc,w.: the position of a poi nt in space and
thereby describe the size of a physical system. Once a stand:lrd unit of
length is defined. one can then use it to define distances and geometric
propcrtks of 3 body as multiples of [his uni t.
Ti me. Time is conceived ,IS a succession of events. Although the
pri nci ples of statics aTC time independent. this quantity plays 311
important role in the study of dynamics.
Mass. MIISS is a measure of a quantity of mailer th at is used to compare
the ac tion of one body wi th th at of another. This propeTly manifests itself
as a gravitational allraction between IWO bodies :lnd provides a measure
of the resistance of mat ter to a change in "elocity.
Force. In genc ral.fim·e is considered as a "push" or "pull"' exerted by
one body on anolher. This interaetion can occur when the re is di reci
oontac[ betwee n the bodies. such as a person pushing on a wall. or it ca n
occur through a distance when the bodies ,Ire physically scparat;;)d.
Examples of the l:lller Iype include gfll\'ilalional. eleclric:ll. and magnetiC"
forces. In any case. a force is completely characterized by its magnitude.
din.:.;:tion. and point of application.
5
Ideal izatio ns . Models or idcalizmions arc used in mechanics in
order \0 si mplify applicat ion of the theory. Here we ..... iIl conside r three
important idca lil~llions.
Particle. A pI/nidI! has a mass. bu t a 5i7£ that ca n be neglected. For
exa mple, the size o f the eart h is insignificant cornpilrcd 10 the size o f its
orbit,a nd therefore the cart h can be modeled as a panicle when stud ying
ils orbital mOl ion. When II body is idca1i7.cd as a particic.lhe principles of
mechanics reduce to a rathe r Simplified foml s ince Ihe geometry of the
body ...iII ' 101 bl! ;m'okf(i in Ihe analysis of the proble m.
Rigid Body. A rigid botly can be considered as a combination of a
large num be r of p(lrIidcs in whi ch all Ihe part icles re main a \ a fixed
dis l.mcc frolll one another. both before and after (Ipplying a Joad. lbis
model is im portanl because the mMerial properties of <l ny body Ihal is
assumed 10 be rigid will nOl han: to be conside red when stud yi ng Ihe
d fecis of forces (ICling on the body. In mOSI eases the llctual dcfomlations
occu rring in structures., machin es. mechanisms. and Ihe like arc relatively
small. and the rigid-body assumption is suitablc for anal ysis.
Concentrated Force. A crmCt'mrllft'(J j"rrt represe nts Ihe erfect of a
load ing which is assumed \0 act at a poi nt on a body. We ca n represent a
load by a concentrated force. prO\.ided the arca o\"cr which the load is
:lpplied is vcry small com pared to the ovcrnll size of the body.An e.~amplc
would bc the oon taet force bctwcen a wheel and the grou nd.
nu ee forces act On tht hook at A . Since Ihesc
f(lrces all mcet at a point. thtn for any ((lrce
analysis.. ""c can assume Ihe hook 10 be
,eprcscnted as a partick.
Sled is a cOmmOn enllineerinll n... tcri~llh~1 docs OOt deform
,"Ct)' mllCh unde.I(l:>d.Thcrcfon:. "'"C can oonsilkr lhis rail.oad
"'hed tobe a rillid body aeled upon b)· thc oollttntrntcd force
of the mil.
6
CH"'PfE~
1
GEN~Ir"" l
PR INCIPLES
Newton 's Thre e l aws of Motion . Engineering mechanics is
formulated on the basis of Newton"s three laws of mOlion. the validit)' of
which is based on I:.~pc rim cn lal obse rvation. These laws appl)' [0 the
motion of a particle as measured from a
frame. They may be bricny Slated as follows.
Firs t law.
/lQIII/cedeNt/ilIS
rcfcrcneo.::
A part ide originally al rest. or moving in a straight line wilh
constant velocit)'. lends to remai n in this Slale prO\ided the particle is 1101
subjected [0 an unbalanced force. Fig. I- la.
"'Y"
"
"
,.)
Equ,hbf;um
Second law. A particle acted upon by an UllIN/lillie!'" force F
cxpcricnC'cs an accele ration a Ih:1\ has the same di r('~ lion as th e force
and a magnitude Ihal is directly proportional 10 the force. Fig. I- l b.If F is applied to a particlc o r mass III, this law mar be exprcssed
malhe rnatil;'a ll yas
F = ilia
( I - I)
Third Law. The mlllu:.1 Cortes of al;'tion au d rcal;'lion belween two
particles arc eqllal. opposite. and collinc;lr. Fig. l- lr.
/ 'pm: of A on H
'~
A
H
F
I... fo",",ofHonA
fijt. I_ I
'Slaled ~n<>lher w~l'. Ihe unb;llan«d force IICIln8 on Ihe parllck;$
Ijme nile of change 0( Ihe pa",ck'5 liMa. momenlum,
Il""ponionallo lhe
1.3
UNITS 01' MEASUREMeNT
Newton's Law of Gravitational Attraction. Shorlly arter
fommlating his three laws of mOl ion. Newton postulated a law governing the
gravita tional attraction betwccn any t".-o j)drliclcs. Stated mathematically.
(1- 2)
where
,.. '" Forcc of gravitmion between the two panicles
G = universal constant of grnvitation; nu:ording to
ellJXrimcntal evidence, G = 66.73( I O-t~) mJ/ (kg· s!)
lilt. III ~
= mass ofcm;h of the lWO panicles
r = diSlal"lcc betwccl"llhc two panicles
Weight.
According 10 Eq. 1- 2. any two particles or bodies have u
mutual attractive (gravi tational ) force acting between thelll. In the casc
of a partide 1000:aled :.t or ncar the surface (If the earth, however, the only
gravitational force having any sizable magni1Ude is thM between the
e;lrt h and the part ide. Consequently. this force. termed the weighl. will be
the only gnll'il(lIional force considered in our study of mechanics.
From Etl. 1- 2. we can develop an appro.~imate expression for finding Ihe
weight IV of a particle having a mass lilt = III. U we assulllc the earlh to be
a nonrotating sphere of cort~tant densit y and having 11 mllSS r112 = M, . lhen
if ris the distancc between the earth's center and the particle. we have
111M,
IV = G- -,-
Th~
astronaut is wcighrl,,~ for all
practical purpos~ ... since she is far
removed from Ihe gravitational r",ld of
Ihe earth.
r
Letting X = GM,I' ! yields
IV - 1118
I
( 1- 3)
By comparison with F '" ilia. we can s..."<: th:lt Ii is the accclemtion due to
gravity. Since il depends on ,.then the weight of a body is 11m an absolute
qu(tntity. lnslead, its magnitude is detennined frOIll where the measurcm~nt
was llIade. For most engineering clliculatiuns, huwever. g is detCmlinoo at
sea level and at a latitude of 45°. which isconsidcr,,"(] the "standard location."
1,3
Units of Measurement
The four basic quan1ities - Ien g1h. time. mass. and force - arc nOI all
imlcpcncJcnt from o ne another:in fact. they arc ,elllled b}' Newton's second
law of mOtion, F = 111>1. Because of this. the IIl1ilS used 10 measure these
qU11llli1ies cannot Ill! be selected arbitrarily. The equalit y F 0: mll is
maintained only if three of the four units. called bal'C Imils. arc deft/It'll
and th e fou rth unit is then derivl'(! fromthc eq uatio n.
7
8
C",APTER 1
,.)
GENERAl PRIN CIPtES
SI Un its. The intcmmiomal System of units. abbrcvi;,h:d SI aftcr the
French "Systemc In!Crn:llionai d'Unitt's," is .. mcxkrn \'crsion of the metric
system which has received worldwide recognition. As shown in Ta bk \- 1.
the SI system defines kngth in IllCICrs (m).limc in seconds (s).and mass in
kilograms (kg). The uni t of force. c;!lied a newton (N). is dl.'ril'clf from
F = maoThUs, 1 newton is l'qua lto a force required \0 gh'e I kil ogram of
mass an acrclcmlioll of J m/ s2 (N :0 kg· m/s2).
If the we ight of .1 body located at .he "standard location" is \0 Ix:
determined in newtons. then Eq. 1-3 must IX: applied. Here measurements
give g = 9.806 6S m/ 52: however. for calculalions..lhc val ue 11 '" 9.8J 111/ ;;2
will be used. Thus.
(g = 9.81
IV = 1118
m/s~)
(1-4)
Thcrdorc. a body of mass I kg has a weight of9.81 N.:I 2· kg bod)' weighs
19.62 N. and so on. Fig. 1- 2a.
fig. 1- 2
U.S. Customary. In the U.S. Customary s),stem of units (FPS) leng.th
is measu red in fec t (fl). time in seconds (s). ;md force in pounds (lb),
Table I-\. 'Illc uni l of mass. e.[!led a .f/ug. is tleril'l.'(i from I' "" mao)-knee.
I slug is equal to Ihe amuunt of mailer accelerated al I fl/ S2 whcn acted
upon bY:I force uf 1 III (slug "" lb· s2/ft).
Therefore. if the measurements arc made at tho.: "standard location:'
where g "" 32.2 f1/52, then from Eq. 1- 3,
I\'
II! "" -
g
And SO;l body weighing 32.2 Ib has a mass of 1 slug. a fi4.4·lb body has a
mass of2 slugs. Hnd so on. Fig. 1- 2b.
TABLE 1- 1
Systems of
Name
Inlcrn~tional
Umt~
Length
Time
Mass
Force
meter
sccond
kilogram
I newton-I
System of Units
51
,
k,
N
m
u.s. Customary
roo'
second
Islug-I
pound
,
(";:.:)
,.
FPS
.I><.., .... ~ WIn
"
(k'/)
1.4
THE INTERNATIONAl S~1€M OF UMlS
Conver sion of Units. Table 1-2 pro\'idcs a sct of direct conversion
factors between FI'S and SI units for the basic quan tities. Also. in the
FI'5 system. recall that I fI ". 12 in. (inches). 5280 fI '" I mi (mile).
1000 Ib =- I Idp (ki lo-pound).;\IId 2000 Ib '" I 1011.
TABLE 1-2
Conversion Filctor s
Quantity
Measurement (FPS)
Unilof
Force
Ib
M,u
slug
LenS1h
1.4
Unit of
Equals
Measurement (SI)
4A48N
14.5<) kg
0.30-1 8 m
"
The International System of Units
The 51syslem or units is uSl!d eXlcnsivcly in this book since il is intended
10 becollle Ihe worldwide standard for measuremcllt.l11ercfore. we will
now present some of the rules for its use ~nd some of ils terminology
relevant to engineering mechanics.
Pref ixes . When a numerical quantity is either very IMge or vcry
small. the units used to define its size may be modified by using a prefix.
Somc of the prefixcs used in thc 51 system are shown in Table 1- 3. Each
represents a multiple or submulliple of a unit which. if applied
successively. moves the decimal point of a numerical quantity to every
third platt.· For cxample. 4 000 000 N = 4 000 kN (kilo-newton) '" 4 MN
(mega-newton). or 0.005 m = 5 mm (milli·meter). Notice th"t the S[
syslCm d~s nOI include the multip!.:: deca ([0) or the submultiple centi
(O.O[). which form part of the melric system. Exce pt for some volume
and area measur,·menls.. the use of these prefi.~es is to be l\1'oidcd in
science and engineering.
TABLE 1-3
Prefu(e~
Prefi)(
51 Symbol
10'
111113
Ill'
Ill'
mella
G
M
10 '
10'
10'
milli
micro
nano
Exponential Form
AlIIWple
I 000 COO COO
1000000
I 000
~ilo
,
Submulliple
0.001
O.COO 001
O.COO 000 00 I
• n..,
k'togrnnl is 'hf onty baloC uni,
,1", i. defined wilh" p,cI;".
m
•
"
9
10
CH"'PfE~ 1
GEN~Ir"' l P~ I N CI PlES
Rules f or Use. Here arc:. few of Ihe imponant rules Ihat describe
the proper usc of the \'l.riOUS SI symbols:
•
•
•
•
Quantitiesdefined by scveml units which arc multiples of onc another
arc scpar:lled by a rim to avoid confusion \lith prefL'I: notation. as
indi cal<:d by N = kg· m/ sl = kg· m· s- 2. Also. m . s (meter·second).
wll\."reas ms (milli·second).
The exponential power on a unil having a prefix refe rs to both the
unit 111111 its prefix. For cxample.~N! = (JLN)t = j.JN· p.N. Likewise.
mm! rcpresents (mOl)! '" 0101· mm.
With the exce ption of the base unit the kilogram. in gene r:.] avoid
the usc of a prefix in the denominator of composite units. For
cxample. do not wri te N/ mm. but rather kN/ m; also. m/ mg should
be written as 1\·lm/ kg.
When pe rforming calculations.. rC'p resent the num bers in terms of
th eir btrst or t/cril't"d IlIIil~' by convcrting all prefix~'$ to pow\'rs of 10.
The final result should then be expressed using 11 singlr PUfi.l . Also.
after calculation. it is best to kee p numerical values betll'een 0.1 and
10Cl0; otherwise. a suitable prefix should be chosen. For CX1,mple,
(50 kN)(60 nm) = [5O(IO l ) NI[6O(1O- 9 ) mJ
"" 3000(10-6) N· m = 3(IO- J ) N· m = 3 mN · m
1.5
Numerical Calculations
Numerical work in engincering practice is mOSt often pcrfonned b)' using
handheld calculators and computers. It is important. hOIl'e\'er. that the
answers to any problem be reported with both justifiable accu racy and
appropria tc significant figu res. In this section we will disellss th\'se topics
together with some other important aspects in\'olved in all engineering
calculations.
CompU!C";3,e ohen uscd in enginee,ing ror
:><.h·anced d... ign and anal)·~is..
Dime nsiona l Ho mog e neity. The lemlS of :lIly etluation used to
deseribe a ph ysical process must be Ilillle/ll'iOllllfly IWlllogelll.'o/ls: Ihat is.
each term mUSI be expressed in the Slime unlls. Provided this is the case.
all the IcmlS of an equalion , an Ihe n Ix: combined if numerical va lues
are subslilnted for the variables. Consider. for example. thc equation
z, where. in SI units.s is the position in me ters. m. t is time in
s '= ~'t +
seconds. s. v is \·docit)' in ml s :md II is acceleration in III/ SZ. Rcgardless of
how this equation is evaluated. it maintains its dimensional homogeneity.
In th e form staled. each of Ihe three terms is expressed in lIleters
[m.(III/ t)3', (m/ sl)sl.J or soll'ing for 11.11 = 2.s/' ~ - lV/ I . the temlS arc
each c~p rcsscd in unilS of III/ S! 1m/s!o m/sz, (III/S)lSJ.
ll1f
12
CH"'PfE~ 1
G EN~Ir"'l P ~ I NC I PlES
1.6
General Procedure for Analysis
The mos t efkeli,'e way of learning the principles of engineering mechanics
is to soll'l! probll'lIu. To be successful at thi s. it is important to alw3Ys
present th e work in a logiclll and oflkrly /lUI",,,:r . as sugges ted b)' the
following sequence of steps:
•
•
•
•
When soh'ing problems. do
ocatly as possible. Being OC3t
lhc work as
w<l l stinlulacc
dear and orderly thinking. and vice \'CI'Sa.
•
Read Ihe problem carefully an d Iry 10 correiate the aclual ph ysical
situation wilh Ihe Iheory studied,
Tabulate Ihc problem dala and draw an)' nl.'cess.u y di agra ms.
Apply the rcle,'ant principles. ge nerally in mathematical form. When
writing any cquations.. be sure the y ;Ire dimcnsionally homogcncous.
Solve th e neceSS'IT)' equalions, and reporlt hc answe r with nO more
th an three signifkant figures.
Study thc answer wi th tcchnic'll judgment and common sensc to
determine whethcr or not it seems reasonable.
Important Points
• Statics is th e study of bodies thaI arc al rest or move with
constant veloci ty.
• A particle has a mass but a size thaI can be neglected .
• A rigid bod)' docs not deform under toad.
• Concentrated forces arc assumed to act at a poi nt on a bod)'.
• Newton's Ihree lawS of motion should be memo rized.
• Mass is measure of a quantit ), of maHer that docs not change
from one location to 3nother.
• Weight refe rs to Ihe gravit ationa l attractiOn of the eanh on a
body or quan tity of mass. Its magnitude depends upon the
elevation at which the mass is located.
• In the SI system th e unit of force. the newton. is a derived unit.
The meter. second, and kilogram are base units..
• Prefixes G, M.I:. m. j.I., and n arc used to re prese nt large and small
numerical <Iuantilies. Their exponen tial size should be known,
along wilh the rules for using thc SI units.
• Perform nume rical calculations ",ilh several Significa nt figures.
and then repon the final answer to three significant figures.
• Al gebraic mani pulations of an equation ca n be checked in part by
verifying that the equation rcmains dimensionally homogeneo us.
• Know the rules for rounding off numbers.
1.6
GENEAAl PROC~OU~E FOR ANAI.'I'SIS
Convert 2 kmj h 10 lU/ S How many rtfs is Ihis?
SOLUTION
Since I kin " lOOOm and 1 h = 36005,lho:: f(lC lors of conversion arc
~rnl1lgcd in Ihc following order. so Ihal <1 canccllation of Ih c un ils C<ln
be applied:
Zk m/ h =
~
Frain Table 1- 2. 1 fl
:0
2~Co:m)(~J
2000m
-- =
3600 S
o. 556 m/'
0,3048 In. Thus.
O.556 m/ S -_
(~)(
S
If'
)
O."'''8
.>Vi m
,. 1.82 fl/S
Ails.
NOTE: Rc m.::m bcr 10 round of[ Ihe final answer 10 Ih re.:: significanl
figu res.
Com'eri Ihe qua ntilics 300 lb· sand 52 sl ug/ ft l to appropri<lle 51 unils.
SOLUTION
Usi ng Table 1- 2. 1 Ib ,. 4.448 2 N.
3OOIb · s "
3OOU1,s(4.~N)
= 1334.5N·s = 1.33kN's
Sinc.:: 1 slug '" 14.5938 kg and I fl
:0
0.304 8 m. then
52slug/rt3 = 52~(14.59kg)(
k'
All$.
1~
J. )'
0.3048m
= 26.8( HY) kg/ m3
= 26.8 Mgjm1
Ails.
13
14
C", APTE R 1
G EN ER A L P RI NCIP L ES
EXAMPLE 1 . 3
EvalulIIc each of the following and express with 51 units having an
appropriate prefix: (a) (50 lllN)(6 GN). (b) (400 mm)(0.6 MN)l .
(c) 45 MN J/ 900 Gg.
SOLUTION
First convert cach nUlllb<':r to bas.:: units., pe rform Ih..: indicated
operations.. tht'n choose an appropriatc prefix.
Part (a)
(50 mN)(6 GN) = [50(10-.1) N][6(1O~) N]
= 300{1d') NZ
~ 300('0') N'(~)(~)
\OJ ~ I!YX
= 300 kN l
Am:.
NOTE: Keep in mind the convcntion kN 2 = (kN)2 = ldi Nl .
Part (b)
(400 rn lll)(O.6 MN)2 = ]400(10- 3) rnIl0.6(10") Nf
= [400(10- J ) rn][0.36(1O!2) N2J
= 144(109)rn·N 2
= 144 Grn-N"
Am:.
We can also write
Am:.
Part (e)
45 MNl
45{ltr N)'
900 Gg = 900(Hf) kg
Am:
15
PlIOIlfMS
•
PROBLEMS
I_ I.
Round of{11\e followmg numbers 10 three signifICant
figures: (a)
(d) 276!! 1;;.
1_2.
4.6S7J.~
Repre~nl
m. (b) 55..S7S
So.
(el 4555 N. and
° 1_1Z. lbc speCIfIC "''Clght ("" ./,'01 .) of br:J.S5 IS 520 lb/ fI J •
Determine lIS denSily
appropriatc prerLl.
(maS$/ ~oI.) In
SI unil" Use an
the corrccI SI form usmg an appropnale pre6x:
(b) N/ ".m. (c) MN/ tr. and (d) kN/ ms.
each oIllIe follo'Mog combinations of units
Cal ",MN.
I - IJ. Con"':rt each 01 the follOl',mg 10 three sigmficant
figures: (a' 20 lb· II to N· m. (b) -150 Ib/ le to kN/ ml. and
(e)]S ft/ h to mm/'"
1-3. Represent each of 11M: following quanli,,,,'s In Ihe
com:<:1SI form using an appropriate prefix: (a) O.OO)..jJI kg.
1- 14. 'Il\c demll)' (mass/ volume) of aluminum IS
5.26 slug/ft). Determine liS dcnsily 10 SI units. Usc an
(b) 35.3(10' ) N. and (el 01WJ2 km .
appropnate prcfi:t.
Represent eath of the following tombinalions of
units In Ihe rom:CI 51 form: (a) MgJms. (b) N/ mm. and
I- IS. Water has a deMlty of 1.9~ slulYft). What IS the
density u preucd in SI units'/ Expre$li the ans"·cr to three
sigmfic:a nl figures.
In
·,_t
(e) mN/ (l.:g· ~).
1 -.~
Represent each of the follov.ing combinations of
I,mils in the OOITOO 51 fonn using an appropriate
prefi~:
(a) kNt ,..!.(b) M YrnN.a nd (e) MNf (kg· msl,
1-6.
Represent each of Ih e fOllowing 10 three significam
figures and express
C~d1
answer in SI units using an
approprilUe prcfi.\: (a) 4S 320 kN. (b) S68(](f') mm . and (e)
0.00563 mg.
1-7. t\ rork':l has a mass of 250( ]()l) s]ugs on eanh.
(D) ilS mass in SI unils and (b) ils weighl in SI uni.s.
S~cify
If Ihe rorket is on the moon .... here Ihe acceleralion due 10
gmvily is r •• 5.30 fl/ Sl . determine 10 Ihree significanl
figures (e) it! " 'Clghl In SI unllS and (d) ils mass In SI units.
If a u r IS .ra,·ehngal 55 mij h.delermine ilssp«d in
kIlome ters per hour and meters ~r serond.
° 1-8.
1_9. 'Ibe P'IJ(u/ (Pa) is aclualt)· a "ery smalt unil of
pressure. To show lllls, con"ert I Pa _ I N/ ml 10 lb/ lt!.
1\lmosphenc pressur~ at sea lel'd is 1-I.7Ib/ inl . How many
p;lS(als IS thIS?
I- Ill. What is the ... eighlln ne ... tonsof an objca Ihal has a
mass of: Ca) \0 kg. (b) 0.5 g. and (e) 4.50 Mg? B:prcss lhe
result to three sigmficant figures. Usc an appropriate prefi:t.
I- I I. Evaluate each of the following to three sigmficant
figurcs and express each ans ... cr in SI units using
an appropri ate prefi~: (a) 3S-1 mg(45 km) / {O.Q356 kN).
(b) (O.rot 53 Mg)(201 nls). and (c) 435 MN/ 23.2 mm.
*1- 16. 1'.... 0 particles hal'e a mass of II kg and 12 kg.
re~pc(,1iwly. If they are 800 mm ~vart. de termine the force
of gm"ity aCllng bet ...·een Ihem. Compare this
the weight of ~lIch pa rticl~ .
r~sult
"ilh
1- 17. Detcrmine Ihe m ~ss in kilograms of an object that
has a .... eight of (a) 20 mN. (b) 150 kN. and (c) 60 MN.
Elc:press the ans .... er to three significant figurcll.
I- III. E"alume each of the fol1o .... ing 10 Ihree SIgni ficant
figurcs and urr~ss ~ach an$ ....er In SI units uSing an
appropriat ~ prerlX: (a) (200 kN)l, (I)) (0.005 mm )! . and
(c) (400 mIl.
1- 19. Using the ba'IC umlS of t he SI system. show that
Eq. 1- 2 is a dImensionally homll&eneous equallon .... hich
gI"c' Fin newtons. Detenntne 10 three signIficant figures
the gravitational forec acting bc' ...·een t....o spheres that
are touch,"! each ot he r. The mass of eKh ~phere is 200 kg
and the radius is 300 mm.
"1- 20. Evaluale each of the foliOll'ing to three sLgniflcant
figu res and express each ans ...u 10 SI units uSing an
appropriate prefix: (a) (0.631 Mm )/ (8.60 kg)!. and
(b) p5 mm)l(4S kg»).
1- 21. £'1Ilu3te (21}.t mm)(O.OOa7 k,)/ (J-' .6 N) to three
significant figures and expr~ss Ihc answ~r in SI umts using
an approprinlc prcfi:t.
This bridge tower is stabilized by cab1es that exert forces at the points of connectIOn.
In thIS chapter we will show how to e><ptess these forces as CarteSIan vectors and then
determine the resultant force
Force Vectors
CHAPTER OBJECTIVES
• To show how to add forces and resolve them into components
using the Parallelogram Law.
• To express force and position in Cartesian vector form and explain
how to determine the vector's magnitude and direction.
• To introduce the dot product in order to determine the angle
between two vectors or the projection of one vector onto another.
2.1
Sca lars and Ve ctors
All physica l quanlitit.-s in engineering mechanics arc measu red using ei ther
scala rs or vectors.
Scalar.
A 5Cf/lllf is any posil h'c or ncgalh'c physical quantity that can
be compklCly spcdficd by lIS 1I111!;lIiIW/", Examples of scalar quantities
include length. mass. and lime.
Vector.
A
I'~C/Q'
is any physical qvamity that requires both a
m(lgl/illl/le and a di(utiQn for its complete description. Exilmples of
,'ectors encountered in statics arc force. position. and moment. A ,'ector
is shown graphically by 3n arrow, The le ngth of th e urrow represents the
IIIl1gllilllll" of the ,'ec tor, an d th e angle 0 betweell the vector and a fh:ed
axis defines the directioll of its lille of OfIiOil, The he,ld or tip of the arrow
indicates the Sl'IISt' of dirn:filJll of thc "eclor, Fig. 2- \'
In print, v\.'(:tor quanlilies are rcpreSo.'nled by bold face ICllers sueh as
A . and ils magnitude of Ihe ,'eelor is italicized. A, For h.mdwriuen work.
it is often oon,'e~i<:!!! 10 denote a ,'cctor quantity by simply drawing.m
arrow on top of It, JI •
Fi~.
!-I
18
CH"'PfE~
2
FO~ CE VE C tO~s
2.2
Vector Operations
Multiplication and Division of a Vector by a Scalar. If a
\'e:c tor is multiplic:d by a positil'c scalar. its magnitude is incrcascd by th~lt
amount. When multiplied b)' a negative scalar it will also change the:
di rectional sense of the ,·ector. Grdphic examples of these opcr.ltions arc
shown in Fig. 2- 2.
Vector Addition. AIl ..eclor qua ntities obey the (JllflIlld(,g,om/ow
of IUfdilimr. To illustrate, the two "compO/rem" " ee/ol'S A and B in
Fig. 2- 311 arc added to form a ",,.SII/'I11"" ,·t'c/or R = A + 8 using the:
following procedure::
• First join the tails of th e components al a point so th at il makes
them concurre:nt. Fig. 2- Jb.
• From the: head of n, draw a line p3rallclto A . Draw anothe: r line
from the he:ad of A that is parallel 10 B. These twO line:s inle:rseet at
point P 10 form the adjacent sides of a paralldogrum.
• The diagonal of this parallelogram that eXlends 10 I' forms R , which
then re: prcsenls the rcsulHlnl l'ector R .. A + B. Fig. 2- 3c.
/
~
(0)
<
(0'
••
"
"
•
K _ ,\
-+
R
l'aml"'l<J&r:I"' Low
,,)
fig. 2- J
We can 81so add B to A. Fig. 2-40. usi ng the 'filllrglt' TIfft'. which is a
special casc of Ihe parallelogram law. wh(:rcby vector H is added to
I'<:Clor A in a "head-Io-Iail"' fashion. i.c.. by connccling the head of A 10
the tail of B. Fig. 2-4b. The resultant R e~le nds from th,;: tail of A to the
he:ld of n . ln a similar manner, R can also Ix! obtained by adding A to n .
Fig. 2-4c. By comp:lrison. il is see:n thai \'ector addition is comrnul1llil'c:
in oll\e: r words. the "cctors can Ix! added in dthcr order. i.e:..
R = A + B = B + A.
2.2
R _ ,\ + 8
,.)
T1i~nKlc
VECtOR OP€RATlONS
1l _ 8 + A
Tr.angle
rule
Ib)
rul~
,<)
As a special case, if the IWO "CCIOn> A and 8 arc W/lill((lf, i.e .. bo lh
h:l\'c the lI:lmc line of aClion. the parallelogram law reduces 10 an
/I/g~b",ir or l"CI,llIr mltfirioll R = A + 8. as shown in fig. 2-5.
•
• ,
:.
Addillon of rolti J\C~' '"«tOB
Fig. 2- 5
Vector Subtraction . The resu ltant of the tfiffrrmCl' between
VCC;[ors A and
n of lh e same
Iwo
I)'PC may be ex presse d as
R' = A - II = A
+
(-8 )
This vtClor sum is shown graph ica ll y in Fig. 2-6. Subtraction is therefore
defined as 11 special case of addition. so the rules of "CClor addition also
apply 10 vector subtract;oll.
I
•
Triangle OOIImunioo
'9
20
CH"'PfE~
2
FOR CE VE C TORS
2.3
Vector Add ition of Forces
Experi ment,tl evidence has shown Ih;lt a force is a ' "CeIOT quanti l)' since
il has a specified magn itude. direction, lind sensc and il adds according [0
the parallelogram Jaw. Two common problems in Sl31ics in\'oJve eithe r
finding the resultant force. knowing its components-or resolving a known
fo rce iOio two co mpon entS- We will now describe how each of th ese
problems is soiv(!d using lhe parallelogram Jaw,
Finding a Resultant Force. The Iwocomponcnl forces F, and Fl
acting on Ihe pin in Fig. '2 - 711 can be added together to form the resultnnt
force FII :;;: FI + F!. 3S shown in Fig. 2_711. From th is tonslruttion. or u~ing
thc tri:mgle rule. Fig. 2- 71'. wc can apply the la w of cosines or the law of
sines to the triangle in order to obtain the magnitude of the resultant
force an d its direction.
The p~talkloglam bw muse he used.o
determine .he rcsullall' <If the ,wo
fOKes at1ln1 On rhe hook.
<
t·,
,.1
"
t·,
'.
y~
'.
.
.'~- .',+ I',
'M
"I
Hg. 2_7
Us,", Ihe parallelogram 13,,· force f
ClIur.cd by rt.c "cnic:ll member can be
resoh'w inlocomponcn15 Dering along
the suspt"llsion ... bles II ~nd b.
Finding th e Components of a Force. Sometimes it is
necess;lry to resolve a force into two COWPOllt'll/S in order to study its
pulling or pushing effect in two specilk direc tions. For example. in
Fig. 2-&1. F is to be resolved in to two components along the two
memhers. defined by the /I and II axes. In ordcr to determine th e
magnitude of each component. a parallelogram is constructed first. by
drawing lines stani ng fro m the tip of )<". one line paral1clto I' . and th e
other line par"Uclto tI. These lines then intersect wi th th e tI and /I axes.
forming a parallelogram. The force components F. and F~ arc then
established by simply joining the tail of F to th e intersection points on
the ,. an d II axes. Fig. 2-8b. This p:.rallelogram can then be reduced 10 a
triangle. which represents the triangle rule. Fig. 2-&. From this. the law of
sines can then be applied to determine th.:: unknown magnitudes of th.::
COmponents.
2 .3
"
21
V{CTOR AOOl110N 01' FO!tC€s
,
"""""-------"
,>,
",
Add ition of Several Forces. If more than twO forces arc to be
added. successive applications of the parallelogmm law can be carried
out in ordt'r to obtain the resultant force. For example. if three forces Fl'
FJ act at a point 0, Hg. 2-9. the resultant of any IWO of the forces is
found. say. •\ + F!-:lIId then this resultant is added to the third (orce.
yielding the resultant of all three forces: i.e.. •"If = (Fl + F2)+FJ . Using
the parallelogram law 10 add morc Ihan t ..... o forces. as shown here. oflen
requires extensive geome tric and trigo nometric c:deulalion 10 determine
Ihe numeric .. l \' ~Iues for the magnitude and direction of the result~nl.
InSlead, problems of Chis type are easily solved by using lhe "reCltlOgularcomponent method." which is e.~plained in Sec. 2.4.
."2.
"b~
r~suhan1 fore<: F. 0 11 Ih" hook
"''lui",. lhe :>ddilion of F, + ..,. Ihen thi~
,esu\1nnl i. added 10 ....
,<,
,.•
t·,
22
C",A PTER 2
F O ~ CE V ecTo Rs
Procedure for Analysis
Proble ms lhat involve the additio n of lWO forces can be solved as
follows:
0
Parallelogram law.
• Two "componenl" forces Fl and F2 in Fig. 2- 106 add according to
the parallelogram law. yielding a re:m/lalll fo rce FRt ha t fonns the
diagonal of the pa ralle logram.
,.,
• If a force )- is to Ix: resolved into ('IJmfl(Jltl'lII$ along two axes 1/
and v, Fig. 2- IOb. Ihen start allhe head of force. F ,lnd construct
lines parallcl to Ihe axes. thereby form ing Ihe paralleIOgr~ m . The
sides of the parallelogram represent the ,omponents. F~ and For
"
""
'.
"I
n.
'
;I
Trigonometry.
• Red raw a half portion of Ihe parallelogram 10 illustrate the
triangular head'lcrtaii addi tion of Ihe componenls.
C
Cosine law:
C · ,' A I .. &
Sinc la,,:
~,o a
• Label ~11 the known ~nd unknown force magnitudes and the
angles on the sketch and identify the two unk nowns as Ihe
magnilude and direction of Fl!. or th e magnitudes of its
components.
ZA lJro< c
--1L""ii~
sio b ~ 'n r
,<,
• From Ihis triangle. the magnitude of the resuhant force can Ix:
determined using the law of cosines. and its dir.::clion is
determi ned from the law of si nes. The magnitudes of two for,e
componen tS arc determined from the law of sines. The formulas
arc given in Fig. 2- lIk
~11:.l- lO
Important Points
• A sc:alar is a posi tive or negaliV!! numbe r.
• A vector is a quantity that h3s a magnitude. direclion. and sense.
• Multiplication or division of a vector by a scalar will change the
magnitude of the \·ccIOT.llle sense of the vector will change if Ihe
SC:lllar is negative.
• As a special case. if Ihe vectors arc eollinl'ar. Ihe resullant is
rormed by an algebraic or scalar addition.
2 .3
V€CfOR AoortION 01' FOItCES
EXAMPLE 2. 1
The screw eye in Fig. 2-11a is subjetled to 11'.'0 forces. . '1 and
Determine the magnitude and direction of the resultant force.
F 2•
<Kr - 25"" 65'
(.,
(>,
SOlUTION
Parallelogram Law. The parallelogram is formed by dr'lwing a line
from the head of FI thai is parallcl lO F!. and anothe r line from Ihe
head of F! Ihat is p,lraUelto Fl' The resultant force FII extends to where
these lines intersect at point A. Fig. 2- 1 lb. The two unknowns arc the
magnitude of .'It and the angle 0 (th eta).
,",
Trigo nometry. From the parallelogram. Ihe vector trian gle is
constructed. Fig. 2-lic. Usi ng the law of eusi nes
F II = '1'(100 N)2
+ (150 Nf
= '1'10000 + 22500
2(100 N)(150 N) cos liS"
30000( 0.4226) = 212.6N
= 2l3N
Fig. l-I I
Applying the Il,w ofsincs to detcrmine 0,
150 N
sinO
212.6 N
sin 115°
sin 0 = 150 N (si n 1[5°)
212.6 N
0 '" 39.So
TIIUs. the direction", (phi) of F,... measured from Ihe horizon tal. is
4>
= 39.8"
+ 15.0" =
54.8"
A m:
NOTE; The results secm reasonable. since Fig. 2-llb shows Fli to ha\'c
a magnitude larger than its components and a directi on that is
between Ihem.
"'"
23
24
C",A PTE R 2
F O ~ CE
VecToRs
EXAMP LE 2 .2
Rcsoll'e the horizontal 6OO-lb force in Fig. 2- 1211 inlU eomponcms
acting along the II and v a~l-S and determine the magnilUdl's ofthesc
components.
"
..,,,
"""
'.
c
(.,
(,'
/'
,/
fig.l-12
'"
SOLUTION
The pa rall elogra m is co nstructcd by extending a line from the /reml of
thl' 600-lb force parallel 10 the u axis until it int ersects th e II axis at
point 8. Fig. 2- 12b. 'The arrow from It to 8 n::prescn ts F.,. Similarly.
the line extended from the head of the 6(X}.lb force drawn parallclto
the /I axis intersec ts the v axis at point C. which gives F<,
1111; \'cctor addition using the triangle rule is ~hown in Fig. 2- 12c. 11lc
two unknowns arc th e magni tud es of Fu and F., Applying the law of
sines.
F"
0:
1O]91b
~= 600l b
si n 300
sin 300
All$.
NOTE: The result for FlO shows that sometimes a componc nt can ha"c
a grea u:r magnitude than the resultant.
2.3 V€CfOR AoortION 01' FOItCES
EXAMPLE 2.3
Determine the magnitude of the component force F in Fig. 2- 1311 and
the magnitude of the resultant force FR if FR is directed along the
positive y axis.
"I
"
"
,.)
'"
re)
fig. 2- 13
SOLUTIO N
The pa rallelogram law of addition is shown in Fig. 2- 13b. and the
triangle rule is shown in Fig. 2- 13c. The magn itudes of FRand .,. are the
two unknowns. They can be determined by applying the Jaw of sines.
F
sin 60"
2001b
sin 45"
F = 2451b
~ = 2001b
sin 75~
sin 45"
Am:
2S
26
C",A PTE R 2
F O ~ CE V ecTo Rs
EXAMPLE 2.4
It is req uired that the rcsultant force acting on th~ eyebolt in
Fig. 2- 14a be dircl:ted along the posHi"e .r axis and thaI F2 have a
minimum magnitude. Determine this magnitudc. the angle O. find the
,orresponding rcsultant for,e.
F,. gooN
F, _ SOON
F, _ SOO N
,
•
,~1-+-'
....
L ":
"'L,;J'---'----
,".
'.
,.,
,
,,,
I
-'-,
,<,
(>,
H I!;. 2-14
SOLUTION
The triangle rule for FI( = F, + F2 is shown in Fig. 2- I-lb. Sine:.:: the
magnitudc~ (lengths) of FI( and Fzarc not specified. then Fz,an a,tually
be any vec tor that has its head touching Ihe linc of action of 1-'1(.
fig. 2- 14c. Howt\'cr. as shown. the magnitude of F2 is., minimtlm or the
sho rt cst length whcn ils line of action is fU"I'~lIllklll(Jr 10 the line of
aClioll of Fl(. that is. when
0 = 90°
An$.
Sin,,, the ve,tor 3ddilion now forms a right tri Hnglc, the two unk nown
magnitudes can be nhl"ined by trigonomctry.
F I( = (800 N)<:os 60" = 400 N
F z = (800 N~in 60" = 693 N
AIlS.
2.3
•
27
V€CfOR AoortION 01' FOItCES
FUNDAMENTAL PROBLEMS'
f"..-I. IXlcrminc tlie magnitude of the r..sultan! force
acting on Ihe 5(rew eye and ils direction measured
clockwise from the.r axis.
HN
t"Z- t Resolve the JO..lb force ;1110 components along Ihe
"and I) a.~cs. and determine the magnitude of each of Ihew
components.
' '1-1
''1-<
P2- l. Two forces a.1 on the hook. Determine Ihe
magnitude o(\he resultant force.
t"!- S. The force F '" 450 lb acts on Ihe (rame. Resolve
this (orcc into componenls aC1ing along member.; AS and
AC. and delcml;nc Ihe mUJ;Ilitudc of cacti companelll.
H-2
t'2-5
F2- J. Determine llie magnimdc of the resultant (ora:
and it~ dire"lion measured counterclockwise from the
posi lh'c .,' axis.
t'2-<i, If force F is 10 have a componen t along the" axisof
1'. _ (>I:N, delermine the maj;l1itudc of .' and lhe
magnitude of its component F" along the !J axis.
t'2- J
• P3,":tl :«,Nul;,,"! 3nd
3n$W~"
10 all
Fund:lm~nl.1
Problems ar~ given in
Ib~
b.ct uf the bQok .
28
•
CH"'PfE~
2
FORCE VECTORS
PROBLEMS
.2-1. If 8 - 30" and T - 6 kN. determine the magnitude
of Ihe rC$ullant force acting on Ihe 1.')'1.'0011 and its dir.::clion
measured clockWise from the posi tive .( axis..
2-2. If (J _ fI.'f' and T _ 5 tN . determine Ihe magniludl.'
of the resultant force acting on the eye boll and its direction
meflSurcd clockwise from Ihe posi li\'c .J axis..
2-7. If foB - 2 kN and the re~ultanl force acts along the
positive" axis. delermine Ihe magnitude of the resultant
force and the angle 9.
"2-3.
If thc resultant force is required to aet along Ihe
posilh'e" uis and ha"c a magnitude of S kN.dclermine the
required magnilude of f Band ils direc.ion 9.
2-3. If the magnitude of the resultant forre is 10 be 9 kN
directed along Ihe posit;"c xaxis,octcrminc the magnitude of
force T ~Iing on the eyebolt and ils :Ingle O.
T
~-IJI-~-··'
Probs. 2-718
('robs. l- l/lIJ
. 2-4. Determine Ihe magniUlde of the resultant forcl.'
acting on
Ihe bracket and ils direction measured
oo\mtcrdod:wisc from Ihe po$it;I'C II
a_~is.
·Z- 5. Rcsoll'c FI inlo components along llie" and v axes.
and determine the magnitudes of these components.
2-6, Rcsol,·c F! ;n10 components along Ihe " and 1/ axes.,
and delCrnl;ne Ihe nlaJllillldcs of Ihcse oo"'pQnenl$.
' 2-9. 1lIc plDIC is subjttled to the t,,·o forces at A and 8
as shown. If 9 - «1'. determine Ihe magnitude of the
resultant of these two forces and its direction measured
clock,,·ij.C from the horizontal.
2-10. Determine the angle of 9 for connecting nlcnlbcr A
to lhe plate 5() Ihat the resultant force of f A and f a is
directed Iiorizontally to.he right. Also.. wll." is the nlagnitude
of lhe rcsUI13n! force?
•
~ F. _
f>kN
l' rob!<.l-4I5I6
I'ml~
2-9/10
2.3
2- 11. If the tellsioll in the table is 400 N. detennine tbe
tn:Igllitudc and di rection of the resultant fon:e acting on
the pulley. This angle is the same angle 0 of lille AB on the
tailboard bind:..
29
V{CTOR AOOl110N 01' FO!tC€s
!-14. [ktennine the design angle 0 to" s 0 s 90") for
strul JI B so that the 4OI).Ib horizontal force has 3
component of 500lb directed from A to ..... ards C. Wh at is the
component of force acting along member AB? Take
<b - 4Q".
!-15. Delennine the design angle <b to" s <b s 90")
bet"'ecn StrulS AB and lie so thai the 4OI).Ib horizoncal
force has a tomponem of 600 lb which aclS up 10 the left. in
Ihe same direction as from B to,,·ardsA.Takc 8 - 30".
,
.100 Ib A
"
c
I' roo. !-Il
!'rob!>. !-14I15
*2- 12. The devke is used for surgical replacement of the
knee jomt, If the force 3"illg along the kg is 360 N.
de tennine its componen ls along the .f and y ' axes.
*2-16. Resolve t'l inca components along the" 8nd u axes
and (\ctermi ne the magnitudes of these tOnlponents.
oZ_IJ. The device is used for surgical replacement of the
knec joint. If the force acting alollg the leg is 360 N.
de termme it5 components along the.t· and axes.
02-17. Resolve f ! into components along Ihe" and u axes
and dete rmine the magllHudes of these componenls.
r
F, - ZSON
!'rob§.. 2- IUIJ
!'robs. 2- 16117
30
CH"'PfE~
2
FORCE VECTORS
'l'he truel: is to be lowed using IWO ropes. Delcnn;nc
the magnitudes of forces f " and FII acting on each rope in
order 10 dC"clop a resultant force of 950 N directed along
2- 1 ~.
lite positive:r axis. ScI 6 .. 50".
2-ZJ. If 0 - 30" and 1-2 - 6 kK (\ctermine the magnitutk
of the resultant fom: acting on the plate and its dircrtion
measured dock"isc from the positi"exaxis;
.·R
2- 19. The truck is \0 be lowed using two ropn. If the
resultant fOIT(: is 10 be 950 N. directed along the posit'I'c:r
axis. determine llie magniludcs of forces FA and FJI aCling
on each rope and the angle II of Fs so thallh c magnitude of
FII is a m;II;"'''"" FII acts at 20" from the ~' :uis as showo.
is directed along a
· 2-24, If the resultant force
line mea su red 75' clockwise from the positi"e x axis and
the magnitude of F2 is to be a minimum. determine the
magnitudes of F R and ••~ and the angk 0 s 90".
,
-
-
...
~
Prob. 2- 18I19
I'robs. 2- lJI24
· 2-20. If 4> _ 45°, Fl .. 5 ~N. and the rcsuhaot (orIX is
6 tN dira:lcd along the Jl'O$ith'c )' axis.. determine lhe required
magnitude of Fz and ils direction 11.
' 1-21. If It> .. 3l)" and the !'e.,ullan! fon:c is 10 be 6 k.N
directed along !he positive y axis.. determine the magnilUOCs
of F , and Fz and the angle 0 if '"1 is required 10 be a minimum.
If <I> - 30", FI - 5 kN. and the resultant force is 10
be direcled along the positil'e y a~i!. delenninc the
magnitude of the resultant force if '"i is to be D minimum.
Also. what is '"1 and the angle 8"!
2- 22.
"robs.. 2-211/1 1/21
°z.-2S, Two forces f j and f t act on the Kreweye. If their
lin..-s of aenon ar..- at an angle 9 apart and the magnitude
of each force is FI - 1-i - F, determine the magnitude of
the resultant force FR and the angle bel,,'eCn . '/1 and .'\'
I'rob. 2-25
2.3
l-16. 'Ibe log is being towcd by two tractors , I and
n,
(}ctcrmine Ihe magniwdes of the \1'10 10"'ing forces t"" and
) "/1 if it is required that the resultant force ha"c a magnitude
1'" '" IOkN and be direCled along the.l axi'!. Sct 0 _ ISO.
2-Z7, 111e resultant .'" of tbe two foren acting on the log is
to be directed along the positive.l a.~is and rum: a magnitude
of I0 kN. determine the angle tl of the cable, aHaehcd 10 B such
Ihal lhe In.1gnitude of foTCt Fa in this (;lblc is a minimum ,
What is tl\o;> magnitude of the foTCt in each (;lblc for this
siluation'!
l-3Il, 'iliree chains aCI OII lhc bracke t slICh thatlhey create
a resultant force ha,ing a magnitude of 500 Ib, If two of thc
chains are subjected to knO"ll forces. as shown. determine
the angle Ii of the third chain measured d oc kwis<: from the
positive A' a~is. so that the magnitude of force .' in this chain
is a ",;",'mllm , All forces lie in the .t-)' plane. What is the
magnitude of .~! lIillr. First find the rcsulta nt of the 11'10
known forces. Force )" acts in this direnion .
,
I
'"
I'robs. 2-21>127
0l-l8. l'hc heam is to be hoisted using twO chain'!. Deter,
mine the magnitudes of foTCts t"" and .'~ acting 00 each ell-1in
in order to de,'clop a re:lultant fom: of 600 N directed along
the posilive)' a.~is. Seltl '" -15°.
02-29. 111(' beam is to be hoisted using t"'O ehains. lf the
resultanl force is to be 600 N directed along the positi,"C "
axis. determine the magnitudcsofforecs F" and .'"acling on
each chain and thc angle Bof . "/1 so thaI the magnitude of ) "/1
is a minimllm. FA aCtS M 3fr from the)' uis. as shown.
,•
31
V{CTOR AOOl110N 01' FO!tC€s
I' rob. 2- 30
""'"
2-J I, Three cableli pull on Ihe pipe such thatlhe}' create a
rcsul\3nt for~e havi ng a magnitude of 900 lb. If two of the
abies arc subjeetcd to known forces. assho"'n in the figure.
determine the angle /I of the third cable so thm Ihc
m.1gn itude of force F in this ~able is a m;lIjlll"'". All forces
lic in the x-y plane. Whm is the magnitude of F? "ml: Firsl
find the resultant of the two known forces.
,,
,
'9-----.
Probs. 2- 2lII!9
l'roh. 2- J I
