.A.
I.
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....
.,~
.,
STATICS P.EVmt1
Review the rree b~
diagr~.
It is ~
major tool ror understanding
and working most problems in engineering mechanics.
The types or Jjroblems are:
rom
THE:R::sULTANT OF A FORCE SYSTEM.
The RESUllXANT
is the s~les~
origiDa~ system without
may be 8 force,
~orce syst~
changing its
8 couple
or 8 force
which can reJ?lace the
effect
on a rigid
1oody. It
and & couple.
B. -Two-dimensional
~ind the direction
or the ;:'es~tant
R
.
a = tan -1 -Z
R '
x
force
--_
R
Q' = sin-l
by:
-I.
.)
R
1R
~
..-cas
R
Find the magnitude of the :~esu1tant force by:
= R sin a
c.
~-dimens
Find
i onal
the direction
of the resultant
force
by:
force
by:
R
= cos-l -z
R .'
e
x
cas
Find
2
e
x
+ cos 2 e
the magnitude
R
x
= L' F ; R
x
Find the length
in the space by:
y
y
+ cas2 e
z
= 1
of .the resultant
= t F ; R
y
.
of a line
z
= t F ; R = (R2 of-R2. + 1~2)1/2
z
or the distance
x
y
z
between two points
:
III.
~
ad
cos
e
.c1
)'"
= d cos e
%
D.
M = Fd
n.
c.
E.
1..
Find
reactions
ot' internal
at the connecting pins.
pins
by separating
1'rame or machine
cross-section.
B.
Ix'
E.
B.
For composite
or ~~.
area
of
~reas
some portion
.to be removed from the total
c.
F = yhA
A = cross-sectional
area.
IV.
I = I
+ Ad2
c
I
= moment of inert~ia
about
centroid
c
A = cross-sectionaJ~
c.
For composite
I = t (I
D.
.area
areas
c
...Ad2)
Product of Inertia
I~
Iy
~ J xydA
Ixy
of area.
"oiJy should
c.
v.
B.
.t'ma,x
= ~N = maximumPO3sible value of static
Types of' problems
.',
I
rriction
j:orce.
~
:
57;fTlcS
.CoJJ~/tA..FY'J 7
h~(,6
,e6.J/~
...rys 7~
2.35 Knowing that a = 35°, determine the resultant of the three
forces shown.
.
[.~,g~~
J
~!:!..E:!:!i
a.:.';)
FIND:
RE:s.~L. TJ.IN1" I
,,/a
/
Of =rU E TII~E!
Fig. P2.35
lIXIN fD~C'ES 5U~.
~o~
F
100-N
l=~ (/00 FOR'!.:
If) cur .3;:
t81.,211,
f = -(JI.()~) $;" J) = ->7.3'
Ii
150
Foc:,cE:
~
Fz':. + (IS'O,J)co.$
,s:~ '1J1~ ~ = -(r./lNjsil: 65"=-1~5'.1SIi
'lO()-N f'DRC;':
.
fs.,:-l1~"~5.:-1,"3.ejN.
r =-(:lrori}si,,~.-': -111f.7£,{
~E,
J
xQ.l"p.ll4)l~
~P.lN)!
Po.: -(18.~H)L
-~ r":~
II~O t4
+ ~1.1l.
-57.3'
I cr.
tro~
+6!.~'
-/3r.~:
I '
-1'3.1)
-11~.1l
I ,
200"
,
,..
i K~= -/B.szl.':-r=-3!'6.0j
«. -::Ifc:Ii: r 1.)'1.
+( .;~
-
.!J ';
-(309.0311})
I
Rt-
)
!::-= 309 N 786.'-
(J'Io~
0 .c A- CCv .P'- G
3.71 T\t,.oparallel60-N forces are applied to a lever as sho\vn. Determine .the mom:nt of the. couple fonned by the hvo forces "
3.71
...,.
~-
fJ)N
~
I..O-).J.,o~t.E.~
~
MOMI;.~' o=-~~
;---/
52)...
Hp-.VE. --M:
Fig. P3.71
-dt:.
W4i:\"\~ cl ~ 1.0.~'-"'\~~lS~-'lQ)
l~eJ-I..
M:-~~~- "i..~\~~.)1l'-ofIl)
oR.
2.0-
.~\ ~ \'2..~~ t\\.~)
~
i~Et-I..
PO/VI
3.148 A 50-lb force is applied as shown to the bracket ABC. Determine the moment of the force about A.
,...
1
~""V~~: F="..:>~ Ec..
SlU, I='Nb: M~"'~Ni
~
-Ec.
P\~~'
A
3.148
(
K
i
I-~~~
w"'~~~
t::1~,:
~~... Et.
I r-.A
~1 :. (S ,~ )~ .lc..
!,
I
Et- --(~II,.
.,
~...
"',
I
~
..- ...ts~~,1\,.)~'w~1;)
(,
j
~
-I
-~
.
---'
-:! ~tl~
Fig. P3.148
-
)~,;)":.'3..::;'~
~
--~
0..
u~,~~)~
-c.-:>~~-l~';:~~.s.)i
51N.~
:So
,~.))
-l
5-
/'
'"
-Cl!-l.
0
S t.c... ~o )""'!:1
~b'Iw)~~
v~ ~\~: (\~ 1'b.,~.)~-tl7..s.Ib-'~.)j-<.7..\l
Supportor Connection
Rocker
~umber ofI
Reac:ion
1
Frictionless
sunace
]~~:::;:~
Short link
t;n-I:nO\VtIS
1
Force with laIO""n
line of action
~~G::::=-
1
Forcewith bown
line of action
1
,.
,.
Collar on
frictionless rod
Fri~onless pin in slot
Fort:e with 1aIown
Une of action
J °:j
,. : .",.
Roughsurface
a
2
t
Folt:eofunbown
dir:ctiOD
JL
LJ
3
~c
{<.16(D
e (;~/E $
:!411~;
t
4.15 The bracketBCD is hinged at C and attachedto a control cable
;it 8. For the loadingsho..m.determine the tensionin the cable..!
a = 0.18 m
L
c
."
.:~
~ON
"i
:
:40N
A'
t
; ..0.18
~:
~0::24~~.O.
4m-l
Fig. P4.15
7; t.e,.,.. ~.!
m
D
1_- ,I
O.~m- -0.4 rnJ
0.14 m
1
k
(l./"~
I
.%.- /-
B
.1.'10"
1.tD
" :;:i1
'
."llc.
R
+
f... T~
~~~
o.¥~
-+)1"~ =0;
I
.aT
8':
~=~
T.. O.z't~
O.'f~ j
7"" (o.l.e.~)-(.2i'O",j".t~)-(.('~,..:lO.8»:-0
Tr~
E(;.~)
-+ /&00
N
7j = ~ f/J~"") : 1200 II
T =v-;;; z;-;:-;
;
4.67
7.ltOH
,
:!~O~
~8
:' V 1 b'00 : .."J ,,~ % -= .2 00" H
7«-2.J}w
For the frame and loading sho\rn, detennine the reactions at .-\
;mdC.
Fig. P4.57
-O.4m
J
c. 017((. D J)Js
~:
I.:
Sift.
i:!
'"
8 in.
t
p.,..t.-e."
8 in.
.~~'W'.JN
,-. :
ii
\,\.p."N~
Sin.
~~
." I
i,t
x. ~W~y
\."
l
'.
.:1
8111.
,-":\-;
i:I
f
12 in.
L
~::
!'!
121ft.
%
Fig. P5.3 I':"f~
.,.
'1
-.++
c.t
~zl./
IT
IC) tN.
~, IOJ.-
'L
],X. I~. -1. ,~- -"
~~.IN
10
!-"ttl4.) 1,.-,-11..111
?'2.".I~
\~~O
-"'!.~\.s~
i\Io~~.A,.~1
\~~~
Iz.'w.
I@
'(
.
14.z.,).Z6~ ,
~ IN1.
;~.
'\.I~
7..~~
I
-"'O~.\~
'l..\~'a.~
~ L.tI.: I. ~ ~
i l u.a,.I~). '1..~'B..4\
OR,.
9'Ep.,rr.;p.,
~.. i. 'l?. ,~
~l-Jb
Y<'7.'t~.,~). z.\Q,I..B
o~ y.~. Sl.:o\~. ~
.--P S".4'~ Determine by direct integrationthe cenb"oidof
he areasho..m.Expressyour anS\verin terms of a and h.
~
'~.--f
10
6 in.
p S. 3 Loc.atethe centroid of the plane areashown.
~:
YI..J'.Ne..
""'~
~~
':t..""'~~ ;
~\l.OW~
\J,,",o.l~ 't)\~~~i
\~~~~
10
r'
,-~z2
~
-
~"'t
ci~
L~
~
LL
Fig. P5.42
IHION..
~
I~=.~
S
h:w.Q. ')
~r.t ~
0.*
~~
-0-
"
~:t*
.1
"
No:.W..d"c )r.d'i
~
L.-~
':'\'
~')..
'J.=a., "\:'"'~
~
-~ ",,"~'
...'l~ d "'0
-,1..E\."
I
-~
r\~
r
b
d~:
-
h
Co
R/~'i
t':'
3
3d'1~4.
"£\.<!~. ~ i-;;::13~
Co
~D'"
:
"'lh
"3.1c.
~f)l
'b
Q:,. ...'I~
"2. '...:t~ '
,
=iL\. ~ '-\
,[
'U t
I '" '0.
L '"
\
.j~t
.
=lo.n:
t
'3
Sj3
d'i :'"2::~~ls'1
J~
0:
I
-~o.'t'h
-I,;:)
!
'
( '"'
~f.l.d~..
(
0.
lo "'\ \. ~/~ 'i
:c
i
.
'"
'r~
0 1:
Q:
('3
"~
h/"3. L , '"'I
'\
30
~0.~1.
i
i~:J'i~d~:
-;~...j;~Ld~:
'i.t{o.~),~o.l'h
~llo.~)~?o.~l
"i"~o.
:.,.~'"'
~.
;
1
~J..r
5.83 DeterminetJ:iereactionsat the beamsupportsfor the given load-
+50 IIlfft I
ill" when Wo = 150Iblft.
~
.
5.83
~ b11
~~M
~:
R~~"'t~~~
rl
Ib
-~
I.
t:---'
fM.w
.C
~ I
I_I
UCt
~~t) 4\..0A.r:.'~
Ct
~OW~.
~:
4.~~
:~J:r:D~
(~
IA ~
~of \bJPII
,
D
-I.t\.-f
I
1~1L
_cJ
I_I
~'"1ft \~I.> \~
~,c
=-~-----~
!
("
."
-\~~O
~F/-:"
:JJZIt?t:-'
\\:.
C>c
::..?
(- ~~/"")+
C'7
I
~t't'F,--;
-
~ /f4-,1..:>+ S~
-'--
\\,.
'§!"\~\bl
6.1 .T;sing
the method of joints, detennine the force in
~:lchmember of the truss sho\m. State whether each member is in tension or
,':.".
'= ,.,..-
'!=:"
r s-2>,~-r- ;7':-0
~
f7 -,.
-~ - ~..">'-~"
-=-
~
~
TI2t1.s:r
8
;-1]':;0
1--
-27:..";:-1?::.~
...~z.SC \\:;0-\:)
~'1 ~ \~
1'-~'2.7...;;(1)-Z7";:t.'rJ
(!~ -S
'l..50:!
't
.:;tJ.~=-:'-:::- :'7:--- :--0:'
C~~~~Sz..S~\b
~..S2.~1\:.'
~
...IL~"~v:
-4..)~ \'\:.-,~S~
\~
~
I,
-
-(4~.\1.,)~ ~~.t+.. "',- t2.I.\'k4,o.)~"b
-l~ "'\)l\~':.~ \\.')4l\"2.tt')~ ~~
I
.",
15.)(;i) -: t-I-:.?ic.~
.Il2.':
(,. ~o
~)LMI'..~v:
.:X-i
Q-'~'l.1
IV,'
z.tt
tll'C.t.~)tl'S~~)
"
."
I
"_,,,!':;;';.i..-'Z.7""~
_.J
.
fGo
it\'Btt.)t4.~c"C+..
'), 4~~~ \~
l'~i.~.. =-Et'~,o:
f1.'
"'l.
-=-,'r-c---
II"O(!II~
\\;1
-\
lft~
\'b
.1"-R1.
Ru"
2ft
I
fIW-l~JE"O('l
~lr)l~
}
: \...0 r\.
~\~"
~1~L-:-\'?.Lt~
~
Fig.ips.83"
r.
~4.\ t,p- tt.(
~...E...
.7'-1:?ft
...~~";.;'O~~
U
J
Wt
rr:"
D
--:~...!:
1
~
~.I~-
~..
:ompression.
1-0
t!r'fl
~
!.t...
A
-
"
~It.~~ ~c.!>y: ~\~"TI~tTRoJ_SS
-r
:-0' ' C _- 0
C :0
~..2.r %.
-1.
~
I
.)..
~"' ,
~
, ~ ~~...~~
I!J; !kN
Fig. P6.~
'-"
'B
§
~m
-+')£1-\
c.
(1.{ll ~..],'y, nl) +-C) (If.~ ",) =()
"--1-
f:.
4.5WI
-!
=D~
to
c~=- -I.Z~
kfJ
~.:1.28 kN~
d'
c.
-~
,.,
...iz. J.:J=o:i5-1.1ZlN-I.~8k.N=().
B = 3.10 J,.N~ ,
~~
If L'..f
fREE
.i
.r
!.~---
F(\!~'r': J"~If-4T..B :
FJ.c
-"'-
.5"
~
3).okH
).8)
~.10:"/oJ .
-=-
J
-
If
~6 = ~.D:J1-..t.J
C .<;&
~
fB'- ='l.lfO coAl
C 41
Fe~E t:O~'r.'!.1ol~IT C.
"2:F=o:-~F: ~.,. ", ...1~Os.N=O
~.
T .,
~
i -
C -0
F~
1:'1r:"=2..,OJcHl
~Z.-
..;1
~C=+:l.I1.l:tI
I
i
I
F;.
.:-2..71iHT~
,.c
.f'"2.f"'f=~(2.72kN)-I.ZZf.H-=O
(CH!'~)",!
5.55 A stadiumroof truss is loaded as sho\vn.Determine the force in
member
ABa'
6.91 Kno'o'-ingthat the pulley has a radius of 0.5 m, detennine the
nponentsof the reactionsat A and E.
c.
F.ea::" []CO:":
Fig. P6.91
fJ>
EN7/~
0
g
-'
~H
E"
if..~-J
dJ '
7.."..
...1$.sE"MIJL'f
.e:".
-I.f"!j
+~rMA.o: E~(7~) -("OOH;(~.r ), 0
L=!j= .:,.s-o
N
T tIF~~o:
.Aq
.j..5"
-.I:.
~=o:
~
.
r-
---7:
..
2,.,
l~
C
- FI2EE
I
-"f
700N
III
IB:)=2".5-c..-1
B E7l .4~' c
-
+trMc =0 :
(700",XI ) -;. .4)o.{3'~)
-(.zl"PHX/~) '= c.
A.~ -IS-ON
or.
j::/Zo.'.-7
L'CI.(,~):
"
/
(I)
801) '1";
-MEM
C
f
0
.~
.4.. ~.rs-o N I
,q" +£")0.":. 0
'>o~
.B
If-
~ = ~"N
4'J -+-~I'{
-;/00"";
~ :Z.S'2>N
""}:.+-Cjr.. =0
-/.sVN-r-E~
~
.4\.-=I.;~rI~
-r
It
I~T
7.37
For the beam and loading sho\m. d.z.~~
sh~
ait.J. /I1l'"~' a.1'".D.
.G!y~
1J~
.~
I
;~ I-j
~~-----
,l
{~(I1l;
I/o r/t{o
sa
.
~.
'" 1!~
Al'--:[:~_.
.~
~
~~
--
"'IH.,=o:
E('.ff)-
.I
-(,t.ifSX2ft)-(IZkif!lXtfrJ
-;~
~
!
.-(~.5t'r)(8f~)=O
E =t/~ ~i~
~ st
f1!~=O;
w.EE 1o1lv:ElJfl~! 8!1I~
I
E:.'~ kipot<3;
.:;!.~=O: 0\:0
AJT/otifl-&'k,"J
:
-(2kip~-'t.;~;~
c.O
4
~=+ 6.,Oki~ !!:E-SOllfSl<:
~.s~ ~Si@
~rj:O
I)
.-~
~ &,5"°,,:':'/2--~=O
.
,,~c-, (. 5".A."~ -.-::J
f'r~=O:
~~-=iltS.t1c t
~ -(",'5c>X/f}-(tXl);-O,ff;';
tJiJ.'F-~
<J
H<tc7toJ
8.14 The coefficients of friction are J.L.= 0.40 and J.L1c
=
0.30 bet\v-eenall surfaces of contact. Detennine the smallest force P re~uired to start the 30-kg block mO\1ngif cable AB
is attacned as sho\\11,
~:
-:fs70,-"",
-'9'~
=~.ZO
.FI...,,:
,-
Fot2t:;;"-P
K£'o",~"O Fc.i'Z. l""'o77~"7'
c:f::;:
-,
~:~;;I
.I-
-2.c=o:
;
F~~YI
,.
r-I;"=-O
~
£,t.e.:=J<
/~,z H
.
r=-&j =78.7:911
RE:"8oO7':~-..PJ.,~
W..:
.('i'
f.J'N-'.
w.,.={ 3'a.8:...17
.8/""/,,~ -m..311
D
.:-z
2o-~
W, ~(20~1.s1-'S'~=-
i '7=-1;.111
= 0,'1[19&,2"")
= .70.'rSN
Nt: /9'.£.2-'(
F~.io.
~JJ,r8
.,.
:lYt.ZH
1-%II = 11&.1J{+29f"..sIi:'
~,.r..y
4- -.--2"1-::0:
P-/j"-;:;;=O
.p= ;8.IfBJ/rl,9&'..vI
:' .?l* 7J1
P=.<7S" 1'1-4-
~I
Fig. P8.14
-1.5kips
E
D1
2 ft
2ft
Fig. P7.37
J
i-u
;
c
AG:=:,,~
.8Eft1-""~ LD~DltI;' .sNI)I/~
~.51:!f11
I.
It
l~
r"""
fi kips 12 lips
M..
.,,:.0:;""'..0
~
2ft
2ft
~
.8.19
A 12.0-1bcabinet is mounted on casters which can be locked to
prevent their rotation. The coefficient of static friction behveen the floor and
each caster is 0.30. If h = 32 in., determine the magnitude of the force P
required to move the cabinet to the right (a) if all castersare locked, (b) if the
casters at B are locked and the castersat A are free to rotate, (c) if the casters
at A are lacked and the casters at B are free to rotate.
.
fa) AL-LCMTe7z:J'" .l.~~
1
+ t ~~
-;
~.
:0:
J/4 ,...hlg-W-=o
.~tFg~~./r'.q+...y~:---f~HlfrN~
1V4+""Q=W=-I.tO/b
~
= 0,10{I.<O/.G)=-36'.to
w
3.1';'.
~
\,
I}~
r f"A-
-~
11
-.4
Ea
~ :-1'$~ = CJ..1'
He
:p
.-.I:r~:c>~
~
P:~=
o.~lIe
(I)
+.:1'V.f ,4:or
-1(12 ;..)-V.?a~y,21;.)t~(~~n,J:o
~
3.z I;',
= (1.<OJ6)(;,z
Ii;,) :0
~
-Ell
I ,.,:- I l.l~
,..' 1.1'il
'!i+~iII._~
(,z:.)c~
!.AI~(3'2J")10NI1(~;:""';)
~
fiT.4
I~.t' Jla: (/~oI6JI'<I;')
!Ill: /~1.6
EQ.{t)
P:O,ZC/oolb)=Zo
f= 3C>/h
-~
~"" Fee AT IS.
I.b
;.:.::.:::.:.
-:.: :
.6T r~~i>i
~tion
.,9~1.i
Determine the moment of inertia and the radius of
of .the'-sn~dedarea sho\m \\ith respect to the !I a.'cis.
"'~~-1 rd1-.
:
1.1
."iz.
~"f 1.~0.. ",,\,' "\1,-b:
""1..
,
b ' ~Q.\.
"'\1.: b.
TH~H
"'"
R -0.1-
t.
t~- c.G.
~~
~
Co
-0.1;
' ~~t.
\.
'1\.-b", 1,- l\.)
"N~w.. dl\. l'1'1.-,\,\d-r. 'tt~l'l..-~) -~'(."~~
'Z.'b
I; 1. ) \
.."'C::'" (,. 0. -~
Q~
a. b
~E.~..
~
0;'
F\' lc.\.\ -~ ~la.1..'I.")d~, ~~ 0."1.-~ 1,.')3:-
, ~ o.b
~
N~w.. dl,,\: "t.d~' 1.'t.t~ta.\.-~'L)d'1.1
\\o\~~.. 1,,\'! d1" ~ j ~ 1,l-l0..
To'(.'L'}d"~~t~o.~~- ~~~1
-0-
~
0.
~ '\
~!')
~
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11- q
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