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2. Objects in Equilibriumpioneer.netserv.chula.ac.th/~mkuntine/42-111/files/ch2_print.pdf · 2 5...
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1 2. Objects in Equilibrium 2142111 Statics, 2011/2 © Department of Mechanical Engineering, Chulalongkorn University 2 Objectives Students must be able to #1 Course Objective Analyze rigid bodies in equilibrium Chapter Objectives State the equations of equilibrium for various bodies (particles, 2D/3D rigid bodies) Draw free body diagrams (FBDs) of various bodies (select the body, draw the isolated body, apply loads and support reactions, add axes and dimensions, use 3 colors to differentiate the body, loads/reactions and other information) Substitute information from FBDs into equation of equilibrium in scalar & vector forms and solve for support reactions 3 Objectives Students must be able to #2 Chapter Objectives State the characteristics and identify 2 and 3-force members Determine from FBDs whether a body in equilibrium is statically determinate, statically indeterminate with redundant supports or statically indeterminate with improper supports 4 Contents Equilibrium of Objects Particles 2D Rigid Bodies 3D Rigid Bodies SD and SI Problems This is the of Statics
Transcript of 2. Objects in Equilibriumpioneer.netserv.chula.ac.th/~mkuntine/42-111/files/ch2_print.pdf · 2 5...
1
2. O
bjec
ts in
Equ
ilibr
ium
2142
111
Stat
ics,
201
1/2
©D
epar
tmen
t of M
echa
nica
l En
gine
erin
g, C
hula
long
korn
Uni
vers
ity
2
Obj
ectiv
es S
tude
nts
mus
t be
able
to #
1
Cou
rse
Obj
ectiv
eA
naly
ze ri
gid
bodi
es in
equ
ilibr
ium
Cha
pter
Obj
ectiv
esS
tate
the
equa
tions
of e
quili
briu
m fo
r var
ious
bod
ies
(par
ticle
s,
2D/3
D ri
gid
bodi
es)
Dra
w fr
ee b
ody
diag
ram
s (F
BD
s) o
f var
ious
bod
ies
(sel
ect t
he
body
, dra
w th
e is
olat
ed b
ody,
app
ly lo
ads
and
supp
ort
reac
tions
, add
axe
s an
d di
men
sion
s, u
se 3
col
ors
to
diffe
rent
iate
the
body
, loa
ds/re
actio
ns a
nd o
ther
info
rmat
ion)
Sub
stitu
te in
form
atio
n fro
m F
BD
sin
to e
quat
ion
of e
quili
briu
m in
sc
alar
& v
ecto
r for
ms
and
solv
e fo
r sup
port
reac
tions
3
Obj
ectiv
es S
tude
nts
mus
t be
able
to #
2
Cha
pter
Obj
ectiv
esS
tate
the
char
acte
ristic
s an
d id
entif
y 2
and
3-fo
rce
mem
bers
Det
erm
ine
from
FB
Ds
whe
ther
a b
ody
in e
quili
briu
m is
sta
tical
ly
dete
rmin
ate,
sta
tical
ly in
dete
rmin
ate
with
redu
ndan
t sup
ports
or
sta
tical
ly in
dete
rmin
ate
with
impr
oper
sup
ports
4
Con
tent
sE
quili
briu
m o
f Obj
ects
Par
ticle
s2D
Rig
id B
odie
s3D
Rig
id B
odie
s
SD
and
SI P
robl
ems
This
is th
e of
Sta
tics
2
5
Equi
libriu
m D
efin
ition
An
obje
ct is
in e
quili
briu
m w
hen
it is
sta
tiona
ry o
r in
stea
dy
trans
latio
n re
lativ
e to
an
iner
tial
refe
renc
e fra
me.
Equ
ilibr
ium
==
==
∑ ∑
vv
v
vv
v0
0OR R
O
FF
MM
Whe
n a
body
is in
equ
ilibriu
m, t
he re
sulta
nt fo
rce
and
the
resu
ltant
cou
ple
abo
ut a
ny p
oint
a
re b
oth
zero
.R
R
F
MO
v
v
6
Equi
libriu
m P
roce
dure
Equ
ilibr
ium
Form
ulat
e pr
oble
ms
from
phy
sica
l situ
atio
ns. S
impl
ify
prob
lem
s by
mak
ing
appr
opria
te a
ssum
ptio
nsD
raw
the
free
body
dia
gram
(FB
D)o
f obj
ects
und
er
cons
ider
atio
nS
tate
the
cond
ition
of e
quili
briu
mS
ubst
itute
var
iabl
es fr
om th
e FB
D in
to th
e eq
uilib
rium
eq
uatio
nsS
ubst
itute
the
num
bers
and
sol
ve fo
r sol
utio
nsD
elay
sub
stitu
te n
umbe
rsU
se a
ppro
pria
te s
igni
fican
t fig
ures
Tech
nica
l jud
gmen
t and
eng
inee
ring
sens
eTr
y to
pre
dict
the
answ
ers
Is th
e an
swer
reas
onab
le?
7
Equi
libriu
m F
ree
Bod
y D
iagr
am (F
BD
)
Equ
ilibr
ium
FBD
is th
e sk
etch
of t
he b
ody
unde
r con
side
ratio
n th
at is
is
olat
edfro
m a
ll ot
her b
odie
s or
sur
roun
ding
s.
The
isol
atio
n of
bod
y cl
early
sep
arat
e ca
use
and
effe
cts
of lo
ads
on th
e bo
dy.
A th
orou
gh u
nder
stan
ding
of F
BD
is m
ost v
italf
or s
olvi
ng
prob
lem
s.
8
Equi
libriu
m F
BD
Con
stru
ctio
n
Equ
ilibr
ium
No
FBD
s
Can
not a
pply
equ
ilibriu
m c
ondi
tions
Sto
p....
......
.→ →
Sel
ect t
he b
ody
to b
e is
olat
edD
raw
bou
ndar
yof
isol
ated
bod
y, e
xclu
ding
sup
ports
Indi
cate
a c
oord
inat
e sy
stem
by
draw
ing
axes
Add
all
appl
ied
load
s (fo
rces
and
cou
ples
)on
the
isol
ated
bo
dy.
Add
all
to s
uppo
rt re
actio
ns (f
orce
s an
d co
uple
s)re
pres
ent
the
supp
orts
that
wer
e re
mov
ed.
Bew
are
of lo
ads
or s
uppo
rt re
actio
ns w
ith s
peci
fic d
irect
ions
due
to
phy
sica
l mea
ning
s A
dd d
imen
sion
s an
d ot
her i
nfor
mat
ion
that
are
requ
ired
in
the
equi
libriu
m e
quat
ion
3
9
Equi
libriu
mH
elps
on
FBD
Use
diff
eren
t col
ours
in F
BD
sB
ody
outli
ne
-blu
eLo
ad (f
orce
and
cou
ple)
-red
Mis
cella
neou
s (d
imen
sion
, ang
le, e
tc.)
-bla
ck
Equ
ilibr
ium
Est
ablis
h th
e x,
yax
esin
any
sui
tabl
e or
ient
atio
n.La
bela
ll th
e kn
own
and
unkn
own
appl
ied
load
and
su
ppor
t rea
ctio
n m
agni
tude
s B
ewar
e of
load
sor
sup
port
reac
tions
with
spe
cific
di
rect
ions
due
to p
hysi
cal m
eani
ngs
Oth
erw
ise,
dire
ctio
ns o
f unk
now
n lo
ads
and
supp
ort
reac
tions
can
be
assu
med
.
10
Equi
libriu
mO
n FB
D A
naly
ses
Obj
ectiv
e: T
o fin
d su
ppor
t rea
ctio
nsA
pply
the
equa
tions
of e
quili
briu
m: s
tate
equ
atio
nsan
d su
bstit
ute
in th
e va
riabl
esfr
om th
e FB
Dun
der
cons
ider
atio
n
Load
com
pone
nts
are
posi
tive
if th
ey a
re d
irect
ed a
long
a
posi
tive
dire
ctio
n, a
nd v
ice
vers
aIt
is p
ossi
ble
to a
ssum
e po
sitiv
e di
rect
ions
for u
nkno
wn
forc
es a
nd m
omen
ts.
If th
e so
lutio
n yi
elds
a n
egat
ive
resu
lt, th
e ac
tual
load
dire
ctio
n is
opp
osite
of t
hat s
how
n in
th
e FB
D.
,,
,
0
or
0,
0,
0
0
or
0,
0,
0x
yz
OO
xO
yO
z
FF
FF
MM
MM
==
==
==
==
∑∑
∑∑
∑∑
∑∑
vv
vv
Equ
ilibr
ium
11
Part
icle
sD
efin
ition
of E
quili
briu
m
A p
artic
le is
in e
quili
briu
m w
hen
it is
st
atio
nary
or i
n st
eady
tran
slat
ion
rela
tive
to a
n in
ertia
l ref
eren
ce
fram
e.
=∑
vv 0
F
==
=∑
∑∑
0,
0,
0x
yz
FF
F
Par
ticle
Equ
ilibr
ium
12
Exam
ple
Hib
bele
rEx
3-1
#1
The
sphe
re h
as a
mas
s of
6 k
g an
d is
sup
porte
d as
sho
wn.
Dra
w a
free
-bod
y di
agra
m o
f the
sph
ere,
the
cord
CE
, the
kno
t at C
an
d th
e sp
ring
CD
.
Par
ticle
Equ
ilibr
ium
4
13
Exam
ple
Hib
bele
rEx
3-1
#2
Par
ticle
Equ
ilibr
ium
Sta
te th
e ph
ysic
al m
eani
ngs
of th
ese
9 fo
rces
Do
any
of th
ese
forc
es h
ave
spec
ific
dire
ctio
ns?
1
2
34
5
6 7 8
9
14
Exam
ple
Hib
bele
rEx
3-2
#1
Det
erm
ine
the
tens
ion
in c
able
s A
Ban
d A
Dfo
r equ
ilibr
ium
of t
he 2
50-k
g en
gine
sho
wn.
Par
ticle
Equ
ilibr
ium
Phy
sica
l mea
ning
s of
forc
esS
peci
fic d
irect
ions
?
15
Exam
ple
Hib
bele
rEx
3-2
#2
Par
ticle
is
in e
quilib
rium
.
0co
s30
0(1
)
0si
n30
0.25
kN
0(2
)
Solv
e (2
) and
sub
st. i
nto
(1)
4.90
35 k
N,
4.24
66 k
N4.
90 k
N,
4.25
kN
Ans
xB
D
yB B
D
BD
A
FT
T
FT
g
TT
TT
⎡⎤
=°−
=⎣
⎦⎡
⎤=
°−=
⎣⎦
==
==
∑ ∑
29.
807
m/s
g=
Par
ticle
Equ
ilibr
ium
16
Exam
ple
Hib
bele
rEx
3-3
#1
If th
e sa
ck a
t Aha
s a
wei
ght o
f 20
lb,
dete
rmin
e th
e w
eigh
t of t
he s
ack
at B
and
the
forc
e in
eac
h co
rd n
eede
d to
ho
ld th
e sy
stem
in th
e eq
uilib
rium
po
sitio
n sh
own.
Par
ticle
Equ
ilibr
ium
5
17
Exam
ple
Hib
bele
rEx
3-3
#2
Par
ticle
Equ
ilibr
ium
18
Exam
ple
Hib
bele
rEx
3-3
#3
⎡⎤
=°−
°=
⎣⎦
⎡⎤
=°−
°−=
⎣⎦
==
==
∑ ∑Con
side
r rin
g in
equ
ilibriu
m
0si
n30
cos4
50
(1)
0co
s30
sin
45(2
0 lb
)0
(2)
Solv
e (1
)&(2
),38
.637
lb,
54.6
41 lb
38.6
lb,
54.6
lbA
ns
xE
GE
C
yE
GE
C
EC
EG
EC
EG
E
FT
T
FT
T
TT
TT
Par
ticle
Equ
ilibr
ium
19
Exam
ple
Hib
bele
rEx
3-3
#4
⎡⎤
=°−
=⎣
⎦⎡
⎤=
+°−
=⎣
⎦
==
==
∑ ∑FBD
of
in e
quilib
rium
0(
lb)c
os45
(45)
0(3
)
0(3
5)(
lb)s
in45
0(4
)
Solv
e (3
)&(4
),
34.1
51 lb
, 47
.811
lb34
.2 lb
, 47
.8 lb
Ans
xE
CC
D
yC
DE
CB
CD
B
CD
B
C
FT
T
FT
TW
TW
TW
Par
ticle
Equ
ilibr
ium
20
Part
icle
sEq
uilib
rium
in 3
D
=
++
=
∑ ∑
vv
v0 ˆ
ˆˆ
()
0x
yz
F Fi
Fj
Fk
= = =
∑ ∑ ∑
0 0 0
x y zF F F
v 1F
v 2F
v 3F
xy
z
Par
ticle
Equ
ilibr
ium
6
21
Exam
ple
Hib
bele
rEx
3-5
#1
A 9
0-N
load
is s
uspe
nded
from
the
hook
. The
load
is s
uppo
rted
by tw
o ca
bles
and
a s
prin
g ha
ving
a
stiff
ness
k=
500
N/m
. Det
erm
ine
the
forc
e in
the
cabl
es a
nd th
e st
retc
h of
th
e sp
ring
for e
quili
briu
m. C
able
AD
lies
in th
e x-
ypl
ane
and
cabl
e A
Clie
s in
the
x-z
plan
e.
Par
ticle
Equ
ilibr
ium
22
Exam
ple
Hib
bele
rEx
3-5
#2
[]
Equ
ilibriu
m o
f FB
D o
f
0si
n30
(45)
0(1
)
0co
s30
0(2
)
0(3
5)(9
0 N
)0
(3)
Solv
e (3
), (1
), (2
):
150
N,
240
N,
207.
85 N
208
NAn
s
/(2
07.8
5 N
)/(5
0
xD
C
yD
B
zC C
DB
sB
AB
AB
B
A
FF
F
FF
F
FF F
FF
Fks
Fks
sF
k
⎡⎤
=°−
=⎣
⎦⎡
⎤=
−°+
=⎣
⎦⎡
⎤=
−=
⎣⎦
=
==
=
== =
=
∑ ∑ ∑
0 N
/m)
0.41
569
m0.
416
mAn
sA
Bs
==
Par
ticle
Equ
ilibr
ium
23
Exam
ple
Hib
bele
rEx
3-7
#1
Det
erm
ine
the
forc
e de
velo
ped
in e
ach
cabl
e us
ed to
sup
port
the
40-
kN c
rate
.
Par
ticle
Equ
ilibr
ium
24
Exam
ple
Hib
bele
rEx
3-7
#2P
ositi
on v
ecto
rsˆ
ˆˆ
34
8 m
ˆˆ
ˆ3
48
m
Forc
e ve
ctor
s
ˆˆ
ˆ(
34
8)
89
ˆˆ
ˆ(
34
8)
89ˆ
ˆ40
kN
AB
AC
AB
BB
BA
B AC
CC
CA
C
DD
ri
jk
ri
jk
rF
FF
ij
kr r
FF
Fi
jk
r
FF
i
Wk
=−
−+
=−
++
==
−−
+
==
−+
+
= =−
v v
vv
vv v v
Par
ticle
Equ
ilibr
ium
7
25
Exam
ple
Hib
bele
rEx
3-7
#3
⎡⎤
=⎣
⎦+
++
=
−−
++
−+
++
+−
=
−−
−+
++
++
−=
∑v
v
vv
vv
v
v v
Equ
ilibriu
m o
f
0
0
ˆˆ
ˆˆ
ˆˆ
ˆˆ
(3
48
)(
34
8)
()
(40
kN
)0
8989
34
83
48
ˆˆ
ˆ(
)(
)(
40 k
N)
089
8989
8989
89
BC
D
CB
D
CC
CB
BB
D
A
F
FF
FW
FF
ij
ki
jk
Fi
k
FF
FF
FF
Fi
jk
Par
ticle
Equ
ilibr
ium
26
Exam
ple
Hib
bele
rEx
3-7
#4
33
dir.
:0
(1)
8989 4
4 d
ir.:
0(2
)89
898
8 d
ir.:
(40
kN)
0(3
)89
89Fr
om (2
),16
From
(3),
(40
kN)
0 or
23
.585
kN
89 6Fr
om (1
),0
or
15.0
kN
8923
.6 k
N,
15.0
kN
Ans
CB
D
CB
CB
BC
CC
CD
D
CB
D
FF
xF
FF
y
FF
z
FF
FF
FF
F
FF
F
−−
+=
−+
=
+−
=
=
−=
=
−+
==
==
=
Par
ticle
Equ
ilibr
ium
27
Equi
libriu
m o
f 2D
Rig
id B
odie
s
Use
sim
ilar a
naly
ses
as th
ose
for p
artic
les
Add
ition
al c
onsi
dera
tion
Act
ion
forc
es in
sup
ports
/con
stra
ints
Free
-bod
y di
agra
m (F
BD
) of 2
D ri
gid
bodi
esE
quili
briu
m e
quat
ions
(sca
lar f
orm
) for
rigi
d bo
dies
Two-
forc
e an
d th
ree-
forc
e m
embe
rs
2D E
quili
briu
m
28
= = =
∑ ∑ ∑
0 0 0
x y OF F M
Equi
libriu
m 2D
Equ
atio
nsS
cala
r For
m
The
sum
of t
he m
omen
t abo
ut a
ny p
oint
Ois
zer
o.
Par
ticle
Rig
id B
ody
2D E
quili
briu
m
8
29
2D S
uppo
rts
Pin
Join
ts
The
pin
supp
ort
prev
ents
the
trans
latio
nby
exe
rting
a re
actio
n fo
rce
on th
e ba
ral
low
s a
free
rota
tion
so it
doe
sno
t exe
rt a
coup
le o
n th
e ba
r.
2D E
quili
briu
m
30
2D S
uppo
rts
Rol
ler S
uppo
rts
The
rolle
r sup
port
prev
ents
the
trans
latio
n in
y-d
irect
ion
by e
xerti
ng a
reac
tion
forc
e to
the
bar.
allo
ws
a tra
nsla
tion
in x
dire
ctio
n an
d a
free
rota
tion
so it
do
es n
ot e
xert
and
a co
uple
to th
e ba
r.
2D E
quili
briu
m
31
2D S
uppo
rts
Fixe
d Su
ppor
ts
The
fixed
or b
uilt-
in s
uppo
rtpr
even
ts a
ll tra
nsla
tion
and
rota
tion
by e
xerti
ng a
reac
tion
forc
e an
d a
coup
le to
the
bar.
2D E
quili
briu
m
32
2D S
uppo
rts
Sum
mar
y #1
2D E
quili
briu
m
9
33
2D S
uppo
rts
Sum
mar
y #2
2D E
quili
briu
m
34
2D S
uppo
rts
Sum
mar
y #3
2D E
quili
briu
m
35
Exam
ple
Hib
bele
rEx
5-1
#1D
raw
the
free-
body
dia
gram
of t
he u
nifo
rm b
eam
sho
wn.
The
beam
has
the
wei
ght o
f 981
N.
2D E
quili
briu
m
36
Exam
ple
Hib
bele
rEx
5-1
#2
2D E
quili
briu
m
10
37
Exam
ple
Hib
bele
rEx
5-1
#3
2D E
quili
briu
m
Find
: ,
, E
quilib
rium
of
00,
0
Ans
0(1
200
N)
(981
N)
0
2181
N2.
18 k
NA
ns
0(1
200
N)(
2 m
)(9
81 N
)(3
m)
0
5343
Nm
5.34
kN
mA
ns
yy
A
xx
x
yy y
AA A
AA
M
AG
FA
A
FA B
MM M
⎡⎤
=−
==
⎣⎦
⎡⎤
=−
−=
⎣⎦
==
⎡⎤
=−
−=
⎣⎦
=⋅
=⋅
∑ ∑ ∑
38
Exam
ple
Hib
bele
rEx
5-3
#1
Two
smoo
th p
ipes
, eac
h ha
ving
the
wei
ght o
f 294
3 N
, are
su
ppor
ted
by th
e fo
rks
of th
e tra
ctor
. D
raw
the
FBD
sfo
r eac
h pi
pe a
nd b
oth
pipe
s to
geth
er.
2D E
quili
briu
m
39
Exam
ple
Hib
bele
rEx
5-3
#2
2D E
quili
briu
m
40
Exam
ple
Hib
bele
rEx
5-3
#3
2D E
quili
briu
m
Pip
es in
equ
ilibriu
mFi
nd
, ,
, T
PF
R
11
41
Exam
ple
Hib
bele
rEx
5-3
#4
2D E
quili
briu
m 0
(294
3 N
)sin
30°
0 14
71.5
N1.
47 k
NAn
s
0
(294
3 N
)cos
30°
0 25
48.7
N2.
55 k
NA
ns
x yF
R R
F
P P⎡⎤
=⎣
⎦−
==
=
⎡⎤
=⎣
⎦−
==
=
∑ ∑
42
Exam
ple
Hib
bele
rEx
5-3
#5
2D E
quili
briu
m 0(2
943
N)s
in30
°0
2943
N2.
94 k
NAn
s
0(2
943
N)c
os30
°0
2548
.7 N
2.55
kN
Ans
x yFR
T
T
FF F
⎡⎤
=−
−+
=⎣
⎦=
=
⎡⎤
=−
=⎣
⎦=
=
∑ ∑
43
Exam
ple
Hib
bele
rEx
5-6
#1
Det
erm
ine
the
horiz
onta
l and
ver
tical
com
pone
nts
of
reac
tion
for t
he b
eam
load
ed a
s sh
own.
Neg
lect
the
wei
ght o
f the
bea
m in
the
calc
ulat
ion.
2D E
quili
briu
m
44
Exam
ple
Hib
bele
rEx
5-6
#2
⎡⎤
=°−
=⎣
⎦=
=
⎡⎤
=+
°−
⎣⎦
°−
=
==
⎡⎤
=−
°−⎣
⎦
∑ ∑ ∑Find
: ,
,
Equ
ilibriu
m o
f
0(6
00 N
)cos
450
424.
26 N
424
NAn
s
0(1
00 N
)(2 m
)(6
00 N
)sin
45(5
m)
(600
N)c
os45
(0.2
m)
(7 m
)0
319.
5031
9 N
Ans
0(6
00 N
)sin
45(1
00
yx
y
xx
x
B
y
y
yy
AB
B AD
B
FB
B
M
A
A
FA
−+
=
==
N)
(200
N)
0
404.
76 N
405
NAn
sy
y
B
B
2D E
quili
briu
m
12
45
Exam
ple
Hib
bele
rEx
5-8
#1
The
link
show
n is
pin
-con
nect
ed a
t Aan
d re
sts
agai
nst a
sm
ooth
sup
port
at B
. C
ompu
te th
e ho
rizon
tal a
nd v
ertic
al
com
pone
nts
of re
actio
ns a
t the
pin
A.
2D E
quili
briu
m
46
Exam
ple
Hib
bele
rEx
5-8
#2
FBD
of l
ink
BA
2D E
quili
briu
m
47
Exam
ple
Hib
bele
rEx
5-8
#3
⎡⎤
=+
−⋅
−+
=⎣
⎦=
⎡⎤
=→
+−
°=
=⎣
⎦⎡
⎤=
↑+
−°−
=⎣
⎦=
=
∑ ∑ ∑Equ
ilibriu
m o
f FB
D o
f lin
k
0 (9
0 N
m)
(60
N)(1
m)
(0.7
5 m
)0
200
NA
ns
0 si
n30
0,10
0 N
Ans
0 co
s30
(60
N)
0
233.
21 N
233
NA
ns
AB
B
xx
Bx
yy
B
y
BA
MN
N
FA
NA
FA
N
A
2D E
quili
briu
m
48
Spec
ial M
embe
r 2-F
orce
Mem
ber #
1
The
obje
ct is
a tw
o-fo
rce
mem
ber
whe
n su
bjec
ted
to tw
o eq
uiva
lent
forc
es
actin
g at
diff
eren
t poi
nts.
2D E
quili
briu
mD
efin
ition
For a
two-
forc
e m
embe
rin
equi
libriu
m, f
orce
sar
e eq
ual i
n m
agni
tude
.ar
e op
posi
te in
dire
ctio
n.ha
ve th
e sa
me
line
of a
ctio
n.Th
e ar
e no
othe
r loa
ds, e
.g.w
eigh
t
The
abili
ty to
reco
gniz
e 2-
forc
e m
embe
rs is
im
porta
nt in
the
anal
yses
of s
truct
ures
.
13
49
Spec
ial M
embe
r 2-F
orce
Mem
ber #
2E
xam
ple
2D E
quili
briu
m
50
Spec
ial M
embe
r 3-F
orce
Mem
ber #
1
The
obje
ct is
a th
ree-
forc
e m
embe
rwhe
n su
bjec
ted
to
thre
e eq
uiva
lent
forc
es a
ctin
g at
diff
eren
t poi
nts.
If a
thre
e fo
rce
mem
ber i
s in
equ
ilibr
ium
, the
thre
efo
rces
are
:co
plan
ar a
ndei
ther
par
alle
l or c
oncu
rren
t.
Def
initi
on2D
Equ
ilibr
ium
51
Spec
ial M
embe
r 3-F
orce
Mem
ber #
2
1Fv
2Fv
3FvO
3Fv
2Fv
1Fv
Con
curr
ent F
orce
sP
aral
lel F
orce
s
2D E
quili
briu
m
52
Exam
ple
Hib
bele
rEx
5-13
#1
The
leve
r AB
Cis
pin
-sup
porte
d at
Aan
d co
nnec
ted
to a
sho
rt lin
k B
Das
sho
wn.
If th
e w
eigh
t of
the
mem
bers
is n
eglig
ible
, de
term
ine
the
forc
e of
the
pin
on
the
leve
r at A
.
2D E
quili
briu
m
14
53
Exam
ple
Hib
bele
rEx
5-13
#2
2D E
quili
briu
m
Equ
ilibriu
m o
f FB
D o
f
0
(400
N)(
0.7
m)
cos4
5(0
.2 m
)si
n45
(0.1
m)
013
19.9
N1.
32 k
NA
ns
0
(400
N)
cos4
50
533.
33 N
533
kNA
ns
0
sin
450
933.
33 N
933
A
BD
BD
BD
x
BD
x
x
y
BD
y
y
AB
C
M
FF
F
F
FA
A
F
FA
A⎡⎤
=⎣
⎦−
+°
+
°=
==
⎡⎤
=⎣
⎦−
°+=
==
⎡⎤
=⎣
⎦−
°+=
==
∑ ∑ ∑
kNA
ns
54
Exam
ple
Hib
bele
rEx
5-13
#3
BD
is a
two
forc
e m
embe
r
AB
Cis
a th
ree-
forc
e m
embe
r.M
ust b
e co
ncur
rent
at O
10.
7ta
n60
.255
0.4
θ−
⎛⎞
==
°⎜
⎟⎝
⎠
2D E
quili
briu
m
55
Exa
mpl
e H
ibbe
ler
Ex
5-13
#4
⎡⎤
=→
+⎣
⎦°−
°+=
⎡⎤
=↑
+⎣
⎦°−
°=
==
==
∑ ∑Equ
ilibriu
m o
f FB
D o
f
0
cos6
0.25
5co
s45
(400
N)
0 (1
)
0
sin
60.2
55si
n45
0 (
2)
Sol
ve (1
) & (2
)10
74.9
N1.
07 k
N13
19.9
N1.
32 k
NA
ns
x
A
y
A A
AB
C
F
FF
F
FF
F F
2D E
quili
briu
m
56
Exam
ple
Hib
bele
rEx
5-4
#1
Dra
w th
e FB
D o
f the
unl
oade
d pl
atfo
rm th
at is
sus
pend
ed o
ff th
e ed
ge
of th
e oi
l rig
sho
wn.
The
pla
tform
has
the
wei
ght o
f 196
2 N
.
2D E
quili
briu
m
15
57
Exam
ple
Hib
bele
rEx
5-4
#2
2D E
quili
briu
m
Pla
tform
in e
quilib
rium
: Fin
d ,
,
0
cos7
0(1
m)
sin7
0(2
.2 m
)(1
962
N)(1
.4 m
)0
1140
.1 N
1.14
kN
Ans
0
cos7
00
389.
92 N
390
NA
ns
0
sin7
0(1
962
N)
0
890.
69 N
89
yy
A x
x x
y
y y
AA
T
M
TT
T
F
AT
A
F
AT
A⎡⎤
=⎣
⎦°
+°
−=
==
⎡⎤
=⎣
⎦−
°=
==
⎡⎤
=⎣
⎦+
°−=
==
∑ ∑ ∑
1 N
Ans
58
Exam
ple
Hib
bele
rEx
5-4
#3
2D E
quili
briu
m
Pla
tform
in e
quilib
rium
: Fin
d ,
,
tan7
0,
2.19
80 m
0.8
m1
mta
n,
66.3
571.
4 m
0
cos
cos7
00
(1)
0
sin
sin7
0(1
962
N)
0(2
)So
lve
(1) &
(2)
1140
.1 N
, 97
2.30
N1.
14 k
A
x
A
y
A
A
FT
dd
d
F
FT
F
FT
TF
T
θ
θθ
θ θ°=
=
+=
=°
⎡⎤
=⎣
⎦−
°=
⎡⎤
=⎣
⎦+
°−=
==
=∑ ∑
N,
972
N,
66.4
Ans
AFθ
==
°
59
Equi
libriu
m o
f 3D
Rig
id B
odie
s3D
Equ
ilibr
ium
60
3D S
uppo
rts
Sum
mar
y #1
3D E
quili
briu
m
16
61
3D S
uppo
rts
Sum
mar
y #2
3D E
quili
briu
m
62
3D S
uppo
rts
Sum
mar
y #3
3D E
quili
briu
m
63
3D S
uppo
rts
Sum
mar
y #4
Com
paris
on w
ith 2
D s
uppo
rts
3D E
quili
briu
m
64
Exam
ple
Hib
bele
rEx
5-14
#1
3D E
quili
briu
m
17
65
Exam
ple
Hib
bele
rEx
5-14
#2
3D E
quili
briu
m
66
Exam
ple
Hib
bele
rEx
5-14
#3
3D E
quili
briu
m
67
Exam
ple
Hib
bele
rEx
5-14
#4
3D E
quili
briu
m
68
3D E
quili
briu
m R
igid
Bod
ies
Act
ion
forc
es in
sup
ports
Equ
ilibr
ium
con
ditio
ns = =
∑ ∑
vv
vv0 0
OF MV
ecto
r for
m
==
==
==
∑∑
∑∑
∑∑
00
0an
d0
00
xx
yy
zz
FM
FM
FM
Sca
lar f
orm
3D E
quili
briu
m
18
69
Exam
ple
Hib
bele
rEx
5-15
#1
The
hom
ogen
eous
pla
te
show
n ha
s a
mas
s of
100
kg
and
is s
ubje
cted
to a
forc
e an
d co
uple
mom
ent a
long
its
edg
es.
If it
is s
uppo
rted
in th
e ho
rizon
tal p
lane
by
mea
ns o
f a ro
ller a
t A, a
ba
ll-an
d-so
cket
join
t at B
, an
d a
cord
at C
, det
erm
ine
the
com
pone
nts
of re
actio
n at
the
supp
orts
.
3D E
quili
briu
m
70
Exam
ple
Hib
bele
rEx
5-15
#2
FBD
of p
late
⎡⎤
==
⎣⎦
⎡⎤
==
⎣⎦
∑ ∑Equ
ilibriu
m o
f pla
te
00
00
xx
yy
AB
C
FB
FB
3D E
quili
briu
m
71
Exam
ple
Hib
bele
rEx
5-15
#3
⎡⎤
=+
+−
−=
⎣⎦
⎡⎤
=−
+=
⎣⎦
⎡⎤
=+
−⎣
⎦−
−⋅
=
∑ ∑ ∑Equ
ilibriu
m o
f pla
te
0(3
00 N
)(9
80.7
N)
0(1
)
0(2
m)
(980
.7 N
)(1 m
)(2
m)
0(2
)
0(3
00 N
)(1.5
m)
(980
.7 N
)(1.5
m)
(3 m
)
(3 m
)(2
00 N
m)
0(3
)
zz
zC
xC
z
yz
z
AB
C
FA
BT
MT
B
MB
A
==
−
= ==
==
=−
Sol
ve (1
), (2
) & (3
)79
0.35
N,
216.
67 N
, 70
7.02
N
790
N,
707
N,
0,
217
N
Ans
zz
C zC
xy
z
AB
T AT
BB
B
3D E
quili
briu
m
72
Exam
ple
Hib
bele
rEx
5-15
#4
′ ′
⎡⎤
=+
+−
−=
⎣⎦
⎡⎤
=+
−=
⎣⎦
⎡⎤
=−
−−
⋅⎣
⎦−
=
∑ ∑ ∑Alte
rnat
ivel
y, fo
r equ
ilibriu
m o
f pla
te
0(3
00 N
)(9
80.7
N)
0(1
)
0(9
80.7
N)(1
m)
(300
N)(2
m)
(2 m
)0
(4)
0(3
00 N
)(1.5
m)
(980
.7 N
)(1.5
m)
(200
Nm
)
(3 m
)0
(5)
zz
zC
xz
y
C
AB
C
FA
BT
MA
M
T
= =−
=
Sol
ve (1
), (4
) & (5
)79
0.35
N,
216.
67 N
, 70
7.02
N
z z CA B T
3D E
quili
briu
m
19
73
Exam
ple
Hib
bele
rEx
5-16
#1
The
win
dlas
s sh
own
is
supp
orte
d by
a th
rust
be
arin
g at
Aan
d a
smoo
th
jour
nal b
earin
g at
B, w
hich
ar
e pr
oper
ly a
ligne
d on
the
shaf
t. D
eter
min
e th
e m
agni
tude
of t
he v
ertic
al
forc
e P
that
mus
t be
appl
ied
to th
e ha
ndle
to m
aint
ain
equi
libriu
m o
f the
100
kg
buck
et.
Als
o ca
lcul
ate
the
reac
tions
at t
he b
earin
gs.
3D E
quili
briu
m
74
Exam
ple
Hib
bele
rEx
5-16
#2
FBD
of w
indl
ass
3D E
quili
briu
m
75
Exa
mpl
e H
ibbe
ler
Ex
5-16
#3
⎡⎤
=−
°=
⎣⎦
=
⎡⎤
=−
++
=⎣
⎦=
⎡⎤
=−
=→
=⎣
⎦
∑ ∑ ∑Equ
ilibriu
m o
f win
dlas
s
0(9
81 N
)(0.1
m)
(0.3
m)c
os30
0
377.
59 N
0(9
81 N
)(0.5
m)
(0.8
m)
(0.4
m)
0
424.
33 N
0(0
.8 m
)0
0
x yz
z
zy
y
MP
P
MA
P
A
MA
A
3D E
quili
briu
m
76
Exam
ple
Hib
bele
rEx
5-16
#4
⎡⎤
==
⎣⎦
⎡⎤
=+
=→
=⎣
⎦⎡
⎤=
−+
−=
⎣⎦
−+
−=
→=
==
==
==
∑ ∑ ∑
00
00
0
0
(981
N)
0
(424
.33
N)
(981
N)
(377
.59
N)
0 934.
26 N
378
N,
0,
424
N,
0,
934
NA
ns
xx
yy
yy
zz
z
z
z
xy
zy
z
FA
FA
BB
FA
BP
BB
PA
AA
BB
3D E
quili
briu
m
20
77
Exam
ple
Hib
bele
rEx
5-18
#1
Rod
AB
sho
wn
is s
ubje
cted
to th
e 20
0-N
forc
e. D
eter
min
e th
e re
actio
ns a
t the
bal
l-and
-soc
ket j
oint
Aan
d th
e te
nsio
n in
cab
les
BD
and
BE
.
3D E
quili
briu
m
78
Exam
ple
Hib
bele
rEx
5-18
#2
FBD
of r
od A
B
=+
+
==
==
+−
=−
=+
−
v vv
v
vv
vv
ˆˆ
ˆ(
) Nˆ N
0.5
ˆˆ
ˆˆ
N(0
.51
) mˆ
ˆˆ
ˆ20
0 N
(2
2) m
xy
z
EE
CB
DD
C B
AA
iA
jA
k
TT
ir
r
TT
jr
ij
k
Fk
ri
jk
3D E
quili
briu
m
79
Exam
ple
Hib
bele
rEx
5-18
#3
⎡⎤
=⎣
⎦+
++
=
++
++
+=
⎡⎤
=+
=⎣
⎦⎡
⎤=
+=
⎣⎦
⎡⎤
=−
=⎣
⎦
∑ ∑ ∑ ∑
vv
vv
vv
v
v
Equ
ilibriu
m o
f bod
y
0
0ˆ
ˆˆ
()
()
()
0
0 0
(1)
00
(2)
0(2
00 N
)0
(3)
ED
xE
yD
z
xx
E
yy
D
zz
AC
B
F
AT
TF
AT
iA
Tj
AF
k
FA
T
FA
T
FA
3D E
quili
briu
m
80
Exam
ple
Hib
bele
rEx
5-18
#4
⎡⎤
=⎣
⎦×
+×
+=
×+
×+
=
×+
+=
+−
×+
+=
⎡⎤
=−
⋅=
⎣⎦
⎡⎤
=−
+⋅
⎣⎦
∑ ∑ ∑
vv
vv
vv
vv
vv
vv
vv
vv
vv
v
vv
vv
Equ
ilibriu
m o
f bod
y
0
()
()
0
(0.5
)(
)0
(0.5
)0
ˆˆ
ˆ(
22
) m(0
.5)
0
0 (2
m)
(200
Nm
)0
(1)
0(
2 m
)(1
00 N
m
A
CB
ED
BB
ED
BE
D
ED
xD
yE
AC
B
M
rF
rT
T
rF
rT
T
rF
TT
ij
kF
TT
MT
MT
=
⎡⎤
=−
=⎣
⎦∑
)0
(2)
0(1
m)
(2 m
)0
(3)
zD
EM
TT
3D E
quili
briu
m
21
81
Exam
ple
Hib
bele
rEx
5-18
#5
= = =−
=−
=
Sol
ve e
qns
(1) t
hrou
gh (6
)10
0 N
50 N 50
N10
0 N
200
NA
ns
The
nega
tive
sign
indi
cate
s th
at
and
h
ave
a di
rect
ion
whi
ch is
opp
osite
to th
ose
show
n in
the
FBD
.
D E x y z
xy
T T A A A
AA
3D E
quili
briu
m
82
Exam
ple
Hib
bele
rEx
5-19
#1
The
bent
rod
is s
uppo
rted
at A
by a
jour
nal b
earin
g, a
t Dby
a
ball-
and-
sock
et jo
int,
at B
by m
eans
of c
able
BC
. U
sing
onl
y on
e eq
uilib
rium
eq
uatio
n, o
btai
n a
dire
ctso
lutio
n fo
r the
tens
ion
in c
able
B
C.
The
bear
ing
at A
is
capa
ble
of e
xerti
ng fo
rce
com
pone
nts
only
in th
e z
and
ydi
rect
ions
sin
ce it
is p
rope
rly
alig
ned
on th
e sh
aft.
3D E
quili
briu
m
83
Exam
ple
Hib
bele
rEx
5-19
#2
FBD
of A
BE
D
3D E
quili
briu
m
84
Exam
ple
Hib
bele
rEx
5-19
#3
=⋅
×=
∑∑
vv
Sum
min
g m
omen
ts a
bout
an
axis
pas
sing
th
roug
h po
ints
a
nd
.ˆ
()
0D
AD
A
DA
Mu
rF
==
−−
v
v
any
poin
tan
y po
int
Rem
inde
r: is
a p
ositi
on v
ecto
r fro
m
on
the
axi
s to
in
the
line
of a
ctio
n of
the
forc
e.1
1ˆ
ˆˆ
22
DA
DA
DA
rD
A
ru
ij
r
3D E
quili
briu
m
22
85
Exam
ple
Hib
bele
rEx
5-19
#4
×=
×+
×
=−
×−
++
−×
−
=−
++
⋅
=⋅
×=
=−
−⋅
=
−
∑ ∑∑
vv
vv
vv
v
v
v
vv
()
0.2
0.3
0.6
ˆˆ
ˆ(
1) m
() N
0.7
0.7
0.7
ˆ(
0.5
) m(
981
) Nˆ
[(0.
8571
449
0.5)
0.28
571
] Nm
Equ
ilibriu
m o
f bod
y ˆ
()
01
1ˆ
ˆ[
][(
0.85
714
22
BB
E BB
B
BB
DA
DA
rF
rT
rW
jT
iT
jT
k
jk
Ti
Tk
AB
ED
Mu
rF
ij
T+
+⋅
=−
−+
⋅=
==
vˆ
490.
5)0.
2857
1] N
m
1(
)(0.
8571
449
0.5)
Nm
02
572.
25 N
572
NA
ns
BB
B
B
iT
k
T
T
3D E
quili
briu
m
86
Sta
tical
ly d
eter
min
ate
(SD
) pro
blem
s ca
n be
ana
lyze
d us
ing
the
equi
libriu
m c
ondi
tion
alon
e.
Stat
ical
ly D
eter
min
ate
Obj
ects
SD
/SI
==
==
∑ ∑
vv
v
vv
v0
0OR R
O
FF
MM
87
An
obje
ct w
ith re
dund
ant s
uppo
rts
An
obje
ct w
ith im
prop
er s
uppo
rts
Stat
ical
ly In
dete
rmin
ate
Obj
ects
SD
/SI S
tatic
ally
inde
term
inat
e (S
I) pr
oble
ms
can
not b
e an
alyz
e w
ith e
quili
briu
m e
quat
ions
.
88
SI O
bjec
tsR
edun
dant
Sup
port
s #1
An
obje
ct h
as re
dund
ant s
uppo
rts, w
hen
it ha
s m
ore
supp
orts
than
the
min
imum
num
ber n
eces
sary
to m
aint
ain
it in
equ
ilibr
ium
.
Res
ult
Mor
e un
know
n fo
rces
or c
oupl
es th
an th
e nu
mbe
r of
inde
pend
ent e
quilib
rium
equ
atio
ns.
SD
/SI
23
89
SI O
bjec
tsR
edun
dant
Sup
port
s #2
5 U
nkno
wns
3 E
quilib
rium
Equ
atio
ns
SD
/SI
90
SI O
bjec
tsR
edun
dant
Sup
port
s #3
8 U
nkno
wns
6 E
quilib
rium
Equ
atio
ns
SD
/SI
91
Exam
ple
Red
unda
nt S
uppo
rts
#1
⎡⎤
==
⎣⎦
⎡⎤
=−
+=
⎣⎦
⎡⎤
=−
−+
=⎣
⎦
∑ ∑ ∑Equ
ilibriu
m o
f bea
m
00
00
0(2
)0
xx
yy
A
AB
FA
FA
FB
MM
FLB
L
3 eq
uatio
ns v
s 4
unkn
own
reac
tions
SD
/SI
92
Exam
ple
Red
unda
nt S
uppo
rts
#2
⎡⎤
=+
=⎣
⎦⎡
⎤=
+−
=⎣
⎦⎡
⎤=
−+
=⎣
⎦
∑ ∑ ∑Equ
ilibriu
m o
f bea
m
00
00
0(2
)0
xx
x
yy
y
Ay
AB
FA
B
FA
BF
MFL
BL
Thre
e eq
uatio
ns a
nd fo
ur u
nkno
wn
reac
tions
:A
yan
d B
yca
n be
foun
d.A
x=
−Bx
SD
/SI
24
93
SI O
bjec
tsR
edun
dant
Sup
port
s #4
Deg
ree
of re
dund
ancy
=#
of u
nkno
wn
reac
tions
−#
of in
depe
nden
t equ
atio
ns
The
unkn
own
reac
tions
can
be
dete
rmin
ed b
y su
pple
men
ting
the
equi
libriu
m e
quat
ions
with
equ
atio
ns re
latin
g fo
rces
/cou
ples
with
de
form
atio
n.
SD
/SI
See
you
in M
echa
nics
of M
ater
ials
I
94
SI O
bjec
tsIm
prop
er S
uppo
rts
#1
An
obje
ct h
as im
prop
er s
uppo
rts w
hen
it ha
s in
adeq
uate
(not
eno
ugh)
sup
ports
to m
aint
ain
it in
equ
ilibriu
m.
Res
ult
The
obje
ct w
ill m
ove
whe
n lo
aded
.
SD
/SI
See
you
in D
ynam
ics
95
SI O
bjec
tsIm
prop
er S
uppo
rts
#2
The
supp
orts
exe
rtpa
ralle
l for
ces
only
.
45
The
supp
orts
exe
rt on
ly
conc
urre
nt fo
rces
.
SD
/SI
96
SI O
bjec
tsIm
prop
er S
uppo
rts
#3
SD
/SI
25
97
SI O
bjec
tsIm
prop
er S
uppo
rts
#4
SD
/SI
98
SI O
bjec
tsIm
prop
er S
uppo
rts
#5
SD
/SI
99
Equi
libriu
m
Her
e en
ds th
e m
ost i
mpo
rtant
ch
apte
r of t
he s
ubje
ct.
100
Con
cept
sW
hen
a bo
dy is
in e
quilib
rium
, the
resu
ltant
forc
ean
d co
uple
abou
t any
poi
nt O
are
both
zer
o. P
robl
ems
can
be
anal
yzed
by
draw
ing
free
bod
y di
agra
ms
(FB
Ds)
and
su
bstit
ute
info
rmat
ion
in th
e FB
D u
nder
con
side
ratio
n in
to
the
equi
libriu
m e
quat
ions
.
Stat
ical
ly d
eter
min
ate
(SD
) pro
blem
s ca
n be
sol
ved
usin
g th
e eq
uilib
rium
con
ditio
ns a
lone
.Th
e st
atic
ally
inde
term
inat
e(S
I) pr
oble
ms
cann
ot d
ue to
to
o fe
w o
r too
man
y su
ppor
ts o
r con
stra
ints
.
Rev
iew
