MO
ME
N D
AN
KO
PE
L
Apa y
ang d
ipela
jari
sekara
ng ?
Me
ng
eta
hu
i d
an m
em
ah
am
i m
aksud d
ari
mom
en g
aya,
mom
en k
ope
l, d
an c
ara
mem
ind
ah g
aya
Apa itu
mo
men
gaya ?
The
mo
men
t of
a fo
rce
about
a poin
t pro
vid
es a
mea
sure
of
the
tenden
cy f
or
rota
tion (
som
etim
es c
alle
d a
torq
ue)
.
MO
ME
NT
IN
2-D
(co
nti
nued
)
In t
he
2-D
cas
e, t
he
mag
nit
ude
of
the
mo
men
t is
Mo =
F d
As
show
n, d
is
the
per
pen
dic
ula
r dis
tance
fro
m p
oin
t O
to t
he
line
of
acti
on o
f th
e fo
rce.
In 2
-D, th
e dir
ecti
on o
f M
O i
s ei
ther
clo
ckw
ise
or
counte
r-cl
ock
wis
e dep
endin
g o
n t
he
tenden
cy f
or
rota
tion.
Mo
me
nt
in 2
-D
b O d
a F
b
O
a
F y
F
F x
As
show
n, d
is
the
per
pen
dic
ula
r dis
tance
fro
m p
oin
t O
to t
he
line
of
acti
on o
f th
e fo
rce.
MO =
F d
and t
he
dir
ecti
on i
s co
unte
r-cl
ock
wis
e.
Oft
en i
t is
eas
ier
to
det
erm
ine
MO b
y u
sing t
he
com
ponen
ts o
f F
as
sho
wn.
MO =
(F
Y a
) – (
FX b
)
CC
W =
(+
)
CW
=
(-)
Exam
ple
1G
iven
: A
40 N
forc
e is
appli
ed t
o t
he
wre
nch
.
Fin
d:
The
mo
men
t of
the
forc
e at
O.
Pla
n:
1)
Res
olv
e th
e fo
rce
along x
and y
axes
.
2)
Det
erm
ine
MO u
sing
scal
ar a
nal
ysi
s.
Solu
tion
: +
� F
y =
- 4
0 c
os
20°
N
+ �
Fx =
- 4
0 s
in 2
0°
N
+ M
O =
{-(
40 c
os
20°)
(200)
+ (
40 s
in 2
0°)
(30)}
N·m
m=
-7107 N
·mm
= -
7.1
1 N
·m
EX
AM
PL
E 2
Giv
en
: A
400 N
forc
e is
appli
ed t
o t
he
fram
e
and �
= 2
0°.
Fin
d:
The
mo
men
t of
the
forc
e at
A.
Pla
n:
1)
Res
olv
e th
e fo
rce
along x
and y
axes
.
2)
Det
erm
ine
MA u
sing s
cala
r an
alysi
s.
EX
AM
PL
E 2
(co
nti
nued
)
Solu
tion
+ �
Fy =
-400 c
os
20°
N
+ �
Fx =
-400 s
in 2
0°
N
+ M
A =
{(4
00 c
os
20°)
(2)
+ (
400 s
in 2
0°)
(3)}
N·m
= 1
160 N
·m
CO
NC
EP
T Q
UE
ST
ION
1.
What
is
the
mo
men
t of
the
10 N
forc
e
about
poin
t A
(M
A)?
A)
10 N
·m
B
) 30 N
·m C
) 13 N
·m
D)
(10/3
) N
·m E
) 7 N
·m
F =
10 N
· A
d=
3m
2. If
a f
orc
e of
mag
nit
ude
F c
an b
e ap
pli
ed i
n f
our
dif
fere
nt
2-D
confi
gura
tions
(P,Q
,R,
& S
), s
elec
t th
e ca
ses
resu
ltin
g i
n t
he
max
imu
m a
nd m
inim
um
torq
ue
val
ues
on t
he
nut.
(M
ax, M
in).
A)
(Q, P
)
B
) (R
, S
)
C)
(P, R
)
D
) (Q
, S
)
P Q
S
R
10 N
3m
P
2m
5N
3. U
sing t
he
CC
W d
irec
tion a
s posi
tive,
the
net
mo
men
t of
the
two f
orc
es a
bout
poin
t P
is
A)
10 N
·m
B
) 20 N
·m
C
) -
20 N
·m
D)
40 N
·m
E
) -
40 N
·m
Apa itu
mo
men
kopel ?
Mo
me
nt
of
a C
ouple
A c
ouple
is
def
ined
as
two
par
alle
l fo
rces
wit
h t
he
sam
e
mag
nit
ude
but
opposi
te i
n
dir
ecti
on s
epar
ated
by a
per
pen
dic
ula
r dis
tance
d.
The
mo
men
t of
a co
uple
is
def
ined
as
MO =
F d
(usi
ng a
sca
lar
anal
ysi
s) o
r as
MO =
r ⋅ F
(usi
ng a
vec
tor
anal
ysi
s).
Her
e r
is a
ny p
osi
tion v
ecto
r fr
om
the
line
of
acti
on o
f
–F
to t
he
line
of
acti
on o
f F
.
Pro
ble
m S
olv
ing
A
B
A t
orq
ue
or
mo
men
t of
12 N
· m
is
requir
ed t
o r
ota
te t
he
whee
l.
Whic
h o
ne
of
the
two g
rips
of
the
whee
l ab
ove
wil
l re
quir
e le
ss
forc
e to
rota
te t
he
whee
l?
Pro
ble
m S
olv
ing (
2-D
)
A
B
M=
Fd
12 =
F 0
.4
F =
30 N
M=
Fd
12 =
F 0
.3
F =
40 N
PR
OB
LE
M S
OL
VIN
G -
SC
AL
AR
Giv
en
: T
wo c
ouple
s ac
t on t
he
bea
m. T
he
resu
ltan
t co
uple
is
zero
.
Fin
d:
The
mag
nit
udes
of
the
forc
es
P a
nd F
and t
he
dis
tance
d.
PL
AN
:
1)
Use
def
init
ion o
f a
couple
to f
ind P
and F
.
2)
Res
olv
e th
e 300 N
forc
e in
x a
nd y
dir
ecti
ons.
3)
Det
erm
ine
the
net
mo
men
t.
4)
Equat
e th
e net
mo
men
t to
zer
o t
o f
ind d
.
Solu
tion
:
Fro
m t
he
def
init
ion o
f a
couple
P =
500 N
and
F =
300 N
.
Res
olv
e th
e 300 N
forc
e in
to v
erti
cal
and h
ori
zonta
l
com
ponen
ts.
The
ver
tica
l co
mponen
t is
(300 c
os
30º)
N a
nd
the
hori
zonta
l co
mponen
t is
(300 s
in 3
0º)
N.
It w
as g
iven
that
the
net
mo
men
t eq
ual
s ze
ro. S
o
+
M =
- (
500)(
2)
+ (
300 c
os
30º)
(d)
+ (
300 s
in 3
0º)
(0.2
) =
0
Now
solv
e th
is e
quat
ion f
or
d.
d =
(1000 –
60 s
in 3
0º)
/ (
300 c
os
30º)
= 3
.96 m
CO
NC
EP
T Q
UE
ST
ION
1. In
sta
tics
, a
couple
is
def
ined
as
__________ s
epar
ated
by a
per
pen
dic
ula
r dis
tance
.
A)
two f
orc
es i
n t
he
sam
e dir
ecti
on.
B)
two f
orc
es o
f eq
ual
mag
nit
ude.
C)
two f
orc
es o
f eq
ual
mag
nit
ude
acti
ng i
n t
he
sam
e dir
ecti
on.
D)
two f
orc
es o
f eq
ual
mag
nit
ude
acti
ng i
n o
pposi
te d
irec
tions.
2. F
1 a
nd F
2 f
orm
a c
ouple
. T
he
mo
men
t F
1
of
the
couple
is
giv
en b
y _
___ .
A)
r 1 ⋅ F
1
B
) r2
⋅ F
1
C)
F2 ⋅ r
1
D
) r2
⋅ F
2
r1
F2
r2
3. A
couple
is
appli
ed t
o t
he
bea
m a
s sh
ow
n. It
s m
om
ent
equal
s _____ N
·m.
50 N
A)
50
B
) 60
1m
2
m
C)
80
D
) 100
3
4
5
Apa itu
me
min
da
h g
aya ?
Sev
eral
forc
es a
nd a
couple
mo
men
t ar
e
acti
ng o
n t
his
ver
tica
l
sect
ion o
f an
I-b
eam
.
Can
you r
epla
ce t
hem
wit
h j
ust
one
forc
e an
d
one
couple
mo
men
t at
poin
t O
that
wil
l hav
e
the
sam
e ex
tern
al
effe
ct?
If y
es, how
wil
l
you d
o t
hat
?
AN
EQ
UIV
AL
EN
T S
YS
TE
M
=
When
a n
um
ber
of
forc
es a
nd c
ouple
mo
men
ts a
re a
ctin
g o
n a
body, it
is
easi
er t
o u
nder
stan
d t
hei
r over
all
effe
ct o
n t
he
body i
f
they
are
com
bin
ed i
nto
a s
ingle
forc
e an
d c
ouple
mo
men
t hav
ing
the
sam
e ex
tern
al e
ffec
t
The
two f
orc
e an
d c
ouple
syst
ems
are
call
ed e
quiv
alen
t sy
stem
s
since
they
hav
e th
e sa
me
exte
rnal
eff
ect
on t
he
body.
Equiv
ale
nt F
orc
e –
Couple
Syste
ms
Movin
g a
forc
e fr
om
A t
o O
, w
hen
both
poin
ts a
re o
n t
he
vec
tors
’ li
ne
of
acti
on, does
not
chan
ge
the
exte
rnal
eff
ect .
Hen
ce, a
forc
e vec
tor
is c
alle
d a
sli
din
g v
ecto
r. (
But
the
inte
rnal
eff
ect
of
the
forc
e on t
he
body d
oes
dep
end o
n
wher
e th
e fo
rce
is a
ppli
ed).
Equiv
ale
nt F
orc
e –
Couple
Syste
ms
Movin
g a
forc
e fr
om
poin
t A
to O
(as
show
n a
bove)
req
uir
es
crea
ting a
n a
ddit
ional
couple
mo
men
t. S
ince
this
new
couple
mo
men
t is
a “
free
” v
ecto
r, i
t ca
n b
e ap
pli
ed a
t an
y p
oin
t P
on t
he
body.
Equiv
ale
nt F
orc
e –
Couple
Syste
ms
If t
he
forc
e sy
stem
lie
s in
the
x-y
pla
ne
(the
2-D
cas
e), th
en t
he
reduce
d e
quiv
alen
t sy
stem
can
be
obta
ined
usi
ng t
he
foll
ow
ing
thre
e sc
alar
equat
ions.
Pro
ble
m S
olv
ing (
2-D
)
Giv
en
: A
2-D
forc
e an
d c
ouple
sy
stem
as
show
n.
Fin
d:
The
equiv
alen
t re
sult
ant
forc
e
and c
ouple
mo
men
t ac
ting a
t
A a
nd t
hen
the
equiv
alen
t
single
forc
e lo
cati
on a
long
the
bea
m A
B.
Pla
n:
1)
Sum
all
the
x a
nd y
co
mponen
ts o
f th
e fo
rces
to f
ind F
RA.
2)
Fin
d a
nd s
um
all
the
mo
men
ts r
esult
ing f
rom
movin
g e
ach
forc
e to
A.
3)
Shif
t th
e F
RA t
o a
dis
tance
d s
uch
that
d =
MR
A/F
Ry
Pro
ble
m S
olv
ing (
2-D
) +
�
FR
x =
25 +
35 s
in 3
0°
= 4
2.5
N
d
FR
+
�
FR
y =
- 2
0 -
35 c
os
30°
= -
50.3
1 N
+ M
RA
= -
35 c
os3
0°
(0.2
)
- 20(0
.6)
+ 2
5(0
.3)
FR =
( 4
2.5
2 +
(-5
0.3
1)2
)1
/2 =
65.9
N
� =
tan
-1 (
50.3
1/4
2.5
) =
49.8
° (
Kw
IV
)
= -
10.5
6 N
.m
The
equiv
alen
t si
ngle
forc
e F
R c
an b
e lo
cate
d o
n t
he
bea
m A
B a
t a
dis
tance
d m
easu
red f
rom
A.
d =
MR
A/F
Ry =
- 1
0.5
6/(
-50.3
1)
= 0
.21 m
.
Recommended
