Force Couples and Their Applications in Mechanics
50
Chapter
2
Force
Systems
2/5
Couple
O
–
F
The
moment
produced
by
two
equal,
opposite,
and
noncollinear
forces
is
called
a
couple
.
Couples
have
certain
unique
properties
and
have
important
applications
in
mechanics.
Conside
r
th
e
actio
n
o
f
tw
o
equa
l
an
d
opposit
e
force
s
F
an
d
-
F
a
dis-
tanc
e
d
apart
,
a
s
show
n
i
n
Fig
.
2/10
a
.
Thes
e
tw
o
force
s
canno
t
b
e
combined int
o
a
singl
e
forc
e
becaus
e
thei
r
su
m
i
n
ever
y
directio
n
i
s
zero
.
Thei
r
only effec
t
i
s
t
o
produc
e
a
tendenc
y
o
f
rotation
.
Th
e
combine
d
momen
t
o
f
the tw
o
force
s
abou
t
a
n
axi
s
norma
l
t
o
thei
r
plan
e
an
d
passin
g
throug
h
any poin
t
suc
h
a
s
O
i
n
thei
r
plan
e
i
s
th
e
coupl
e
M
.
Thi
s
coupl
e
ha
s
a
magnitude
a
F
d
(
a
)
r
B
O
B
–
F
r
r
A
M
=
F
(
a
+
d
)
-
Fa
F
or
A
M
=
Fd
(
b
)
Its
direction
is
counterclockwise
when
viewed
from above
for
the
case
il-
lustrated.
Note
especially
that
the
magnitude
of
the
couple
is
indepen-
dent
of the
distance
a
which
locates
the
forces
with
respect
to
the
moment
center
O
.
It
follows
that the
moment
of
a
couple
has
the
same
value
for
all
moment
centers.
M
O
Vector
Algebra
Method
We
may
also
express
the
moment
of
a
couple
by
using
vector
alge-
bra.
With the
cross-product
notation
of
Eq.
2/6,
the
combined
moment
about
point
O
of
the
forces
forming
the
couple
of
Fig.
2/10
b
is
(
c
)
M
=
r
A
x
F
+
r
B
x
(
-
F
)
=
(
r
A
-
r
B
)
x
F
where
r
A
and
r
B
are
position
vectors
which
run
from
point
O
to
arbi-
trary
points
A
and
B
on
the
lines
of
action
of
F
and
-
F
,
respectively.
Be-
cause
r
A
-
r
B
=
r
,
we
can
express
M
as
M
=
r
x
F
M
M
M
M
Here
again,
the
moment
expression
contains
no
reference
to
the
mo-
ment
center
O
and,
therefore,
is
the
same
for
all
moment
centers.
Thus,
we
may
represent
M
by
a
free
vector,
as
shown
in
Fig.
2/10
c
,
where
the
direction
of
M
is
normal
to
the
plane
of
the
couple
and
the
sense
of
M
is
established
by
the
right-hand
rule.
Because
the
couple
vector
M
is
always
perpendicular
to
the
plane
of the
forces
which
constitute
the
couple,
in
two-dimensional
analysis
we can
represent
the
sense
of
a
couple
vector
as
clockwise
or
counterclock- wise
by
one of
the
conventions
shown
in
Fig.
2/10
d
.
Later,
when
we
deal with
couple
vectors
in
three-dimensional problems,
we
will
make
full use
of
vector
notation
to
represent
them,
and
the
mathematics
will
au- tomatically
account
for
their
sense.
Counterclockwise
couple
Clockwise
couple
(
d
)
Figure
2/10
Equivalent
Couples
Changin
g
th
e
value
s
o
f
F
an
d
d
doe
s
no
t
chang
e
a
give
n
coupl
e
as
lon
g
a
s
th
e
produc
t
F
d
remain
s
th
e
same
.
Likewise
,
a
coupl
e
i
s
no
t
af-
fecte
d
i
f
th
e
force
s
ac
t
i
n
a
differen
t
bu
t
paralle
l
plane
.
Figur
e
2/11
Article
2/5
Couple
51
M
M
M
M
–
2
F
d
/2
–
F
d
F
≡
–
F
2
F
F
≡
≡
–
F
d
d
F
Figure
2/11
show
s
fou
r
differen
t
configuration
s
o
f
th
e
sam
e
coupl
e
M
.
I
n
eac
h
of
th
e
fou
r
cases
,
th
e
couple
s
ar
e
equivalen
t
an
d
ar
e
describe
d
b
y
the
sam
e
fre
e
vecto
r
whic
h
represent
s
th
e
identica
l
tendencie
s
t
o
rotat
e
the
bodies.
Force–Couple
Systems
The
effect
of
a
force
acting
on
a
body
is
the
tendency
to
push
or
pull
the
body
in
the
direction
of
the
force,
and
to
rotate the
body
about
any
fixed
axis which
does
not
intersect
the
line
of
the
force.
We
can
repre-
sent
this
dual
effect
more
easily
by
replacing
the
given
force
by
an
equal
parallel
force
and
a
couple
to
compensate
for
the
change
in
the
moment
of
the
force.
The
replacement
of
a
force
by
a
force
and
a
couple
is
illustrated
in
Fig.
2/12,
where
the
given
force
F
acting
at
point
A
is
replaced by
an
equal
force
F
at
some
point
B
and
the
counterclockwise
couple
M
=
Fd
.
The
transfer
is
seen
in
the
middle
figure,
where
the
equal
and opposite
forces
F
and
-
F
are
added
at
point
B
without
introducing
any
net
exter-
nal
effects
on
the
body.
We
now
see
that
the
original
force
at
A
and
the
equal
and
opposite
one
at
B
constitute
the
couple
M
=
Fd
,
which
is
counterclockwise
for
the
sample
chosen,
as
shown
in
the
right-hand
part of
the
figure.
Thus,
we
have
replaced
the
original
force
at
A
by
the
same force
acting
at
a
different
point
B
and
a
couple,
without
altering
the
ex- ternal
effects
of
the
original force
on
the
body.
The
combination
of
the force
and
couple
in
the
right-hand part
of
Fig.
2/12
is
referred
to
as
a
force–couple system
.
By
reversing
this
process,
we
can
combine
a
given
couple
and
a
force
which
lies
in
the
plane of
the
couple
(normal
to
the
couple
vector)
to
produce a
single,
equivalent
force.
Replacement
of
a
force
by
an
equiva-
lent
force–couple system,
and
the
reverse
procedure,
have
many
applica-
tions
in
mechanics
and
should
be
mastered.
B
F
F
≡
≡
B
B
–
F
d
F
F
A
A
M
=
Fd
Figure
2/12
52
Chapter
2
Force
Systems
SAMPL
E
PROBLE
M
2/7
The
rigid
structural member
is
subjected
to
a
couple
consisting
of
the
two
100-N
forces.
Replace
this
couple
by
an
equivalent
couple
consisting
of
the
two
forces
P
and
+
P
, each
of
which
has
a
magnitude
of
400
N.
Determine
the
proper
angle
O
.
0
Solution.
The
original
couple
is
counterclockwise
when
the
plane
of
the
forces
is
viewed
from
above,
and
its
magnitude
is
[
M
=
Fd
]
M
=
100(0.1)
=
10
N
m
The
forces
P
and
+
P
produce
a
counterclockwise
couple
M
=
400(0.040)
cos
O
O
Equating
the
two
expressions
gives
10
=
(400)(0.040)
cos
O
O
=
cos
+
1
10
=
51.3
°
Ans.
16
Helpful
Hint
O
Since
the
two
equal
couples
are
parallel free
vectors,
the
only
dimensions
which
are
relevant
are
those
which
give
the
perpendicular
distances
between the
forces
of
the
couples.
SAMPL
E
PROBLE
M
2/8
Replace
the
horizontal
80-lb
force
acting
on
the
lever
by
an
equivalent
sys-
tem
consisting
of
a
force
at
O
and
a
couple.
Solution.
We
apply
two
equal
and
opposite
80-lb
forces
at
O
and
identify
the
counterclockwise
couple
[
M
=
Fd
]
M
=
80(9
sin
60
°
)
=
624
l
b
-
in.
Ans.
O
Thus,
the
original
force
is
equivalent
to
the
80-lb
force
at
O
and
the
624-lb-in.
couple
as
shown
in
the
third
of
the
three
equivalent
figures.
Helpful
Hint
O
Th
e
revers
e
o
f
thi
s
proble
m
i
s
ofte
n
encountered
,
namely
,
th
e
replacement
o
f
a
forc
e
an
d
a
coupl
e
b
y
a
singl
e
force
.
Proceedin
g
i
n
revers
e
i
s
th
e
sam
e
as
replacin
g
th
e
coupl
e
b
y
tw
o
forces
,
on
e
o
f
whic
h
i
s
equa
l
an
d
opposit
e
t
o
the
80-l
b
forc
e
a
t
O
.
Th
e
momen
t
ar
m
t
o
th
e
secon
d
forc
e
woul
d
b
e
M/
F
=
624/8
0
=
7.7
9
in.
,
whic
h
i
s
9
si
n
60
°
,
thu
s
determinin
g
th
e
lin
e
o
f
actio
n
of
th
e
singl
e
resultan
t
forc
e
o
f
8
0
lb.
80
lb
9″
60°
0
80
lb
80
lb
≡
≡
0
0
0
80
lb
80
lb
80
lb
624
lb-in.
M
40
P
0
-
P
100
100
100
60
100
N
100
N
Dimensions
in
millimeters
P
=
400
N
0
40
mm
d
0
0
P
=
400
N
Article
2/5
Problems
53
PROBLEMS
Introductory
Problems
2/59
Compute
the
combined
moment
of
the
forces
about
(
a
)
point
O
and
(
b
)
point
A
.
2/61
Replace
the
force–couple
system
at
point
O
by
a
single
force.
Specify
the
coordinate
y
A
of
the
point
on
the
y
-axis
through
which
the
line
of
action
of
this resultant
force
passes.
two
90-lb
y
y
x
x
Problem
2/61
2/62
The
top
view
of
a
revolving
entrance
door
is
shown.
Two
persons
simultaneously
approach
the
door
and
exert
force
of
equal
magnitudes as
shown.
If
the
resulting
moment
about
the
door
pivot
axis
at
O
is
25
N
r
m,
determine
the
force
magnitude
F.
9
0
lb
Problem
2/59
2/60
Replace
the
12-kN
force
acting
at
point
A
by
a
force–couple system
at
(
a
)
point
O
and
(
b
)
point
B
.
y
0.
8
m
O
x
Problem
2/62
Problem
2/60
12
kN
30°
4
m
A
O
5
m
B
15
°
F
0.
8
m
–
F
15
°
9
0
lb
A
8
″
4
″
O
4
″
200
N
O
80
N·m
54
Chapter
2
Force
Systems
2/63
Determine
the
moment
associated
with
the
couple
applied to
the
rectangular
plate.
Reconcile
the
results
with those
for
the
individual
special
cases
of
O
=
0,
b
=
0,
and
h
=
0.
2/65
The
7-lb
force
is
applied
by
the
control
rod
on
the
sector
as
shown.
Determine
the
equivalent
force–
couple
system
at
O.
T
=
7
lb
30°
A
h
15°
O
15°
3
"
θ
Problem
2/63
Problem
2/65
2/64
As
part
of
a
test,
the
two
aircraft
engines
are
revved
up
and
the
propeller
pitches
are
adjusted
so
as
to
re-
sult
in
the
fore
and
aft
thrusts
shown.
What
force
F
must
be
exerted
by
the
ground
on
each
of
the
main
braked
wheels
at
A
and
B
to
counteract
the
turning
effect
of
the
two
propeller thrusts?
Neglect
any
ef-
fects
of
the
nose
wheel
C
,
which
is
turned
90
°
and
unbraked.
2/66
Replace
the
10-kN
force
acting
on
the
steel
column
by
an
equivalent force–couple
system
at
point
O.
This
replacement
is
frequently
done
in
the
design
of
structures.
75
mm
1
m
500
8′
14′
Problem
2/66
C
500
Problem
2/64
lb
A
lb
B
10
kN
75
O
1
F
θ
b
F
Article
2/5
Problems
55
2/67
Each
propeller
of
the
twin-screw
ship
develops
a
full-speed
thrust
of 300
kN.
In
maneuvering
the
ship,
one
propeller
is
turning
full
speed
ahead and
the
other
full
speed
in
reverse.
What
thrust
P
must
each
tug
exert
on
the
ship
to
counteract
the
effect of
the
ship’s
propellers?
2/69
A
lug
wrench
is
used
to
tighten
a
square-head
bolt.
If
50-lb
forces
are
applied
to
the
wrench
as
shown,
determine the
magnitude
F
of
the
equal
forces
exerted on
the
four
contact
points
on
the
1-in.
bolt
head
so that
their
external
effect
on
the
bolt
is
equivalent to
that
of
the
two
50-lb
forces.
Assume
that
the
forces
are
perpendicular
to
the
flats
of
the
bolt
head.
1″
50
lb
A
C
V
iew
C
Detail
(
c
learances
exaggerated)
14″
50
lb
B
Problem
2/67
14″
Representative
Problems
2/68
The
force–couple
system
at
A
is
to
be
replaced
by a
single
equivalent
force
acting
at
a
point
B
on
the
vertical
edge
(or
its
extension)
of
the
triangular
plate.
Determine
the
distance
d
between
A
and
B.
Problem
2/69
2/70
A
force–couple
system
acts
at
O
on
the
60
°
circular
sector.
Determine
the
magnitude of
the
force
F
if
the
given
system
can
be
replaced by
a
stand-alone
force
at
corner
A
of
the
sector.
M
A
A
R
F
F
h
60°
b
8
0
N·m
O
0.
4
m
A
Problem
2/68
Problem
2/70
B
d
5
0
m
F
1
2
m
F
12
0
m
56
Chapter
2
Force
Systems
2/7
1
Durin
g
a
stead
y
righ
t
turn
,
a
perso
n
exert
s
th
e
forces
show
n
o
n
th
e
steerin
g
wheel
.
Not
e
tha
t
eac
h
force
consist
s
o
f
a
tangentia
l
componen
t
an
d
a
radially-
inwar
d
component
.
Determin
e
th
e
momen
t
exerted
abou
t
th
e
steerin
g
colum
n
a
t
O.
2/73
The
tie-rod
AB
exerts
the
250-N
force
on
the
steer-
ing
knuckle
AO
as
shown.
Replace
this
force
by
an
equivalent
force–couple
system
at
O
.
10
°
F
=
25
0
N
23
5
mm
8
N
y
30
°
B
x
O
A
5
0
m
m
15
°
15
°
B
30
°
8
N
37
5
mm
Problem
2/73
2/74
The
250-N
tension
is
applied
to
a
cord
which
is
se-
curely
wrapped
around
the
periphery
of
the
disk.
Determine
the
equivalent
force–couple
system
at
Problem
2/71
2/72
A
force
F
of
magnitude
50
N
is
exerted
on
the
point
C.
Begin
by
couple
system
at
A.
finding
the
equivalent
force–
automobile
parking-brake
lever
at
the
position
x
=
250
mm.
Replace
the
force
by
an
equivalent
force–couple system
at
the
pivot
point
O.
120
mm
F
20°
45°
15°
200
mm
A
10°
B
250
N
O
400
mm
Problem
2/72
60°
60°
D
C
Problem
2/74
100
mm
x
5
0
m
O
A
Article
2/5
Problems
57
2/75
The
system
consisting
of
the
bar
OA
,
two
identical
pulleys,
and
a
section
of
thin
tape
is
subjected
to
the
two
180-N
tensile
forces
shown in
the figure.
Deter-
mine
the
equivalent
force–couple
system
at
point
O.
2/77
The
device
shown
is
a
part
of
an
automobile
seat-
back-release mechanism.
The
part
is
subjected
to
the
4-N
force
exerted
at
A
and
a
30
0
-
N
r
mm
restor- ing
moment
exerted
by
a
hidden
torsional
spring. Determine
the
y
-intercept
of
the
line
of
action
of
the single
equivalent
force.
18
0
N
r
y
A
F
=
4
N
r
=
2
5
mm
15
°
A
4
0
mm
10
0
mm
r
18
0
N
x
O
30
0
N·mm
45°
O
5
0
mm
Problem
2/77
2/78
The
force
F
acts
along
line
MA
,
where
M
is
the
mid-
point
of
the
radius
along
the
x
-axis.
Determine
the
equivalent
force–couple
system
at
O
if
O
=
40
°
.
Problem
2/75
2/76
Points
A
and
B
are
the
midpoints
of
the
sides
of
the
rectangle.
Replace
the
given
force
F
acting
at
A
by
a
force–couple system
at
B.
y
y
F
x
O
B
x
–
R
–
–
R
–
Problem
2/76
2
2
Problem
2/78
F
A
θ
M
A
h
O
b
1
0
mm
Force couples, generated by equal and opposite forces, produce rotational effects without any net translation. They possess unique properties and find significant applications in mechanics. Couples have a constant magnitude regardless of the distance between forces, making them independent of the location of the moment center. The concept of equivalent couples and force couple systems are explored, illustrating how couples can be used to represent the combined effect of a force and a tendency to rotate a body.
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50 Chapter 2 Force Systems 2/5 Couple O F The moment produced by two equal, opposite, and noncollinear forces is called a couple. Couples have certain unique properties and have important applications in mechanics. Consider the action of two equal and opposite forces F and -F a dis- tance d apart, as shown in Fig. 2/10a. These two forces cannot be combined into a single force because their sum in every direction is zero. Their only effect is to produce a tendency of rotation. The combined moment of the two forces about an axis normal to their plane and passing through any point such as O in their plane is the couple M. This couplehas amagnitude M =F(a +d) -Fa or M =Fd a F d (a) rB O B F r rA F A (b) Its direction is counterclockwise when viewed from above for the caseil- lustrated. Note especially that the magnitude of the couple is indepen- dent of the distance a which locates the forces with respect to the moment center O. It follows that the moment of a couple has the same valuefor all moment centers. M O Vector Algebra Method We may also express the moment of a couple by using vector alge- bra. With the cross-product notation of Eq. 2/6, the combined moment about point O of the forces forming the couple of Fig. 2/10bis (c) M =rAxF +rBx(-F) =(rA-rB) xF M where rAand rBare position vectors which run from point O to arbi- trary points A and B on the lines of action of F and -F, respectively. Be- causerA-rB=r, we can expressM as M M =r xF M M Here again, the moment expression contains no reference to the mo- ment center O and, therefore, is the same for all moment centers. Thus, we may represent M by a free vector, as shown in Fig. 2/10c, where the direction of M is normal to the plane of the couple and the sense of M is established by the right-handrule. Because the couple vector M is always perpendicular to the plane of the forces which constitute the couple, in two-dimensional analysis we can represent the sense of a couple vector as clockwise or counterclock- wise by one of the conventions shown in Fig. 2/10d. Later, when we deal with couple vectors in three-dimensional problems, we will make full use of vector notation to represent them, and the mathematics will au- tomatically account for their sense. Counterclockwise couple Clockwise couple (d) Figure 2/10 Equivalent Couples Changing the values of F and d does not change a given couple as long as the product Fd remains the same. Likewise, a couple is not af- fected if the forces act in a different but parallel plane. Figure 2/11
Couple 51 Article 2/5 M M M M 2F F d /2 F d 2F F F d d F F Figure 2/11 shows four different configurations of the same couple M. In each of the four cases, the couples are equivalent and are described by the same free vector which represents the identical tendencies to rotate the bodies. Force Couple Systems The effect of a force acting on a body is the tendency to push or pull the body in the direction of the force, and to rotate the body about any fixed axis which does not intersect the line of the force. We can repre- sent this dual effect more easily by replacing the given force by an equal parallel force and a couple to compensate for the change in the moment of the force. The replacement of a force by a force and a couple is illustrated in Fig. 2/12, where the given force F acting at point A is replaced by an equal force F at some point B and the counterclockwise couple M = Fd. The transfer is seen in the middle figure, where the equal and opposite forces F and -F are added at point B without introducing any net exter- nal effects on the body. We now see that the original force at A and the equal and opposite one at B constitute the couple M = Fd, which is counterclockwise for the sample chosen, as shown in the right-hand part of the figure. Thus, we have replaced the original force at A by the same force acting at a different point B and a couple, without altering the ex- ternal effects of the original force on the body. The combination of the force and couple in the right-hand part of Fig. 2/12 is referred to as a force couple system. By reversing this process, we can combine a given couple and a force which lies in the plane of the couple (normal to the couple vector) to produce a single, equivalent force. Replacement of a force by an equiva- lent force couple system, and the reverse procedure, have many applica- tions in mechanicsand should be mastered. B F F B B F d F F A A M = Fd Figure 2/12
52 Chapter 2 Force Systems SAMPLE 2/7 PROBLEM M The rigid structural member is subjected to a couple consisting of the two 100-N forces. Replace this couple by an equivalent couple consisting of the two forces P and +P, each of which has a magnitude of 400 N. Determine the proper angle O. 40 P 0 0 -P 100 Solution. The original couple is counterclockwisewhen the plane of the forces is viewedfrom above, and its magnitude is 100 [M =Fd] M =100(0.1) =10 N m The forces P and +P produce a counterclockwisecouple 100 60 M =400(0.040) cos O 100 N O Equating the two expressions gives 100 N 10 =(400)(0.040) cos O Dimensions in millimeters O =cos+110=51.3 Ans. P = 400 N 16 0 Helpful Hint O Since the two equal couples are parallel free vectors, the only dimensions which are relevant are those which give the perpendicular distances between the forces of the couples. 40 mm d0 0 P = 400 N SAMPLE PROBLEM 2/8 80 lb Replace the horizontal 80-lb force acting on the lever by an equivalent sys- tem consistingof a force at O and a couple. 9 60 0 Solution. We apply two equal and opposite 80-lb forces at O and identify the counterclockwisecouple [M =Fd] M =80(9 sin 60 ) =624 lb-in. Ans. 80 lb 80 lb O Thus, the original force is equivalent to the 80-lb force at O and the 624-lb-in. couple as shown in the third of the three equivalent figures. Helpful Hint O The reverse of this problem is often encountered, namely, the replacement of a force and a couple by a single force. Proceeding in reverse is the same as replacing the couple by two forces, one of which is equal and opposite to the 80-lb force at O. The moment arm to the second force would be M/F = 624/80 = 7.79 in., which is 9 sin 60 , thus determining the line of action of the single resultant force of 80 lb. 0 0 0 80 lb 80 lb 80 lb 624 lb-in.
Article 2/5 Problems 53 PROBLEMS 2/61 Replace the force couple system at point O by a single force. Specify the coordinate yAof the point on the y-axis through which the line of action of this resultant force passes. Introductory Problems 2/59 Compute the combined moment of the forces about (a) point O and (b) point A. two 90-lb y y 90 lb x 200 N O 80 N m x A O Problem 2/61 8 4 4 The top view of a revolving entrance door is shown. Two persons simultaneously approach the door and exert force of equal magnitudes as shown. If the resulting moment about the door pivot axis at O is 25 N r m, determine the force magnitude F. 2/62 90 lb Problem 2/59 2/60 Replace the 12-kN force acting at point force couple systemat (a) point O and (b) point B. A by a 15 F y 0.8 m 12 kN 0.8 m O 4 m 30 x A O F 5 m 15 Problem 2/62 B Problem 2/60
54 Chapter 2 Force Systems Determine the moment associated with the couple applied to the rectangular plate. Reconcile the results with those for the individual special cases of O =0, b =0, and h =0. The 7-lb force is applied by the control rod on the sector as shown. Determine the equivalent force couple systemat O. 2/63 2/65 T = 7 lb F 30 A h 15 15 O b 3" F Problem 2/63 Problem 2/65 As part of a test, the two aircraftenginesare revved up and the propeller pitches are adjusted so as to re- sult in the fore and aft thrusts shown. What force F must be exerted by the ground on each of the main braked wheels at A and B to counteract the turning effect of the two propeller thrusts? Neglect any ef- fects of the nose wheel C, which is turned 90 and unbraked. 2/64 Replace the 10-kN force acting on the steel column by an equivalent force couple system at point O. This replacement is frequently done in the design of structures. 2/66 10 kN 75 mm 75 1 m 1 500 lb O A 8 14 Problem 2/66 C B lb 500 Problem 2/64
Article 2/5 Problems 55 2/67 Each propeller of the twin-screw ship develops a full-speed thrust of 300 kN. In maneuvering the ship, one propeller is turning full speed ahead and the other full speed in reverse. What thrust P must each tug exert on the ship to counteract the effect of the ship s propellers? A lug wrench is used to tighten a square-headbolt. If 50-lb forces are applied to the wrench as shown, determine the magnitude F of the equal forces exerted on the four contact points on the 1-in. bolt head so that their external effect on the bolt is equivalent to that of the two 50-lb forces. Assume that the forces are perpendicular to the flats of the bolt head. 2/69 50 m 1 F 12 m 50 lb F A C View C Detail (clearancesexaggerated) 120 m 14 50 lb Problem 2/67 B 14 Representative Problems Problem 2/69 2/68 The force couple system at A is to be replaced by a single equivalent force the vertical edge (or its extension) of the triangular plate.Determinethe distance d betweenA and B. acting at a point B on 2/70 A force couple system acts at O on the 60 circular sector. Determine the magnitude of the force F if the given system can be replaced by a stand-alone force at corner A of the sector. MA A R d F F h B 60 b 80 N m 0.4 m O A Problem 2/68 Problem 2/70
56 Chapter 2 Force Systems 2/71 During a steady right turn, a personexertsthe forces shown on the steering wheel. Note that each force consists of a tangential component and a radially- inward component. Determine the moment exerted about the steeringcolumn at O. The tie-rod AB exerts the 250-N force on the steer- ing knuckle AO as shown. Replace this force by an equivalent force couple systemat O. 2/73 10 F = 250 N 235 mm 8 N y 30 B x O A 50 mm 50 m 15 A 15 B O 30 8 N 375 mm Problem 2/73 The 250-N tension is applied to a cord which is se- curely wrapped around the periphery of the disk. Determine the equivalent force couple system at point C. Begin by couple systemat A. 2/74 Problem 2/71 2/72 A force F of magnitude 50 N is automobile parking-brake lever at the position x =250 mm. Replace the force by an equivalent force couple systemat the pivot point O. exerted on the finding the equivalent force 120 mm F 20 45 15 200 mm A 10 B 250 N O 400 mm Problem 2/72 60 60 D C Problem 2/74
Article 2/5 Problems 57 2/75 The system consisting of the bar OA, two identical pulleys, and a section of thin tape is subjected to the two 180-N tensile forces shown in the figure. Deter- mine the equivalent force couple systemat point O. The device shown is a part of an automobile seat- back-release mechanism. The part is subjected to the 4-N force exerted at A and a 300- N r mm restor- ing moment exerted by a hidden torsional spring. Determine the y-intercept of the line of action of the singleequivalent force. 2/77 180 N r y F = 4 N A 10 mm r = 25 mm 15 A 40 mm 100 mm r 180 N x O 300 N mm 45 O 50 mm Problem 2/77 2/78 The force F acts along line MA, where M is the mid- point of the radius along the x-axis. Determine the equivalent force couple systemat O if O =40 . Problem 2/75 2/76 Points A and B are the midpointsof the sidesof the rectangle.Replacethe given force F acting at A by a force couple systemat B. y F y F A A x M O h B x O b R 2 2 R Problem 2/76 Problem 2/78
