Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
3/1 Moment Of a Force About a Point
Consider a force F acting on a rigid body (Fig. 3-1a). The magnitude of the
moment or tendency of the force to rotate the body about the axis O-O perpendicular
to the plane of the body is proportional both to the magnitude of the force and to the
moment arm d, which is the perpendicular distance from the axis to the line of action
of the force. Therefore, the magnitude of the moment is defined as
. … … … … … … … … … … … … … … . . 3.1
The sense of the moment may be determined by the right
hand rule (Fig. 3-1b).
Note: If the force tends to rotate the structure clockwise,
the sense of the moment vector is out of the plane
(a)
of the structure and the magnitude of the moment is
positive and vice versa (Fig. 3-2).
(b)
Fig. (3-1)
Fig. (3-2)
Note: In some two-dimensional and many of the three-dimensional problems to
follow, it is convenient to use a vector approach for moment calculations. The
moment of F about point A of Fig. (3-1a) may be represented by the cross-product
expression
… … … … … … … … … … … … … … … … … … … … … … … … . 3 2
32
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
Where r is a position vector which runs from the moment reference point A to any
point on the line of action of F.
Finally, denoting by α the angle between the lines of action of the position vector r
and the force F , we find that the magnitude of the moment of F about A is
. … … … … … … … … … … … … … … … … … … … … … . . 3 3
Where d represents the perpendicular distance from A to the line of action of F.
3/2 Varignon’s Theorem
The distributive property of vector products can be used to determine the
moment of the resultant of several concurrent forces . If several forces F 1 , F 2 , . . .
are applied at the same point A ( Fig. 3.3 ), and if we denote by r the position vector
of A. The moment about a give point O of the resultant of several concurrent forces is
equal to the sum of the moments of the various moments about the same point O.
(
)
r r r
r r r r
r × F1 + F2 + L = r × F1 + r × F2 + L
Fig. (3-3)
3/3 Rectangular Components of the Moment of a Force
The moment of F about O,
r r
r
r
r r r
M O = r × F, r = x i + y j + zk
r
r
r
r
F = Fx i + Fy j + Fz k
33
Chapter three
A.Lecturer Saddam K. Kwais
r
r
r
r
MO = M x i + M y j + Mzk
r
i
= x
Fx
r
j
y
Fy
r
k
z
Fz
r
r
r
= (yFz − zFy ) i + (zFx − xFz ) j + (xFy − yFx )k
The moment of F about B,
r
r
r
M B = rA / B × F
r
r r
rA / B = rA − rB
r
r
r
= (x A − x B ) i + (y A − y B ) j + (z A − z B ) k
r
r
r
r
F = Fx i + Fy j + Fz k
r
i
r
M B = (x A − x B )
Fx
(y
r
j
A
− yB )
(z
r
k
A
Fy
− zB )
Fz
For two-dimensional structures,
r
r
M O = (xFy − yFz )k
MO = M Z
= xFy − yFz
r
r
M O = [(x A − x B )Fy − (y A − y B )Fz ]k
MO = M Z
= (x A − x B )Fy − (y A − y B )Fz
34
Rigid Bodies:Equivalent Systems Of Forces
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
EXAMPLE 1. A 100-lb vertical force is applied
to the end of a lever which is attached to a shaft at O.
Determine:
a) moment about O,
b) horizontal force at A which creates the same moment,
c) smallest force at A which produces the same moment,
d) location for a 240-lb vertical force to produce the same moment,
e) whether any of the forces from b, c, and d is equivalent to the original force.
SOLUTION:
a) moment about O,
M O = Fd
d = (24 in.)cos 60° = 12 in.
M O = (100 lb)(12 in.)
M O = 1200 lb ⋅ in
b) horizontal force at A which creates the same moment,
d = (24 in.)sin 60° = 20.8 in.
M O = Fd
1200 lb ⋅ in. = F(20.8 in.)
F=
1200 lb ⋅ in.
20.8 in.
F = 57.7 lb
c) The smallest force at A occurs when the perpendicular distance is a maximum or
when F is perpendicular to OA.
M O = Fd
1200 lb ⋅ in. = F(24 in.)
F=
1200 lb ⋅ in.
24 in.
F = 50 lb
35
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
d) To determine the point of application of a 240 lb
force to produce the same moment,
M O = Fd
1200 lb ⋅ in. = (240 lb )d
1200 lb ⋅ in.
= 5 in.
240 lb
OB cos60° = 5 in. ⇒ OB = 10 in.
d=
e) Although each of the forces in parts b), c), and d) produces the same moment as
the 100 lb force, none are of the same magnitude and sense, or on the same line of
action. None of the forces is equivalent to the 100 lb force.
EXAMPLE 2. A force of 800 N acts on a bracket as shown.
Determine the moment of the force about B.
SOLUTION:
1. Cross product method
r
r
r
M B = rA / B × F
Where rA/B is the vector drawn from B to A . Resolving
rA/B and F into rectangular components, we have
r
r r
rA / B = rA − rB
r
r
= (x A − x B ) i + ( y A − y B ) j
r
r
= − (0.2 m ) i + (0.16 m ) j
r
r
r
F = Fx i + Fy j
r
r
= (800 N) cos 60 o i + (800 N) sin 60 o j
r
r
= (400 N ) i + (693) j
r
i
r
M B = (x A − x B )
Fx
(y
r
j
A
− yB )
Fy
36
Chapter three
r
i
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
r
j
r
r
r
r
M B = − 0.2 0.16 = [(−0.2 × 693) − (0.16 × 400)]k = [−138.6 − 64]k = −202.6 k N .m
400 693
M = −202.6 N .m = 202.6 N .m cw.
F=800 N
2. Another method
Fy
800 60° 400
A
60
◦
Fx
800 60° 693
160 mm
The moment of Fx is Clockwise
400 0.16 64. 64. The moment of Fy is Clockwise
693 0.2 138.6. 138.6. ! 64 138.6 202.6. 202.6. .
EXAMPLE 3. The rectangular plate is supported by the
brackets at A and B and by a wire CD. Knowing that
the tension in the wire is 200 N, determine the
moment about A of the force exerted by the wire at C.
SOLUTION:
The moment MA of the force F exerted by the wire
is obtained by evaluating the vector product,
r
r
r
M A = rC A × F
r
r
r
r r
rC A = rC − rA = (0.3 m )i + (0.08 m ) j
37
B
200 mm
Chapter three
A.Lecturer Saddam K. Kwais
r
r
r
r
F = Fλ = (200 N ) C D
rC D
r
r
Rigid Bodies:Equivalent Systems Of Forces
r
− (0.3 m )i + (0.24 m ) j − (0.32 m )k
= (200 N )
0.5 m
r
r
r
= −(120 N ) i + (96 N ) j − (128 N )k
r
r
r
i
j
k
r
M A = 0.3
0
0.08
− 120 96 − 128
r
r
r
v
M A = −(7.68 N ⋅ m ) i + (28.8 N ⋅ m ) j + (28.8 N ⋅ m )k
3/4 Moment of a Couple
Couple: Two forces F and -F having the
same magnitude, parallel lines of action, and
opposite sense are said to form a couple.
From Fig. (3-4)
" # "
∴ . Fig. (3-4)
Note : Two couples lie in parallel planes, and have
the same sense or the tendency to cause rotation
in the same direction (Fig.( 3-5)). Thus, Two couples
will have equal moments if
F1d1 = F2d 2
Fig. (3-5)
38
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
3/5 Equivalent Couple: Figure (3-6) shows three couples which act successively on
the same rectangular box. Since each of the three couples shown has the same
moment M (same direction and same magnitude M = 120 lb . in.), we can expect the
three couples to have the same effect on the box.
Fig. (3-6)
3/6 Addition of Couples: Two Couple may be replaced by asingle Couple of momentequal to algabric sum of the moment of the given couple is illustrated in Fig. (3-7).
s
-s
d
-s
(-p)+(-s)
d
-p
d
s
d
p+s
-p
p
p
Fig. (3-7)
3/7 Force-Couple system: The replacement of a force by a force and a couple is
illustrated in Fig. (3-8), 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.
Fig. (3-8)
39
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
EXAMPLE 1. A 50 Ib force is applied to
B
3 in.
a corner plate as shown. Determine
A
a) An equivalent force couple system at A.
5 in.
b) An equivalent system consisting of
a 150 Ib force at B and another force at A.
30◦
10 in.
SOLUTION:
50 Ib
Force-couple system at (A) we resolve the 50 Ib force into x and y components.
Fy
50 sin 30° 25)* →
50 cos 30° 43.3)* ↓
A
Fx
Rx= 25 Ib
5 in.
Fx
Ry= 43.3 Ib
/ 25 5 43.3 10
10 in.
Fy
308)*. MA= 308 Ib.in
A
Rx= 25 Ib
/ 308)*. ↷
Ry= 43.3 Ib
b) Force at (A) and (B)
p=150 Ib
/ 1 cos 2 3
308 150 cos 2 3
A
308
cos 2 ⇒ 2 ±46.8°
450
P'=150 Ib
5 # 1́ 43.3 150 sin 2 152.645)*
1 150)*
46.8◦
Rx= 25 Ib
θ
5 # 1́ 25 150 cos 2 77.682)*
5 85 9 # 5 9 171.3)* 63.0◦
θ
B
at A
Q==171.3
40
150 Ib
θ
B
A
θ
at B
Ry= 43.3 Ib
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
EXAMPLE 2. Replace the couple and force shown
by an equivalent single force applied to the lever.
Determine the distance from the shaft to the point
of application of this equivalent force.
SOLUTION:
B
B
400 N
400 N
400 N
24 N.m
24 N.m
150 mm
0
150 mm
0
400 N
C
≡
24 N.m
400 N
60◦
60 N.m
84 N.m
0
0
400 N
400 N
: 400 0.15 60. : 60. ↷
84. 400 ⇒ 0.210 210
;< cos 60° 210 ⇒ ;< 420=.
41
0
d
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
3/8 Reduction of a System of Coplanar Forces to one Force and one Couple.
Resultant of System of Coplanar Forces.
Coplanar forces F1 , F2 , and F3 act in the same plane as shown in Fig. (3-9a), They
can be reduced to an equivalent force – couple system acting at a given point 0 as
shown in Fig. (3-9b).
≡
≡
≡
Fig. (3-9)
When R is equal to zero, the force – system reduces to the Couple M. Thus, the
given system of forces is reduced to a single Couple M, called the resultant couple
of the system.
When both R and M are zero, then the system is an equilibrium.
Steps Solution of (x-y) plane
1. Determined >? ∑ AandD? ∑ E
2. Compute ? F?9 # ?9
3. Find 2 tanHI J? ⁄? L
42
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
4. Find ∑ M (about point O)
5. For calculated the distance d apply the equation M ?
EXAMPLE 1. For the beam, reduce the system of forces shown to (a) an equivalent
force-couple system at A, (b) an equivalent force couple system at B, and (c) a single
force or resultant.
SOLUTION:
a) Compute the resultant force and the resultant couple at A.
? 0
? 150 600 # 100 250 600 600 ↓
↺ ∑ / 600 1.6 # 100 2.8 250 4.8 1880. 1880. ↻
∴ ? 600 ↓ , / 1880. ↷
b) Force – Couple system at B based on the force-couple system at A.
600 4.8 2880. ↶
! 2880 1880 1000. ↶
∴ ? 600 ↓ , ! 1000. ↶
43
Chapter three
A.Lecturer Saddam K. Kwais
Rigid Bodies:Equivalent Systems Of Forces
600 N
c) Single force or resultant
x
1880 600 ∗ S
A
B
S 3.13
? 600 ↓ ,S 3.13
EXAMPLE 2. Determine the resultant of the four forces and one couple which act on
the O as shown in Fig. below.
? 40 # 80 cos 30° 60 cos 45°
66.9 →
? 50 # 80 sin 30° # 60 sin 45°
132.4 ↑
? 8?9 # ?9 F66.99 # 132.49 148.3=.
IV9.W
2 tanHI J? ⁄? L tanHI U
[M \]
XX.Y
Z 63.2°
M 140 505 60 sin 45° 7 # 60 cos 45° 4
273. 273. ↷
(a)
The force–couple system consisting of R and MO is shown
in Fig. a. We now determine the final line of action of R such
that R alone represents the original system.
M ? ⇒ 273 148.3 ⇒ 1.6=.
44