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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
MOMENT OF A FORCE
Moment of aforce
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APPLICATIONS
What is the net effect of the
two forces on the wheel?
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APPLICATIONS (continued)
What is the effect of the 30 N
force on the lug nut?
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MOMENT IN 2-D
The moment of a force about a point provides a measure of thetendency for rotation (sometimes called a torque).
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MOMENT IN 2-D (continued)
In the 2-D case, the magnitude of the moment is
Mo = F d
As shown, d is the perpendicular distance from point O to the
line of action of the force.
In 2-D, the direction of MO is either clockwise or
counter-clockwise depending on the tendency for rotation.
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MOMENT IN 2-D (continued)
For example, MO = F d and the
direction is counter-clockwise.
Often it is easier to determine MO by using the components of F
as shown.
Using this approach, MO = (FY a) – (FX b). Note the different
signs on the terms! The typical sign convention for a moment in2-D is that counter-clockwise is considered positive. We can
determine the direction of rotation by imagining the body pinned
at O and deciding which way the body would rotate because of
the force.
Fa b
d
O
a bO
F
F x
F y
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Moment of a Force About a Point
• A force vector is defined by its magnitude and
direction. Its effect on the rigid body also depends
on it point of application.
• The moment of F about O is defined as F r M O ×=
• The moment vector M O
is perpendicular to the
plane containing O and the force F.
• Any force F’ that has the same magnitude and
direction as F, is equivalent if it also has the same line
of action and therefore, produces the same moment.
• Magnitude of M O
measures the tendency of the force
to cause rotation of the body about an axis along M O.
The sense of the moment may be determined by theright-hand rule.
Fd rF M O == θ sin
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Moment of a Force About a Point
• Two-dimensional structures have length and breadth but
negligible depth and are subjected to forces contained in
the plane of the structure.
• The plane of the structure contains the point O and the
force F. M O, the moment of the force about O is
perpendicular to the plane.
• If the force tends to rotate the structure clockwise, the
sense of the moment vector is out of the plane of the
structure and the magnitude of the moment is positive.
• If the force tends to rotate the structure counterclockwise,
the sense of the moment vector is into the plane of the
structure and the magnitude of the moment is negative.
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Varignon’s Theorem
• 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.
• Varigon’s Theorem makes it possible to
replace the direct determination of the
moment of a force F by the moments of two
or more component forces of F.
L
rr
rr
L
rrr
+×+×=++× 2121 F r F r F F r
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MOMENT IN 3-D
Moments in 3-D can be calculated using scalar (2-D) approach but
it can be difficult and time consuming. Thus, it is often easier to
use a mathematical approach called the vector cross product.
Using the vector cross product, M O
= r F .
Here r is the position vector from point O to any point on the line
of action of F
.
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Rectangular Components of the Moment of a Force
( ) ( ) ( )k yF xF j xF zF i zF yF
F F F
z y x
k ji
k M j M i M M
x y z x y z
z y x
z y xO
rrr
rrr
rrrr
−+−+−=
=
++=
The moment of F about O,
k F jF iF F
k z j yi xr F r M
z y xO rrrr
rrrrr
rr
++=
++=×= ,
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Rectangular Components of the Moment of a Force
The moment of F about B,
F r M B A B
rr
r
×= /
( ) ( ) ( )
k F jF iF F
k z z j y yi x x
r r r
z y x
B A B A B A
B A B A
rrrr
rrr
rrr
++=
−+−+−=
−=/
( ) ( ) ( )
z y x
B A B A B A B
F F F
z z y y x x
k ji
M −−−=
rrr
r
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Rectangular Components of the Moment of a Force
For two-dimensional structures,
z y
Z O
z yO
yF xF M M
k yF xF M
−=
=
−=rr
( ) ( )
( ) ( ) z B A y B A
Z O
z B A y B AO
F y yF x x
M M
k F y yF x x M
−−−=
=
−−−=rr
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Sample Problem 3.1
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.
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Sample Problem 3.1
a) Moment about O is equal to the product of the
force and the perpendicular distance between the
line of action of the force and O. Since the force
tends to rotate the lever clockwise, the momentvector is into the plane of the paper.
( )
( )( )in.12lb100
in.1260cosin.24
=
=°=
=
O
O
M
d
Fd M
inlb1200 ⋅=O M
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Sample Problem 3.1
b) Horizontal force at A that produces the same
moment,
( )
( )
in.8.20
in.lb1200
in.8.20in.lb1200
in.8.2060sinin.24
⋅=
=⋅
=
=°=
F
F
Fd M
d
O
lb7.57=F
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Sample Problem 3.1
c) The smallest force A to produce the same moment
occurs when the perpendicular distance is a
maximum or when F is perpendicular to OA.
( )
in.42
in.lb1200
in.42in.lb1200
⋅=
=⋅
=
F
F
Fd M O
lb50=F
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Sample Problem 3.1
d) To determine the point of application of a 240 lb
force to produce the same moment,
( )
in.5cos60
in.5lb402
in.lb1200
lb240in.lb1200
=°
=⋅
=
=⋅
=
OB
d
d Fd M O
in.10=OB
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Sample Problem 3.1
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 the100 lb force.