MOMENT OF A FORCE – VECTOR
FORMULATION
• Moment of force F about point O can be expressed
using cross product
MO = r X F
Magnitude
• For magnitude of cross product,
MO = rF sinθ
• Treat r as a sliding vector.
Since d = r sinθ,
MO = rF sinθ = F (rsinθ) = Fd
axis
MOMENT OF A FORCE – VECTOR
FORMULATION (cont)
Direction
• Direction and sense of MO
are determined by righthand rule
*Note:
- “curl” of the fingers
indicates the sense of
rotation
- Maintain proper order of r
and F since cross product is
not commutative
MOMENT OF A FORCE – VECTOR
FORMULATION (cont)
Principle of Transmissibility
• For force F applied at any point A, moment created
about O is MO = rA x F
• F has the properties of a
sliding vector, thus
M O = r1 X F = r2 X F = r3 X F
MOMENT OF A FORCE – VECTOR
FORMULATION (cont)
Cartesian Vector Formulation
• For force expressed in Cartesian form,

 
M O  r F  rx

j

k
ry
rz
Fx
Fy
Fz

i
• With the determinant expanded,
MO = (ryFz – rzFy)i
– (rxFz - rzFx)j + (rxFy –ryFx)k
112
3.5 Moment of a Force about a Specified Axis
• For moment of a force about a point, the moment and
its axis is always perpendicular to the plane
• A scalar or vector analysis is used to find the
component of the moment along a specified axis that
passes through the point
113
114
Scalar Analysis
• According to the right-hand rule,
My is directed along the
positive y axis
• For any axis, the moment is
M a  Fd a
• Force will not contribute a moment
if force line of action is parallel or
passes through the axis
115
MOMENT OF A FORCE ABOUT A
SPECIFIED AXIS (cont)
Vector Analysis
• For magnitude of MA,
MA = MOcosθ = MO·ua
where ua = unit vector
• In determinant form,
uax uay uaz

  
M a  ua  (r F )  rx
ry
rz
Fx
Fy
Fz
i
M a  (uax i  ua y j  uaz k )  rx
Fx
j
ry
Fy
k
rz
Fz
Ma  M aua  ua  (r  F)ua
M a  ua  (r  F)  ua   (r  F)
117
118
119
120
121



rC  0.6i  0.3k
M AB
0.894 0.447
 
 u B  rC  F   0.6
0
 600
200
0
0.3  53.64
 300
122
Homework:
3-63
123
3.6 Moment of a Couple
• Couple
– two parallel forces
– same magnitude but opposite direction
– separated by perpendicular distance d
• Resultant force = 0
• Tendency to rotate in specified direction
• Couple moment = sum of moments of both couple
forces about any arbitrary point
124
MOMENT OF A COUPLE (cont)
Scalar Formulation
• Magnitude of couple moment
M = Fd
• Direction and sense are
determined by right hand rule
• M acts perpendicular to
plane containing the forces
MOMENT OF A COUPLE (cont)
Vector Formulation
•
For couple moment,
M=rXF
• If moments are taken about point A, moment of –F is
zero about this point
• r is crossed with the force F to which it is directed
M = rB X F + rA X –F
= (rB – rA) X F
rB = rA + r
A couple moment is a free vector, i.e., it can act at any point
since M depends only on the position vector r directed
between the forces not the position vectors rA and rB.
127
128
M R   (r  F)
129
MOMENT OF A COUPLE (cont)
Resultant Couple Moment
• Couple moments are free vectors and may be applied
to any point P and added vectorially
• For resultant moment of
For more than 2
moments,
two couples at point P,
MR = ∑(r X F)
MR = M1 + M2
Equivalent Couples
• 2 couples are equivalent if they produce the same
moment
• Forces of equal couples lie on the same plane or plane
parallel to one another
131
132
133
134
135
136
Homework:
3-70, 3-90
137
Equivalent System
Point O is on the line of action of the force:
Point O is not on the line of action of the force:
138
139
SIMPLIFICATION OF A FORCE AND
COUPLE SYSTEM
• An equivalent system is when the external effects are
the same as those caused by the original force and
couple moment system
• External effects of a system is the translating and
rotating motion of the body
• Or refers to the reactive forces at the supports if the
body is held fixed