ME 2560 Statics
Moments of Forces and the Cross Product
Moment of a Force – Torque
o The moment (or torque) of a force about a point O is defined as the magnitude of the
force  F  multiplied by the perpendicular distance from the point to the line of
action of the force  d  r sin( )  . So, | M O | | F | r sin( )
o The direction of the moment is defined by the right-hand-rule. Let the fingers of your
right hand show the direction of the circulation of F around O, and your thumb
shows the direction of the moment. So, M O | F | r sin( ) k .
θ
A
O
θ
Line of
Action
o The moment of F about O can also be calculated by first breaking the force into
components, and then summing the moments of the individual components.
o As an example, consider the force F shown in the diagram. The line of action of the
X-component of F passes through O and, hence, has no moment about O. The line of
action of the Y-component is perpendicular to the position vector r . So the moment
of F may be calculated as




M O  | F |cos( )   0  k  | F |sin( )   r  k  | F |sin( ) r  k




 X component

 Y component

Example l:
Given: Force F  300 i  100 j (lb) is applied at point B.
Find: M A the moment of F about point A.
Solution:
M A   4  300   8 100  k  2000 k (ft-lb)
5 ft.
B
4 ft.
A
3 ft.
Kamman – ME 2560: page 1/3
The Cross Product
Geometric Definition
o The cross product of two vectors is defines as
A  B   A B sin( )  n 
o Here, n  is a unit vector perpendicular to the plane
formed by the two vectors A and B .
o The sense of n  is defined by the right-hand-rule, that is, the right thumb points in
the direction of n  when the fingers of the right hand point from A to B .
Properties of the Cross Product
o Product is not commutative: A  B    B  A
o Product is distributive over addition: A   B  C    A  B    A  C 
o Multiplication by a scalar  :   A  B    A   B 
Calculation
o Given two vectors A and B expressed in terms of
mutually perpendicular unit vectors i , j and k , we
calculate the cross product as

 
A  B  ax i  a y j  az k  bx i  by j  bz k

  a ybz  az by  i   axbz  azbx  j   a xby  a ybx  k
o The cross product is zero if the two vectors are parallel.
o The cross product may be calculated using determinant form.
i
j
k
A  B  ax
ay
bx
by
i
j
k
az  ax
ay
bz
by
bx
i
j
k
az  ax
ay
bz
by
bx
i
j
k
az  ax
ay
az
bz
by
bz
bx
  a ybz  az by  i   axbz  azbx  j   axby  a ybx  k
Kamman – ME 2560: page 2/3
Moment of a Force – Using the Cross Product
o The moment of a force about a point O may be calculated using the cross product
MO  r  F
o Here, r is a position vector from O to any point on the line of action of F .
Example 2:
Given: Force F  100 i  50 j  200 k (lb)
Point A : (3,4,5) (ft)
Find: M O the moment of F about O
Solution:
M O  r A/ O  F

i
j
k
3
4
5
100 50 200
  800  250  i   600  500  j  150  400  k
 550 i  1100 j  550 k (ft-lb)
Resultant Moment
o As we did with forces, we can define a resultant moment about a point O. This is
defined as the sum of the moments of all the forces about point O.
 M O  R    M O i    r i  F i 
i
i
Kamman – ME 2560: page 3/3