Moment of a Force about an Axis

ME 2560 Statics
Moment of a Force about an Axis
Two Dimensional Systems
o The moment of a force about a point O may be calculated using a cross product.
MO  r  F
o Here, M O is perpendicular to the plane formed by the vectors r and F , and it has
magnitude | M O |  | F | d  | F | r sin( ) .
o In the two dimensional system shown, M O represents the moment of the force about an
axis perpendicular to the page (in the k direction) and passing through O.
A
O
θ
θ
Line of
Action
Three Dimensional Systems
o To find the moment of a force about an axis in three dimensional analysis, we first
calculate the moment about any point on that axis, say O, then we project that
moment onto the axis using the dot product.
M n -axis  M O  n   r  F   n
o Here, M n -axis is the scalar moment of F about the axis passing through O and parallel
to the unit vector n . In vector form, we write M n-axis   M O  n  n
o M n -axis can be positive or negative depending on the angle between n and M O . If it is
positive, point your right thumb in the direction of n and your right fingers will show
the circulation of F about the axis. If it is negative, point your right thumb opposite
the direction of n and your right fingers will show the circulation of F about the axis.
o As before, r is a position vector from O to any point on the line of action of F .
Kamman – ME 2560: page 1/2
Example:
Given: F  100 i  50 j  200 k (lb) ; A : (3,4,5) (ft)
Find: a) M x , M y , and M z the moments of F
about the X, Y, and Z axes, b) M n -axis the
scalar moment of F about an axis in the X-Y
plane that makes an angle of 30 (deg) with
the X-axis, and c) M n -axis
Solution:
a) M O  r A/O  F

i
j
k
3
4
5   800  250  i   600  500  j  150  400  k
100 50 200
 550 i  1100 j  550 k (ft-lb)
M x  M O  i  550 (ft-lb) , M y  M O  j  1100 (ft-lb) , M z  M O  k  550 (ft-lb)
b) n  cos(30) i  sin(30) j
M n-axis  M O  n   550cos(30)    1100  sin(30)    550  0  73.686  73.7 (ft-lb)
c) M n -axis   M O  n  n  73.686 n  63.8 i  36.8 j (ft-lb)
Note on Calculation of the Scalar Moment: M n -axis
o Calculation of the scalar moment can also be done in determinant form. Simply
replace the first row of the determinant by the components of n and expand as usual.
nx
ny
nz
M n -axis  rx
ry
rz  nx  ry Fz  rz Fy   n y  rx Fz  rz Fx   nz  rx Fy  ry Fx 
Fx
Fy
Fz
o So, for the above example, we have
cos(30) sin(30)
M n -axis 
3
100
4
50
0
5  cos(30) 800  250   sin(30)  600  500 
200
 73.7 (ft-lb)
Kamman – ME 2560: page 2/2