Page 194 - Dynamics of Mechanical Systems

0593_C06_fm Page 175 Monday, May 6, 2002 2:28 PM

Forces and Force Systems 175

*
These expressions show that if there is some point Q such that the component of M Q*
along n is zero, then M  will be smaller or, at most, equal to the magnitude of the
⊥
Q*
moment about any other point. That is, if Q is any point, then:
M ≥ M = M * (6.6.8)
Q Q *

*
The question that remains, however, is do such points Q exist? To answer this question,
observe that if Q exists, then for any point O we have:
*
×
*
*
M = M + Q O R = M n (6.6.9)
Q * O R
If we consider the vector product of n with the terms of this equation, we have:
R
×
n × M + n ×( Q O R) = 0 (6.6.10)
*
R
O
R
By expanding the triple product and by dividing by R, we obtain:
R (
n × M ) + QO −( n ⋅ Q O n )
*
*
O
R R R (6.6.11)
*
Solving for OQ , we obtain:
×
RM +( OQ n n )
OQ =− Q O = O * ⋅ (6.6.12)
*
*
R 2 R R
2
Observe that the terms (R × M )/R and (OQ • n )n are perpendicular (R × M is
*
O
O
R
R
perpendicular to R, and n is parallel to R). Hence, let Q be selected such that:
*
R
OQ = ( R M ) 2
×
*
O R (6.6.13)
Then, from Eq. (6.6.9), we have:
×
M = M + Q O R = M − OQ × R
*
*
*
Q O O
( RM )
×
= M − O × R
O 2
R
(6.6.14)
M ⋅ R
= M − M + O
O O 2
R
= ( M ⋅ ) M * n
n n =
O R R R
Therefore, if Q is located according to Eq. (6.6.13), then  M *  is smaller than (or at most
*
Q
equal to) the magnitude of the moment about any other point. That is, M * is a minimum
Q
moment. Moreover, upon examination of Eqs. (6.6.11) to (6.6.14) we see that if L is a line
through Q and parallel to n , then all points on L produce minimum moments.
*
R