4569846498

52 Multibody Systems Approach to Vehicle Dynamics
Body 5
A
{F A } 1
{F A52 } 1
{F A54 } 1
{F A25 } 1
{F A53 } 1
{F A45 } 1
A
{F A35 } 1
A
Body 2
A
Body 3
Body 4
B
D {F D41 } 1
{F B21 } 1
{F D14 } 1
C
{F B12 } 1
{F C31 } 1
B
D
Body 1
{F C13 } 1
C
Body 1
Fig. 2.23
Body 1
Free-body diagram of tripod structure
For the action–reaction forces shown here Newton’s third law would apply
so that for the interaction of Body 5 and Body 2 we can say that {F A52 } 1 and
{F A25 } 1 are equal and opposite or
{F A52 } 1 {F A25 } 1 (2.135)
In this example we are looking at linkages that are pin-jointed, or have spherical
joints at each end. An example of a similar linkage would be a tie rod
in a suspension system. As both ends of the linkage are pin-jointed the
force by definition must act along the linkage. In this case we could use this
information to reduce the number of unknowns by using a scale factor as
shown in equations (2.136) to (2.138):
{F A25 } 1 f 2 {R AB } 1 (2.136)
{F A35 } 1 f 3 {R AC } 1 (2.137)
{F A45 } 1 f 4 {R AD } 1 (2.138)
In this case the solution would be assisted as the three unknowns F A25x , F A25y
and F A25z , associated with equation (2.136) for example, would be reduced
to a single unknown f 2 . To maintain rigour the choice of position vector, for
example {R AB } 1 rather than {R BA } 1 , has also been used so that comparing