If the force and couple of an equivalent force-couple system are perpendicular
to one another, we have the equation
FR MRP
= 0
The system can be reduced to a single force FR located
at a point Q. The moments of the two systems can be equated
about the point P giving,
MRP = rPQ
× FR
(4)
By substituting Eqs. 1, 2, and 3 into Eq. 4; performing the cross products;
and equating the i, j, and k components; the three scalar equation are,
Σ(rPFy
Fz - rPFz Fy ) = rPQy ΣFz
- rPQz ΣFy
Σ(rPFz
Fx - rPFx Fz ) = rPQz ΣFx
- rPQx ΣFz
Σ(rPFx
Fy - rPFy Fx ) = rPQx ΣFy
- rPQy ΣFx
Solving this system of three equations for the three unknowns rPQx,
rPQy, and rPQz gives the position vector rPQ
of the force FR,
rPQ = rPQxi
+ rPQyj + rPQzk
FR = ΣF
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