When the resultant doesn't act on the same object as the components I can understand what the resultant means when it acts on the same object as the components. What I don't understand is situations like this: If two oppositely parallel forces (red arrows in diagram) act at the ends of a beam then according to my textbook the resultant will act in thin air (blue arrow in diagram) outside of the beam. This doesn't make sense to me as it seems to be saying that if I apply two unequal forces to a beam, the beam won't move (because the resultant doesn't act on the beam!)
A thorough explanation of this phenomenon would be highly appreciated!
Diagram:

Edit: Here is what the book says.
Page 1:

Page 2:

 A: 
What I don't understand is situations like this: If two oppositely parallel forces (red arrows in diagram) act at the ends of a beam then according to my textbook the resultant will act in thin air 

That is incorrect. A force can't act 'in thin air': it would have no effect on the beam.
In order to find the resultant force and its position, the net force and net torque of the resultant has to be the same as the individual forces combined:



*

*Net force: as there are only forces acting in the vertical direction (with 'up' designated as the positive sense), the net resultant force $F_R$ is:


$$\boxed{F_R=F_2-F_1}$$


*Net torque: we can take the torque about the point $O$, so we get:


$$F_R\times x=F_2\times L$$
$$\boxed{x=\frac{F_2L}{F_2-F_1}}$$
With this magnitude and position of $F_R$ the resultant has the same effect as the combined forces $F_1$ and $F_2$.
A: As Gert stated, the resultant is defined as a single force on the system that has the same global effect as the forces really acting on the system. In this case, we have indeed that $F_R = F_2-F_1$ for the magnitude of the resultant and $x = \frac{F_2L}{F_2-F_1}$ the position from O the resultant should apply.
However, as you can see, it is possible to have $\frac{F_2}{F_2-F_1} =2$ (for $F_1/F_2 = 1/2$) in which case the resultant should act at $x = 2L$, which is outside your physical bar. There is however no problem because the resultant is a fictitious force which is useful for calculations, but do not reflect any physical reality.
So yes, it is possible that the resultant action be not on the object.
