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I'm self-studying Lanczos book The Variational Principles of Mechanics and in the chapter on the principle of virtual work there's a problem

Show that any given system of forces acting on a rigid body can be replaced by a single force if, and only if, the resultant moment $\bf{\bar{M}}$ and the resultant force $\bf{\bar{F}}$ are perpendicular to each other

$$\bf{\bar{F}} \cdot \bf{\bar{M}} = 0$$ where $\bf{M}=R \times F$ and $\bf{\bar{F}}=\sum F_k$

I'm pretty miffed on how to approach this. Can I just assume a single force and a single moment that point in the same directions as the problem, and then use the fact that two systems of forces which have the same resultant force and resultant moment are mechanically equivalent?

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  • $\begingroup$ Something missing form $F \cdot M$? Maybe it's equal to zero? $\endgroup$
    – user93146
    Commented Jul 10, 2018 at 20:27
  • $\begingroup$ sorry yes fixed now $\endgroup$
    – DS08
    Commented Jul 10, 2018 at 20:38
  • $\begingroup$ The proposition as stated is false. Consider two opposite forces of equal magnitude creating a non-zero torque on the body. $\mathbf F=0$ but $\mathbf M$ is not zero, so it cannot be written as $\mathbf R\times \mathbf F$ no matter what $\mathbf R$ is. $\endgroup$
    – Ján Lalinský
    Commented Jul 10, 2018 at 22:23

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With that equation you get the net change in angular momentum of the system to be zero. In that case you can resolve the net force into an applied force that acts on the total mass, and does not produce any rotation. Think snooker. If you hit the cue against the ball dead center then you don't get spin. But if you are off center then you get spin, and the force is partially going to produce motion of the mass of the ball, and partially produce rotation.

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  • $\begingroup$ How can I analytically show this? $\endgroup$
    – DS08
    Commented Jul 10, 2018 at 21:20

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