0
$\begingroup$

An extract from a school maths mechanics textbook reads:

To deal with cases in which the body is not rotating freely under gravity, we need to take into account also the work done by other external forces acting on the body. Suppose that $F$ is one such force, and let $OA$ be the perpendicular from the fixed axis to the line of action of $F$. By the principle of transmissibility we can take $A$ as the point of application of $F$.

Why is this case? This is a chapter on rotation about a fixed axis, following on from Moments of Inertia. Seeing as I previous been considering a body as a collection of masses, each with a force acting on them or their own energy, it seems odd to make this generalisation.

I have never encountered 'transmissibility' before so it is a foreign concept to me.

$\endgroup$
2
  • $\begingroup$ Under gravity a body is not freely rotating, but freely falling. $\endgroup$
    – Sofia
    Dec 26, 2014 at 13:47
  • $\begingroup$ Is that not what the extract says, 'not rotating freely under gravity'? Regarding the other case the book says 'If the rigid body is rotating freely under gravity about a fixed axis then the only external forces doing work are the weights of the component particles'. Is this wrong? $\endgroup$ Dec 26, 2014 at 13:58

1 Answer 1

1
$\begingroup$

If I understand your question correctly, the principle of transmissibility is just the fact that, when you consider momenta, that is $\mathbf r\times\mathbf F$, you can decompose $\mathbf r$ into a component $\mathbf r_\perp$ which is perpendicular the line of action of $\mathbf F$, and another component $\mathbf r_\parallel$ which is parallel to it. Now $\mathbf r\times\mathbf F = \mathbf r_\perp\times \mathbf F$ and therefore you can assume that $\mathbf F$ is applied at the point $A$ rather than at $OA+\mathbf r_\parallel=\mathbf r_\perp + \mathbf r_\parallel$ (here $OA = \mathbf r_\perp$), where the force actually is.

$\endgroup$
2
  • $\begingroup$ So the point can be considered to by applied anywhere along the line of action of the force? Also do you therefore assume that the force only acts on the particles of the body along the line of action, and the rest of the particles are unaffected? Does rotation only occurred when the line of action is not through the centre of mass then? Sorry for inundating you with questions. I am self-teaching M5 dynamics which is completely new to me. $\endgroup$ Dec 26, 2014 at 14:19
  • $\begingroup$ Yes, the force can be thought as being applied anywhere on the line of force, but this is just mathematical convenience. Physically this would make no sense as a force must be applied to some material component of the body so it's best to consider that the force is still applied to the point of the rigid body. Rigidity (which translates into an inertia tensor) will then give you the correct motion for the body. If your rigid body is not constrained to revolve around a fixed axis then any force applied at the centre of mass won't cause the body to rotate, otherwise it will start rotating. $\endgroup$
    – Phoenix87
    Dec 26, 2014 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.