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In my book, its written that:

When we produce torque on a pivoted body, we are always producing a couple as the reaction force from the pivot provides an equal, opposite and parallel force.

So suppose I have a body like this:enter image description here

Here, two parallel, opposite but unequal forces $A$ and $B$ are acting on a body where magnitude of $A$>$B$.

  • Will the body rotate?
  • If yes, how will it do so?
  • Where is the couple here, and the pivot?
  • The net force of $A$ and $B$ is acting downwards outside the body towards $A$, so what kind of motion will be exhibited by the body?
Qmechanic
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1 Answers1

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Add forces $A'$, $A''$ and $B'$, $B''$ acting at the centre of mass $G$ of the rod.

enter image description here

Forces $A$ and $A'$ act as a couple of magnitude $Aa$ in a counter-clockwise direction. Forces $B$ and $B'$ act as a couple of magnitude $Bb$ in a counter-clockwise direction.
So the net couple on the rod is $Aa+Bb$ counter-clockwise and the rod will rotate about the centre of mass $G$ under the influence of the net couple.

You are left with a net force of $A''- B''$ downwards which will accelerate the centre of mass.

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Using the symbols above one can write two equations, one for linear acceleration of the centre of mass $G$, $\dot v_{\rm G}$, and one for the angular acceleration about the centre of mass, $\dot \omega_{\rm G}$.

$B-A= m\,\dot v_{\rm G} $ and $B\, b+A\, a= I_{\rm G}\, \dot \omega_{\rm G}$.

Now the constraint is that $A$ is a pivot and so does not move, thus the upward linear acceleration is equal to the downward linear acceleration, $\dot v_{\rm G}=a\, \dot \omega_{\rm G}$.

This gives the relationship between the magnitude of the applied force, $B$, and the magnitude of the force exerted by the pivot, $A$.

$B(I_{\rm G} -m\,a\,b) = A(I_{\rm G} +m\,a^2) $.

Thus $B>A$ as suggested by @tryingtobeastoic.

Farcher
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    Where do the forces $A'$ and $B'$ come from in the first place? –  Nov 13 '21 at 13:38
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    The four extra forces are chosen so that the system of forces can be converted into two couples and two forces acting at the centre of mass. Have a look at Appendix 20A Chasles’s Theorem: Rotation and Translation of a Rigid Body. – Farcher Nov 13 '21 at 13:56
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    Alright, but why do we consider the two forces $A'$ and $B'$ seperately when both of them, acting on the same point, should cause a net force to act upwards as $A'-B'$? –  Nov 13 '21 at 14:07
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    Just two show that it can be considered that two couples are acting on the bar. – Farcher Nov 13 '21 at 17:21
  • But your answer doesn't seem to address the quote from the book cited by the OP which involves I believe a pivot point (restraint). Instead of G being the COM, suppose it is a frictionless pin so that the only forces applied to the rod are B and the vertical reaction B'' at G. Would they constitute a "couple" in your view? – Bob D Mar 28 '22 at 19:36
  • @tryingtobeastoic actually I mean B’, and only B’ – Bob D Mar 29 '22 at 05:48
  • If the bar is restrained by a pivot at A, then how does G undergo linear acceleration? Shouldn't in only rotate about A without translation? BTW thanks for taking the time to respond. – Bob D Mar 30 '22 at 11:46
  • @Farcher, Sir, if the bar is pinned/constrained at one end of the bar (let's call that point A), and if a force $F$ acts at the other end, then the reaction force at A, $F_A$, will be equal to $F$ if I understand you correctly. According to you, $F$ and $F_A$ will constitute a couple and the bar will rotate about A (not G, which is the center of mass). My question is that how is that possible? Doesn't a couple always make a bar rotate about the bar's center of mass? Why is the couple making the bar rotate about A then? – tryingtobeastoic Mar 30 '22 at 13:40
  • @tryingtobeastoic I believe the COM (and every point on bar) will rotate about A while at the same time the bar will rotate about the COM. I prepared a figure to show this. Farcher, FYI for the example tryingtobeastoic and I are working on the force B in your diagram is continually perpendicular to the bar. – Bob D Mar 30 '22 at 14:51
  • @BobD "I prepared a figure to show this"-where can I find that figure, sir? – tryingtobeastoic Mar 30 '22 at 15:19
  • @Farcher, I and BobD are working on this question. See BobD's answer to the question and the room that we are in. – tryingtobeastoic Mar 30 '22 at 15:22
  • @tryingtobeastoic I have added to my answer but now with the left hand end of the rod as a pivot. – Farcher Apr 01 '22 at 03:21